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This book is written by STEPHEN H. HALL who is a guru in Signal Integrity an EMC EMI subjects.A must have book for each PCB Engineer.Enjoy it !

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ADVANCED SIGNAL INTEGRITY FOR HIGH-SPEED DIGITAL DESIGNS ADVANCED SIGNAL INTEGRITY FOR HIGH-SPEED DIGITAL DESIGNS STEPHEN H. HALL HOWARD L. HECK A JOHN WILEY & SONS, INC., PUBLICATION Copyright 2009 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and speciﬁcally disclaim any implied warranties of merchantability or ﬁtness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of proﬁt or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress Cataloging-in-Publication Data: Hall, Stephen H. Advanced signal integrity for high-speed digital designs / Stephen H. Hall, Howard L. Heck. p. cm. Includes bibliographical references and index. ISBN 978-0-470-19235-1 (cloth) 1. Digital electronics. 2. Logic designs. I. Heck, Howard L. II. Title. TK7868.D5H298 2009 621 .381—dc22 2008027977 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 CONTENTS Preface xv 1. Introduction: The Importance of Signal Integrity 1 1.1 Computing Power: Past and Future, 1 1.2 The Problem, 4 1.3 The Basics, 5 1.4 A New Realm of Bus Design, 7 1.5 Scope of the Book, 7 1.6 Summary, 8 References, 8 2. Electromagnetic Fundamentals for Signal Integrity 9 2.1 Maxwell’s Equations, 10 2.2 Common Vector Operators, 13 2.2.1 Vector, 13 2.2.2 Dot Product, 13 2.2.3 Cross Product, 14 2.2.4 Vector and Scalar Fields, 15 2.2.5 Flux, 15 2.2.6 Gradient, 18 2.2.7 Divergence, 18 2.2.8 Curl, 20 2.3 Wave Propagation, 23 2.3.1 Wave Equation, 23 2.3.2 Relation Between E and H and the Transverse Electromagnetic Mode, 25 2.3.3 Time-Harmonic Fields, 27 v vi CONTENTS 2.3.4 Propagation of Time-Harmonic Plane Waves, 28 2.4 Electrostatics, 32 2.4.1 Electrostatic Scalar Potential in Terms of an Electric Field, 36 2.4.2 Energy in an Electric Field, 37 2.4.3 Capacitance, 40 2.4.4 Energy Stored in a Capacitor, 41 2.5 Magnetostatics, 42 2.5.1 Magnetic Vector Potential, 46 2.5.2 Inductance, 48 2.5.3 Energy in a Magnetic Field, 51 2.6 Power Flow and the Poynting Vector, 53 2.6.1 Time-Averaged Values, 56 2.7 Reﬂections of Electromagnetic Waves, 57 2.7.1 Plane Wave Incident on a Perfect Conductor, 57 2.7.2 Plane Wave Incident on a Lossless Dielectric, 60 References, 62 Problems, 62 3. Ideal Transmission-Line Fundamentals 65 3.1 Transmission-Line Structures, 66 3.2 Wave Propagation on Loss-Free Transmission Lines, 67 3.2.1 Electric and Magnetic Fields on a Transmission Line, 68 3.2.2 Telegrapher’s Equations, 73 3.2.3 Equivalent Circuit for the Loss-Free Case, 76 3.2.4 Wave Equation in Terms of LC, 80 3.3 Transmission-Line Properties, 82 3.3.1 Transmission-Line Phase Velocity, 82 3.3.2 Transmission-Line Characteristic Impedance, 82 3.3.3 Effective Dielectric Permittivity, 83 3.3.4 Simple Formulas for Calculating the Characteristic Impedance, 85 3.3.5 Validity of the TEM Approximation, 86 3.4 Transmission-Line Parameters for the Loss-Free Case, 90 3.4.1 Laplace and Poisson Equations, 91 3.4.2 Transmission-Line Parameters for a Coaxial Line, 91 3.4.3 Transmission-Line Parameters for a Microstrip, 94 3.4.4 Charge Distribution Near a Conductor Edge, 100 3.4.5 Charge Distribution and Transmission-Line Parameters, 104 CONTENTS vii 3.4.6 Field Mapping, 107 3.5 Transmission-Line Reﬂections, 113 3.5.1 Transmission-Line Reﬂection and Transmission Coefﬁcient, 113 3.5.2 Launching an Initial Wave, 116 3.5.3 Multiple Reﬂections, 116 3.5.4 Lattice Diagrams and Over- or Underdriven Transmission Lines, 118 3.5.5 Lattice Diagrams for Nonideal Topologies, 121 3.5.6 Effect of Rise and Fall Times on Reﬂections, 129 3.5.7 Reﬂections from Reactive Loads, 129 3.6 Time-Domain Reﬂectometry, 134 3.6.1 Measuring the Characteristic Impedance and Delay of a Transmission Line, 134 3.6.2 Measuring Inductance and Capacitance of Reactive Structures, 137 3.6.3 Understanding the TDR Proﬁle, 140 References, 140 Problems, 141 4. Crosstalk 145 4.1 Mutual Inductance and Capacitance, 146 4.1.1 Mutual Inductance, 147 4.1.2 Mutual Capacitance, 149 4.1.3 Field Solvers, 152 4.2 Coupled Wave Equations, 153 4.2.1 Wave Equation Revisited, 153 4.2.2 Coupled Wave Equations, 155 4.3 Coupled Line Analysis, 157 4.3.1 Impedance and Velocity, 157 4.3.2 Coupled Noise, 165 4.4 Modal Analysis, 177 4.4.1 Modal Decomposition, 178 4.4.2 Modal Impedance and Velocity, 180 4.4.3 Reconstructing the Signal, 180 4.4.4 Modal Analysis, 181 4.4.5 Modal Analysis of Lossy Lines, 192 4.5 Crosstalk Minimization, 193 4.6 Summary, 194 References, 195 Problems, 195 viii CONTENTS 5. Nonideal Conductor Models 201 5.1 Signals Propagating in Unbounded Conductive Media, 202 5.1.1 Propagation Constant for Conductive Media, 202 5.1.2 Skin Depth, 204 5.2 Classic Conductor Model for Transmission Lines, 205 5.2.1 Dc Losses in Conductors, 206 5.2.2 Frequency-Dependent Resistance in Conductors, 207 5.2.3 Frequency-Dependent Inductance, 213 5.2.4 Power Loss in a Smooth Conductor, 218 5.3 Surface Roughness, 222 5.3.1 Hammerstad Model, 223 5.3.2 Hemispherical Model, 228 5.3.3 Huray Model, 237 5.3.4 Conclusions, 243 5.4 Transmission-Line Parameters for Nonideal Conductors, 244 5.4.1 Equivalent Circuit, Impedance, and Propagation Constant, 244 5.4.2 Telegrapher’s Equations for a Real Conductor and a Perfect Dielectric, 246 References, 247 Problems, 247 6. Electrical Properties of Dielectrics 249 6.1 Polarization of Dielectrics, 250 6.1.1 Electronic Polarization, 250 6.1.2 Orientational (Dipole) Polarization, 253 6.1.3 Ionic (Molecular) Polarization, 253 6.1.4 Relative Permittivity, 254 6.2 Classiﬁcation of Dielectric Materials, 256 6.3 Frequency-Dependent Dielectric Behavior, 256 6.3.1 Dc Dielectric Losses, 257 6.3.2 Frequency-Dependent Dielectric Model: Single Pole, 257 6.3.3 Anomalous Dispersion, 261 6.3.4 Frequency-Dependent Dielectric Model: Multipole, 262 6.3.5 Inﬁnite-Pole Model, 266 6.4 Properties of a Physical Dielectric Model, 269 6.4.1 Relationship Between ε and ε , 269 6.4.2 Mathematical Limits, 271 CONTENTS ix 6.5 Fiber-Weave Effect, 274 6.5.1 Physical Structure of an FR4 Dielectric and Dielectric Constant Variation, 275 6.5.2 Mitigation, 276 6.5.3 Modeling the Fiber-Weave Effect, 277 6.6 Environmental Variation in Dielectric Behavior, 279 6.6.1 Environmental Effects on Transmission-Line Performance, 281 6.6.2 Mitigation, 283 6.6.3 Modeling the Effect of Relative Humidity on an FR4 Dielectric, 284 6.7 Transmission-Line Parameters for Lossy Dielectrics and Realistic Conductors, 285 6.7.1 Equivalent Circuit, Impedance, and Propagation Constant, 285 6.7.2 Telegrapher’s Equations for Realistic Conductors and Lossy Dielectrics, 291 References, 292 Problems, 292 7. Differential Signaling 297 7.1 Removal of Common-Mode Noise, 299 7.2 Differential Crosstalk, 300 7.3 Virtual Reference Plane, 302 7.4 Propagation of Modal Voltages, 303 7.5 Common Terminology, 304 7.6 Drawbacks of Differential Signaling, 305 7.6.1 Mode Conversion, 305 7.6.2 Fiber-Weave Effect, 310 Reference, 313 Problems, 313 8. Mathematical Requirements for Physical Channels 315 8.1 Frequency-Domain Effects in Time-Domain Simulations, 316 8.1.1 Linear and Time Invariance, 316 8.1.2 Time- and Frequency-Domain Equivalencies, 317 8.1.3 Frequency Spectrum of a Digital Pulse, 321 8.1.4 System Response, 324 8.1.5 Single-Bit (Pulse) Response, 327 8.2 Requirements for a Physical Channel, 331 8.2.1 Causality, 331 x CONTENTS 8.2.2 Passivity, 340 8.2.3 Stability, 343 References, 345 Problems, 345 9. Network Analysis for Digital Engineers 347 9.1 High-Frequency Voltage and Current Waves, 349 9.1.1 Input Reﬂection into a Terminated Network, 349 9.1.2 Input Impedance, 353 9.2 Network Theory, 354 9.2.1 Impedance Matrix, 355 9.2.2 Scattering Matrix, 358 9.2.3 ABCD Parameters, 382 9.2.4 Cascading S-Parameters, 390 9.2.5 Calibration and Deembedding, 395 9.2.6 Changing the Reference Impedance, 399 9.2.7 Multimode S-Parameters, 400 9.3 Properties of Physical S-Parameters, 406 9.3.1 Passivity, 406 9.3.2 Reality, 408 9.3.3 Causality, 408 9.3.4 Subjective Examination of S-Parameters, 410 References, 413 Problems, 413 10. Topics in High-Speed Channel Modeling 417 10.1 Creating a Physical Transmission-Line Model, 418 10.1.1 Tabular Approach, 418 10.1.2 Generating a Tabular Dielectric Model, 419 10.1.3 Generating a Tabular Conductor Model, 420 10.2 NonIdeal Return Paths, 422 10.2.1 Path of Least Impedance, 422 10.2.2 Transmission Line Routed Over a Gap in the Reference Plane, 423 10.2.3 Summary, 434 10.3 Vias, 434 10.3.1 Via Resonance, 434 10.3.2 Plane Radiation Losses, 437 10.3.3 Parallel-Plate Waveguide, 439 References, 441 Problems, 442 CONTENTS xi 11. I/O Circuits and Models 443 11.1 I/O Design Considerations, 444 11.2 Push–Pull Transmitters, 446 11.2.1 Operation, 446 11.2.2 Linear Models, 448 11.2.3 Nonlinear Models, 453 11.2.4 Advanced Design Considerations, 455 11.3 CMOS receivers, 459 11.3.1 Operation, 459 11.3.2 Modeling, 460 11.3.3 Advanced Design Considerations, 460 11.4 ESD Protection Circuits, 460 11.4.1 Operation, 461 11.4.2 Modeling, 461 11.4.3 Advanced Design Considerations, 463 11.5 On-Chip Termination, 463 11.5.1 Operation, 463 11.5.2 Modeling, 463 11.5.3 Advanced Design Considerations, 464 11.6 Bergeron Diagrams, 465 11.6.1 Theory and Method, 470 11.6.2 Limitations, 474 11.7 Open-Drain Transmitters, 474 11.7.1 Operation, 474 11.7.2 Modeling, 476 11.7.3 Advanced Design Considerations, 476 11.8 Differential Current-Mode Transmitters, 479 11.8.1 Operation, 479 11.8.2 Modeling, 480 11.8.3 Advanced Design Considerations, 480 11.9 Low-Swing and Differential Receivers, 481 11.9.1 Operation, 481 11.9.2 Modeling, 482 11.9.3 Advanced Design Considerations, 483 11.10 IBIS Models, 483 11.10.1 Model Structure and Development Process, 483 11.10.2 Generating Model Data, 485 11.10.3 Differential I/O Models, 488 11.10.4 Example of an IBIS File, 490 11.11 Summary, 492 References, 492 xii CONTENTS Problems, 494 12. Equalization 499 12.1 Analysis and Design Background, 500 12.1.1 Maximum Data Transfer Capacity, 500 12.1.2 Linear Time-Invariant Systems, 502 12.1.3 Ideal Versus Practical Interconnects, 506 12.1.4 Equalization Overview, 511 12.2 Continuous-Time Linear Equalizers, 513 12.2.1 Passive CTLEs, 514 12.2.2 Active CTLEs, 521 12.3 Discrete Linear Equalizers, 522 12.3.1 Transmitter Equalization, 525 12.3.2 Coefﬁcient Selection, 530 12.3.3 Receiver Equalization, 535 12.3.4 Nonidealities in DLEs, 536 12.3.5 Adaptive Equalization, 536 12.4 Decision Feedback Equalization, 540 12.5 Summary, 542 References, 545 Problems, 546 13. Modeling and Budgeting of Timing Jitter and Noise 549 13.1 Eye Diagram, 550 13.2 Bit Error Rate, 552 13.2.1 Worst-Case Analysis, 552 13.2.2 Bit Error Rate Analysis, 555 13.3 Jitter Sources and Budgets, 560 13.3.1 Jitter Types and Sources, 561 13.3.2 System Jitter Budgets, 568 13.4 Noise Sources and Budgets, 572 13.4.1 Noise Sources, 572 13.4.2 Noise Budgets, 579 13.5 Peak Distortion Analysis Methods, 583 13.5.1 Superposition and the Pulse Response, 583 13.5.2 Worst-Case Bit Patterns and Data Eyes, 585 13.5.3 Peak Distortion Analysis Including Crosstalk, 594 13.5.4 Limitations, 598 13.6 Summary, 599 References, 599 Problems, 600 CONTENTS xiii 14. System Analysis Using Response Surface Modeling 605 14.1 Model Design Considerations, 606 14.2 Case Study: 10-Gb/s Differential PCB Interface, 607 14.3 RSM Construction by Least Squares Fitting, 607 14.4 Measures of Fit, 615 14.4.1 Residuals, 615 14.4.2 Fit Coefﬁcients, 616 14.5 Signiﬁcance Testing, 618 14.5.1 Model Signiﬁcance: The F -Test, 618 14.5.2 Parameter Signiﬁcance: Individual t-Tests, 619 14.6 Conﬁdence Intervals, 621 14.7 Sensitivity Analysis and Design Optimization, 623 14.8 Defect Rate Prediction Using Monte Carlo Simulation, 628 14.9 Additional RSM Considerations, 633 14.10 Summary, 633 References, 634 Problems, 635 Appendix A: Useful Formulas, Identities, Units, and Constants 637 Appendix B: Four-Port Conversions Between T- and S-Parameters 641 Appendix C: Critical Values of the F-Statistic 645 Appendix D: Critical Values of the T-Statistic 647 Appendix E: Causal Relationship Between Skin Effect Resistance and Internal Inductance for Rough Conductors 649 Appendix F: Spice Level 3 Model for 0.25 µm MOSIS Process 653 Index 655 PREFACE Technology has progressed to a point where digital design has entered a new realm that requires design techniques and concepts that bafﬂe even some of the most seasoned digital system designers. The fact is that state-of-the-art digital systems such as personal computers cannot be designed without a thorough understanding of advanced signal integrity. As computer technology evolves, highspeed interconnect phenomena that designers historically have ignored begin to dominate performance, and unforeseen problems arise that dramatically increase the complexity of design. Consequently, every new generation of computer design requires an understanding of new signal integrity issues that were previously not critical and new design techniques that were previously not necessary. In modern and future systems, an incomplete understanding of high-speed interconnect phenomena will literally result in the progress of digital computing coming to a standstill. In this book we leverage theory and techniques from ﬁelds such as applied physics, communications, and microwave engineering and apply them to the ﬁeld of high-speed digital design, creating an optimal combination between theory and practical applications. Although some basic material is covered, we assume that readers are well acquainted with basic electromagnetic theory, vector calculus, differential equations, statistics, and transmission-line analysis. In this book we build on the traditional knowledge base and discuss advanced topics ranging from electromagnetic theory for signal integrity to equalization methods that compensate for signal integrity problems with circuitry as required to design modern and future digital systems. Detailed theory is presented in the context of real-life design examples so that it can be applied immediately by practicing engineers, yet provides more than enough technical content to facilitate complete understanding of the concepts. xv xvi PREFACE Features 1. Visual description of theoretical concepts wherever possible, so each chapter includes numerous ﬁgures to help reinforce the concepts discussed. 2. Rigorous coverage of theory and use of practical examples to demonstrate how to use the theory in practical, real-world applications. 3. Summary of the electromagnetic theory concepts required to comprehend signal integrity. 4. Rigorous development of transmission-line and crosstalk theory to build a fundamental understanding and then apply the theory to real-world problems. 5. Development of physically consistent dielectric and conductor models to account for frequency-dependent properties, surface roughness, and physical anomalies due to manufacturing and environmental effects. 6. Description of differential signaling at a practical and a theoretical level. 7. Explanation of the mathematical limits of models such as causality, pas- sivity, stability, and reality that must be obeyed to ensure that simulations remain consistent with nature. 8. Full description of network theory, including S -parameters and frequencydomain analysis. 9. Coverage of topics such as nonideal current return paths, tabular modeling, and via resonance. 10. Covers the basics of I/O design and channel equalization. 11. Methods for modeling and budgeting of timing jitter and noise. 12. System analysis techniques for handling large numbers of variables using response surface modeling. Contemporary signaling systems continue to offer new problems to solve. Engineers who can solve these problems will deﬁne the future. This book will equip readers with the necessary practical understanding to contend with contemporary problems in high-speed digital design and with enough theory to see beyond the book to solve problems that the authors have not yet encountered. ACKNOWLEDGMENTS Many people have contributed directly or indirectly to the writing of this book. We have been fortunate to keep the company of excellent engineers, scientists, and peers, whose technical knowledge, wisdom, and mentorship have truly inspired us throughout our careers. Without these contributions, this book would not have been possible. Our gratitude is immeasurable. PREFACE xvii Among the direct contributors are: Guy Barnes of Ansoft Corporation, who reviewed our mathematics, provided simulation data for Chapter 10, and provided the graphics for the front cover. Stephen Hall, Sr., Stephen’s father, for educating us on how molecules interact with externally applied electric ﬁelds and for writing that section in Chapter 6. Paul Huray of the University of South Carolina, who educated us on complicated aspects of electromagnetic theory and acted as a role model for academic excellence. Yun Ling of Intel Corporation, who patiently reviewed our mathematics and politely corrected our mistakes. Kevin Slattery of Intel Corporation, for reviewing Chapter 2 and strongly encouraging us to write the book. Chaitanya Sreerama Burt of Intel Corporation, who both reviewed chapters and double-checked all of our calculations in Chapter 2. Steven Pytel of Ansoft Corporation, who graciously reviewed Chapter 5, correcting errors and making excellent recommendations for improvements. Daniel Hua, for checking results and helping to solve the difﬁcult differential equations. Luis Armenta of Intel Corporation, for reviewing the dielectric models in Chapter 6. Paul Hamilton of Tyco, for reviewing Chapter 6. Gerardo Romo of Intel Corporation, who reviewed Chapters 4 and 12. Michael Mirmak of Intel Corporation and current chair of the IBIS open forum, who reviewed Chapter 11. Richard Allred of Intel Corporation, who reviewed Chapters 13 and 14. Pelle Fornberg and Adam Norman, who reviewed the jitter and peak distortion material in Chapter 13. Stephen would also like to thank: Dorothy Hall , Stephen’s mother, for a lifetime of encouragement and for instilling the enthusiasm, discipline, and drive needed not only to become an engineer, but to write this book. Garrett Hall , not only for reviewing chapters and double-checking results, but for providing much of the material and motivation for the introduction and for being the most encouraging, devoted, trustworthy, and principled friend that a big brother could ever have. Howard would like thank Eric Dibble of Lockheed-Martin Corporation and Martin Rausch of Intel Corporation for supporting his efforts to break into the signal integrity ﬁeld and for guiding his professional and personal growth. He also thanks Dr. Ricardo Suarez-Gartner of Intel Corporation for encouraging him to write the book. Finally, Stephen dedicates his contributions to this book to his lifelong companion, best friend, wife, and jewel of his life, Jodi , with all his love and gratitude, as well as his little girls, Emily and Julia, who bring him more joy than he could have thought possible. Howard dedicates his contributions to his beautiful wife, Lisa, and sons, Tyler and Nick , whose support and love provide him the day-to-day happiness that makes each day a gift. Hillsboro, OR July 2008 STEPHEN H. HALL HOWARD L. HECK 1 INTRODUCTION: THE IMPORTANCE OF SIGNAL INTEGRITY 1.1 Computing power: past and future 1 1.2 The problem 4 1.3 The basics 5 1.4 A new realm of bus design 7 1.5 Scope of the book 7 1.6 Summary 8 References 8 1.1 COMPUTING POWER: PAST AND FUTURE It is estimated that sometime between the years 2025 and 2050, commonplace personal computers will exceed the calculation power of a human brain. Further extrapolation based on historical trends indicates that a single commonplace computer could exceed the computational power of the human race sometime between 2060 and 2100. Are such vast increases in computational power possible in less than 100 years? We cannot say for certain because it is impossible to predict the future. However, hindsight is always 20/20, and if we subscribe to the notion that history tends to repeat itself, we can look at the progress of computational capabilities over the last century to see if historical data support a rate sufﬁcient to achieve such performance. Hans Moravec, a researcher from the Robotics Institute at Carnegie Melon University, estimated that a computer would require 100 million megainstructions per second (MIPS) to mimic sufﬁciently closely the behavior of a human brain [Moravec, 1998]. Based on the Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 1 2 INTRODUCTION: THE IMPORTANCE OF SIGNAL INTEGRITY MIPS PC extrapolation Computational capacity of a human brain 1.00E+08 100,000,000 MIPS 1.00E+05 1.00E+02 Pentium 4 Cray 1 Historical extrapolation Human Monkey Lizard 486DX Spider 386DX 1.00E−01 1.00E−04 Eniac Univac Collosus Monroe Calculator IBM PC Worm Apple II Commodore 64 1.00E−07 1.00E−10 1870 Hand IBM Tabulator Calculation 1890 1910 1930 1950 1970 Year 1990 2010 2030 2050 Figure 1-1 Historical computational power and extrapolation into the future. (Adapted from Moravec [1998].) number of neurons, he was also able to compare the current state of computer technology to the estimated computational power of animal brains. These data points outline a particularly interesting way to examine the history of computational power while level-setting computer performance against the brainpower of common animals. Figure 1-1 plots the computational power for mechanical and electrical computers used over the last 100 years. Some of the more interesting data points are labeled on the plot, ranging from hand calculation (ca. 1/100,000,000 MIPS) to the Pentium 4 processor of 2002 (10,000 MIPS), which is only two orders of magnitude away from the estimated brain power of a monkey (1,000,000 MIPS). As the plot indicates, computers comparable to the human brain could appear as early as the 2020s based on the extrapolation of personal computer (PC) performance over the last three decades. If the historical data for the entire twentieth century are used, the time frame is extended to 2050. The predictions get even more outrageous if we extend the extrapolation to the estimated computational power of all humans presently on Earth (ca. 6 billion), which would require 6 × 1017 MIPS. Such a computer could exist by 2060, as shown in Figure 1-2. The question is: Can the historical pace of development be sustained? Observation of the data indicates that the historical trend shows no sign of slowing. In fact, the rate appears to be increasing. However, one often encounters articles by knowledgeable people in the computer industry who believe that the trend cannot be sustained and that the decades of exponential growth must stop. In 1998 it was estimated in such an article that COMPUTING POWER: PAST AND FUTURE 3 MIPS PC extrapolation 1.00E+17 1.00E+14 1.00E+11 1.00E+08 1.00E+05 Computational capacity of the human race 600,000,000,000,000,000 MIPS Extrapolation of digital computers 1.00E+02 1.00E−01 1.00E−04 1.00E−07 1.00E−10 1880 1900 1920 1940 1960 1980 2000 2020 2040 2060 2080 2100 Year Figure 1-2 Historical computational power and extrapolation into the future. commonplace printed circuit boards (PCBs) built on an FR4 dielectric could not support bus speeds faster than 300 MHz [Porter, 1998]. Current designs using FR4 substrates exceed that bus speed in commonplace personal computers by almost 10-fold (PCI Express Gen 2 buses run at 5 giga-transfers per second, which has a fundamental frequency of 2.5 GHz). History is ﬁlled with “experts” who mispredicted the future: Heavier-than-air ﬂying machines are impossible. —Lord Kelvin, British mathematician and physicist, president of the British Royal Society, 1895 Fooling around with alternating current is just a waste of time. Nobody will use it, ever. —Thomas Edison, American inventor, 1889 Rail travel at high speed is not possible because passengers, unable to breathe, would die of asphyxia. —Dr Dionysys Larder (1793–1859), professor of natural philosophy and astronomy, University College London In the mid-1970s, integrated circuits held approximately 10,000 components, which was enough to construct an entire computer with devices as small as 3 µm. Experienced engineers worried that semiconductor technology had reached its pinnacle. Three micrometers was barely larger than the wavelength of the light 4 INTRODUCTION: THE IMPORTANCE OF SIGNAL INTEGRITY used to sculpt the chip. Interactions between ever-closer wires were about to ruin the signals. Chips would soon generate so much heat that they would be impossible to cool without a refrigeration unit. The list goes on [Moravec, 1998]. A look at the computer growth graph shows that the industry found solutions to all those problems. Chip progress not only continued—it sped up. Technology companies, motivated by the potential of high proﬁts, dedicated tremendous resources to making the “impossible” possible: developing more efﬁcient transistor designs, better heat sinks, new manufacturing processes, and more advanced analysis techniques. History indicates that the rate of performance will continue to grow at exponential rates. Historically, the mechanism for advancing computation has been to miniaturize components, allowing more devices to ﬁt in and operate in a smaller space, thus producing more performance per unit volume. First, the gears in mechanical calculators shrunk, which allowed them to spin faster. Then the relays in electromechanical machines became smaller, which allowed them to switch faster. Next, the switches in digital machines evolved from shotgun shell–sized vacuum tubes, to pea-sized transistors, to tiny integrated-circuit chips [Moravec, 1998]. Each of these technological advancements came with a price: New problems that were never before considered arose that needed to be solved . How does this relate to signal integrity? The ﬁeld of signal integrity arose directly from the exponential growth of computing power. A computer system is comprised of many integral components in addition to the processor, such as the memory, cache, and chip set. The interconnections between these parts within a computer system are known collectively as system buses. Essentially, a bus is an integrated set of interconnections used to transfer data between different parts of a digital system. Accordingly, to capitalize on the beneﬁts of increased processor power, system buses must also operate at higher data transfer rates. For example, if the memory bus fails to transmit data at a sufﬁciently fast rate, the processor simply sits idle until data are available. This bottleneck would negate much of the performance gained from a more powerful processor. Subsequently, it is vital that the bus performance scale correspondingly with processor performance. 1.2 THE PROBLEM The two mechanisms used historically to scale bus design to feed the growing performance of computer processors have been speed and width. Speed facilitates higher information transfer rates by sending more bits in a given amount of time. Width facilitates more information transfer by sending more bits in parallel. From now on, the rate of information transfer on a bus will be referred to as the bus bandwidth . Increasing the bus speed to overcome bandwidth limitations becomes problematic for many reasons. As bus frequencies increase in speed, the pathways that comprise the bus, called interconnects, begin to exhibit high-frequency behavior, which thoroughly puzzles many conventional digital designers. What is THE BASICS 5 required is complete comprehension of the relevant analog techniques and theory commonly used in microwave system and radio designs applied carefully to the digital realm. As the operating frequencies of digital systems increase, these analog effects become more prevalent and severely impede the overall performance if not resolved properly. Furthermore, increased bus speed usually requires more power, which is a precious commodity, especially in mobile designs such as laptops, which rely on battery power. Increasing the bus width to overcome bandwidth roadblocks is self-limiting. Practical mechanical limitations arise quickly due to increased pin counts on packages, sockets, connectors, and the shortcomings of PCB technology. Furthermore, interactions between closely spaced interconnects lower the signal-to-noise level, making clean data transmission more difﬁcult. Since Moore’s law results in computer performance doubling every 18 months, and the bus bandwidth must scale in proportion, doubling the number of signals in the bus to facilitate the required bandwidth provides a solution to the problem that lasts less than two years. Increasing the width of the bus is simply a short-term “band-aid.” At some point, faster bus speeds will be required. The problem is that as bus designs become both wider and faster and form factors shrink to provide more computational power per unit volume, the assumptions used for past designs become outdated and new techniques must be developed. As a result, the ﬁeld of signal integrity is evolving continuously to encompass new effects that were not relevant to earlier designs. Modern bus designs have become so fast that the designer must calculate the voltage and timing numbers to a resolution as small as a few picoseconds and a few millivolts. This degree of resolution was unheard of in computer designs just a few years ago. Just to put this problem in perspective, the light that is reﬂected off your nose takes a little over 85 ps to travel to the surface of your eye, which is well over 10 times the required timing resolution of some modern bus designs. This dramatic decrease in bus timing requirements leads to several problems. First, the number of effects that must be accounted for in the design stage increases. This is because effects that were either second order, or ignored completely in previous designs, begin to dominate the performance. Consequently, the total number of variables that must be accounted for increases, which makes the problem more difﬁcult. Furthermore, the new variables are often very difﬁcult or impossible to model using conventional methods. So we have not only many more variables to worry about, but most of the new variables are very difﬁcult to model correctly. Finally, to top it off, the laboratory equipment currently available is often insufﬁcient to resolve these small timing variations, making it difﬁcult or impossible to verify the completed design or to correlate the models to reality. 1.3 THE BASICS As the reader undoubtedly knows, the basic idea in digital design is to communicate information with signals representing 1’s or 0’s. Typically, this involves 6 INTRODUCTION: THE IMPORTANCE OF SIGNAL INTEGRITY sending and receiving series of trapezoidal voltage signals in which a high voltage is a 1 and a low voltage is a 0. The conductive paths carrying the digital signals are known as interconnects. The interconnect includes the entire electrical pathway from the chip sending a signal to the chip receiving the signal. This includes the chip packages, connectors, sockets, transmission lines, and vias. A group of interconnects is referred to as a bus. The region of voltage where a digital receiver distinguishes between a high and a low voltage is known as the threshold region. Within this region, the receiver will either switch high or switch low. On the silicon, the actual switching voltages vary with temperature, supply voltage, silicon process, and other variables. From the system designer’s point of view, there are usually high- and low-voltage thresholds, known as Vih and Vil, associated with the receiving silicon, above and below which a high or low value is guaranteed to be received under all conditions. Thus, the designer must guarantee that the system can, under all conditions, deliver high voltages that do not, even brieﬂy, fall below Vih, and low voltages that remain below Vil, to ensure the integrity of the data. To maximize the speed of operation of a digital system, the timing uncertainty of a transition through the threshold region must be minimized. This means that the rise or fall time of the digital signal must be as fast as possible. Ideally, an inﬁnitely fast edge rate would be used, although there are many practical problems that prevent this. Realistically, edge rates as fast as 35 ps are encountered in real systems. The reader can use Fourier analysis to verify that the quicker the edge rate, the higher the frequencies that are found in the spectrum of the signal. Herein lies a clue to the difﬁculty. Every conductor has a frequency-dependent capacitance, inductance, conductance, and resistance. At a high-enough frequency, none of these things are negligible. Thus, a wire is no longer a wire but a distributed, frequency-dependent parasitic element that has delay and a transient impedance proﬁle that can cause distortions and glitches to manifest themselves on the waveform propagating from the driving chip to the receiving chip. The wire is now an element that is coupled to everything around it, including power and ground structures, heat sinks, other traces, and even the wireless network. The signal is not contained in the conductor itself but is, instead, carried in the local electric and magnetic ﬁelds around the conductor. The signals on one interconnect will affect, and be effected by, the signals on another. The inductance, capacitance, and resistance of all the structures in the vicinity of the interconnect have vital roles in the simple task of guaranteeing proper signaling transitions with appropriate timing at the receiver. One of the most difﬁcult aspects of high-speed design is the fact that there are many codependent variables that affect the outcome of a digital design. Some of the variables are controllable, and others force the designer to live with the random variation. One of the difﬁculties in high-speed design is how to handle the many variables, whether they are controllable or uncontrollable. Often, simpliﬁcations can be made by neglecting or assuming values for variables, but this can lead to unknown failures down the road for which it will not be SCOPE OF THE BOOK 7 possible after the fact to locate the root cause. As timing becomes more constrained, the simpliﬁcations of the past are rapidly dwindling in utility to the modern designer. In this book we also show how to incorporate a large number of variables that would otherwise make the problem intractable. Without a methodology for handling the large number of variables, a design ultimately incorporates some guesswork, no matter how much the designer understands the system physically. The ﬁnal step in handling all the variables is often the most difﬁcult part and the one most readily ignored by designers. A designer crippled by the inability to handle large numbers of variables will ultimately resort to proving a few “point solutions” instead and hope that they plausibly represent all known conditions. Although such methods are sometimes unavoidable, this can be a dangerous guessing game. Of course, a certain amount of guesswork is always present in design, but the goal of the system designer should be to minimize uncertainty. 1.4 A NEW REALM OF BUS DESIGN Technology has progressed to a point where digital design has entered a new realm, where new design techniques and concepts are required that bafﬂe even the most seasoned digital system designers. The fact is that present and future state-of-the-art digital systems, such as personal computers, cannot be designed without a thorough understanding of the principles outlined in this book. Why hasn’t this been a problem before? The answer is that digital designers didn’t need to understand these things. But digital circuits are reaching speeds where design will not be possible without an understanding of this subject. Seasoned engineers face the threat of becoming a legacy if they do not adapt to the new design space. This book will help practicing engineers adapt. From the Monroe calculator to the Pentium, from punch cards to ﬂash memory, from vacuum tubes to integrated circuits, computer performance is increasing at an exponential rate. In this book we address the needs of the contemporary digital designer as he or she encounters the numerous new challenges with modern and future high-speed digital systems and is forced to learn material previously not needed. As the conventional digital designer transitions to faster designs, he or she will indeed experience a completely different view of logic signals at high speeds. This book will help to make sense of the ugly, distorted, and smeared waveforms produced in a high-speed digital system. 1.5 SCOPE OF THE BOOK This book was written to be an advanced study in signal integrity. Although some basic material is covered, it is assumed that the reader is well acquainted with basic electromagnetic theory, vector calculus, differential equations, statistics, and transmission-line analysis. The book builds on the traditional knowledge base and covers topics required to design present and future digital systems. 8 INTRODUCTION: THE IMPORTANCE OF SIGNAL INTEGRITY 1.6 SUMMARY All of this leads to the present situation: There are new problems to solve. Engineers who can solve these problems will deﬁne the future. This book will equip readers with the necessary practical understanding to contend with contemporary problems of modern high-speed digital design with enough theory to see beyond this book and solve problems that the authors have not yet encountered. ERRATA Inevitably, some errors will slip past the layers of review. Although they will be remedied in subsequent printings, it is useful to summarize the corrections in one place. The errors, along with the corrections, will be posted at ftp://ftp.wiley.com/public/sci tech med/high speed design. REFERENCES Moravec, Hans, 1998, When will computer hardware match the human brain?, Journal of Evolution and Technology, vol. 1. Porter, Chris, 1998, High chip speeds spell end for FR4, Electronic Times, Mar. 30. 2 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY 2.1 Maxwell’s equations 10 2.2 Common vector operators 13 2.2.1 Vector 13 2.2.2 Dot product 13 2.2.3 Cross product 14 2.2.4 Vector and scalar fields 15 2.2.5 Flux 15 2.2.6 Gradient 18 2.2.7 Divergence 18 2.2.8 Curl 20 2.3 Wave propagation 23 2.3.1 Wave equation 23 2.3.2 Relation between E and H and the transverse electromagnetic mode 25 2.3.3 Time-harmonic fields 27 2.3.4 Propagation of time-harmonic plane waves 28 2.4 Electrostatics 32 2.4.1 Electrostatic scalar potential in terms of an electric field 36 2.4.2 Energy in an electric field 37 2.4.3 Capacitance 40 2.4.4 Energy stored in a capacitor 41 2.5 Magnetostatics 42 2.5.1 Magnetic vector potential 46 2.5.2 Inductance 48 2.5.3 Energy in a magnetic field 51 2.6 Power flow and the poynting vector 53 2.6.1 Time-averaged Values 56 2.7 Reﬂections of electromagnetic waves 57 Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 9 10 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY 2.7.1 Plane wave incident on a perfect conductor 57 2.7.2 Plane wave incident on a lossless dielectric 60 References 62 Problems 62 Much of signal integrity is based heavily in electromagnetic theory. Various aspects of this theory are found in numerous books on a variety of topics, such as microwaves, electromagnetics, optics, and mathematics. To rely on these books to form a basis of the fundamental understanding of signal integrity would result in a confusing disarray of conﬂicting assumptions, notations, and conventions. Although it is assumed that readers have a basic understanding of electromagnetics, the presentation of Maxwell’s equations and subsequent solutions in the form most often used in signal integrity will minimize confusion and help readers extract the relevance from the haze of mathematical calculation often encountered in generalized electromagnetic textbooks. It is also convenient to summarize, in one place, the underlying physics that forms the basis of succeeding chapters. In this section we present Maxwell’s equations and the underlying electromagnetic theory needed for signal integrity. The concepts are used and expanded on in several subsequent chapters. This analysis does not constitute a complete theoretical study; however, it does present the fundamental electromagnetic concepts needed to develop the basis of signal integrity theory. As the book progresses, this material will be built on to describe more advanced concepts as they are applied to real-world examples. Initially, the most common vector operators are reviewed brieﬂy. This is important because Maxwell’s equations will be presented in differential form and a fundamental understanding of the vector operators will allow readers to visualize the behavior of electromagnetic ﬁelds. Next, the equations that govern a plane wave propagating in free space are derived directly from Maxwell’s equations. Then the concepts of wave propagation, intrinsic impedance, and the speed of light are derived. Next, the theory of electrostatics and magnetostatics is covered to explain the physical meaning of an electric and a magnetic ﬁeld, the energy they contain, and how they relate to speciﬁc circuit elements, such as inductance and capacitance, used in later chapters. Finally, we discuss the power carried by electromagnetic waves and how they react when propagating into different materials, such as metal or other dielectric regions. Other aspects of electromagnetic theory are covered in later chapters, but the basis of that analysis is deﬁned here. 2.1 MAXWELL’S EQUATIONS Electromagnetic theory is described by Maxwell’s equations, published originally in 1873. In this section we outline the fundamentals of electromagnetic theory MAXWELL’S EQUATIONS 11 that we will need for the remainder of the book. Since a broad study of Maxwell’s equations is beyond our scope in this text, we present only the necessary information that applies directly to the speciﬁc problems addressed here. Note that it is assumed that readers have completed basic electromagnetic theory classes as a prerequisite to this material. The differential form of Maxwell’s equations in SI units is summarized in equations (2-1) through (2-4). The equivalent integral forms of Maxwell’s equations are presented throughout when necessary; however, emphasis is placed on the differential forms when convenient because they lend themselves to better intuitive understanding: ∇ × E + ∂B = 0 ∂t ∇ × H = J + ∂D ∂t ∇ ·D =ρ ∇·B =0 (Faraday’s law) (Ampe`re’s law) (Gauss’s law) (Gauss’s law for magnetism) (2-1) (2-2) (2-3) (2-4) where E = electric ﬁeld intensity (V/m) H = magnetic ﬁeld intensity (A/m), where B H = −M µ0 M = magnetization density (A/m) B = magnetic ﬂux intensity (Wb/m2) J = current density (A/m2) ρ = charge density (C/m3) D = electric ﬂux density (C/m2) where (2-5) D = ε0E + P (2-6) P = electric polarization density (C/m2) ε0 = permittivity of free space (8.85 × 10−12 F/m) µ0 = permeability of free space (4π × 10−7 H/m) Electromagnetic ﬁelds are derived from the movement of charge, so J and ρ are the ultimate sources that induce the electric and magnetic ﬁelds, while the other quantities are responses. Note that the magnetization density, M, does not exist in nature, as it is a mere mathematical convenience. Realistic sources 12 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY of magnetic current are caused by an electric current loop, as opposed to the ﬂow of a magnetic charge, which is discussed in detail in Section 2.5. The magnetization density is included here only for completeness, as it is typically not used for the applications covered in this book. The constants ε0 and µ0 dictate the electromagnetic properties of free space, such as the speed of light and the intrinsic impedance, both of which are discussed in more detail in Section 2.3. Equations (2-1) through (2-6) are not sufﬁcient to describe the electromagnetic properties of general materials; they must be supplemented with relations that comprehend the properties of media other than free space. Speciﬁcally, equation (2-7) comprehends the ﬁnite conductivity of metal, equation (2-8) accounts for the magnetic properties of a material, and equation (2-9) describes how the dielectric will respond to an applied electric ﬁeld: J = σE B = µr µ0H = µH P = ε0(εr − 1)E (2-7) (2-8) (2-9) where J is the current density (A/m2), σ the conductivity of a medium (e.g., a metal) (S/m), µr the relative permeability, and εr the relative permittivity (also known as the relative dielectric constant). Note that both µr and εr are unitless quantities. The convention used in this book is to represent the equivalent relative permittivity and permeability as µ = µr µ0 ε = εr ε0 (2-10a) (2-10b) For use in high-speed digital design, materials included typically have descriptive coefﬁcients σ , µr , and εr that are linear, meaning that they do not change as a function of the applied ﬁeld. However, for many realistic dielectric materials used to construct circuit boards in digital designs, the descriptive coefﬁcients are not homogeneous (independent of position) or isotropic (independent of direction), meaning that extreme care must be taken to ensure that the material properties are accounted for properly, which is explored in detail in Chapter 6. Furthermore, the descriptive coefﬁcients often exhibit strong frequency dependence, the nature of which is explored throughout the book. Although this brief review of Maxwell’s equations may initially seem intimidating, throughout the book the theory will be simpliﬁed and applied directly to the solution of practical real-world problems, allowing readers to extract the important concepts from the haze of mathematical calculations so that an intuitive understanding can be conveyed. COMMON VECTOR OPERATORS 13 2.2 COMMON VECTOR OPERATORS Maxwell’s equations are presented in differential form using many vector operators, which simpliﬁes their representation. The vector operators describe the behavior of the ﬁelds and how they interact, so a full understanding of the meaning behind these operators will allow readers to visualize the problem, which will lead to an intuitive understanding. In engineering, the most valuable tool is a comprehensive understanding of the concept. In electromagnetics and signal integrity, the concept is often best understood through visualization of the ﬁelds. The vector operators used in this book are reviewed here, with an emphasis on how they affect the ﬁelds visually. 2.2.1 Vector In physics and in vector calculus, a vector is a concept characterized by a magnitude and a direction. A component of a vector is the inﬂuence of that vector in a given direction. A vector is often described by a ﬁxed number of components that sum uniquely to the total vector. When used in this role, the choice of directions is dependent on the particular coordinate system being used: Cartesian coordinates, spherical coordinates, or polar coordinates. A common example of a vector is force, because it has a magnitude and a direction. Whenever possible, the problems and analysis are presented in rectangular (Cartesian) coordinates. When the geometry of the problem dictates coordinate transformation into a spherical or cylindrical coordinate system, the relevant transformations are given in Appendix A. Recall that if the vector A was located at the point in space P (x,y,z), with the components axA1, ayA2, and azA3, where am is a unit vector along the axis m, the expression for A could be written A = ax Ax + ay Ay + azAz (2-11a) where the magnitude of the vector is shown as A = A2x + A2y + A2z This vector is shown graphically in Figure 2-1. (2-11b) 2.2.2 Dot Product The dot product of two vectors A and B is a metric of how much parallelism exists between vectors: A · B ≡ AB cos φ (2-12) where φ is the angle between A and B. Note that if φ is 90◦, then A · B = 0, and if φ is 0◦ (the vectors are parallel), then A · B = AB, which is simply the product of their magnitudes. 14 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY x P(Ax, Ay, Az) A axAx ay Ay z azAz y Figure 2-1 Graphical representation of a vector in rectangular coordinates. If the vectors A and B are expressed in terms of their generalized orthogonal components as in (2-11) and the expression is expanded, the dot product can be calculated as A · B = Ax Bx + Ay By + AzBz (2-13) 2.2.3 Cross Product Similarly, the cross product of A and B is a measure of how orthogonal the vectors are, as shown by A × B ≡ (AB sin φ)an (2-14) where an is a unit vector normal to the plane containing A and B. If φ is 0◦, then A × B = 0, and if φ is 90◦ (the vectors are at a right angle to each other), then A × B = ABan with a direction perpendicular to both A and B, with the ambiguity of direction resolved by means of the right-hand rule, as shown in Figure 2-2. If the angle between A and B is something other than 0◦ or 90◦, the following determinant form of the cross product applies: which simpliﬁes to ax ay az A × B = Ax Ay Az Bx By Bz (2-15) A × B = ax(Ay Bz − AzBy) + ay(AzBx − Ax Bz) + az(Ax By − AyBx ) (2-16) COMMON VECTOR OPERATORS 15 B = axBx + ayBy + azBz B×A A×B A = axAx + ayAy + azAz Figure 2-2 Graphical representation of the cross product. 2.2.4 Vector and Scalar Fields In electromagnetic theory, a ﬁeld is deﬁned as a mathematical function of space and time. Fields can be classiﬁed as either scalar or vector. A scalar ﬁeld has a speciﬁc value (magnitude) at every point in a region of space at each instance in time. Figure 2-3 shows two examples of a scalar ﬁeld, temperature in a block of material and the voltage across a resistive strip. Note that each point P (x,y,z), there exists a corresponding temperature T (x,y,z) or voltage v(x) at any instant in time. Other examples of scalar ﬁelds are pressure and density. A vector ﬁeld has a variable magnitude and direction at any point in time, as illustrated with Figure 2-4. Note that the velocity and direction of the ﬂuid inside the pipe changes in the vicinity of the neck-down region, so the magnitude and direction (phase) of the vectors that describe the motion of the ﬂuid at a given instant in time are a function of the position in space. Other examples of vector ﬁelds are acceleration and electric and magnetic ﬁelds. 2.2.5 Flux A vector ﬁeld, F (x,y,z,t), can be represented graphically by depicting a large number of individual vectors with a speciﬁc magnitude and phase (direction); however, this is cumbersome. A more useful method for representation of a vector ﬁeld is to use the concept of ﬂux. Flux is a measure of how many ﬁeld vectors pass though a surface in space, as depicted in Figure 2-5a. The net ﬂux of vector ﬁeld F through surface S is shown as ψ = F · ds S (2-17) 16 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Block of metal x = x′ T = 250ο T = 300ο T = 350ο T = 400ο Temperature field at x = x ′ 1V 0.75 V 0.5 V 0.25 V 0V Heat (a) x Resistive strip 1V +− (b) Figure 2-3 Examples of scalar ﬁelds: (a) temperature; (b) voltage. P1 n1 P2 n2 Figure 2-4 Example of a vector ﬁeld: ﬂuid velocity. A ﬂux plot can replace vectors with a system of ﬂux lines that are created in accordance with the following rules: 1. The transverse density of the ﬂux lines agrees with the magnitudes of the vectors. So, in Figure 2-5b, the velocity of the ﬂuid is slower near the pipe wall, necessitating that the ﬂux lines be drawn farther apart than in the middle, where the ﬂuid ﬂow is faster. 2. The direction of the ﬂux lines must agree with the direction of the vectors. COMMON VECTOR OPERATORS 17 Surface (S) Vector field Flux is a measure of how many vector field lines pass through a surface S (a) Fluid velocity flux field (b) Figure 2-5 (a) Deﬁnition of ﬂux; (b) example of a ﬂux ﬁeld. Flux is, however, useful for more than just simplifying a vector ﬁeld. If a surface S is drawn in a region of space that includes ﬂux lines, the number of ﬂux lines passing through that surface is a measure of several physical quantities, such as current or power ﬂow. Note that if (2-17) is integrated over a closed surface, the net ﬂux will always be zero, assuming that no sources exist within the volume of the closed surface. This is because the same number of ﬂux lines enter the volume as exit it. To illustrate the utility of the ﬂux concept with an example, consider current ﬂow in a wire. Suppose that a wire contains electric charges of density ρ(C/m3) in a region and the charges have a velocity ν(m/s). The current density in the region is calculated as J = ρν A/m2 (2-18) the instantaneous rate of charge ﬂow per unit cross-sectional area at point P in space. For n points in space with charge densities ρi and velocities νi, the current density becomes n J = ρiνi A/m2 (2-19) i=1 Therefore, the total current ﬂowing through a surface (e.g., the cross section of the wire) is the sum of all the current density functions within the area of the surface times the surface area. This calculates the total number of vectors (J ) passing though the cross-sectional surface S of the wire, which is ﬂux. Therefore, the ﬂux of the current density function is the current ﬂowing through area S and is calculated as ψi = i = J · ds A S (2-20) 18 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY az Current Density Vector J ds l = 1 mA 5mm Current Flux Figure 2-6 Current ﬂux through a wire. Example 2-1 If a current of 1 mA is measured ﬂowing through a wire with a radius of 5 mm, calculate the current density. See Figure 2-6. SOLUTION Assume that the current density is constant in the cross section so that J = azJ = Jz and A is the cross-sectional area of the wire. Therefore, i = Jz · az ds = Jz ds = JzA = Jz(π r2) S S Jz = i πr2 ≈ 12.7 A/m2 2.2.6 Gradient The vector operator ∇, pronounced del , is shorthand for the gradient of a scalar ﬁeld. In simple terms, the gradient is the space rate of change of a scalar ﬁeld . In rectangular coordinates, the gradient of a function f is ∇f = ∂f ax ∂x + ay ∂f ∂y + az ∂f ∂z (2-21) Subsequently, the gradient constructs a vector ﬁeld from a scalar ﬁeld. 2.2.7 Divergence The divergence of a vector ﬁeld F is a measure of the outward ﬂux per unit volume. For example, if F is represented by a continuous system of unbroken ﬂux lines in a volume region, the region is said to be source-free and divergenceless. However, if F is discontinuous through the volume region or contains broken ﬂux lines, the region contains sources of ﬂux ﬁelds and has a nonzero divergence. The divergence of F (x,y,z,t) is ∇ · F (x, y, z, t) = ∂Fx + ∂Fy + ∂Fz ∂x ∂y ∂z (2-22) COMMON VECTOR OPERATORS 19 y x Test Surface (a) (b) Figure 2-7 (a) Example of a divergence-free ﬂux plot (incoming ﬂux = outgoing ﬂux); (b) ﬂux plot with nonzero divergence (outgoing ﬂux is greater than incoming ﬂux, indicating that there are sources in the test region). Figure 2-7a is an example of the ﬂux plot for a divergence-free ﬁeld. Inspection reveals why the divergence is zero because the vectors do not seem to converge or emerge from any source points. Additionally, a test closed surface (the box) placed in the region will have zero net ﬂux emanating from it, because the ﬂux lines going into the test region equal the ﬂux lines leaving it. Figure 2-7b has nonzero divergence. Note that the emergence or convergence of the ﬁeld vectors from source points is a common characteristic of a ﬁeld with ﬁnite divergence. Simply put, if a region contains a source of a ﬁeld, it will have a positive divergence. In Figure 2-7b, the nonzero divergence is evident from inspection because discontinuous ﬂux lines are required to represent the increasing density with x, yielding a net nonzero ﬂux. In other words, there are more ﬂux lines emanating from the closed surface than are entering it, meaning that a source of the ﬁeld must exist inside the test region. An understanding of the meaning of divergence allows us to gain some insight into Gauss’s laws: ∇ ·D = ρ ∇·B =0 (Gauss’s law) (Gauss’s law for magnetism) (2-3) (2-4) Note that the divergence of D, which equals εE, is nonzero and equal to the charge density, which implies that the source of the electric ﬁeld is an electrical charge. Equivalently, if the electric ﬁeld terminates abruptly, the termination must be an electric charge. Conversely, the divergence of B is zero, indicating that there is no magnetic equivalent to the electric charge and that the magnetic ﬁeld is always source-free. For a test surface, the number of ﬂux lines entering the surface must equal the ﬂux leaving it, and there are no abrupt terminations of 20 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY the magnetic ﬁeld. Therefore, the ﬂux lines of a magnetic ﬁeld consist of closed lines. 2.2.8 Curl Historically, the concept of curl comes from a mathematical model of hydro- dynamics. Early work by Helmholtz studying the vortex motion of ﬂuid led ultimately to Maxwell’s and Faraday’s conceptions of electric ﬁelds induced by time-varying magnetic ﬁelds, which is shown in equation (2-1) [Johnk, 1988]. To visualize the concept of the curl, consider a paddle wheel immersed in a stream of water, with a velocity ﬁeld as shown in Figure 2-8. In Figure 2-8a, the paddle is oriented along the z-axis perpendicular to the water ﬂow, and since the velocity of the ﬂuid is larger on the top of the paddle, the paddle will rotate clockwise, and therefore has a ﬁnite curl along the z-axis, with a direction pointing into the page as determined using the right-hand rule. Similarly, if the paddle is rotated so that it is oriented along the x-axis, as in Figure 2-8b, the paddle will not rotate, and the curl is zero. The curl of F (x,y,z,t) in determinant form (in rectangular coordinates) is shown as ax ay az ∂∂∂ ∇ × F (x, y, z, t) = ∂x ∂y ∂z (2-23) Fx Fy Fz which simpliﬁes to ∇ × F = ax ∂Fz − ∂Fy ∂y ∂z + ay ∂Fx − ∂Fz ∂z ∂x + az ∂Fy − ∂Fx ∂x ∂y (2-24) y n y n x x z z (a) (b) Figure 2-8 (a) The ﬂuid velocity ﬁeld causes the paddle wheel to rotate when it is oriented orthogonal to the ﬁeld, giving it a nonzero curl with a direction pointing into the page (−z ); (b) when the paddle wheel is parallel to the ﬁeld it has a zero curl because the ﬁeld will not make it rotate. COMMON VECTOR OPERATORS 21 Simply put, if the curl is ﬁnite, a ﬁeld will be induced that possesses circulation. This allows us intuitively to understand Faraday’s and Ampe`re’s laws: ∇ × E + ∂B = 0 ∂t ∇ × H = J + ∂D ∂t (Faraday’s law) (Ampe`re’s law) Faraday’s law says that a time-varying magnetic ﬁeld will induce an electric ﬁeld that possesses circulation around B. More intuitively, if we examine Ampe`re’s law for a steady-state current, it reduces to ∇×H =J (2-25) Equation (2-25) implies that a current ﬂowing in a wire will induce a magnetic ﬁeld that circulates around the wire, which is consistent with Gauss’s law for magnetism (2-4), which implies that the ﬂux lines of a magnetic ﬁeld must consist of closed lines. Example 2-2 Calculate the magnetic ﬁeld of a current I ﬂowing through an inﬁnitely long wire of radius a. Show that the current ﬂowing in the wire induces a magnetic ﬁeld that circulates around the z-axis. See Figure 2-9. SOLUTION To solve this problem it is necessary to present the integral form of Ampe`re’s law for static ﬁelds: B · dl = J · ds = i l µ0 S (2-26) Switching to a cylindrical coordinate system, B = aφBφ and dl = aφrdφ, yielding 2π Bφ rdφ = 2π rBφ = i 0 µ0 µ0 Bφ (r >a ) = iµ0 2π r for r > a To calculate the magnetic ﬁeld inside the conductor, only the amount of current passing through a percentage of the wire area must be considered. This is achieved by expressing the current in terms of an area ratio: 2π 0 Bφ r dφ = µ0 2π rBφ µ0 = i πr2 πa2 Bφ (r a) r Bf(r >a) Bf(r >a) Bf(ra) r Figure 2-9 How the magnetic ﬁeld will rotate around a wire carrying current. The analysis above shows that the magnetic ﬁeld has only a φ-component that is perpendicular to the current ﬂow, proving that the magnetic ﬁeld will wrap around the wire. Since the current ﬂow is inducing the magnetic ﬁeld, its intensity will increase until r becomes greater than the wire radius a. When examining ﬁelds outside the wire radius (where no current is ﬂowing), the magnetic ﬁelds will decrease, as shown in Figure 2-9. We can conﬁrm that B circulates around the wire by calculating the curl. The curl of the magnetic ﬁeld inside the wire can be calculated using the differential form of Ampe`re’s law for the static case: ∇×H =∇× B =J µ0 The curl of F in cylindrical coordinate is (from Appendix A) ∇ × F = ar 1 ∂Fz − ∂Fφ r ∂φ ∂z + aφ ∂Fr − ∂Fz ∂z ∂r + az 1 ∂(rFφ) − 1 ∂Fr r ∂r r ∂φ The solution of the integral form of Ampe`re’s law shows that the only component of the magnetic ﬁeld in the φ-direction is a function of r. Consequently, the curl WAVE PROPAGATION 23 becomes az 1 ∂(rBφ) r ∂r ∇ × B = az 1 ∂ r(µ0ir/2π a2) r ∂r = µ0i πa2 az for r < a which says that the magnetic ﬁeld with a φ-component will be induced that circulates around a wire when current I is ﬂowing in the z-direction. For r > a, the curl becomes ∇ × B = az 1 ∂ (r(µ0i/2π r)) r ∂r =0 The curl of the magnetic ﬁeld outside the wire is zero. This does not mean that the magnetic ﬁeld does not circulate around the wire outside the conductor (it certainly does). The zero curl result is simply due to the fact that the area outside the conductor does not contain any current density (J = 0), and therefore Ampe`re’s law states that the curl of the magnetic ﬁeld must be zero. 2.3 WAVE PROPAGATION When studying Maxwell’s equations, it becomes apparent that Faraday’s and Ampe`re’s laws (the two curl equations), which state, respectively, that a changing magnetic ﬁeld will produce an electric ﬁeld and a changing electric ﬁeld will produce a magnetic ﬁeld, are responsible for the propagation of an electromagnetic wave. In this section we derive equations that regulate electromagnetic wave propagation in a simple source-free medium. In the study of signal integrity, the propagation of waves in packages, on printed circuit boards, though cables, and between power and ground planes constitutes a very large portion of the discipline. In fact, communication between components in a high-speed digital design necessitates the intentional propagation of electromagnetic waves guided by transmission lines and the prevention of energy propagation across unintentional pathways (such as crosstalk) or in unwanted signal propagation modes. Without a detailed study of wave propagation, the study of signal integrity would become impossible. 2.3.1 Wave Equation In subsequent chapters it will become necessary to analyze electromagnetic wave propagation only in terms of magnetic or electric ﬁelds because they are related directly to the voltage and current propagating on transmission lines, through vias, or across planes. The wave equation forms the basis for calculating critical 24 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY electrical properties, such as crosstalk, reﬂections, standing waves, and different modes of propagation in multiconductor systems (e.g., a bus). We begin by manipulating Faraday’s and Ampe`re’s laws using some useful vector identities: ∇ × E + ∂B = 0 ∂t ∇ × H = J + ∂D ∂t Taking the curl of (2-1) produces (Faraday’s law) (Ampe`re’s law) (2-1) (2-2) ∇ × (∇ × E) = −∇ × ∂B ∂t Since B = µr µ0H [from (2-8)], the equation above can be written in terms of the electric ﬁeld by substituting (2-2) into the right-hand part: ∇ × (∇ × E) = −∇ × ∂B = − ∂(∇ × µH ) = −µ ∂ J + ∂D ∂t ∂t ∂t ∂t where µ = µr µ0. If it is assumed that the region of wave propagation is source-free, the current density J is zero. Combining equations (2-6) and (2-9) yields the relation D = εr ε0E = εE and allows the equation to be expressed only in terms of E: ∇ × (∇ × E) = −µε ∂2E ∂t2 The formula can be simpliﬁed further by using the following vector identity (see Appendix A): ∇ × (∇ × E) = ∇(∇ · E) − ∇2E Since we have assumed a source-free medium, the charge density is zero (ρ = 0), Gauss’s law reduces to ∇ · E = 0, yielding equation (2-27), which is known as the wave equation for the electric ﬁeld : ∇2E ∂2E − µε ∂t2 = 0 (2-27) Using the identical technique, the wave equation for the magnetic ﬁeld can be derived: ∇2H − εµ ∂2H ∂t2 =0 (2-28) Note that equations (2-27) and (2-28) are similar except that the order of multiplication of µε is reversed. The order of multiplication for this derivation was preserved because it will become important when using matrices to calculate the solution for waves propagating on multiple transmission lines in Chapter 4. WAVE PROPAGATION 25 2.3.2 Relation Between E and H and the Transverse Electromagnetic Mode The wave equations (2-27) and (2-28) are presented in their most general form, where the ﬁelds have components in four dimensions: x, y, z, and time. However, for the vast majority of signal integrity analysis, the wave equations (and all of Maxwell’s equations) can be simpliﬁed so that the ﬁelds have only one nonzero component that varies with one spatial coordinate. For example, the electric ﬁeld E(x, y, z, t) can be reduced to axEx(z, t). A good example of this is the magnetic ﬁeld that was calculated in Example 2-2, where the magnetic ﬂux intensity B had only one component in the φ-direction. Although the wave equations (2-27) and (2-28) were derived separately, they are coupled and interdependent. For example, if the electric ﬁeld is restricted so that E = axEx(z, t), similar restrictions on the magnetic ﬁeld cannot be chosen arbitrarily. In this case, since (2-27) was derived using the entire set of Maxwell’s equations, once the electric ﬁeld is restricted, the magnetic ﬁeld is already determined. Thus, the proper way to calculate the magnetic ﬁeld is to derive it from the electric ﬁeld. For example, if a wave is propagating in a source-free medium in the z-direction and its electric ﬁeld only has a component in the x-direction [E = axEx (z,t)], the magnetic ﬁeld can be calculated from(2-1) and (2-2): ∇ × E + ∂B = 0 ∂t (2-1) Since we have restricted E so that it varies only with z (∂E/∂x = ∂E/∂y = 0), equation (2-24) shows that the curl of E can only produce components in the x- and y-directions. ay ∂ Ex ∂z + ax ∂ Bx ∂t + ay ∂ By ∂t =0 Since B = µH , ay ∂ Ex ∂z + ax µ ∂ Hx ∂t + ay µ ∂ Hy ∂t =0 Grouping into vector components yields ay ∂ Ex ∂z + ay µ ∂ Hy ∂t =0 ax µ ∂ Hx ∂t =0 From Ampe`re’s law (2-2), ∇ × H = J + ∂D ∂t 26 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY x Ex y z Hy Figure 2-10 How the electric and magnetic ﬁelds are related as a TEM electromagnetic pulse propagates through space along the z -axis. where J = 0 for a source-free medium and D = εE (for now, we assume that P = 0): ay ∂ Hx ∂z − ax ∂ Hy ∂z = 0 + ax ε ∂ Ex ∂t Grouping into vector components yields −ax ∂ Hy ∂z = ax ε ∂ Ex ∂t ay ∂ Hx ∂z =0 The nonzero components of the equations above can be grouped to see the contributions in both the x- and y-directions. ay ∂Ex = −µ ∂Hy ∂z ∂t ax ε ∂Ex = − ∂Hy ∂t ∂z (2-29) (2-30) Equations (2-29) and (2-30) symbolize an important concept used throughout signal integrity analysis, which is that the electric and magnetic ﬁelds are orthogonal and there are no components in the z-direction. When waves propagate in this manner, it is called the transverse electromagnetic mode (TEM). Figure 2-10 depicts the relationship between the electric and magnetic ﬁelds as the TEM electromagnetic pulse propagates along the z-axis. It should be noted that waves are not always restricted to propagate only in TEM mode because some structures (such as a microstrip) distort the waveform so that a small portion of the ﬁelds will have a z-component that lies WAVE PROPAGATION 27 outside the x –y plane. However, the relationship between the wavelength and the structure sizes in practical systems allows us to assume that the waves are propagating in TEM mode until very high frequencies. Measurements have conﬁrmed that the TEM assumption remains valid to at least 50 GHz for typical transmission-line structures used in contemporary digital designs. The validity of the TEM assumption for transmission lines is discussed further in Chapter 3. 2.3.3 Time-Harmonic Fields A simpliﬁcation of Maxwell’s equations can be made if the time variation is assumed to be steady-state sinusoidal or time harmonic in nature. Although perfect sinusoidal waveforms are rarely encountered in digital design, the trapezoidal digital pulses usually employed can be constructed from a series of sinusoidal waveforms via the Fourier transform, making this general simpliﬁcation particularly useful. Time-harmonic electromagnetic ﬁelds will be generated whenever their charge and current sources also have densities that have a sinusoidal variation with time. Assuming that the sinusoidal sources are steady state permits the assumption that both E and B also reach steady state and vary according to cos(ωt + θE) and cos(ωt + θB ), where ω = 2πf and θ is the phase of either the electric or the magnetic ﬁeld. Generally, a sinusoidal waveform can be represented as cos φ + j sin φ = ejφ (2-31) so the sinusoidal form of a time-harmonic ﬁeld will vary according to the complex exponential factor ejωt , which leads to a reduction of Maxwell’s equations from a function space and time to simply space: E(x, y, z, t) = E(x, y, z)ejωt B(x, y, z, t) = B(x, y, z)ejωt (2-32a) (2-32b) Equations (2-32) allow Maxwell’s equations to be rewritten as ∇ × (Eejωt ) + ∂B(ejωt ) = 0 ∂t ∇ × (H ejωt ) = J ejωt + ∂(Dejωt ) ∂t ∇ · (Dejωt ) = ρejωt ∇ · (Bejωt ) = 0 The curl and the gradient affect only space-dependent functions, and the ejωt is operated on only by the partial time derivatives. Therefore, after canceling all 28 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY the extraneous ejωt terms, the time-harmonic form of Maxwell’s equations, where time has been eliminated, is shown. ∇ × E + jωB = 0 ∇ × H = J + jωD ∇ ·D = ρ ∇·B =0 (2-33) (2-34) (2-35) (2-36) Note that the time variation of the ﬁelds can be restored by multiplying by ejωt and taking the real part: F (x, y, z, t) = Re[F (x, y, z)ejωt ] (2-37) 2.3.4 Propagation of Time-Harmonic Plane Waves As will be demonstrated in subsequent chapters, the propagation of time-harmonic plane waves is of particular importance for the study of transmission-line or other guided-wave structures. This allows us to study a simpliﬁed subset of Maxwell’s equations where propagation is restricted to one direction (usually along the z-axis) and time is removed as described in Section 2.3.3. A plane wave is deﬁned so that propagation occurs in only one direction (z) and the ﬁelds do not vary with time in the x- and y-directions. If the ﬁelds were observed at an instant in time, they would be constant in the x –y plane for any given point z and would change for different values of z or t. Figure 2-11 depicts a plane wave propagating in the z-direction. To study the behavior of time-harmonic plane waves, it is necessary to re-derive the wave equation from the time-harmonic form of Maxwell’s x Direction of propagation z y Figure 2-11 Plane wave propagating in the z -direction. WAVE PROPAGATION 29 equations using the procedure employed in Section 2.3.1. Again, assume a source-free, linear, homogeneous medium: ∇ × (∇ × E) = −j ωµ(∇ × H ) The formula can be further simpliﬁed by using the following vector identity (Appendix A): ∇ × (∇ × E) = ∇(∇ · E) − ∇2E Since we have assumed a source-free medium, the charge density is zero (ρ = 0) and Gauss’s law reduces to ∇ · E = 0, yielding ∇2E + j 2ω2µεE = ∇2E − ω2µεE = 0 (2-38) Substituting γ 2 = ω2µε yields ∇2E − γ 2E = 0 (2-39) which is the time-harmonic plane-wave equation for the electric ﬁeld, where γ is known as the propagation constant. If the solution is limited to plane waves propagating in the z-direction that have an electric ﬁeld component only in the x-direction, the wave equation becomes (see Appendix A) ∇2E = ∂ 2 Ex ∂x2 + ∂ 2 Ex ∂y2 + ∂ 2 Ex ∂ z2 = ∂ 2 Ex ∂ z2 − γ 2Ex =0 ∂ 2 Ex ∂ z2 − γ 2Ex = 0 (2-40) which is a second-order ordinary differential equation with the general solution Ex = C1e−γ z + C2eγ z (2-41) where C1 and C2 are determined by the boundary conditions of the particular problem. As discussed in Chapter 3, equation (2-41) and its magnetic ﬁeld equiva- lent will prove to be particularly important for signal integrity because they describe the propagation of a signal on a transmission line. The ﬁrst term, C1e−γ z, describes completely the forward-traveling part of the wave propagating in the z-direction (i.e., down the length of the transmission line), and the second term, C2e+γ z, describes the propagation of the rearward-traveling wave in the −z-direction. Observing equation (2-41) allows the deﬁnition of an important term, the propagation constant: γ = α + jβ (2-42) 30 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY The terms in (2-42) have special meanings used throughout the book to describe the medium where the electromagnetic wave is propagating, whether it is in free space, is an inﬁnite dielectric, or is a transmission line. Speciﬁcally, α is the loss term, which describes signal attenuation as it propagates through the medium. The loss term accounts for the fact that real-world metals are not inﬁnitely conductive (except superconductors) and dielectrics are not perfect insulators (except free space), both of which are discussed in detail in Chapters 5 and 6. The imaginary portion of (2-42), β, called the phase constant, essentially dictates the speed at which the electromagnetic wave will travel in the medium. To visualize these waves propagating as described in (2-41), it is necessary ﬁrst to recover the time dependency removed in Section 2.3.3. Considering only the forward-propagating component of a wave in a vacuum, replacing C1 with the magnitude of the electric ﬁeld, restoring the time dependency as in (2-37), and applying the identity of equation (2-31) yields E(z, t) = Re(Ex+e−γ zejωt ) = Re(Ex+e−αze−jβzejωt ) = e−αzEx+ cos(ωt − βz) (2-43) Assuming that the loss term is zero (α = 0), Figure 2-12 depicts successive snapshots of a wave propagating though space. To determine how fast the wave is propagating, it is necessary to observe the cosine term for a small duration of time t. Since the wave is propagating, a small change in time will be proportional to a small change in distance z, which means that an observer moving with the wave will experience no phase change because she is moving at the phase velocity (νp). Setting the term inside the cosine of (2-43) to a constant (ωt − βz = constant) and differentiating allows the deﬁnition of the phase velocity from the cosine term in (2-43): νp = dz dt = ω β m/s (2-44) E + x cos (wt − bz) at t = t0 at t = t1 at t = t2 ∆t ∆z ∆z ∆t = νp l Figure 2-12 Snapshots in time of a plane wave propagating along the z -axis, showing the deﬁnition of phase velocity. WAVE PROPAGATION 31 The relationship between the frequency and its wavelengths is calculated based on the speed of light, which is the phase velocity (νp) in a vacuum: f=c λ hertz (2-45) Since ω = 2πf and c is the speed of light in a vacuum (ca. 3×108 m/s), equation (2-45) can be substituted into (2-44) to obtain a useful formula for β in terms of the wavelength λ: c = ω = 2πc → β = 2π β βλ λ rad/m (2-46) The speed of light in a vacuum is deﬁned as the inverse of the square root of the product of the permeability and the permittivity of free space: c≡ √1 m/s µ0ε0 (2-47) Calculation of λ in terms of (2-47) allows the phase constant β to be rewritten in terms of the properties of free space: β = 2πf √ µ0ε0 = √ ω µ0ε0 rad/m (2-48) This is expanded on later in this chapter to include propagation of a wave in a dielectric medium. Now that the propagation constant has been deﬁned, (2-43) can be rewritten in physical terms, assuming free space (which is lossless, so α = 0): E(z, t) = Re(Ex+ e−j √ ωz µ0ε0 ej ωt ) = Ex+ cos(ωt − √ ωz µ0ε0) (2-49) Since (2-49) is a solution to the wave equation, the magnetic ﬁeld is found simply by using Faraday’s law (∇ × E + j ωB = 0): ∂ ∂z (Ex+e−j √ ωz µ0ε0 )ej ωt = −j ωµ0Hy+ √ Hy+ = µ0ε0 µ0 Ex+ e−j √ ωz µ0 ε0 ej ωt = 1 η0 Ex+e−j √ ωz µ0 ε0 ej ωt = 1 η0 Ex+ cos(ωt − √ ωz µ0ε0) (2-50) where η0 is the intrinsic impedance of free space and has a value of 377 : η0 ≡ µ0 = 377 ε0 (2-51) 32 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Equations (2-49) and (2-50) describe how a plane wave propagates in free space. The intrinsic impedance and the speed of light are constants that describe how the electromagnetic wave will propagate through the medium. The speed of light deﬁnes the phase delay of the wave, and the intrinsic impedance describes the relationship between the electric and magnetic ﬁelds. However, for wave propagation in other media, such as the dielectric of a printed circuit board (PCB), the speed of light and the intrinsic impedance are calculated using the relative permittivity εr and relative permeability µr , which simply describe the properties of the material relative to free-space values. Note that both µr and εr are unitless values that are real numbers for loss-free media but become complex for lossy media, as described in Chapters 5 and 6. The speed of light (referred to as the phase velocity for media other than free space) and the intrinsic impedance in a medium is calculated as νp = 1 √ µr µ0εr ε0 = c √ µr εr η ≡ µr µ0 = µ = E εr ε0 εH m/s ohms (2-52) (2-53) Note that for free space, µr and εr are both deﬁned to be unity. Equations (2-54) and (2-55) summarize the TEM plane waves of both the electric and magnetic ﬁelds in general form, with the time dependency removed: Ex (z) = Ex+e−γ z + Ex−eγ z Hy (z) = 1 η (Ex+e−γ z − Ex−eγ z) (2-54) (2-55) where γ = α + jβ is the propagation constant; α describes how the signal is attenuated by conductor and dielectric losses, described in full detail in Chapters 5 and 6; and β is the phase constant, as deﬁned by (2-46) when the phase velocity in (2-52) is substituted for the speed of light in a vacuum (c). Note that the second term in (2-55) is negative. This is because the sign of the exponent for the reverse traveling wave does not cancel the negative sign in Faraday’s law as it did for the forward-traveling wave in equation (2-50) when the derivative with respect to z was calculated. The terms Ex+ and Ex− describe the directions of each component of a propagating wave. For example, the total propagating wave could have a portion of the electric ﬁeld propagating in the +z-direction and another propagating in the −z-direction. Figure 2-13 depicts a time-harmonic TEM plane wave propagating along the z-axis. 2.4 ELECTROSTATICS Electrostatics is the study of stationary charge distributions. A complete understanding of electrostatics is essential for the high-speed digital designer because ELECTROSTATICS 33 x Ex(z,t ) = E +xe−gz y Hy(z,t ) = 1 h E +xe−gz z Figure 2-13 Time-harmonic TEM plane wave propagating down the z -axis. it forms the basis of fundamental signal integrity theory and promotes an intuitive understanding of how electric ﬁelds behave. In this chapter we (1) deﬁne the electric ﬁeld, (2) describe how energy is stored in the electric ﬁeld, and (3) deﬁne capacitance, which is the circuit element used in circuit models to represent the energy stored in an electric ﬁeld. The vast majority of signal integrity analysis performed in the industry today uses electrostatic techniques to calculate critical design variables such as transmission-line impedance, phase velocities, and effective dielectric permittivities. The concepts introduced in this section are expanded on in later chapters to describe a myriad of concepts. When a weightlifter hoists a barbell over his head, the energy expended (or the work done against the gravitational ﬁeld) is stored in the form of potential energy. The gravitational potential energy can be recovered by lowering the barbell to the ground. Similarly, when two inﬁnitely separated charges of the same polarity are brought together, the charges will experience a repulsive force whose magnitude depends on the distance between the charges. The existence of this force is described by saying that a charge q with units of coulombs (C) produces an electric ﬁeld in the region surrounding it. When the electric ﬁelds of two charges of the same polarity begin to interact, a force will be generated that will push the charges apart. Therefore, the region surrounding a charge is permeated by a force ﬁeld known as the electric ﬁeld , deﬁned fundamentally as force per unit charge, with units of newtons per coulomb (N/C). Note that because a volt is deﬁned as joules per coulomb (J/C), newtons per coulomb is equivalent to volts per meter (V/m), which are the units commonly used to describe an electric ﬁeld. V= J →C= J C V 1J = 1kg · m2/s2 1N = 1kg · m/s2 N C = N·V J = 1kg · m/s2 · V 1kg · m2/s2 = V m 34 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY The description above makes the argument that an electric ﬁeld is a ﬁeld of force. If that force ﬁeld acts upon a body to move it, work is done. The energy used to perform the work must either be dissipated by losses (friction in mechanical systems) or stored in the form of kinetic or potential energy. For example, if a test charge q is moved to different positions in an electric ﬁeld produced by a ﬁxed charge Q, either work must be performed to keep the charges separated (in the case of opposite charges) or work must be performed to bring the charges closer together (in the case of the same polarity charges). In this case, no energy is dissipated (in a loss-free system) because the energy in stored in the separated conﬁguration and the energy can be recovered if the charges are allowed to return to the initial positions. The stored energy is potential energy because it depends on the position of the charges within the ﬁeld. The concept of scalar electric potential , which will now be derived, provides a metric to describe the work or energy required to move charges from one point to another inside an electrostatic ﬁeld. The discussion above describes how an electric ﬁeld is produced when two charges are brought into the vicinity of each other. If we assume that one charge (Q) is stationary and the other charge (q) is moved toward the stationary charge from point a to point b, the work can be calculated as force × distance. To calculate the work done while moving the charge along a path, the following line integral is used: b Wa→b = − F · dl a joules (2-56) Note that the minus sign is necessary because (2-56) represents work being done against the ﬁeld. The dot product accounts for the fact that it takes zero work to move the charge perpendicular to the ﬁeld because there is no opposing force in that direction. Since a fundamental unit of an electric ﬁeld is newtons per coulomb (described above), the force ﬁeld in (2-56) can be rewritten in terms of the electric ﬁeld as F E= q a→b N/C or V/m W b = v(b) − v(a) = − E · dl V q a→b a (2-57) (2-58) where q is a test charge being moved from point a to point b within the vicinity of a ﬁxed charge Q, with units of coulombs. To calculate the electric ﬁeld generated by a point charge, the integral form of Gauss’s law is used: εE · ds = ρ dV = Qenc S V (2-59) ELECTROSTATICS 35 E = arEr Q Er = Q 4pe0 r 2 r Figure 2-14 Electric ﬁeld generated by a point charge Q. where is the volume density of the charge within volume V in C/m3 and Qenc is the total charge enclosed by the surface S and contained in the volume V in units of coulombs. As shown in Figure 2-14, a point charge has an electric ﬁeld that radiates out in all directions, necessitating the use of spherical coordinates. In spherical coordinates, the φ and θ omponents of the electric ﬁeld are zero, leaving only an r component directed outward from the point charge. Therefore, since the surface area of a sphere is 4πr 2, the electric ﬁeld around a point charge in free space is derived from (2-59): E = ar Er = Q 4π ε0r2 V/m (2-60) Substituting (2-60) into (2-58) allows the calculation of the work done per unit charge when a charge is moved along r from point a to point b in an electrostatic ﬁeld. Note that if the test charge was moved along φ or θ , there would be no work done because E · dl = 0; however, along the radial component, E · dl = Er dr. W b bQ q = − E · dl = − a→b a a 4π ε0r2 dr = Q 1 − 1 = (a) − (b) V 4π ε0 rb ra (2-61) Therefore, we can deﬁne the electrostatic potential as work done to move a charge in an electrostatic ﬁeld from point a to point b: ab = Q 4π ε0rab V (2-62) 36 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY where rab denotes the radial distance that the stationary charge moved and ab is known as the electrostatic potential or voltage between points a and b (don’t confuse the symbol for potential with the polar coordinate variable φ). 2.4.1 Electrostatic Scalar Potential in Terms of an Electric Field For electrical engineers, it is often convenient to think of ﬁelds in terms of more familiar circuit concepts, which are usually described in terms of potential differences (i.e., voltage) between points in a circuit. Therefore, a relationship between the electrostatic potential and the electric ﬁelds that is described in terms of a scalar function will prove to be useful in the study of signal integrity. To derive this relationship, it is convenient to sidestep the complexities of spherical coordinates and think in terms of simple rectangular coordinates. From equations (2-61) and (2-62) it is obvious that the differential amount of work done in moving a charge in an electrostatic ﬁeld is directly proportional to the potential difference: dW = q (x + x, y + y, z + z) − q (x, y, z) =q ∂ x+∂ y+∂ z ∂x ∂y ∂x (2-63) However, from equation (2-58), b W = − qE · dl → dW = −qE · dl a In rectangular coordinates, dl = ax x + ay y + az z. Thus, (2-63) can be simpliﬁed using the deﬁnition of the dot product in (2-13). dW = q ∂ x + ∂ y + ∂ z = −qE · dl ∂x ∂y ∂x ∂ ∂ ∂ = q ax ∂x + ay ∂y + az ∂z · (ax x + ay y + az z) = −qE · (ax x + ay y + az z) E=− ∂ ax ∂x + ay ∂ ∂y + az ∂ ∂ z (2-64) Note that the form of the last term in (2-64) is identical to (2-21), which is the gradient, which gives a very useful relation between the electric ﬁeld and the electrostatic potential: E = −∇ (2-65) ELECTROSTATICS 37 Also note that (2-65) can be derived using vector identities, which is signiﬁcantly simpler but does not provide any intuition. The alternative derivation is shown here because we employ a similar technique when studying magnetostatics in the next section. For an electrostatic ﬁeld, Ampe`re’s law is reduced to ∇ × E = 0 because the ﬁeld does not vary with time. For any differentiable scalar function, the following vector identity holds true (from Appendix A): ∇ × ∇ψ = 0 Therefore, since ∇ × E = 0, E must be derivable from the gradient of a scalar function. Since (2-61) shows a relationship between the electrostatic potential and the electric ﬁeld, a leap of logic says that the scalar function in the vector identity must be the electrostatic potential: ∇ × E = ∇ × (−∇ ) = 0 (2-66) Equation (2-65), known as the electrostatic scalar potential , is used often when solving electrostatic problems such as transmission-line impedance or calculating the effective dielectric constant of a microstrip, as we demonstrate in Chapter 3. 2.4.2 Energy in an Electric Field To calculate the energy stored in an electric ﬁeld, it is necessary to begin with a stationary charge (q1) in free space that is inﬁnitely far way from any other charges. It takes no work to move the ﬁrst charge into position (W1 = 0) because there are no other nearby charges to provide electrostatic repulsion. Then we calculate how much work it takes to bring another charge (q2) into the vicinity of the ﬁrst charge using (2-61) and (2-62): W2 = q2 12 = q1q2 4π ε0r12 (2-67) If we bring another charge, q3, into the vicinity of q1 and q2, the work is calculated as W3 = q3( 13 + 23) = q3 1 4π ε0 q1 + q2 r13 r23 Thus, the total work done to bring the three charges together is Wtot = W1 + W2 + W3 = 0+ q1q2 4π ε0r12 + q3 4π ε0 = 1 q1q2 + q1q3 + q2q3 4π ε0 r12 r13 r23 q1 + q2 r13 r23 38 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY which can be generalized into the following double summation to account for n charges: Wtot = 1 4π ε0 n i=1 n j =1 qi qj rij (2-68) j >i The limit on the j term exists to ensure that terms are not counted twice. For example, without the limit the term 12 would be counted twice because 12 = 21, which simply means that it takes the same amount of work to move q1 into the vicinity of q2 as it does to move q2 into the vicinity of q1. Equation (2-68) can be simpliﬁed by allowing the terms to be counted twice and dividing by 2 to compensate for the double counting of terms: Wtot = 1 2 1 4π ε0 n i=1 n j =1 qi qj rij = 1 2 n i=1 n j =1 qi ij j =i j =i (2-69) Equation (2-69) is the work it takes to assemble n charges. It is interesting to consider where the energy is stored in an accumulation of charges. It is analogous to a mechanical system of a compressed spring with a weight on each end. If the weights are forced together, the energy is stored in the stressed state of the spring. Therefore, just like the spring example, the charges, which will tend to repel each other, will have stored energy that is a function of the proximity of the charges and the properties of an electric ﬁeld. To calculate the energy stored in a continuous charge distribution, and thus an electric ﬁeld, it is more convenient to express the charge in terms of the unit volume, the potential in terms of a continuous function, and to take the limit as n → ∞, which allows us to write (2-69) in terms of an integral, Wtot = We = 1 2 V ρ(r) (r) dV joules (2-70) where ρ (r) is the charge density in units of C/m3. It is useful to express (2-70) in terms of the electric ﬁeld. To do this, Gauss’s law is used to express the charge density in terms of the electric ﬁeld: ∇ · D = ∇ · εE = ρ → We = 1 2 V (∇ · εE) (r) dV (2-71) This equation can be simpliﬁed using the ﬂowing vector identity (Appendix A): ∇ · ψa = a · ∇ψ + ψ(∇ · a) → ψ(∇ · a) = ∇ · ψa − a · ∇ψ ELECTROSTATICS 39 The identity can be rewritten in terms of the electrostatic vector potential, substituting ψ = and a = εE: We = ε 2 (∇ · V E−E·∇ ) dV Since E = −∇ , the equation can be simpliﬁed further: ε We = 2 (∇ · V E + E · E) dV The divergence theorem of vector calculus states (Appendix A) that (2-72) (∇ · F ) dV = F · ds V S allowing further simpliﬁcation of (2-72): We = ε 2 S E · n ds + ε E · E dV 2V (2-73) where n is a unit vector normal to the surface. Equation (2-73) describes the total energy in an electric ﬁeld that is induced by a volume of charges. If we expand the volume of integration to include all space, any contribution of (2-73) outside the charge distribution will con- tribute nothing to the work done since there are no charges in that space. Also note that if the volume of integration is chosen to be inﬁnity, the sur- face integral disappears. To understand why, remember that the surface integral sums all the contributions evaluated at the surface. Since ∝ 1/r [equation (2-62)], E ∝ 1/r2 [equation (2-60)], and ds ∝ r2, the limit of E · nds is pro- portional to (1/r)(1/r2)r2, whose limit goes to zero when r is inﬁnity and the surface integral disappears. The volume integral includes contributions over the entire volume, not only on the surface. Subsequently, equation (2-73) can be reduced to (2-74), which is the work done by accumulating charges to create an electric ﬁeld: We = ε 2 (E2) dV all space joules (2-74) This leads to the deﬁnition of the volume energy density, which expresses the stored energy in a charge distribution in terms of the electric ﬁeld: we = ε E2 2 joules/m3 (2-75) 40 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY 2.4.3 Capacitance In circuit terms, the quantity associated with storing energy in an electric ﬁeld is capacitance. To deﬁne capacitance, imagine two conductors, with a charge of +Q on one and of −Q on the other. If we assume that the voltage is constant over each conductor, the potential difference (voltage) between them is calculated as b v(b) − v(a) = − E · dl V a (2-58) We show that E is proportional to Q: E = ar Er = Q 4π ε0r2 V/m (2-60) Since E is proportional to both Q and v, we can deﬁne a constant of proportion- ality that relates Q and v. The constant of proportionality is deﬁned to be the capacitance: C≡Q v farads (2-76) where Q is the total charge in coulombs and v is the voltage potential between the conductors, given in units of farads, deﬁned as 1 coulomb per volt. Capacitance depends purely on the geometry of the structures and the value of the dielectric permittivity. Note that v is deﬁned as the potential of the positive conductor minus the negative conductor and that Q is the charge on the positive conductor. Therefore, capacitance is always a positive value. Example 2-3 Consider the case where two conductive plates of area A are oriented parallel to each other separated by a distance d. Assume that we place a charge of +Q on the top plate and −Q on the bottom plate and assume that the charges will spread out evenly (a reasonable assumption, assuming a good conductor). Then the surface charge density becomes ρ = Q/A (C/m2). Calculate the capacitance. SOLUTION Using the integral form of Gauss’s law (2-59), we can calculate the electric ﬁeld: εE · ds = ρ dV s V where dV in this case refers to the volume. Since we are considering the charge distribution on a surface, ds = dV = nA (where n is the unit normal vector to the plate), we can write the electric ﬁeld in terms of the area and dielectric permittivity: εEA = QA → E = Q A εA ELECTROSTATICS 41 The voltage is evaluated with equation (2-58), where dl = x(a) − x(b) = d, which is the distance between the plates: b Q v = E · dl = d a εA Therefore, the capacitance between two parallel plates is C = Q = Q = εA v (Q/εA) d d farads where A is the area of the parallel plates, v the voltage, and d the distance between them. 2.4.4 Energy Stored in a Capacitor The process of storing energy in a capacitor involves electric charges of equal magnitude, but opposite polarity, building up on each plate. As long as the capacitor holds a charge, it is storing energy. To calculate how much energy is stored in a capacitor, consider how much energy it would take to transport a single charge from the positive plate to the negative plate. From equation (2-58) we know that voltage is the work done to move a charge from point a to b, and capacitance is deﬁned as W =v q a→b C≡Q v farads (2-76) Therefore, the amount of work needed to move one charge q of the total charge Q from plate a to plate b is dW = v dq = q dq C To calculate the total work done to charge up the capacitor to a value of Q, all charges must be moved: W = Q q dq = 1 Q2 0C 2C However, from (2-69), Q = Cv. Therefore, the energy stored in a capacitor with a ﬁnal potential of v is W = 1Cv2 2 joules (2-77) 42 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY 2.5 MAGNETOSTATICS In Section 2.4 we discussed the problem of classical electrostatics, where we deﬁned the electric ﬁeld in terms of the force that a collection of charges exert on each other. In the case of electrostatics, we considered only cases where the charges are at rest. Now it is time to consider the forces that exist between charges in motion. To begin this discussion, let’s consider an experiment that most people performed in high school physics class. If you recall, direct current (dc) (from a battery) driving through a coil will produce an electromagnet. The magnetic ﬁeld produced by such a conﬁguration can be descried by Ampe`re’s law, ∇ × H = J . Note that the time dependence of the electric ﬁeld (∂D/∂t) has been eliminated because we are considering only a dc ﬂow. Ampe`re’s law tells us that a steady-state current J will induce a magnetic ﬁeld H that circulates around the wire. As described in Example 2-2, the direction of the circulation can be determined using the right-hand rule. If the thumb points in the direction of the current ﬂow, the ﬁngers of the right hand will curl around in the direction of the magnetic ﬁeld. Subsequently, it is easy to imagine the form of the magnetic ﬁeld from a single loop of current in our electromagnet, as shown in Figure 2-15a. Now consider a tiny elemental loop of current circulating around a point that will induce a small magnetic ﬁeld as shown in Figure 2-15b. This small current loop will produce a small magnetic ﬁeld that is analogous to an electric charge. In fact, historically, scientists initially speculated that there was a magnetic charge analogous to the electric charge described earlier. However, experimental evidence suggests overwhelmingly that magnetic charges do not exist. Magnetic ﬁelds are not generated by the forces that magnetic charges exert on each other; rather, magnetic ﬁelds are generated by current loops. Similar to our description of the forces between charges in electrostatics, consider an isolated tiny current loop (l0) which will induce a magnetic ﬁeld and behave like a small electromagnet. Now, bring another tiny current loop (l1) from inﬁnity into the magnetic ﬁeld of l0. If the orientations of the current loops are similar, it will take work to push them together because magnets exert forces on one another similar to electric charges. Like poles will repel each other, and unlike poles will attract. Each “electromagnet” has its own north and south B B l dl B (a) (b) Figure 2-15 (a) Magnetic ﬁeld generated by a loop of current; (b) elemental current loop analogous to an electric charge. MAGNETOSTATICS 43 poles. It is interesting to note that the fundamental source of the magnetic ﬁeld is the moving charge Q that constitutes the steady-state current. Subsequently, when l1 is moved into the proximity of l0, the force induced between the two electromagnets is caused by the charge (Q) of l1 moving in the magnetic ﬁeld of l0 and is described by the Lorenz force law: Fm = Q(ν × B) (2-78a) A charge moving in the presence of both an electric and a magnetic ﬁeld produces a force calculated as Fm = Q(E + ν × B) (2-78b) The implications of (2-78) are that the force is perpendicular to both the velocity ν of the charge q and the magnetic ﬁeld B. The magnitude of the force is F = qvB sin θ , where θ is the angle between the velocity vector and the magnetic ﬁeld. Because sin(0) = 0, this implies that the magnetic force on a stationary charge or a charge moving parallel to the magnetic ﬁeld is zero. From the force relationship in (2-78) it can be deduced that the units of magnetic ﬁeld are newton · seconds/coulomb · meter or newtons/ampere · meter. This unit is named the tesla. It is a large unit, and the smaller unit gauss is used for small ﬁelds such as Earth’s magnetic ﬁeld. A tesla is 10,000 G. Earth’s magnetic ﬁeld is on the order of 0.5 G. To make this concept more apparent, the force can be deﬁned in terms of the current, which is the ﬂow of 1 C of charge per second: 1A = 1C/s (2-79) If we consider the current ﬂowing along on a differential slice of a wire (dl ), we can write (2-78) in terms of the current. Since Qν has units of C (m/s), which is the same as A · m, Qν can be simpliﬁed to I dl : Qν = Q l = dl Q → Idl s s This allows us to write the force in terms of both the current and the magnetic ﬁeld: Fm = (I × B) dl (2-80) Equation (2-80) says that the force caused by the magnetic ﬁeld will be perpendicular to the current ﬂow and the magnetic ﬁeld. When deriving the energy in an electric ﬁeld in Section 2.4, we calculated the amount of work done against the electric ﬁeld to bring an accumulation of charges together. However, when calculating the energy in the magnetic ﬁeld, a different approach must be taken because of a unique feature of equation (2-78). 44 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY If a charge Q moves an amount dl = ν dt in a magnetic ﬁeld, the amount of work done is dWm = Fm · dl = Q(ν × B) · ν dt = 0 (2-81) Note that ν × B is perpendicular to the ﬂow of the current, which follows the path of dl. In other words, the work done by the magnetic ﬁeld is zero because the force is perpendicular to the moving charge. Therefore, the magnetic ﬁeld can change the direction of the moving particle, but it cannot speed it up or slow it down. This concept may be confusing, especially when considering the simple electromagnets that we all played with in high school, because we all know that we can use an electromagnet to pick up a paper clip. Since we are moving the mass of the paper clip against Earth’s gravitational ﬁeld, we know that work is being done. However, if the work is not being done by the magnetic ﬁeld, what is doing the work? The answer is demonstrated in the following example. Example 2-4 Consider a long wire carrying current I1 in the presence of a rigid rectangular loop carrying current I2, as shown in Figure 2-16. The long wire will generate a magnetic ﬁeld as calculated in Example 2-2: B1 = I1µ0 2π r Calculate the magnetic force. B F F l2 b C x F z y F A D a F1 l1 Figure 2-16 Forces generated on a wire loop in the vicinity of a magnetic ﬁeld generated from a wire. MAGNETOSTATICS 45 SOLUTION According to equation (2-80), the force exerted on the loop carrying current I2 in the presence of B1 is Floop = (I2 × B1) dl = I2 I1µ0 2π B A ax dx × ay x + C B az dz × ay b + D C −ax dx × ay x + A D −az dz × ay a Note that the segments AB and CD are equal but opposite, so they cancel. Floop = I2 I1µ0 2π C B az dz × ay b + A D −az dz × ay a From the right-hand rule, the cross products are as follows: az × ay = −ax −az × ay = ax Therefore, the force is reduced to Floop = I2 I1µ0 2π −ax 1 (B b − C) + ax 1 a (D − A) Since segments BC = DA, we can call this length d: Floop = ax I2 I1µ0 2π d 1−1 ab Therefore, the loop will be pushed away from the wire in the direction of ax. Note that the magnetic force has caused the wire loop to move. Since work is force × distance, it would be easy to conclude that the magnetic force has performed work. However, equation (2-81) explicitly states that the magnetic ﬁeld can do no work. What is performing work? To answer this question, consider the force vector on a single segment of the loop as soon as it begins to move. Remember that the force is perpendicular to the direction of the current ﬂow, and the current ﬂow is deﬁned by the movement of charge. When the loop moves, the direction of the current ﬂow I2 will be altered. To understand this, Figure 2-17 shows that the direction in which a single charge in the loop will travel when the loop is moved in the +x-direction. Instead of moving from right to left, it is moved up and to the left because the loop is moving in the +x-direction. This will cause the force vector, which must remain perpendicular to the current ﬂow, to tilt to the right, as shown in Figure 2-17. When the force vector tilts, the component azFz opposes the charge ﬂow of the current I2 in the loop. For I2 to remain constant, the source of the current must overcome this force. This leads us to the conclusion that the power source is performing the work! The magnetic ﬁeld simply alters the direction of the force vector. 46 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Direction of loop movement I −azIz F axIx Q (t = t 2) Q (t = t1) l2 Q (t = t0) B1 l1 Figure 2-17 Forces generated on a wire loop in the vicinity of a magnetic ﬁeld generated from a wire. 2.5.1 Magnetic Vector Potential Because magnetic ﬁelds can do no work, we cannot calculate the energy stored in a magnetic ﬁeld in the same manner as we did for the electric ﬁeld (i.e., by calculating the work done to accumulate a distribution of elemental current loops), so another approach is needed. In this section we develop the concept of the magnetic vector potential . The magnetic vector potential is used to calculate inductance, which is in turn used to calculate the energy stored in a magnetic ﬁeld. The basic laws that rule magnetostatics are the time-invariant forms of Ampe`re’s and Gauss’s laws for magnetism: ∇×H =J ∇·B =0 (2-82a) (2-82b) If the following vector identity is applied (from Appendix A), ∇ · (∇ × A) = 0 it implies that B can be written in terms of the curl of a vector A. This is called the magnetic vector potential , which sometimes helps to simplify calculations such as inductance: B =∇×A (2-83) To calculate the form of A, it is ﬁrst necessary to introduce the most basic law of magnetostatics, the Biot–Savart law , which describes how the B ﬁeld at a given point is produced by the moving charges in the vicinity of that point. During MAGNETOSTATICS 47 application of this law, we consider currents that are either static or very slowly varying with time. The Biot–Savart law is given by [Inan and Inan, 1998] dBp = µ0I dl × 4π R2 R (2-84) where I is the current, R a unit vector pointing from the location of the differential current element I dl to the point P , and R = |r − r | the distance between the current element and point P . Note that a primed quantity represents the position vector or the coordinates of the source points and the unprimed quantities represent the position vector or the coordinates of the point where B is being evaluated. Note that the cross product in (2-84) indicates that dB is perpendicular to both I dl and R, with the orientation being described by the right-hand rule. Also note that the ﬁeld generated by I dl will fall off with the square of the distance. Note that the current element I dl is a small part of a closed current loop and that an arbitrarily shaped loop can be constructed by the superposition of many such closed elemental loops of current, as shown in Figure 2-18. Subsequently, the Biot–Savart law can be written in terms of an integral: B = µ0 4π Idl × R C R2 (2-85) Careful observation of (2-85) allows us to calculate the form of the magnetic vector potential A. Equation (2-83) states that the magnetic ﬁeld is the curl of A. The trick is to manipulate (2-85) into the form of a curl so that A can be found. The right-hand term of (2-85) has a term that can be equated to the gradient of 1/R in spherical coordinates (see Appendix A): ∇ 1 R =R ∂ ∂R 1 R +θ 1 ∂ R ∂θ 1 R +φ 1 ∂ R sin θ ∂φ 1 R = −R 1 R2 (2-86) I dl Figure 2-18 Any current loop can be constructed from many elemental current loops dI . 48 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Substituting (2-86) into (2-85) gives B = µ0 −I dl × ∇ 1 4π C R (2-87) The del operator (∇) does not operate on source variables and therefore does not affect I dl . Therefore, it is acceptable to rewrite (2-87) in the form of a curl: B = ∇ × µ0I dl 4π C R (2-88) Also note that the negative sign has been eliminated because it indicates current direction, which is comprehended by the vector I dl . Comparing (2-88) to the deﬁnition of the magnetic vector potential in (2-83), we can deduce the form of A: A = µ0I dl 4π C R (2-89) 2.5.2 Inductance Suppose that two current loops are in close proximity to each other as shown in Figure 2-19. If a steady-state current (I1) is ﬂowing in loop 1, it produces a magnetic ﬁeld B1, as predicted by (2-85). If some of the magnetic ﬁeld lines pass though loop 2, the ﬂux passing through loop 2 is calculated using the form of equation (2-17): ψ2 = B1 · ds2 (2-90) Note that the ﬂux in loop 2 is proportional to B1 and is therefore also proportional to I1. This allows us to deﬁne a constant of proportionality, more commonly known as the mutual inductance: L21 ≡ ψ2 I1 (2-91) Substituting the (2-83) into (2-90) gives the ﬂux passing through loop 2 in terms of the magnetic vector potential, where s2 is the surface enclosed by loop 2: ψ2 = (∇ × A1) · ds2 (2-92) Now we can simplify using Stokes’ theorem and substitute (2-89) for A [Jackson, 1999]: ψ2 = (∇ × A1) · ds2 = µ0I1 dl1 4π C R · dl2 = µ0I1 4π dl R · dl2 (2-93) MAGNETOSTATICS 49 y2 s1 B1 s2 I I1 Figure 2-19 Mutual inductance caused by magnetic ﬂux from loop 1 passing though loop 2. Since ψ2 = L21I1 [from (2-91)], the mutual inductance between loops 1 and 2 is determined by dividing (2-93) by the current in loop 1: L21 = µ0 4π dl1 · dl2 R (2-94) Equation (2-94), called the Neumann formula, involves integration around both loops 1 and 2. Note two very important concepts that can be derived from equation (2-94). 1. The mutual inductance is a function of the size, shape, and distance between the two loops. 2. The mutual inductance from loop 1 to loop 2 (L21) is identical to the mutual inductance from loop 2 to loop 1 (L12). If we consider the implications of Faraday’s law (∇ × E + ∂B/∂t = 0), another very important concept used throughout signal integrity can be surmised. For simpliﬁcation, let’s assume that we only have a component of the electric ﬁeld in the x-direction and it is propagating along z so that E = axEx(z, t). Faraday’s law is then reduced to ∇ ×E+ ∂B ∂t = ay ∂ Ex ∂z + ∂B ∂t = 0 (2-95) Equation (2-95) says that a time-varying magnetic ﬂux B generated from loop 1 and passing through loop 2 will induce an electric ﬁeld and subsequently a 50 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY +− L12 dI2 dt L21 dI1 dt −+ I2 +− L2 dI2 dt −+ B1 L1 dI1 dt −+ V1 I1 Figure 2-20 Voltage sources induced by the time-varying currents in loop 1. voltage in loop 2. Faraday’s law can be rewritten in terms of circuit parameters: v2 = − dψ2 dt = −L21 d I1 dt (2-96) Therefore, every time you change the current in loop 1, an electromotive force (i.e., a voltage) is induced in loop 2, which causes current to ﬂow. In addition to inducing a voltage on loops in the vicinity, changing current in loop 1 will change the magnetic ﬂux ﬂowing through itself and, consequently, induce a voltage in itself. This is called the self-inductance: L11 ≡ ψ1 I1 (2-97) Similar to the mutual inductance, if the current changes a voltage is induced in the loop: v = −L11 d I1 dt (2-98) By observing (2-98), the units of inductance are determined to be volt · seconds per ampere, also known as henries. Figure 2-20 shows the voltage elements superimposed on the magnetically coupled loops corresponding to the mutual and self-inductance values calculated with (2-96) and (2-98). Note that the voltage sources were chosen to be positive and the negative sign is accounted for with the direction of current ﬂow. Example 2-5 Equivalent Circuit of a Magnetically Coupled System The magnetically coupled circuit can be understood more easily if conventional circuit theory is applied. Figure 2-21 shows the circuit for the magnetically coupled MAGNETOSTATICS 51 Rw I1 I2 Rw L1ddIt1 + − + − L2 dI2 dt V1 + − RL L12 dI2 dt + − + − L21 dI1 dt Figure 2-21 Circuit for Example 2-5, showing magnetically coupled loops. loops, assuming that the second loop is terminated in a resistor RL and each wire has a resistance of Rw. Kirchhoff’s voltage relations can be written for each loop: Loop 1: v1 = Rw I1 + L1 d I1 dt + L12 d I2 dt Loop 2: (Rw + RL)I2 + L2 d I2 dt + L21 d I1 dt = 0 Note that inductance is always a positive quantity that can be compared to a mass in a mechanical system. Large masses are difﬁcult to move, making it difﬁcult to accelerate in any direction. Similarly, the greater the inductance, the more difﬁcult it is to change the current because of the back emf (back voltage) generated in a direction to oppose the current, which is enforced by the negative sign in (2-98). This is called Lenz’s law . 2.5.3 Energy in a Magnetic Field If a circuit has a ﬁnite amount of inductance, it takes energy to make a current ﬂow because it requires work to overcome the back EMF voltage described by Lenz’s law. The work done on a charge against this back EMF is —v , from (2-98). The negative sign dictates that it is work being done against the EMF, not work done by the EMF. Since current is charge ﬂow per unit time, the work done per unit time is derived from (2-58) and given by W = −vq → dW = −vI = − dI L I dt dt (2-99) 52 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Integrating the second half of (2-99) will provide the total work done and thus energy in the magnetic ﬁeld in terms of the current and the inductance: W = − LI 2 joules 2 (2-100) Note that (2-100) only depends on the loop geometry (which is where the induc- tance comes from) and the steady-state current. It is often useful to represent the total energy in terms of the magnetic ﬁeld. To translate inductance, which is a circuit quantity to a ﬁeld quantity, equation (2-97) is used. L = ψ → ψ = LI I (2-101) Recalling equations (2-90) and (2-92) and invoking Stoke’s theorem as in (2-93), the ﬂux can be equated as follows: ψ = LI = B · ds = (∇ × A) · ds = A · dl (2-102) Therefore, the work calculated in (2-100) can be rewritten W = 1 2 LI 2 = 1 2 I A · dl = 1 2 (A · I ) dl (2-103) Expressing (2-103) in a volume integral allows us to use Ampe`re’s law (∇ × B/µ0 = J ) in place of I : W = 1 (A · I ) dl = 1 (A · J ) dV = 1 A · ∇ × B dV (2-104) 2 2V 2V µ0 Conjuring up more mathematical trickery from Jackson [1999] allows us to rewrite (2-104) in a simpliﬁed form using the following vector identity: ∇ · (A × B) = B · (∇ × A) − A · (∇ × B) (2-105) Rearranging the terms and utilizing the deﬁnition of the vector magnetic potential, A · (∇ × B) = B · (∇ × A) − ∇ · (A × B) ∇ ×A = B A · (∇ × B) = B · B − ∇ · (A × B) allows the work to be expressed in terms of the magnetic ﬁelds: Wm = 1 2µ0 [B · B − ∇ · (A × B)] dV V POWER FLOW AND THE POYNTING VECTOR 53 which can be simpliﬁed: Wm = 1 2µ0 B2 dV all space (2-106) This leads to the deﬁnition of the volume energy density, which expresses the stored energy in a magnetic ﬁeld: wm = B2 2µ0 joules/m3 2.6 POWER FLOW AND THE POYNTING VECTOR (2-107) As electromagnetic waves propagate through space, they carry power. The goal of this section is to derive a relationship between the electric and the magnetic ﬁeld vectors and the power transferred. To begin this derivation, the energy stored in both the electric and magnetic ﬁelds must be quantiﬁed within a volume of space. Consider Figure 2-22, which depicts a cubic volume of space with an electromagnetic plane wave propagating through it. To quantify the total power, which is the transfer of energy over time, propagating through the cube, all sources of energy must be accounted for. The power balance equation can be expressed as PA = PS − PL − PEM (2-108) where PA is the power ﬂowing through the surface of the far end of the cube of space with area A, as shown in Figure 2-22, PS represents any sources of power within the cube, PL represents the losses within the cube that dissipate power as heat (such as resistive losses), and PEM represent the power contained within the electromagnetic waves that propagate into the cube. If we assume a source-free and loss-free medium, the power ﬂowing into the cube must equal the power ﬂowing out of the cube, where the minus sign convention is chosen because the power is ﬂowing out of the surface: PA = −PEM (2-109) Power in PA x z1 z2 Power out z PEM y Surface area=A Figure 2-22 Volume of space used to calculate power ﬂow and the Poynting vector. 54 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY To quantify (2-109) in terms of the electric and magnetic ﬁelds, we begin with the volume energy densities for an electric and a magnetic ﬁeld, which were derived earlier and listed here for convenience. we = ε E2 2 wm = B2 2µ0 joules/m3 joules/m3 (2-75) (2-107) Integrating the sum of (2-75) and (2-107) over the volume of the cube will give total energy: WEM = A z2 1 z1 2 ε0E2 + 1 B2 µ0 dz (2-110) Since power is energy transfer per unit time interval, PEM is calculated by taking the time derivative: ∂ ∂t WEM = ∂ ∂t A z2 1 z1 2 ε0E2 + 1 µ0 B2 dz (2-111) Deﬁning the ﬁelds in time-harmonic form lets us express (2-111) in terms of the partial time derivatives: E = E0e−jωt ∂ ∂t E = −j ωE0e−jωt ∂ E2 ∂t = −j 2ωE02e−j2ωt = 2E0e−jωt (−j ωE0e−jωt ) = 2E0 e−j ωt ∂ E2 = 2E ∂ E ∂t ∂t ∂ ∂t E0 e−j ωt Similarly, ∂ B2 = 2B ∂ B ∂t ∂t which allows (2-111) to be rewritten as ∂ ∂t WEM = A z2 z1 ∂ ε0E ∂t E + 1 B µ0 ∂ ∂t B dz (2-112) Now, for a plane wave propagating in the z-direction, the contributions of the ﬁelds in the y-direction were calculated earlier with equation (2-29): ay ∂Ex = −µ ∂Hy ∂z ∂t = ay ∂Ex = − ∂By ∂z ∂t (2-29) POWER FLOW AND THE POYNTING VECTOR 55 and the contributions in the x-direction are given by ax ε ∂Ex = − ∂Hy ∂t ∂z = ax µε ∂Ex = − ∂By ∂t ∂z (2-30) Note that J = 0 since it is assumed that there are no sources in the volume under consideration. Substituting (2-29) and (2-30) into (2-112) allows us to simplify the equation: ∂ ∂t WEM z2 =A ε0Ex z1 z2 = −A Ex z1 − 1 ∂By µ0ε0 ∂z ∂ Hy ∂z + Hy + 1 µ0 By − ∂Ex ∂z ∂Ex dz ∂z dz (2-113) Note that the form of the integrand in (2-113) looks like a curl multiplied by a vector. This tells us that (2-113) might be simpliﬁed if we use the vector identity introduced earlier: ∇ · (E × H ) = H · (∇ × E) − E · (∇ × H ) (2-105) Note that from (2-24), ax(∂Hy/∂z) = −∇ × H and ay(∂Ex/∂z) = ∇ × E. Therefore, E · (∇ × H ) = − ∂Hy ∂z H · (∇ × E) = ∂Ex ∂z This allows us to rewrite (2-113) in terms of the cross product of the electric and magnetic ﬁelds: z2 z2 PEM = −A (H · ∇ × E − E · ∇ × H ) dz = −A ∇ · (E × H ) dz z1 z1 (2-114) Note that the ﬁnal term in (2-114) is equivalent to a volume integral, where A dz = dV . This allows us to use the divergence theorem to eliminate the del operator: (∇ · F ) dV = F · ds (divergence theorem) V S z2 PEM = −A (∇ · (E × H )) dz = − ∇ · (E × H ) dV z1 V = − (E × H ) · ds S 56 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY This allows us to deﬁne the Poynting vector, which represents the ﬂow of power per unit area through a surface S at an instant in time. S = E × H W/m2 (2-115) Note that the direction of the power ﬂow is perpendicular to both E and H . 2.6.1 Time-Averaged Values When considering the electromagnetic power delivered by a sinusoidal time-varying ﬁeld, practical measurement considerations tend to favor the time-averaged value of the power rather than the instantaneous value described in (2-115). This is because the time-averaged power entering a passive network, as measured with a watt-meter, is a measure of the power dissipated by heat in all the resistive circuit elements. In the laboratory, the time average of a time-harmonic function is taken over an interval of many periods. For a steady-state sinusoidal function, the average of one period will be the same as the average over many periods, since each period looks identical. The time average of the Poynting vector is deﬁned as the area under the function for one period, divided by the duration of the cycle: Save = Aperiod Tperiod = 1 T T S(x, y, z, t) dt 0 (2-116) Although the math is not shown here, Jackson [1999] shows the derivation of the time average value for the Poynting vector: Save = 1 2 Re(E × H ∗) W/m2 (2-117) where the star represents the complex conjugate. It is sometimes useful to represent the magnitude of the Poynting vector in terms of the electric or magnetic ﬁelds and the intrinsic impedance deﬁned in equation (2-53). Integrating sinusoidal functions as described in (2-116) will yield the time-averaged Poytning vector in terms of the E and H ﬁelds [Johnk, 1988]: E = axE+ cos(ωt − βz) H = ay H+ η cos(ωt − βz) S = E × H = [axE+ cos(ωt − βz)] × ay E+ η cos(ωt − βz) = az (E+) 2 η cos2(ωt − βz) (2-118) (2-119) (2-120) REFLECTIONS OF ELECTROMAGNETIC WAVES 57 To obtain the time-averaged value of (2-120), remember that sin2 θ + cos2 θ = 1. Therefore, over a complete cycle, the average of cos2 θ is equal to the average of sin2 θ and sin2 θ = cos2 θ = 1 2 . Subsequently, the time-averaged value of a cosine-squared term over a complete cycle is 1 2 : (E+)2 Save = az 2η (2-121) Similarly, the Poynting vector can be expressed in terms of the magnetic ﬁeld: S = E × H = [axηE+ cos(ωt − βz)] × [ayH + cos(ωt − βz)] = azη(H +)2 cos2(ωt − βz) (2-122) Save = (H +)2 azη 2 2.7 REFLECTIONS OF ELECTROMAGNETIC WAVES So far, we have considered the propagation of electromagnetic waves in a simple, inﬁnitely large medium. However, most practical problems involve waves propagating in multiple dielectric media. Since each medium will have different electric characteristics, it is essential to understand how a propagating electromagnetic wave will behave when it enters a region where the properties of the medium change. Generally, when an electromagnetic plane wave propagating in medium A enters region B, where the properties of the dielectric change, two things happen: (1) a portion of the wave is reﬂected away from region B, and (2) a portion of the wave is transmitted into region B. When these plane waves encounter planar interfaces, both the reﬂected and transmitted waves are also planar, so their directions, amplitudes, and phase constants can easily be calculated. The simultaneous existence of both the transmitted and reﬂected waves is a direct result of the boundary conditions that must be satisﬁed when solving Maxwell’s equations at the interface between the two regions. We begin with a plane wave incident on a perfect conductor. 2.7.1 Plane Wave Incident on a Perfect Conductor Consider a plane wave propagating in medium A in the z-direction. Assume that the medium in region A is a simple, loss-free medium and medium B is a perfectly conducting metal plane, as shown in Figure 2-23. Assume that the electric ﬁeld is oriented in the x-direction, necessitating that the magnetic ﬁeld 58 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Region A (Dielectric media) x z y Region B (Perfect conductor) Ei Incident wave Hi Direction of propagation Figure 2-23 Incident electromagnetic wave propagating in dielectric region A and impinging on a perfect conductor. be oriented in the y-direction. The incident ﬁelds propagating in medium A are given by Ei (z) = ax Ei e−j √ ωz µ0 ε0 Hi (z) = ay 1 η Ei e−j √ ωz µ0ε0 ej ωt (2-123) (2-124) Since the boundary is normal to the incident waves, the reﬂected portion will be reﬂected back in the −z-direction. Subsequently, the reﬂected electric ﬁeld will have the form Er (z) = ax Er e−j ω(−z)√µ0ε0 = ax Er e+j √ ωz µ0 ε0 (2-125) Since the electric ﬁeld of the reﬂected wave is propagating in the −z-direction, the magnetic ﬁeld must also ﬂip and point in the −y-direction, to maintain the proper relationship between the electric and magnetic ﬁelds: Hr (z) = −ay 1 η Er e+j √ ωz µ0ε0 ej ωt (2-126) To calculate the values of the reﬂected and transmitted waves we must consider what happens when an electromagnetic ﬁeld impinges on a perfect conductor. Equation (2-7) implies that an electric ﬁeld impinging on a conductor with conductivity σ will produce a current density J : J = σE (2-7) REFLECTIONS OF ELECTROMAGNETIC WAVES 59 For a perfect conductor, σ = ∞. However, since the real world limits currents to ﬁnite densities, equation (2-7) implies that inﬁnite conductivity will produce inﬁnite currents inside a perfect conductor, which is impossible; therefore, the electric ﬁeld inside the perfect conductor must be zero. If the electric ﬁeld is zero, the magnetic ﬁelds must also be zero. This allows us to deduce the boundary conditions for both E and H at the interface between a dielectric medium and a perfect electrical conductor (PEC). Since the ﬁelds in region A are ﬁnite and the ﬁelds in region B (the PEC) are zero, we can deduce that the wave impinging on the conductor must induce a wave equal but opposite to the incident wave at the surface, so the ﬁelds in the conductor are zero. The boundary condition at the surface (z = 0) of the PEC requires that the tangential electric ﬁeld must vanish for all x and y to ensure that the electric ﬁeld inside the conductor is zero. Applying the boundary conditions to both the incident and the reﬂected portions of the electric ﬁeld gives E(z = 0) = Ei(z = 0) + Er (z = 0) = axEi + ax Er = 0 which produces the relationship between the incident and reﬂected electric ﬁelds for a electromagnetic wave impinging on a PEC: ax Ei = −ax Er (2-127) This means that when an electromagnetic wave is incident normal to a perfect electrical conducting plane traveling in the +z-direction, it will experience a 100% reﬂection back toward the −z-direction with the same magnitude as the incident wave with a negative amplitude. This is the same as saying that the magnitude of the reﬂected wave will remain constant but the phase will be shifted by 180◦. This is shown in Figure 2-24. Region A (Dielectric media) Ei Incident wave x z y Region B (Perfect conductor) Hi Hr Reflected wave Er Figure 2-24 Incident electromagnetic wave propagating in dielectric region A and impinging on a perfect conductor, showing that 100% of the wave is reﬂected back in the −z -direction. 60 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY Region A Incident wave Region B x z y Reflected wave Transmitted wave Figure 2-25 Incident electromagnetic wave propagating in dielectric region A and impinging on dielectric region B , showing that a portion of the wave is reﬂected back in the −z -direction and a portion is transmitted in the +z -direction. 2.7.2 Plane Wave Incident on a Lossless Dielectric In Section 2.7.1 we considered the special case where a propagating electromagnetic plane wave was incident on a plane boundary where the second medium was a perfect conductor. Now we consider the more general case where the second medium is a lossless dielectric. When a plane wave impinges on an area with a different dielectric, a portion of the wave is reﬂected and a portion of the wave in transmitted into the new medium, where it continues to propagate, as shown in Figure 2-25. If we assume that the incident ﬁelds are represented by phasors, we can separate the total plane wave into three parts: 1. Incident wave: 2. Reﬂected wave: Ei (z) = ax Ei e−jβAz Hi (z) = ay 1 ηA Ei e−jβAz Er (z) = ax Er e+jβAz Hr (z) = −ay 1 ηA Er e+jβAz REFLECTIONS OF ELECTROMAGNETIC WAVES 61 3. Transmitted wave: Et (z) = ax Et e−jβB z Ht (z) = ay 1 ηB Et e−jβB z where βA = √ ω µAεA, βB = √ ω µB εB , ηA = √ ω µA/εA, and ηB = √ ω µB /εB (where µx = µrxµ0 and εx = εrxε0) are the phase constants and intrinsic impedances of regions A and B, respectively. When a wave intersects a boundary between two lossless dielectric regions, the tangential components of both the electric and magnetic ﬁelds across the interface must remain continuous. In other words, the tangential component of the ﬁelds cannot change instantaneously. In our particular scenario, the plane wave is propagating in TEM mode in the z-direction, so both the electric and magnetic ﬁelds are oriented parallel (i.e., tangent) to the boundary of the dielectric interface. Subsequently, we can say that at the interface (z = 0), the sum of the incident and reﬂected waves must equal the transmitted wave: Et (z = 0) = Ei(z = 0) + Er (z = 0) → Et = Ei + Er Ht (z = 0) = Hi(z = 0) + Hr (z = 0) → Et η2 = Ei η1 − Er η1 (2-128) (2-129) Since the incident waves are known, we can solve (2-128) and (2-129) simultaneously for the transmitted and reﬂected portions of the wave: Et = Ei 2ηB ηB + ηA Er = Ei ηB ηB − ηA + ηA (2-130) (2-131) Equations (2-130) and (2-131) lead to the deﬁnition of the reﬂection and transmission coefﬁcients: ≡ Er = ηB − ηA Ei ηB + ηA T ≡ Et Ei = 2ηB ηB + ηA =1+ (2-132) (2-133) The reﬂection coefﬁcient is a measure of how much of the wave is reﬂected back off the intersection between the two media, and the transmission coefﬁcient tells how much of the wave is transmitted. If the reﬂection coefﬁcient is zero, it means that the intrinsic impedance in the two regions is identical. If the intrinsic impedances are not equal, the reﬂection coefﬁcient will be ﬁnite. These 62 ELECTROMAGNETIC FUNDAMENTALS FOR SIGNAL INTEGRITY equations are used heavily in Chapter 3 when analyzing multiple reﬂections on transmission-line structures. REFERENCES Huray, Paul G., 2009, Foundations of Signal Integrity, Wiley, Hoboken, NJ. Inan, Umran S., and Aziz S. Inan, 1998, Engineering Electromagnetics, Addison Wesley Longman, Reading, MA. Jackson, J. D., 1999, Classical Electrodynamics, 3rd ed., Wiley, New York. Johnk, Carl T. A., 1988, Electromagnetic Theory and Waves, Wiley, New York. PROBLEMS 2-1 For a TEM plane wave traveling in the z-direction where Ex = 60 V/m, if λ = 20 cm and ν = 1.5 × 108, determine the frequency of the wave, the relative dielectric permittivity (εr ), the phase constant, and the intrinsic impedance. 2-2 For the wave in Problem 2-1, write formulas for the time-domain expressions for the E and H ﬁelds, and plot them. 2-3 For the plane wave in Problem 2-1, how much power is being transported? 2-4 If a wire with a diameter of 1 in. is carrying a dc current of 1000 A, how much power would be delivered to the 100- resistive load in the loop shown in Figure 2-26? R = 100 ohms 1 meter 1 meter I = 1000 A 10 meter Figure 2-26 2-5 If two identical point charges are separated by 10 in. in free space and are experiencing a repulsion force of 1 N, what is the magnitude of each charge? PROBLEMS 63 2-6 Assume that two inﬁnitely long parallel wires are separated by a distance of 5 in. with currents that are equal but opposite. Find the current if the force per unit length is 0.25 N/m. 2-7 Determine the energy needed to assemble three point charges in free space from inﬁnity to the following (x,y,z) coordinates: (0,0,0), (1 mm, 0, 0), (−1 mm, 0, 0). 2-8 For a parallel-plate capacitor in free space (εr = 1) with total change Q on one plate and—Q on the other with plate areas of A separated by the distance d, calculate the external force required to keep the plates stationary. 2-9 If a plane wave is propagating in the z-direction in free space and it impinges on a planar dielectric medium where εr = 9.6, how much of the wave is propagated? How much is reﬂected? If the electric ﬁeld incident on the boundary has a value of 60 V/m, how much power is carried in the initial wave? How much power is reﬂected and transmitted? Show that power is conserved. 2-10 If the electric ﬁeld in an area of space is known to be E = br3ar (where b is a constant) in spherical coordinates, ﬁnd the charge density ρ and the total charge within a sphere of diameter d. 3 IDEAL TRANSMISSION-LINE FUNDAMENTALS 3.1 Transmission-line structures 66 3.2 Wave propagation on loss-free transmission lines 67 3.2.1 Electric and magnetic fields on a transmission line 68 3.2.2 Telegrapher’s equations 73 3.2.3 Equivalent circuit for the loss-free case 76 3.2.4 Wave equation in terms of LC 80 3.3 Transmission-line properties 82 3.3.1 Transmission-line phase velocity 82 3.3.2 Transmission-line characteristic impedance 82 3.3.3 Effective dielectric permittivity 83 3.3.4 Simple formulas for calculating the characteristic impedance 85 3.3.5 Validity of the TEM approximation 86 3.4 Transmission-line parameters for the loss-free case 90 3.4.1 Laplace and poisson equations 91 3.4.2 Transmission-line parameters for a coaxial line 91 3.4.3 Transmission-line parameters for a microstrip 94 3.4.4 Charge distribution near a conductor edge 100 3.4.5 Charge distribution and transmission-line parameters 104 3.4.6 Field mapping 107 3.5 Transmission-line reﬂections 113 3.5.1 Transmission-line reﬂection and transmission coefﬁcient 113 3.5.2 Launching an initial wave 116 3.5.3 Multiple reﬂections 116 3.5.4 Lattice diagrams and over- or underdriven transmission lines 118 3.5.5 Lattice diagrams for nonideal topologies 121 3.5.6 Effect of rise and fall times on reﬂections 129 3.5.7 Reﬂections from reactive loads 129 3.6 Time-domain reﬂectometry 134 3.6.1 Measuring the characteristic impedance and delay of a transmission line 134 Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 65 66 IDEAL TRANSMISSION-LINE FUNDAMENTALS 3.6.2 Measuring inductance and capacitance of reactive structures 137 3.6.3 Understanding the TDR proﬁle 140 References 140 Problems 141 Transmission lines are simple electrical structures that consist of an insulating layer of dielectric material sandwiched between two layers of metal (usually, a nonferrous metal such as copper). In high-speed digital design, transmission lines are used for communicating between electrical components, such as a microprocessor and a memory module, or chip set. In modern designs such as computer systems, the data transfer rates are so high that the width (in time) of the digital pulses are small compared to the time it takes to propagate a signal from one end of a transmission line to another. As a result, it is quite common to have more than one bit of information propagating on the transmission line at any given instant in time, where the transmission line is essentially “storing” information before the receiving circuitry can latch the data. Consequently, to preserve the quality of the digital pulse stream to a degree where it can be captured with no errors at the receiving agent, great attention must be given to the construction and design of the transmission lines so that the electrical characteristics are controlled and predictable. Designing a successfull bus for high-data-rate information transfer requires a thorough understanding of how the signals propagate on transmission lines. In this chapter we introduce basic transmission-line structures typically used in digital systems and present fundamental transmission-line theory for the ideal case. 3.1 TRANSMISSION-LINE STRUCTURES Transmission lines come in many shapes and sizes. One of the most common transmission lines is the coaxial cable, which is used with cable television. When designing a high-speed digital system, transmission lines are usually manufactured on a printed circuit board (PCB) or a multichip module (MCM), which typically consists of conductive traces buried in or attached to a dielectric with one or more reference power and ground planes. The metal typically used is copper (although the copper is often electroplated with silver or nickel to prevent corrosion) and the dielectric is often FR4, a ﬁberglass–resin composite discussed in detail in Chapter 6. The two most common types of transmission lines used in digital designs are microstrips and striplines. A microstrip is routed on an outside layer of the PCB and has only one reference plane. There are two types of microstrips, buried and nonburied. A buried (sometimes called embedded ) microstrip is simply a transmission line that is embedded into the dielectric but WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 67 Microstrip PCB substrate Packaged integrated circuit Stripline t Cross-sectional view taken here w Via Transmission line cross sections t Copper plane w Signal (microstrip) Ground/power Signal (stripline) Signal (stripline) Ground/power Signal (microstrip) Figure 3-1 Typical application of transmission lines in a digital design. still has only one reference plane. A stripline is routed on an inside layer and has two reference planes. Figure 3-1 represents a PCB with traces routed between the various components on both internal (stripline) and external (microstrip) layers. The accompanying cross section is taken at the given mark so that the position of the transmission lines relative to the ground and power planes can be seen. Transmission lines in the book are often represented in cross-section form because they are useful for calculating and visualizing the various transmission-line parameters described. Multiple-layer PCBs such as the one depicted in Figure 3-1 can provide a variety of stripline and microstrip structures. Control of the conductor and dielectric layers (which is referred to as the stackup) is required to make the electrical characteristics of the transmission line predictable. These basic electrical characteristics are deﬁned in this chapter and referred to as transmission-line parameters. Figure 3-2 shows several transmission-line structures discussed in the text [Hall, 2000]. 3.2 WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES Transmission lines are designed to guide electromagnetic waves from one point to another for the purpose of information transfer. For practical applications, the electromagnetic wave can be approximated as planar because the wavelength is usually much larger than the electrical delay across the width of the transmission line. This means that there will be no variation in the electric or magnetic 68 IDEAL TRANSMISSION-LINE FUNDAMENTALS (a) (b) (c) (d) (e) (f) Figure 3-2 Common transmission-line structures: (a) balanced stripline; (b) asymmetrical stripline; (c) microstrip line; (d) buried microstrip line; (e) coaxial line; (f) slotline. ﬁelds across the width of the transmission line. The approximation of a planar electromagnetic wave will become inadequate when the transmission-line width is no longer small compared to the wavelength. For a typical transmission line on a common PCB used in motherboard designs for personal computers, the bandwidth of the plane-wave approximation can be estimated using (2-45) and (2-52) by calculating the frequency where the wavelength equals the width of the conductor. For typical applications, the dielectric constant of FR4 (εr = 4) and a trace width of 5 mils (127 µm) can be used to calculate the frequency where the wavelength equals the width of the transmission-line signal conductor: fplane = √ c/ εr λ = √ 3.0 × 108 m/s/ 4 127 × 10−6 m = 1181 GHz The fundamental frequency of the fastest digital design in a product at the time of this writing is less than 20 GHz. Consequently, the analysis of electromagnetic plane waves described in Section 2.3 applies for digital transmission lines because f fplane. 3.2.1 Electric and Magnetic Fields on a Transmission Line On a transmission line, information is transferred from one component to the other by guiding electromagnetic energy from point A to point B. To gain an intuitive understanding of how a signal propagates on a transmission line, we WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 69 must understand how the electric and magnetic ﬁeld patterns are distributed. To do this, we ﬁrst derive the boundary conditions of the electric ﬁeld at the interface between a dielectric and a perfect conductor, which allows us to draw the electric ﬁeld patterns for typical transmission line cross sections in a uniform dielectric. Next, we explore how the electric ﬁeld behaves at a dielectric boundary, such as the interface between the air and the board material in a microstrip transmission line. Finally, we use the relationships for a TEM wave derived in Chapter 2 to obtain the magnetic ﬁeld pattern from the electric ﬁeld. When a voltage is applied between the signal conductor and the reference plane on one end of a transmission line, an electric ﬁeld is established as described in Section 2.4. The voltage is calculated by the line integral of the electric ﬁeld between the signal conductor and the reference plane: b v = − E · dl a (3-1) As discussed in Section 2.3.2, once the electric ﬁeld is established, the properties of the magnetic ﬁeld can be calculated using Faraday’s and Ampe`re’s laws (2-1) and (2-2). From Section 2.3.2, the relationship between the electric and magnetic ﬁelds was shown always to remain orthogonal: ay ∂Ex = −µ ∂Hy ∂z ∂t ax ε ∂Ex = − ∂Hy ∂t ∂z (2-29) (2-30) Note that since we are applying a voltage source to generate the electric ﬁeld, and voltage is deﬁned in terms of the work done per unit charge, equation (2-58) implies that charges must be present on both the signal and the reference conductors: W b = v(b) − v(a) = − E · dl q a→b a (2-58) Furthermore, Gauss’s law (∇ · εE = ρ) states that the divergence of εE is nonzero and equal to the charge density, which implies that the sources of the electric ﬁeld are the electrical charges. Equivalently, if the electric ﬁeld terminates abruptly, the termination must be an electric charge. To understand how the electric ﬁeld behaves at the interface between the dielectric and a conducting surface, we use the integral form of Gauss’s law to calculate the boundary conditions: εE · ds = ρdV = Qenc S V (2-59) 70 IDEAL TRANSMISSION-LINE FUNDAMENTALS n A Dielectric h (Region 1) Conductor (Region 2) n Figure 3-3 Surface used to calculate the boundary conditions at the dielectric– conductor interface. First, we must choose a surface to integrate and compare the normal components of εE on both sides of the conductor–dielectric boundary. A convenient surface is a cylinder, as depicted in Figure 3-3. Since we are observing the behavior of ﬁelds at the surface, the height (h) of the cylinder can be made inﬁnitely small, reducing (2-59) to εE · ds = ε1E1 · ds1 + ε2E2 · ds2 = (n · ε1E1)A + (−n · ε2E2)A = ρA S S S (3-2) where ε1 is the dielectric permittivity of the dielectric in region 1 and ε2 describes the dielectric permittivity of region 2, which in this example is a perfect con- ductor. To interpret this equation, recall the discussion in Section 2.7.1, where it was shown that the electric ﬁeld (E2 in this case) must be zero inside a perfect conductor. Setting E2 = 0 allows us to simplify (3-2) for the special case of a boundary between a dielectric and a perfect conductor: n · εE = ρ C/m2 (3-3) Equation (3-3) means that the electric ﬁeld must emanate normal from and terminate normal to the conductor surface. Since equations (2-29) and (2-30) say that the magnetic ﬁeld must be orthogonal to the electric ﬁeld, we can conclude that the magnetic ﬁeld must be tangential to the conductor surface. These two rules make it easy to visualize electric and magnetic ﬁelds on transmission-line structures with perfect electrical conductors. Simply draw the electric ﬁeld lines so that they are always perpendicular to the conductor surface, emanating from the high and terminating in the low potential conductor, and then draw the magnetic ﬁeld lines so that they are always perpendicular to the electric ﬁeld lines WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 71 (a) Electric field Magnetic field (b) Figure 3-4 Electric and magnetic ﬁeld patterns for homogeneous dielectrics in (a) a stripline and (b) a microstrip. er = 1 er > 1 Figure 3-5 How the electric ﬁeld behaves at a dielectric boundary. and tangential to the conductor surfaces. If we assume that the voltage is applied with the positive value on the signal conductors as shown in Figure 3-4, we can draw the electric and magnetic ﬁelds for various transmission-line structures assuming a homogeneous dielectric. When the dielectric is not homogeneous, as is almost always the case with microstrip transmission lines, the ﬁeld lines are distorted as they cross dielectric boundaries. Figure 3-5 illustrates how the electric ﬁeld lines are bent away from the normal to the dielectric boundary when the relative dielectric permittivity on the top half of the structures is smaller than that on the bottom. To understand why the ﬁeld lines are distorted, we must examine how electrostatic ﬁelds behave at the boundaries between two dielectric regions. First, 72 IDEAL TRANSMISSION-LINE FUNDAMENTALS we can recycle equation (3-2) to derive the boundary conditions of the normal components of the electric ﬁeld at the interface between two dielectrics: (n · ε1E1) − (n · ε2E2) = ρ (3-4) If the surface charge density between the dielectric layers is assumed to be zero (usually, a valid assumption), the relationships between the electric ﬁelds in both regions is described by n · ε1E1 = n · ε2E2 (3-5) Equation (3-5) means that the normal component of the electric ﬁeld is not continuous across a dielectric boundary. In Section 2.7.2 it was mentioned that the tangential component of the electric ﬁeld must remain continuous across a dielectric boundary. This can be shown with the integral form of Faraday’s law for the electrostatic case: E · dl = 0 l (3-6) If we integrate (3-6) around a closed differential contour that encompasses the dielectric boundary such as that shown in Figure 3-6, we can calculate the tangential components of the electric ﬁeld: b c d a E · dl = E · dl + E · dl + E · dl + E · dl l a b c d (3-7) Since we are considering the behavior at the surface ( h → 0), segments da and bc can be eliminated. Furthermore, the tangential segments ab and cd are equal but opposite, which means that (3-7) can be simpliﬁed to (E1t − E2t ) l = 0 → E1t = E2t (3-8) Equation (3-8) means that the tangential components of the electric ﬁeld across a dielectric boundary must remain continuous. Assume that an electric ﬁeld E1 is incident on a boundary between two dielectrics as shown in Figure 3-7. The change in the orientation of the electric ∆l a b e1 ∆h e2 d c Figure 3-6 Differential contour encompassing a dielectric boundary. WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 73 e1E1 q1 e1 e2 q2 e2E1 Figure 3-7 Change in the orientation of an electric ﬂux line at a dielectric boundary. ﬂux lines across the interface can be calculated by using the boundary conditions for the normal and tangential components of the electric ﬁeld derived in (3-5) and (3-8). When θ1 = 0, the boundary conditions of (3-5) apply. Consequently, we need a function of θ1 that will satisfy (3-5) when the electric ﬁeld is normal to the boundary. Since cos(0) = 1, the following equation will satisfy the boundary conditions for the normal components of the electric ﬁeld: ε1E1 cos θ1 = ε2E2 cos θ2 (3-9) Similarly, when θ1 = 90◦, the boundary conditions of (3-8) apply, and since sin(90◦) = 1, the following equation will satisfy the boundary conditions for the tangential components of the electric ﬁeld: E1 sin θ1 = E2 sin θ2 (3-10) If E1 is calculated from (3-10) and substituted into (3-9), the change in orientation that the electric ﬁelds experience at a dielectric boundary can be calculated: θ2 = tan−1 ε2 ε1 tan θ1 (3-11) Equation (3-11) means that the ﬁeld lines will be bent farther away from the normal to the dielectric interface in the medium with the higher permittivity. 3.2.2 Telegrapher’s Equations So far, we have concentrated primarily on the derivation and calculation of the electric and magnetic ﬁelds. However, since this is a book targeted primarily at 74 IDEAL TRANSMISSION-LINE FUNDAMENTALS engineers, we want to express the effect of the ﬁelds in terms that allow us to use circuit theory. The telegrapher’s equations express the properties of the electric and magnetic ﬁelds of transmission lines with simple equivalent circuits. This simpliﬁes the analysis dramatically and provides a well-known mechanism for engineers to understand transmission-line behavior. Since a transmission line is a structure intended to guide a planar electromagnetic wave down a speciﬁc path, we begin the derivation of the telegrapher’s equations by observing the relationships described in Section 2.3.2 for a plane wave propagating in the z-direction in the TEM mode. First, let’s examine equation (2-29), which says that a time-varying magnetic ﬁeld will produce an electric ﬁeld: ay ∂Ex = −µ ∂Hy ∂z ∂t (2-29) Equation (2-29) is a written in general terms assuming that the electromagnetic wave is propagating in an inﬁnite space. In reality, the electric and magnetic ﬁelds are conﬁned to an area dictated by the transmission-line geometry, so (2-29) can be simpliﬁed by calculating the equivalent-circuit parameters. First, we calculate the voltage between the signal conductor and the reference plane from the electric ﬁeld using equation (3-1), assuming that point a is on the signal conductor and point b is the position on the reference plane directly below. b v = Ex · dl a Consequently, for given transmission-line geometry, the left-hand part of (2-29) is equivalent to the partial derivative of the voltage between the signal conductor and the reference plane with respect to z: ∂Ex → ∂ E · dl = ∂v(z, t) ∂z ∂z ∂z Similarly, the right-hand side of equation (2-29) can be related to the inductance as discussed in Section 2.5.2. Since B = µH and the magnetic ﬂux is a function of B, ψ = B · ds (where ds is deﬁned by the transmission-line geometry), equation (2-97) can be used to write the magnetic ﬂux in terms of the current and inductance (ψ = IL): L11 ≡ ψ1 I1 (2-97) WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 75 Therefore, for a given transmission-line geometry, the right-hand side of (2-29) can be written in terms of circuit parameters, where L is the series inductance of the transmission line per unit length: ∂µHy → ∂ By · ds = ∂ψ = L ∂i ∂t ∂t ∂t ∂t Don’t get confused by the fact that the units of ∂µHy/∂t are HA/m2 (henry · amperes per square meter) and those of L(∂i/∂t) are H · A/m. Equation (2-29) is in terms of generalized electric and magnetic ﬁelds. When the ﬁelds are repre- sented by circuit parameters, the units change. Therefore, the circuit equivalent of (2-29) is ∂v(z, t) = −L ∂i(z, t) ∂z ∂t (3-12) Note that equation (3-12) is simply the classic response of an inductor from circuit theory. Similarly, equation (2-30) says that a time-varying electric ﬁeld will produce a magnetic ﬁeld: ax ε ∂Ex = − ∂Hy ∂t ∂z (2-30) Note that the form of the numerator on the left-hand side of equation (2-30) (εE) has units of F · V/m2 (farad · volts per square meter), which indicates that the equivalent-circuit form should be in terms of a capacitance and a voltage. From the integral form of Gauss’s law (2-59) we can calculate the left-hand side of (2-30) in terms of the charge, assuming that the signal conductor is an inﬁnite thin strip: QA Q εE · ds = S ρdV → εExA = V A → εEx = A (3-13a) where dV refers to volume. For a given transmission-line geometry, the voltage between the conductors from point a on the signal conductor to point b on the reference plane is calculated with (3-1), which has units of volts: b Q v= a Ex · dl = d εA = Ex d (3-13b) where A is the area of the conductors where the electric ﬁeld is established and d is the distance between the signal conductor and the reference plane. Therefore, since Q = Cv, εEx = Q A → Cv A = C b Ex · dl a 1 = CExd = C v → Cv AA l (3-14) 76 IDEAL TRANSMISSION-LINE FUNDAMENTALS where C is normalized to length with units of farads per meter. As a result, for a speciﬁc transmission-line geometry, we can write the left-hand side of (2-30) in terms of the capacitance: ε ∂Ex → ∂v C ∂t ∂t (3-15) To derive the circuit equivalent of the right side of (2-30), recall from equation (2-79) that current is the rate of charge ﬂow per second: i = dQ dt (3-16) Substituting Q = Cv into equation (3-16) produces the current in terms of the voltage and capacitance: i = dv C dt (3-17) which is equal to equation (3-15). Therefore, (2-30) can be rewritten in terms of circuit parameters: ∂i(z, t) = −C ∂v(z, t) ∂z ∂t (3-18) Note that (3-18) is the classic response of a capacitor from circuit theory. Equations (3-12) and (3-18) are the loss-free forms of the telegrapher’s equations, which describe the electrical characteristics of a transmission line. 3.2.3 Equivalent Circuit for the Loss-Free Case Although signal integrity is largely a study in electromagnetic theory, application of the discipline is performed almost entirely using circuit parameters because they are more intuitive to most engineers. Consequently, it is necessary to derive a model for the transmission line in terms of the equivalent inductance L and the capacitance C per unit length. In this section the equivalent circuit of a transmission line is developed for the loss-free case. The model is reﬁned in Chapters 5 and 6 to include loss from nonperfect dielectrics and ﬁnite conductivity conductors. To begin, consider a differential element of transmission line with a length of z as shown in Figure 3-8, which represents a section of a transmission line with a signal conductor and a reference conductor. If we assume that current is traveling down the signal conductor and returning on the reference conductor (remember from circuit theory that current must always complete a loop along a return path), it can be represented by a series of differential current elements, WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 77 B I ∆I B E I ∆I E V+ Signal Conductor ∆I E Vref Reference Conductor Figure 3-8 Segment of a transmission line showing the magnetic and electric ﬁelds, the associated differential current loops, and the voltage potentials at each conductor. as was shown in Figure 2-18. When representing the total current as a series of small loops, the adjacent vertical components of the current elements cancel each other out, leaving a net current of I on the signal conductor and (−I ) on the reference conductor. As described in equation (2-97), a current change in a loop will change the magnetic ﬂux and will thus induce a self-inductance. Consequently, the magnetic ﬁeld of a transmission line is represented by a series inductor in the circuit model . The value of the equivalent-circuit inductance is given by L z = zL (3-19) where z is the length of the differential section of transmission line and L is the inductance per unit length. Similarly, when a voltage is applied to the transmission line between the signal conductor and the reference plane as shown in Figure 3-8 (v = V + − Vref), an electric ﬁeld is established in units of volts per meter: b v(b) − v(a) = − E · dl a (2-58) The presence of an electric ﬁeld implies that charges exist on the conductors and consequently implies the existence of a capacitance (as described in Section 2.4.3). Therefore, the electric ﬁeld of a transmission line is represented by a shunt capacitor in the circuit model . The value of the equivalent-circuit capacitance is given by C z = zC (3-20) 78 IDEAL TRANSMISSION-LINE FUNDAMENTALS L∆z C∆z 1 L∆z C∆z (a) 2 L∆z C∆z Ns L∆z C∆z (b) Figure 3-9 (a) Model for a differential element of a transmission line; (b) full model. where z is the length of a differential section of transmission line and C is the capacitance per unit length. Figure 3-9a shows the equivalent-circuit model of a differential element of transmission line. However, to make use of the model, we need a practical methodology to represent transmission lines that are much longer than z. Simply increasing z to scale the inductance and capacitor values for longer lengths will not produce a legitimate model unless the resonance of the LC circuit is much higher than the maximum frequency to be simulated: √1 2π C zL z fsimulation (3-21) If equation (3-21) is not satisﬁed, the equivalent circuit will behave like an LC ﬁlter circuit instead of like a transmission line. The equivalent circuit of a transmission line can be represented by a series inductance and a shunt capacitor only if the LC resonant frequency is signiﬁcantly greater than the maximum frequency of interest. The correct method to scale the transmission-line model for long lengths is to cascade a sufﬁcient number of small LC segments together until the correct overall length is achieved, as shown in Figure 3-9b. Each LC segment represents a small transmission-line section of length z. The trick is to choose the correct value of z to achieve adequate model accuracy. If the value of z is too small, it will take an inordinate amount of time to perform SPICE simulations. However, if it is too large, the model will not exhibit realistic transmission-line properties. A good “rule of thumb” is to choose z so that the delay of each segment is WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 79 approximately one-tenth the signal rise time if working in the time domain, z ≤ t√r c 10 εr (3-22a) or one-tenth the wavelength that corresponds to the maximum frequency of interest if working in the frequency domain, z ≤ λf,max 10 (3-22b) where tr m/s), εr is the signal rise or fall time, c the speed the dielectric permittivity, and λf,max = of light c/(fmax √in a εr vacuum (3 × 108 ) the wavelength that corresponds to the highest frequency of interest in the simulation. When using a distributed LC model for modeling transmission lines, the num- ber of segments for time-domain simulations is determined by √ Ns = l = 10l εr z tr c (3-23a) For frequency-domain simulations, Ns = l = 10l z λf,max (3-23b) where Ns is the minimum number of segments required to model a transmission line of length l. Therefore, the capacitance and inductance per segment are given by C z= lC Ns L z= lL Ns (3-24a) (3-24b) where C and L are the per unit length values of the capacitance and inductance. Example 3-1 Create a transmission-line model for the 20-cm transmission line shown in Figure 3-10a assuming the following inductance and capacitance values and a dielectric permittivity of εr = 4.5. L = 3.54 × 10−7 H/m C = 1.41 × 10−10 F/m 80 IDEAL TRANSMISSION-LINE FUNDAMENTALS 20 cm 100 ps 100 ps 0.498 nH er = 4.5 (a) 0.498 nH Signal Conductor Reference Conductor 0.498 nH 0.198 pF 0.198 pF 0.198 pF 1 2 142 (b) Figure 3-10 Equivalent circuit of a transmission line. SOLUTION Since the digital waveform has rise and fall times of 100 ps (tr = tf = 100 ps), equations (3-22) and (3-23) give the required values for the model: z = 100 ps(3√.0 × 108) = 1.41 × 10−3 m 10 4.5 Ns = l z = 0.2 1.41 × 10−3 = 141.8 segments Since it is inconvenient to build an equivalent-circuit model with 141.8 segments, Ns is rounded up to 142. The inductance and capacitance values for each segment are calculated with equations (3-24): C z= (0.2)(1.41 × 10−10) 142 = 1.98 × 10−13 F L z= (0.2)(3.54 × 10−7) 142 = 4.98 × 10−10 H Figure 3-10b shows the equivalent circuit for this transmission line. 3.2.4 Wave Equation in Terms of LC The wave equations, which were used as the basis for analyzing propagating electromagnetic ﬁelds, were derived in Section 2.3.1. To analyze transmission lines in terms of circuit parameters, we rederive the wave equation from the WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES 81 telegrapher’s equations (3-12) and (3-18): ∂v(z, t) = −L ∂i(z, t) ∂z ∂t ∂i(z, t) = −C ∂v(z, t) ∂z ∂t (3-12) (3-18) Assuming that the digital signals can be decomposed into sinusoidal harmon- ics using the Fourier transform, the telegrapher’s equations can be expressed in time-harmonic form where the voltage and current have the forms v(t) = V0ejωt and i(t) = I0ejωt . Similar representations for the time-harmonic ﬁelds were discussed in Section 2.3.3. Consequently, the time-harmonic forms of the telegra- pher’s equations are dv(z) = −j ωLi(z) dz di(z) = −j ωCv(z) dz Taking the derivative of (3-25) with respect to z produces (3-25) (3-26) d2v(z) = −j ωL di(z) d z2 dz (3-27) and substituting (3-26) into (3-27) allows us to write an equation only in terms of voltage, d 2 v (z) d z2 + ω2LCv(z) = 0 (3-28) which is the loss-free transmission-line wave equation for voltage. Equa- tion (3-28) is a second-order differential equation with the general solution given by √ √ v(z) = v(z)+e−jzω LC + v(z)−ejzω LC (3-29) √ The term v(z)+e−jzω LC describes the vo√ltage propagating down the transmission line in the +z-direction and v(z)−ejzω LC describes the voltage propagating in the −z-direction. Note the similarity of (3-29) and (2-41), which is the solution equivalent to the wave equation for the electric ﬁeld. In Section 2.3.4, a propagation constant for a wave traveling in an inﬁnite medium was deﬁned that completely describes the medium where the electromagnetic wave is propagating: γ = α + jβ (2-42) 82 IDEAL TRANSMISSION-LINE FUNDAMENTALS Comparing the solution to the electromagnetic wave equation (2-41) to equation (3-29) allows us to draw direct parallels for the propagation constant of a wave propagating on a loss-free transmission line: √ √ γ = α + jβ = 0 + j ω LC → β = ω LC (3-30) Equation (3-30) is the phase constant of a transmission line. Notice that for the loss-free transmission line, the attenuation constant (α) is zero. In Chapters 5 and 6 we describe how to calculate the losses for a transmission line and subsequently, α. 3.3 TRANSMISSION-LINE PROPERTIES With the phase constant deﬁned by (3-30), we can calculate the transmission-line properties using the techniques described in Section 2.3.4 for the propagation of a time-harmonic plane wave. 3.3.1 Transmission-Line Phase Velocity Substitution of the propagation constant deﬁned in equation (3-30) into equation (2-46) allows us to calculate the phase velocity of a harmonic component traveling on a transmission line in terms of L and C: ω1 νp = β = √ LC m/s (3-31) were L and C are the per unit length values. Note that phase velocity can also be calculated from the dielectric properties: νp = c √ µr εr (2-52) where c is the speed of light in a vacuum (3 × 108 m/s), µr the relative magnetic permeability of the dielectric (it is almost always unity for practical dielectrics), and εr the relative permittivity of the dielectric. 3.3.2 Transmission-Line Characteristic Impedance In Chapter 2 we showed that the electric and magnetic ﬁelds were interdependent. Therefore, if the values of one ﬁeld were known, the other could be calculated using Maxwell’s equations. Similarly, for particular transmission-line geometry, if the voltage is known, the current can be calculated using the telegrapher’s equations. For example, if we assume that a voltage is propagating along an TRANSMISSION-LINE PROPERTIES 83 inﬁnitely long transmission line along the z-axis, the current can be calculated by differentiating the voltage and substituting into (3-25): √ v(z) = v(z)+e−jzω LC dv(z) = −j √ ω LC e−j √ zω LC v(z)+ = −j ωLi(z)+ dz i(z)+ = √ LC e−j √ zω LC v(z)+ = C v(z)+ e−j √ zω LC L L (3-32) Since√the units of C are F/m = A · S/m and L is in units of H/m = V · S/A, the term√ C/L has units of A/V, which is the same as siemens. Therefore, the quantity L/C must be similar to the intrinsic impedance deﬁned in Section 2.3.4. The characteristic impedance of a loss-free transmission line in ohms is given by Z0 = L C (3-33) Just as the intrinsic impedance describes the relationship between the electric and magnetic ﬁelds, the characteristic impedance describes the relationship between the voltage and current on a transmission line. Thus, we can write the transmission-line equation for the current in terms of the voltage equation: i(z) = 1 √ v(z)+e−jzω LC + 1 √ v(z)−ejzω LC Z0 Z0 (3-34) 3.3.3 Effective Dielectric Permittivity As described in Section 3.3.1, electrical signals propagating along a transmission line will travel at a speed that is dependent on the dielectric permittivity of the surrounding medium and the geometry of the transmission-line cross section. Although this concept is relatively simple, it becomes complicated when a transmission line is built with a nonhomogeneous dielectric. The most common example of this is a microstrip transmission line. For example, consider Figure 3-11a, which shows the electric ﬁeld of a signal on a microstrip transmission line built on an FR4 dielectric, which has a relative dielectric permittivity of approximately εr = 4.0. Note that a percentage the electric ﬁeld is propagating simultaneously in the dielectric and in the air, where the latter has a relative dielectric permittivity of approximately εr = 1.0. Consequently, the propagation velocity will depend on an “effective” dielectric permittivity, which is a weighted average of how much electric ﬁeld is fringing through the air versus how much stays contained within the dielectric material. When the electric ﬁeld is contained completely within the board, as in the case of a stripline as shown in Figure 3-11b, the effective dielectric constant will be equal to the dielectric permittivity of the insulating material, and the signals will propagate more slowly than microstrip traces. When signals are routed on the external layers of the board, as in the case 84 IDEAL TRANSMISSION-LINE FUNDAMENTALS er = 1 er = 4 (a) er = 4 (b) Figure 3-11 Electric ﬁelds of (a) a microstrip and (b) a stripline. of a microstrip line, the electric ﬁeld fringes through both the dielectric material and the air, which lowers the effective dielectric constant; thus, the signals will propagate more quickly than those on an internal layer. Note that this description assumes that the relative magnetic permeability of the dielectric material under consideration is unity (µr = 1). In Sections 3.4.3 and 3.4.4 we describe how to derive the effective dielectric permittivity of a microstrip using quasistatic approximations of Maxwell’s equations. Furthermore, there are many commercially available two-dimensional electromagnetic ﬁeld solvers available which will produce accurate results. However, in the absence of a ﬁeld solver, equation (3-35) has been shown to produce results of reasonable accuracy [Hammerstad and Jensen, 1980] for structures where the conductor thickness t is much smaller than the dielectric thickness h. 1 u4 + (u/54)2 1 u3 a = 1 + ln 49 u4 + 0.432 + ln 1 + 18.7 18.1 b = 0.564 εr − 0.9 0.053 εr + 3 εeff(u, εr ) = εr + 2 1 + εr − 2 1 1 + 10 u −ab (3-35) where u = w/ h and the dimensions are deﬁned in Figure 3-12a. TRANSMISSION-LINE PROPERTIES 85 w t er = 1 h er (a) er t w h2 h1 (b) Figure 3-12 Dimensions used in the impedance and effective dielectric constant formulas: (a) microstrip line; (b) stripline. 3.3.4 Simple Formulas for Calculating the Characteristic Impedance For maximum accuracy it is necessary to use one of the many commercially available two-dimensional electromagnetic ﬁeld solvers to calculate the impedance of the PCB or MCM traces for design purposes. The solvers will typically provide the impedance, propagation velocity, and the L and C elements per unit length using many of the concepts that will be presented in Section 3.4. In the absence of a ﬁeld solver, the formulas presented here will provide approximations to the impedance values of typical transmission lines as a function of the trace geometry and the dielectric constant εr , where the dimensions are as shown in Figure 3-12. Microstrip: Inﬁnitely Thin Conductors (t h) [Hammerstad and Jensen, 1980] Z0 = η 2π ln ξ h w + 1+ 2h 2 w ξ = 6 + (2π − 6)e−(30.666h/w)0.7528 η = 377 √ εeff (3-36a) Microstrip: Finite Thickness [Collins, 1992] The formulas below are accurate for 1 < εr ≤ 16 and 0.25 ≤ w/ h ≤ 6. Note that εeff in this equation set accounts for the ﬁnite thickness of the signal conductor when calculating the effective dielectric constant for the microstrip. The term we is an effective width that accounts for the extra capacitance caused by the ﬁnite thickness of the signal conductor. Since electric ﬁeld lines will be established between the edge of the conductor and the reference plane, thicker signal conductors will exhibit increased 86 IDEAL TRANSMISSION-LINE FUNDAMENTALS capacitance. The effective width we is slightly wider than the physical width w, to account for the extra capacitance. Z0 = ε0µ0 1 εeff Ca w + 0.398t 1 + ln 4πw t w≤ 1 h 2π we = w + 0.398t 1 + ln 2h w1 > Ca = t 2π ε0 ln(8h/we + we/4h) ε0 we + 1.393 + 0.667 h ln h 2π we + 1.444 h we ≤ 1 h we > 1 h εeff = εr + 2 1 + εr − 2 1 h 1 + 12 we 0.02(εr − 1) ξ = 0 1− w h 2 −1/2 t + ξ − 0.217(εr − 1) √ weh w <1 h w >1 h (3-36b) Symmetrical Stripline (h1 = h2) [IPC, 1995] Zo = 60 √ εr ln 1.9(2h + t) 0.8w + t w 0.1 < < 2.0 h t < 0.25 h 1 < εr < 15 (3-36c) Asymmetrical Stripline (h1 > h2) [IPC, 1995] Z0 = 80 √ εr ln 1.9(2h2 + t) 0.8w + t 1 − h2 4h1 w 0.1 < < 2.0 h2 t < 0.25 h2 1 < εr < 15 (3-36d) 3.3.5 Validity of the TEM Approximation The assumptions of how the electric and magnetic ﬁelds are related for a propagating electromagnetic wave were discussed in Section 2.3.2. One of TRANSMISSION-LINE PROPERTIES 87 t=0 er = 1 t = t1 t = t2 Ex_air er > 1 Ex Ez Ex_dielectric Direction of signal propagation (+z) Figure 3-13 The electric ﬁeld develops a z -component when propagating down a microstrip transmission line, due to the nonhomogeneous dielectric. the fundamental concepts used throughout signal integrity analysis is that the electric and magnetic ﬁelds are orthogonal and there are no components in the z-direction. When waves propagate in this manner, it is called the transverse electromagnetic mode (TEM). However, when discussing the concept of an effective dielectric permittivity in Section 3.3.3, where the component of the electric ﬁeld propagating through the air travels faster than the component propagating in the board material, it becomes obvious that the electric ﬁeld is no longer restricted to a single component. For example, consider Figure 3-13, which depicts the side view of an electric ﬁeld established in the x-direction at t = 0 on a microstrip transmission line between the signal conductor and the reference plane. As the signal begins to propagate down the line, the electric ﬁeld lines in the air will travel at a faster speed than those in the board, effectively tilting the electric ﬁeld in the z-direction. Consequently, the electric ﬁeld develops a component in the z-direction which violates the assumption of the TEM approximation. Furthermore, as the frequencies increase, the electric ﬁeld will become more conﬁned to the region between the microstrip and the reference plane, resulting in less fringing through the air, causing the effective dielectric permittivity to increase. To understand this, refer to Figure 3-14. When a dc voltage is applied between the signal conductor and the reference plane, the charge will be distributed uniformly across the cross section of the signal conductor. As the frequency of the signal is increased, the charge will tend to concentrate at the bottom of the signal conductor closest to the reference plane because that is the area of highest ﬁeld concentration. This means that for a microstrip transmission line, the electric ﬁeld will tend to concentrate in the board material, which increases the effective dielectric permittivity with increasing frequency. The charge distribution in transmission lines is discussed in more detail in Sections 3.4.4 and 5.1.2. 88 IDEAL TRANSMISSION-LINE FUNDAMENTALS Increased charge density er = 1 er > 1 Figure 3-14 At high frequencies, the proximity of the ﬁelds concentrates the charge in the bottom of the strip nearest the reference plane. The frequency-dependent nature of the effective permittivity in a microstrip will cause the spectral components of the digital waveform (as calculated with a Fourier transform) to travel at different speeds, which will distort the waveform. This is known as dispersion. A relatively simple formula for calculating how the effective dielectric permittivity for a microstrip changes with frequency due to the nonhomogeneous nature of the dielectric was developed empirically by [Collins 1992], and is given by εeff(f ) = εr − εr − εeff(f = 0) 1 + (f/fa)m (3-37) where fa = 0.75 + (0.75 − fb 0.332εr−1.73)(w/ h) fb = √ h εr 47.746 − εeff(f = 0) tan−1 εr εeff(f = 0) − 1 εr − εeff(f = 0) m = m0mc ≤ 2.32 m0 = 1 + 1 + √1 w/ h + 0.32 1+ w −3 h mc = 1 1 + 1 1.4 + w/h 0.15 − 0.235e−0.45(f/fa) w ≤ 0.7 h w > 0.7 h where εeff(f = 0) is calculated with (3-35), f is in gigahertz, and the units of w and h are millimeters.∗ ∗Fortunately, a frequency-dependent effective dielectric permittivity does not pose signiﬁcant obstacles to modeling transmission lines. In Chapter 10, techniques that employ frequency-dependent equivalent circuits using tabular SPICE models are described that allow a unique value of the transmission-line parameters to be described at every frequency point. TRANSMISSION-LINE PROPERTIES 89 er,eff w = 2h 4 er = 1 3.9 er = 4.5 h 3.8 3.7 h = 2.5 mils h = 25 mils 3.6 Quasi-TEM approximation 3.5 3.4 3.3 3.2 3.1 3 1 10 100 Frequency (GHz) Figure 3-15 Effective dielectric permittivity compared to the quasi-TEM approximation for a 12-in. microstrip. For most practical applications encountered in high-speed digital design, the TEM approximations are valid and the frequency-dependent nature described by (3-37) is ignored. For microstrip lines and other nonhomogeneous structures, when TEM propagation is assumed, it is referred to as the quasi-TEM approximation. To demonstrate the conditions when the quasi-TEM approximation breaks down, equation (3-37) was used to calculate the frequency variation of the relative effective dielectric permittivity εeff for two cases, a thick and a thin dielectric, as shown in Figure 3-15. The thin dielectric example (h = 2.5 mils) represents the transmission-line dimensions typically used to design buses on conventional motherboards for personal computers. The thick dielectric example (h = 25 mils) is an exaggerated case chosen to demonstrate when the TEM approximation breaks down (however, similar dimensions are sometimes used in radio-frequency (RF) applications where density is not as much of a concern). Equation (3-35) was used to calculate the frequency-invariant quasistatic TEM value for εeff, and equation (3-37) was used to estimate the variation of εeff with frequency. Note that for the thin case, the deviation from the quasi-TEM approximation is small up to very high frequencies, but the thick case begins to deviate much earlier. To evaluate when the quasi-TEM approximation breaks down, we choose a metric of 1% error in total delay. Since the changes in εeff will alter the velocity as shown in equation (2-52), the errors will accumulate for longer line lengths. If we choose a line length, the valid frequency range of the TEM approximation can be calculated. Figure 3-16 shows the percent error in propagation delay caused by the quasi-TEM approximation for a 12-in. microstrip. Note that for the thick case, the quasi-TEM will induce a 1% error in delay at about 8 GHz, and the thin case remains accurate to about 80 GHz. Consequently, for typical transmission-line 90 IDEAL TRANSMISSION-LINE FUNDAMENTALS Quasi-TEM error (%) 10 h = 2.5 mils 9 h = 25 mils 8 1% error threshold 7 6 5 4 3 2 1 0 2 20 40 60 80 100 Frequency (GHz) Figure 3-16 Bandwidth where the quasi-TEM approximation induces a 1% error in the delay for a 12-in. microstrip. dimensions used in contemporary digital design, the quasi-TEM approximation for the transmission-line parameters is valid. 3.4 TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE Earlier, we discussed how an electromagnetic wave propagates on a transmission line by drawing parallels to a plane wave in free space and developed an equivalent circuit that was constructed from inductance and capacitance elements. In this section we discuss a variety of methods to calculate the equivalent inductance and capacitance values per unit length for various transmission-line structures. Although in practice, a commercial two-dimensional ﬁeld solver is usually employed to calculate the equivalent-circuit parameters, it is vital for the engineer to understand the approximations and methodologies typically used. Note that the solutions in this chapter assume that the electrostatic solutions to Maxwell’s equations are adequate approximations for high-speed digital design purposes. This is known as the quasistatic approximation. This technique allows signiﬁcant simpliﬁcation over the full-wave solution, where no approximations are assumed. Indeed, at high frequencies (above a few hundred megahertz), the quasistatic approximation breaks down because the equivalent-circuit parameters have signiﬁcant frequency dependence. Nonetheless, virtually all two-dimensional ﬁeld solvers used in the industry are quasistatic in nature. Full-wave solvers, which do not assume quasistatic ﬁelds are in theory more accurate. However, in practice they require too many computational resources and too much time to be of practical use for modeling transmission lines in a digital design. Fortunately, the errors caused by the quasistatic approximation can be corrected relatively painlessly with excellent accuracy up to very TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 91 high frequencies (at least 50 GHz), as discussed in Chapters 5, 6, 8, and 10. Consequently, the methodologies in this book for modeling transmission lines are based on a quasistatic approach, and corrections are applied in later chapters to recover the frequency-dependent behavior. 3.4.1 Laplace and Poisson Equations In Section 2.4 we show that the behavior of an electrostatic ﬁeld can be described by the differential equations ∇ ·D = ρ E = −∇ (2-3) (2-65) the latter coming from the fact that Ampe`re’s law is zero for an electrostatic ﬁeld (∇ × E = 0). If (2-65) is substituted into (2-3), ∇ · (−∇ ) = ρ ε we arrive at Poisson’s equation: ∇2 = −ρ ε (3-38) In a medium that lacks any charge density, Poisson’s equation is reduced to Laplace’s equation, ∇2 = 0 (3-39) Poisson’s and Laplace’s equations are particularly useful for solving quasistatic transmission-line problems, as demonstrated in the remainder of this chapter. 3.4.2 Transmission-Line Parameters for a Coaxial Line Assume that a pair of long, coaxial, circular conductors are statically charged with the inner conductor at the potential = V relative to the outer conductor, which is held at zero potential (ground), where the cross section is as shown in Figure 3-17. The region between the conductors has a dielectric permittivity of ε = ε0εr and a relative magnetic permeability of µr = 1. The ﬁrst step is to calculate the capacitance per unit length. If we make the assumption that the dielectric is charge free (very reasonable; see Chapter 6), the capacitance can be derived by solving Laplace’s equation (3-39) in cylindrical coordinates. From Appendix A we get ∇ = ar ∂ ∂ r + aφ 1 r ∂ ∂ φ + az ∂ ∂ z 92 IDEAL TRANSMISSION-LINE FUNDAMENTALS Φ=0 b a Φ=V e = e0er mr = 1 Figure 3-17 Cross section of a coaxial transmission line. Assuming TEM, there will be no variation of the electric ﬁeld in the z-direction, and because of the symmetry, there should be no variation with φ. Therefore, ∇2 = ∇ · (∇ ) = 1 ∂ ∂ r =0 r ∂r ∂r Integrating ∂[(r(∂ /∂r)]/∂r produces r(∂ /∂r) = C1, and a second integration yields the solution: (r) = C1 ln(r) + C2 (3-40) The boundary conditions must be applied to (3-40) to determine the values of C1 and C2. The boundary conditions are: • When r = b, = 0. • When r = a, (a) = V . Applying the ﬁrst boundary condition yields (b) = 0 = C1 ln(b) + C2 C2 = −C1 ln(b) which is substituted back into (3-40): (r ) = C1 ln(r ) − ln(b)C1 = C1 ln r b Applying the second boundary condition produces the following: (a) = V = C1 ln a b → C1 = V ln(a/b) → C1 = −V ln (b/a) TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 93 Note that ln(a/b) = − ln(b/a), which is done here as a convenience: (r) = − V ln(r) + V ln(b) = V b ln ln(b/a) ln(b/a) ln(b/a) r The electric ﬁeld is computed using (2-65): E = −∇ = −∇ V b ln ln(b/a) r = −ar ∂ ∂r V b ln ln(b/a) r = ar r V ln(b/a) (3-41) The next step is to ﬁnd the total surface charge on the signal conductor using (3-3), which describes the boundary conditions between a dielectric and a perfect conductor: n · εE = ρ C/m2 (3-3) At r = a (the inner conductor surface), the surface charge is calculated with εE = εV = ρ a ln(b/a) C/m2 (3-42) Multiplying (3-42) by the circumference of the inner conductor will give the equivalent surface charge per unit length: Q = ρ2πa = 2πεV C/m l ln(b/a) (3-43) Using (2-76) we can get the capacitance per unit length of a coaxial transmission line: C = Q/ l = 2πε F/m (3-44) V ln(b/a) To ﬁnd the inductance per unit length, we exploit the relationship between the magnetic and electric ﬁelds: 1 √ µr µ0εr ε0 c =√ µr εr m/s νp = ω = √1 m/s β LC (2-52) (3-31) which allows us to write (assuming µr = 1) c √ = √1 εr LC (3-45) Since µr = 1, the dielectric properties do not inﬂuence the inductance. Consequently, since the speed of light in a vacuum is constant, the inductance can be 94 IDEAL TRANSMISSION-LINE FUNDAMENTALS calculated from (3-45) if the capacitance is calculated with εr = 1: 1 L= c2Cεr =1 (3-46) where c is the speed of light in a vacuum and Cεr=1 is the capacitance with the relative dielectric permittivity εr set to 1. Substituting (3-44) into (3-46) with εr = 1 gives the value of the inductance per unit length for a coaxial line: L = ln(b/a) c22π ε0 H/m (3-47) 3.4.3 Transmission-Line Parameters for a Microstrip To derive the impedance of a microstrip transmission line, we begin with the solution to Laplace’s equation. To begin the derivation, consider Figure 3-18, which shows the boundary conditions that we must satisfy when solving the differential equations. Note that the microstrip is surrounded by a box with conducting walls at x = ±d/2, y = 0, and y = ∞. This is a valid placement of the boundary conditions only if d h, where h is the height of the dielectric. In this case, the partial differential equations are solved using a method called separation of variables, as described by Jackson [1999]. The rectangular nature of the microstrip geometry allows us to remain in rectangular coordinates, which gives the two-dimensional Laplace equation the form ∇2 ∂2 ∂2 = ∂x2 + ∂y2 = 0 (3-48) y ∞ e0 −w w 2 2 h −d 2 er d 2 x Figure 3-18 Dimensions used to derive microstrip transmission-line parameters from Laplace’s equation. TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 95 The solution to this partial differential equation can be found in terms of two ordinary differential equations, assuming that the potential can be represented by the product of a function for each coordinate [Jackson, 1999]: (x, y) = X(x)Y (y) (3-49) If equation (3-49) is substituted back into (3-48) and then divided by , the result is 1 XY d2XY dx2 +1 XY d2XY dy2 = 1 d2X X dx2 + 1 Y d2Y dy2 =0 (3-50) where the partial derivatives are replaced by total derivatives because each term involves only one variable. This allows us to write two separate ordinary differential equations: 1 X d2X dx2 = −β 2 1 d2Y Y dy2 = β2 (3-51) (3-52) The constant β must satisfy the conditions β2 + (−β2) = 0 to satisfy Laplace’s equation. The differential equations in (3-51) and (3-52) are solved by ﬁnding the roots of the characteristic equations: X(x) = C1nejβx + C2ne−jβx = C1n(cos βx + j sin βx) + C2n(cos βx − j sin βx) Y (y) = C1neβy + C2ne−βy = C1n(cosh βy + sinh βy) + C2n(cosh βy − sinh βy) (3-53) (3-54) The potential is then calculated by substituting (3-53) and (3-54) into (3-49): (x, y) = X(x)Y (y) = [C1n(cos βx + j sin βx) + C2n(cos βx − j sin βx)] · [C1n(cosh βy + sinh βy) + C2n(cosh βy − sinh βy)] (3-55) Note that since the solution is periodic, β must be an integer. The boundary conditions that must be applied to solve (3-55) are (x, y) = 0 when x = ± d 2 which is the potential at the sidewalls and (3-56) (x, y) = 0 when y = 0, ∞ (3-57) 96 IDEAL TRANSMISSION-LINE FUNDAMENTALS which is the potential at the ground plane and at a point inﬁnitely far away. This means that we must consider two different solutions: region 1, which exists from y = 0 to h (in the dielectric), and region 2, which exists from y = h to inﬁnity (in the air), where the potentials must be equal at the boundary of these regions. Looking at X(x) in (3-55), the boundary condition in (3-56) is met if C1n = C2n and β is chosen so that the cosine term equals zero when x = ±d/2. When C1n = C2n, X(x) is reduced to X(x) = C1n cos βx and when β = nπ/d for odd n and x = ±d/2, X(x) = 0 yielding an appropriate form for X(x): X(x) = C1n cos nπ x d (3-58) The second term Y (y) in (3-55) will satisfy the boundary condition in (3-57) at y = 0 if C1n = −C2n, yielding nπ Y (y) = C2n sinh βy = C2n sinh y d (3-59) At y = ∞, the boundary condition (3-57) is satisﬁed when C1n = 0: Y (y) = C2n(cosh βy − sinh βy) = C2ne−βy (3-60) This allows us to write equations for the potential that satisﬁes the boundary conditions by multiplying X(x) and Y (y) for the appropriate boundary conditions, where the product of the constants have been renamed An and Bn for clarity. ∞ n=1 odd An cos nπ x d sinh nπ y d (x, y) = ∞ n=1 odd Bn cos nπ x d e−(nπ /d )y when 0 ≤ y < h (region 1) (3-61a) when h ≤ y < ∞ (region 2) (3-61b) The potential at the signal conductor (y = h) must be continuous, so An cos nπ x d sinh nπ h d = Bn cos nπ x d e−(nπ /d )h TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 97 yielding Ane(nπ/d)h sinh nπ h d = Bn which allows the equations for the potential to be written in terms of An alone: ∞ n=1 odd (x, y) = ∞ n=1 odd nπ An cos x sinh d nπ An sinh h cos d nπ y d nπ x d e−(nπ /d )(y −h) when 0 ≤ y < h (3-62a) when h ≤ y < ∞ (3-62b) To get the electric ﬁeld between the signal conductor and the ground plane, we apply equation (2-65), Ey = −∇φ = −∂ /∂y. Since d(sinh ax)/dx = a cosh ax and d(eax)/dx = aeax, the electric ﬁelds become Eyn = − ∂ ∂y An cos nπ x d sinh nπ y d = − nπ An cos d nπ x d cosh nπ y d for region 1 and Eyn = − ∂ ∂y An sinh nπ h d cos nπ x d e−(nπ /d )(y −h) = nπ An sinh nπ h cos nπ x e−(nπ /d )(y −h) d d d for region 2, yielding Ey (x, y) = − ∞ nπ An cos nπ x cosh d d n=1 odd ∞ nπ An sinh nπ h d d n=1 odd × cos nπ x e−(nπ /d )(y −h) d nπ y d when 0 ≤ y < h (3-63a) when h ≤ y < ∞ (3-63b) 98 IDEAL TRANSMISSION-LINE FUNDAMENTALS To calculate the coefﬁcient An, we must assume a charge distribution on the signal conductor. As a ﬁrst-order approximation, it can be assumed that the charge (ρ) is spread out uniformly across the signal conductor at y = h. Consequently, there will be no variation of ρ with x on the strip, but it will be zero off the strip. 1 ρ(x) = 0 when |x| < w/2 when |x| > w/2 (3-64) From equation (3-4), the electric ﬁelds above and below the signal conductor are equated to the charge density: ρ(x) = (ε0Ey1) − (ε0εr Ey2) = ∞ ε0nπ An cos nπ x d d n=1 odd × sinh nπ h d + εr cosh nπ h d (3-65) If we temporarily assign all terms in (3-65) that are not a function of x (except An) to a constant term k as in (3-66), it is easy to see that it takes the form similar to the Fourier series shown in (3-68) when a0 and bm are zero: ρ(x) = ∞ Ank cos n=1 odd nπ x d k = ε0 nπ d sinh nπ h d + εr cosh nπ h d f (x) = 1 2 a0 + ∞ am cos mπ x d ∞ + bm cos n=1 n=1 mπ x d (3-66) (3-67) (3-68) This allows the use of Fourier series techniques to solve for the coefﬁcient, An: w/2 ρ(x) cos −w/2 mπ x d dx = d /2 Ank cos −d /2 nπ x d cos mπ x d dx Setting m = n, we arrive at w/2 ρ(x) cos −w/2 nπ x d dx = d /2 Ank cos2 −d /2 nπ x d dx (3-69) TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 99 After substituting (3-67) for k, integration of both sides of (3-69) yields 2d sin nπ w = Ankd nπ 2d 2 An = ε0[(nπ )2/d] 4 sin(nπw/2d) sinh((nπ/d)h) + εr cosh ((nπ/d)h) (3-70) The voltage of the signal conductor can now be calculated with respect to the ground plane at x = 0 using (3-1) and (3-63a): b h ∞ nπ v = − E · dl = − Ey(x = 0, y) dy = a 0 An sinh h d n=1 odd (3-71) The total charge is calculated with w/2 Q= ρ(x)dx = w −w/2 (3-72) So the capacitance per unit length is calculated from (3-71) and (3-72) using (2-76): C=Q= v w ∞ n=1 An sinh[(nπ/d)h] odd (3-73) and the inductance per unit length is calculated using (3-46) by ﬁrst calculating the capacitance with (3-73) with a relative dielectric permittivity of unity (εr = 1): L = 1 c2Cεr =1 (3-46) With (3-73) and (3-46), we can calculate several useful quantities, such as the phase constant (3-30), the characteristic impedance (3-33), and the effective relative dielectric permittivity (εeff). To understand how to calculate εeff, we can examine the formulation of a parallel-plate capacitor, as derived in Example 2-3: Cεr,eff Cεr =1 = ε0εr,effA/d ε0A/d = εr,eff (3-74) Consequently, the effective permittivity is determined by calculating the capacitance of the microstrip with (3-73) using the correct value of the dielectric constant and dividing by the capacitance calculated with (3-73) when εr = 1. Since the dimensions that determine the capacitance remain identical, all that remains is the effective dielectric permittivity. 100 IDEAL TRANSMISSION-LINE FUNDAMENTALS The accuracy of (3-73) and (3-74) is reasonable, but two approximations made during the derivation degrade the accuracy. First, we assumed that the signal conductor is inﬁnitely thin, which is not realistic because microstrip transmission lines manufactured on printed circuit boards often have conductor thicknesses of dimensions similar to the dielectric height. There is little beneﬁt of deriving a ﬁnite thickness formula here because the most useful method would employ numerical solutions that are not covered in this book. Second, the charge distribution on the signal conductor is not uniform. The charge distribution near sharp edges is derived in Section 3.4.4 and applied to this formulation. 3.4.4 Charge Distribution near a Conductor Edge The approximation made in the analysis of Section 3.4.3 assumed that charge distribution on the signal conductor on the microstrip was uniform. Realistic charge distributions tend to increase signiﬁcantly near the corners on the conductors, which will alter the capacitance and subsequently the effective permittivity of the transmission line. To begin the derivation of the charge distribution, consider Figure 3-19, which shows two intersecting planes at an angle of θ . The two cases of interest are when θ = 270◦, which will yield the charge distribution near a sharp corner, and when θ = 360◦, which will yield the charge distribution near the edge of a very thin strip. The derivation assumes that the conducting planes are kept at a potential V with respect to a point far away. Since we are interested y P r f q x Φ=V Edges of a conductor Figure 3-19 Two conducting planes held at a potential V with respect to a far-off point intersecting at an angle. TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 101 in the behavior of the ﬁelds only close to the edge, the details of the reference conductor far away are not considered. We begin the derivation with Laplace’s equation in cylindrical coordinates: ∇2 =1 ∂ ∂ r r ∂r ∂r + 1 ∂2 r2 ∂φ2 =0 (3-75) Similar to the solution of the microstrip transmission line, the solution to this partial differential equation is found using separation of variables. The problem is reduced to two ordinary differential equations, assuming that the potential can be represented by the product of a function for each coordinate [Jackson, 1999]: (r, φ) = R(r)Y (φ) (3-76) Substitution of (3-76) into (3-75) and dividing each side by R(r)Y (φ)/r2 produces r2 ∂ RY r ∂r ∂RY r ∂r r2 ∂2RY r ∂ + RY r2 ∂φ2 = R ∂r ∂R r ∂r + 1 Y ∂2Y ∂φ2 =0 (3-77) Since the sum of the ordinary differential equations must equal zero, we can solve each one separately: d r dR r = β2R dr dr d2Y dφ2 = −β 2 Y (3-78) (3-79) For the solution of (3-78), try R = krl: d r d (kr l ) r = r d (rklrl−1) = r d klrl = kl2rl dr dr dr dr Setting the solution above to β2R yields β2R = β2(krl) = l2(krl) l = ±β Hence, the solution to (3-78) is shown: R(r) = arβ + br−β (3-80) The solution to (3-79) is found by solving the roots of the characteristic equation Y (D2 + β2) = 0 102 yielding IDEAL TRANSMISSION-LINE FUNDAMENTALS Y (φ) = A cos βφ + B sin βφ (3-81) Since (3-81) is composed of periodic functions, β must be an integer. Substituting (3-80) and (3-81) into (3-76) yields the general solution for the potential when β = 0: (r, φ) = arβA cos βφ + arβB sin βφ + br−β A cos βφ + br−β B sin βφ (3-82) The boundary conditions that must be satisﬁed are: • = V when φ = θ . • = V when φ = 0. If β = nπ/θ , the sin terms in (3-82) become zero when n is odd. Since sin(0) = 0, the choice of β allows us to write a function for the potential that satisﬁes the boundary conditions by setting the constant A = 0, and combining the constants αn = aB and αn = bB: (r, φ) = V + ∞ αnrnπ/θ sin n=1 odd nπ φ θ + αnr−nπ/θ sin nπ φ θ (3-83) The φ component of the electric ﬁeld is calculated from equation (2-65) (E = −∇ ). From Appendix A we get ∇ = ar ∂ ∂r + aφ 1 r ∂ ∂ φ + az ∂ ∂ z yielding Eφ = −1 ∂ r ∂φ = ∞ n=1 odd − nπ αn r(nπ/θ)−1 cos nπ φ θ θ + nπ αn r−(nπ/θ+1) cos nπ φ θ θ (3-84) TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 103 Finally, the charge can be calculated with (3-3): n · εE = ρ ∞ ρ(r, φ) = εEφ = n=1 odd − nπ εαn r(nπ/θ)−1 cos nπ φ θ θ + nπ εαn r−(nπ/θ+1) cos nπ φ θ θ Since we are interested in the behavior of the charge in the immediate vicinity of the edge, r is small and therefore only the ﬁrst term, the summation, is important (n = 1). Also, since for a ﬂat plane the dependence on r should vanish and the last term shows dependence on r for θ = π and n = 1, it is necessary that αn = 0. ρ(r, φ) = − π εα1 r0 cos π φ + π εα1 r−2 cos π φ θ θ θ θ This yields the surface charge density as a function of the angle between two conducting planes in the near vicinity of the corner: ρ(r, φ) = − π εα1 r(π/θ)−1 cos π φ θ θ (3-85) For a sharp corner, θ = 3π/2, as shown in Figure 3-20, 3.85 reduces to ρ(r, φ = 3π /2) = 2εα1 r−1/3 3 (3-86) Since we are concerned with the charge density on the metal planes, φ will be either θ or 0, meaning that the cosine term will equal 1 or -1. Equation (3-86) means that for a transmission line built with a square con- ductor, the charge density increases dramatically near the corner, as shown in Figure 3-20. In fact, for a perfectly sharp edge, the charge density becomes singular as r approaches zero. For the edge of a thin strip, θ = 2π and (3-85) reduces to ρ(r, φ = 2π ) = εα1 r−1/2 2 (3-87) Equation (3-87) means that for an inﬁnitely thin metal strip, as was assumed in the derivation for the microstrip line in Section 3.4.3, the charge density also approaches inﬁnity near the edge. Since inﬁnitely thin sheets or inﬁnitely sharp corners do not really exist in nature, the trends of (3-86) and (3-87) are what must be understood. The charge density increases dramatically near sharp corners on metal conductors. The coefﬁcients αn are dependent on the initial conditions remote from the corner, which are not solved here. 104 IDEAL TRANSMISSION-LINE FUNDAMENTALS Charge density, r r r=0 Φ=V q = 270° = 3p 2 Figure 3-20 Charge distribution near a sharp corner. 3.4.5 Charge Distribution and Transmission-Line Parameters Now that the behavior of the charge in the near vicinity of the edge is understood, it is easy to imagine what the total charge distribution should look like across the signal conductor of a transmission line. The charge density should be almost uniform near the center of the strip but will increase dramatically near the edges, as predicted by (3-85). A useful formula to predict the current distribution for a microstrip, which works well for a variety of dimensions, was derived by Collins [1992] using conformal mapping techniques: ρ(x) = 2Q πw 1− x2 w2 2 (3-88) where Q is the total charge, x the distance from the conductor center, and w the width of the signal conductor. Figure 3-21 shows an example of a realistic current distribution in the signal conductor of a microstrip transmission line, as calculated with (3-88) and normalized so that the center of the conductor has a charge density of 1. To understand how the charge distribution will inﬂuence the transmission-line parameters such as delay and impedance, the equations for the microstrip transmission line derived in Section 3.4.3 can be improved by applying a more realistic charge distribution to the signal conductor and solving (3-69) for An: w/2 ρ(x) cos −w/2 nπ x d dx = d /2 Ank cos2 −d /2 nπ x d dx (3-69) TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 105 charge density, r 5 4 3 2 1 0 x −0.5 −0.3 −0.1 0.1 0.3 0.5 w −w2 w 2 Figure 3-21 Charge distribution across a microstrip signal conductor. The problem arises that (3-88) is singular at x = w/2, so it cannot be integrated. Fortunately, we can circumvent this problem by choosing a well-behaved function to approximate (3-88) that can be integrated: ρ(x) = 2Q ∼= Q(axm + c) πw 1− x2 w2 2 (3-89) where a, m, and c should be chosen so that the polynomial approximates the real- istic charge distribution. If the total charge is to remain the same as in (3-72), and the width of the transmission line is normalized to w = 1, the charge distribution must satisfy w/2 Q= ρ(x)dx = 1 (3-90) −w/2 Furthermore, the order of the polynomial must be high to approximate the sharp increase in the charge density near the edges. If the order of (3-89) is chosen to get a reasonable ﬁt (m = 6), the polynomial will satisfy (3-90) when a = 100 and c = 0.77 for a strip width of 1, ρ(x) ≈ 100x6 + 0.77 (3-91) where 0.5 (100x6 + 0.77) dx ≈ 1. −0.5 106 IDEAL TRANSMISSION-LINE FUNDAMENTALS 5 w = 1, a = 100, 4 m = 6, c = 0.77 3 charge density, r 2 1 0 −0.5 −0.3 −0.1 0.1 0.3 0.5 w Equation (3-88) Equation (3-91) Figure 3-22 Comparison of the charge distribution calculated analytically and the nonsingular polynomial approximation. TABLE 3-1. Effect of Charge Distribution on the Impedance and Effective Relative Permittivitya Z0( )/εeff 1 when |x| < w/2 w/ h ρ(x) = 0 when |x| > w/2 ρ(x) = 100x6 + 0.77 Two-Dimensional Numerical Field Solver Results 4 2 1 0.667 0.5 37.6/3.26 59.7/3.02 84.8/2.86 99.9/2.81 110.8/2.78 31.6/3.37 52.6/3.08 77.0/2.90 92.1/2.83 99.1/2.76 ad = 100h, widths normalized to 1, εr = 4.0, n = 5000. 32.5/3.38 51.4/3.13 74.6/2.96 89.3/2.89 99.9/2.84 Figure 3-22 shows a comparison between the charge distribution approximated with (3-91), which is not singular, and (3-88). Plugging (3-91) into (3-69) and solving for An will provide a formula for the microstrip transmission line that accounts for a realistic charge distribution on the signal conductor. Unfortunately, the integration gets messy and the ﬁnal form of An is ungainly and therefore is not shown here. Mathematica was programmed to perform the integration and solve (3-69) for An when the charge distribution was approximated with (3-91). Comparisons between the calculated impedance and the effective dielectric constant assuming a uniform charge distribution using (3-64) and a realistic distribution calculated with (3-88) and approximated with (3-91) is shown in Table 3-1. Note that when a realistic charge distribution is used, the quasistatic approximations are signiﬁcantly more accurate when compared to a commercial two-dimensional numerical ﬁeld solver. TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 107 3.4.6 Field Mapping So far, we have used Laplace’s equation to calculate the transmission-line parameters. Next we explore a graphical method for solving general two-dimensional electrostatic problems. The chief merit of this method is that it provides unique insight into the ﬁeld patterns for complex geometries. Additionally, the method is applicable to systems where the analytical approach is either extremely difﬁcult or impossible. It also demonstrates a methodology where the structure is meshed, which is similar to techniques used by many numerical ﬁeld solving techniques. A full understanding of this method is useful because it will allow the signal integrity engineer to develop an intuitive understanding of how ﬁelds are distributed and how transmission-line geometries affect impedance and the effective dielectric permittivity. For any two-dimensional system possessing a uniform charge distribution along the z-axis, the same electrostatic ﬁeld sketch can be applied to every cross section. An example of the ﬁeld sketch for an arbitrary cross section possessing the charges q and −q in every length l is shown in Figure 3-23. Equation (3-3) tells us that the electric ﬁeld lines must terminate on a perfect conductor normal to the surface. This means that if a line is drawn on a two-dimensional ﬁeld sketch so that it is always perpendicular to the electric ﬁelds, and that line is extended into the page, the surface created must be a constant potential, as shown in Figure 3-23. These constant potential surfaces can be treated as Equal potential surface Φ=V Φ=0 Electric field flux Figure 3-23 Electrostatic ﬁeld sketch of two arbitrary conductors. 108 IDEAL TRANSMISSION-LINE FUNDAMENTALS inﬁnitely thin sheets of “virtual” conducting foil. Using the constant potential surfaces, the capacitance per unit length of a two-dimensional system can be found with reasonable accuracy from a careful ﬁeld sketch. For example, consider the cross section of a coaxial line as shown in Figure 3-24, where electric ﬁeld ﬂux lines originate from a charge q distributed over a length l on the inner conductor and terminate on −q in the same length on the outer conductor. If equal potential lines are drawn between the inner and outer conductors, the total −q C0 C0 C0 +q Φ=V Φ=0 (a) ∆C Φ=V Φ=0 (b) Figure 3-24 Field map of a coaxial line showing (a) series capacitance between equal potential surfaces and (b) capacitance per unit cell. TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 109 capacitance is simply the series combination of the capacitance between equal potential lines as shown in Figure 3-24a. If the equal potential lines are chosen so that each capacitance is the same and ns denotes the number of elements in series, the total capacitance is C = C0 ns (3-92) Furthermore, each of the series capacitors in Figure 3-24a can be subdivided into parallel capacitance values with a value of C for each cell, as shown in Figure 3-24b. Assuming np parallel elements, C0 = np capacitance C = np C ns C, yielding a total (3-93) In order to use (3-93), we need to calculate the cell capacitance C. Assuming that the charge q and − q are present on the top and bottom of each cell wall, we can write the cell capacitance in terms of the potential difference between the boundaries using (2-76) and (3-1): C= q E · dl (3-94) If the problem is simpliﬁed by drawing the ﬁeld lines such that each cell is approximately the same size, the capacitance can be written in terms of the average electric ﬁelds, q C= Eave have (3-95) where have is the average height of a cell and Eave is the average electric ﬁeld for a cell. Equation (3-95) can be simpliﬁed to eliminate the electric ﬁeld using the integral form of Gauss’s law, (2-59), S εE · ds = S D · ds = q: C= q = εq = εq = ε wave l (3-96) Eave have Dave have ( q/ l wave) have have where wave is the average cell width and ds = l wave is the area of the cell surface for the transmission line length l. If the cells can be drawn so that wave ≈ have, the total capacitance is calculated by substituting (3-96) into (3-93): C = ε np F/m l ns (3-97) Example 3-2 Use ﬁeld mapping techniques to calculate the impedance of a coaxial transmission line shown in Figure 3-25, where b/a = 2 and the permittivity of the dielectric is εr = 2.3. 110 IDEAL TRANSMISSION-LINE FUNDAMENTALS a b = 2a 1 1 1 1 1 1 1 1 V Figure 3-25 Field map used to calculate the capacitance in Example 3-2, showing one quadrant divided up into two series sections and four parallel sections. SOLUTION Since the cross section of the coaxial line is symmetrical, we can take advantage of symmetry and draw the ﬁeld lines for only one quadrant, as shown in Figure 3-25. In this sketch, the quadrant was divided into two series sections using a single equal potential line, and four parallel sections yielding ns = 2 and np = (4)(4) = 16. Therefore, equation (3-97) yields C = εr ε0 np ns = (2.3)(8.85 × 10−12) 16 2 = 162.8 × 10−12 F/m To calculate the impedance of the coaxial line, we must ﬁrst recalculate the capacitance for εr = 1 and calculate the inductance from (3-46), Cεr =1 = ε0 np ns = 8.85 × 10−12 16 2 = 70.78 × 10−12F /m L = 1 c2Cεr =1 = 1 (3 × 108)2(70.78 × 10−12) = 156.9 × 10−9H /m yielding the characteristic impedance calculated from (3-33): Z0,ﬁeldmap = 156.9 × 10−9 162.8 × 10−12 = 31.0 This can be compared directly to the results derived in Section 3.4.2. C = 2π ε = 2π(2.3)(8.85 × 10−12) = 184.5 × 10−12 F/m ln(b/a) ln(2) TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE 111 Again, to calculate the impedance, we must ﬁrst recalculate the capacitance for εr = 1. C = 2π ε = 2π(1)(8.85 × 10−12) = 80.2 × 10−12 F/m ln(b/a) ln(2) L = 1 c2Cεr =1 = 1 (3 × 108)2(80.2 × 10−12) = 138.5 × 10−9 H/m Z0 = 138.5 × 10−9 184.5 × 10−12 = 27.3 Comparing this result to Z0,ﬁeldmap shows reasonable accuracy but demonstrates the inherent errors that are inevitable when using hand sketches. Example 3-3 Use ﬁeld mapping techniques to calculate the impedance and the effective dielectric permittivity of the microstrip transmission line shown in Figure 3-26, where w/ h = 1 and the permittivity of the dielectric material is εr = 4.0. SOLUTION It is inherently more difﬁcult to apply mapping techniques to a microstrip because the ﬁelds are difﬁcult to draw accurately and the dielectric is not homogeneous. Figure 3-26 shows a drawing of the ﬁeld lines. Note 1 1 1 1 1 1 1 w 1 0.5 0.5 1 0.5 1 0.5 1 1 1 1 11 1 1 1 1 11 1 1 1 111 1 11 1 1 1 er = 1 er = 4 h Figure 3-26 Field map used to calculate the effective dielectric permittivity and the impedance of a microstrip. 112 IDEAL TRANSMISSION-LINE FUNDAMENTALS that the authors were not able to draw the cells to be all the same size; however, we will still assume an average electric ﬁeld strength for each cell, which will induce some error. Since some cells are small and some cells are large, the errors induced by the cell size discrepancy will partially average out and should give reasonable results. For this example, symmetry was used and only half of the ﬁeld lines were drawn. The distortion of the electric ﬁeld lines at the dielectric boundary was ignored, which will induce an additional, although small error. Initially, we must calculate the effective dielectric permittivity. In Section 3.3.3 we explained that the effective permittivity was due to the weighted average of the ﬁelds fringing through the air and the dielectric. By counting the number of cells in the air and in the board, we can use a weighted average to calculate the effective permittivity. Counting the cells in Figure 3-26 yields nine cells in the air and 27 cells in the dielectric. Note that some of the cells were divided up between the air and the dielectric material. Therefore, the effective dielectric permittivity is calculated: εeff ≈ 9 9 + 27 (1) + 9 27 + 27 (4) = 3.25 The results from the two-dimensional ﬁeld solver in Table 3-1 shows εeff = 2.96 for w/ h = 1, so the result above is not a bad approximation. The ﬁeld plot was divided into ﬁve series sections, and seven parallel sections for one-half of the structure, yielding ns = 5 and np = (2)(7) = 14 (due to the symmetry). Using, εr = εeff = 3.25, equation (3-97) yields C = εr ε0 np ns = (3.25)(8.85 × 10−12) 14 5 = 80.53 × 10−12 F/m and the impedance is calculated as in Example 3-2: Cεr=1 = ε0 np ns = (8.85 × 10−12) 14 5 = 24.77 × 10−12 F/m L= 1 c2Cεr=1 = 1 (3 × 108)2(24.77 × 10−12) = 448.5 × 10−9 H/m Z0 = 448.5 × 10−9 80.53 × 10−12 = 74.6 Comparison to the numerically calculated results from the two-dimensional ﬁeld solver in Table 3-1 show Z0 = 74.6 . The fact that the exact answer was obtained as the two-dimensional solver is serendipity. TRANSMISSION-LINE REFLECTIONS 113 3.5 TRANSMISSION-LINE REFLECTIONS So far, we have considered the propagation of signals on a single transmission line. However, most practical problems involve signals propagating on a series of cascaded transmission lines, such as those built on a motherboard, a daughtercard, and a chip package. In addition, digital designs often utilize transmission-line topologies that branch out from one driver to multiple receivers. Since each transmission line will have different electric characteristics, and topologies will affect signal integrity dramatically, it is essential to understand how a propagating electromagnetic wave will behave when it enters a region where the properties of the transmission line and/or the topology of the interconnection change. 3.5.1 Transmission-Line Reﬂection and Transmission Coefﬁcient Generally, when an electromagnetic plane wave propagating on transmission line A transitions to line B, where the characteristic impedance changes, two things happen: (1) a portion of the wave is reﬂected away from the impedance discontinuity back toward the source, and (2) a portion of the wave is transmitted onto transmission line B, as shown in Figure 3-27. The simultaneous existence of both the transmitted and reﬂected waves is a direct result of the boundary conditions that must be satisﬁed when solving Maxwell’s equations at the interface between the two regions. These boundary conditions are identical to those derived in Section 2.7 for a plane wave impinging on a region with different properties. The parallels between a plane wave propagating in free space and a wave propagating on a transmission line were shown in Section 3.2, where the wave equation was derived in terms of the transmission-line circuit parameters (L and C). If we assume that the values are represented by phasors, we can separate the total plane wave into three parts: the incident, the reﬂected, and the transmitted. Incident wave Transmitted wave ∼R Reflected wave Z01 Line A Z02 Line B Figure 3-27 When a signal propagating on transmission line A encounters an impedance discontinuity, part will be reﬂected back toward the source and part will be transmitted onto transmission line B . 114 IDEAL TRANSMISSION-LINE FUNDAMENTALS 1. Incident wave: vi (z) = vi+e−jβ1z ii (z) = 1 Z01 vi+ e−jβ1 z 2. Reﬂected wave: vr (z) = vr−e+jβ1z ir (z) = − 1 Z01 vr−e+jβ1z 3. Transmitted wave: vt (z) = vt+e−jβ2z it (z) = 1 Z02 vt+ e−jβ2 z √ √ √ √ where β1 = ω L1C1, β2 = ω L2C2, Z01 = L1/C1, and Z02 = L2/C2 are the phase constants and characteristic impedances of transmission line A and region B, respectively. When the wave intersects a boundary between the transmission lines, the tan- gential components of both the electric and magnetic ﬁelds across the interface must remain continuous, as shown with equation (3-8). In other words, the tan- gential component of the ﬁelds cannot change instantaneously. In our particular scenario, the plane wave is propagating in the TEM mode in the z-direction, so both the electric and magnetic ﬁelds are oriented parallel (i.e., tangent) to the boundary of the dielectric interface. Subsequently, we can say that at the interface (z = 0), the sum of the incident and reﬂected waves must equal the transmitted wave. vt (z = 0) = vi(z = 0) + vr (z = 0) → vt = vi + vr it (z = 0) = ii (z = 0) + ir (z = 0) → it Z02 = ii Z01 − ir Z01 (3-98) (3-99) Since the incident waves are known, we can solve (3-98) and (3-99) simultaneously for the transmitted and reﬂected portions of the wave: vt = vi 2Z02 Z02 + Z01 vr = vi Z02 Z02 − + Z01 Z01 (3-100) (3-101) TRANSMISSION-LINE REFLECTIONS 115 Following the method used to derive equations (2-132) and (2-133) leads to the deﬁnition of the reﬂection and transmission coefﬁcients: ≡ vr vi = Z02 − Z01 Z02 + Z01 T ≡ vt vi = 2Z02 Z02 + Z01 =1+ (3-102) (3-103) The reﬂection coefﬁcient is a measure of how much is reﬂected back off the intersection between the two impedance regions, and the transmission coefﬁcient tells how much of the wave is transmitted . If the reﬂection coefﬁcient is zero, it means that the characteristic impedances in the two regions are identical. If the characteristic impedances are not equal, the reﬂection coefﬁcient will be ﬁnite. If the impedance discontinuity is inﬁnite, such as an open circuit, the signal propagating on transmission line A will be reﬂected 100%, as shown in Figure 3-28a. This is easy to show simply by taking the limit of (3-102) as Z02 goes to inﬁnity: open = 1 (3-104) Incident wave ∼R Reflected wave Z0 Γ=1 (a) Incident wave ∼R Reflected wave Z0 Γ = −1 (b) Figure 3-28 Reﬂections caused by (a) open- and (b) short-circuit termination. 116 IDEAL TRANSMISSION-LINE FUNDAMENTALS If the impedance discontinuity is shorted to ground, as shown in Figure 3-28b, the signal propagating on transmission line A will also be reﬂected 100%, but the reﬂected wave will be out of phase from the incident by 180◦, yielding short = −1 (3-105) The reason for the phase shift is explained in Section 2.7.1, for a plane wave incident on a perfect conductor. 3.5.2 Launching an Initial Wave When a driver launches a signal onto a transmission line, the magnitude of the initial voltage (vi) seen on the transmission line at t = 0 will be governed by the voltage divider between the source resistance and the line impedance (see Figure 3-29): vi = vs Z0 Z0 + Rs (3-106) If the end of the transmission line is terminated with an impedance that exactly matches the characteristic impedance of the line, the signal with amplitude vi will be terminated to ground and the voltage vi will remain on the line until the signal source switches again. If the end of the transmission line is terminated with a resistance different than the characteristic impedance of the line, a portion of the signal will be terminated to ground and the remainder of the signal will be reﬂected back down the transmission line toward the source. 3.5.3 Multiple Reﬂections As described above, when a signal is reﬂected from an impedance discontinuity at the end of the line, a portion of the signal will be reﬂected back toward the source. The amount of signal reﬂected back is determined by the reﬂection coefﬁcient between the transmission-line impedance (Z0) and the termination resistance (Rt ) and calculated with (3-102), where Z02 = Rt . When the incident wave hits the termination, Rt , a portion of the signal, vi t , is reﬂected back υs Rs l Rt zo t = 0 Vi vi Γt Γs t = 2td vi Γt t = td v (t = τd) = vi + viΓt v (t = 2td) = vi + viΓt + vi Γt Γs Figure 3-29 Initial voltage launched onto a transmission line and subsequent reﬂections. TRANSMISSION-LINE REFLECTIONS 117 toward the source and is added to the incident wave to produce a total magnitude on the line of vi t + vi, as shown in Figure 3-29. The reﬂected component (vi t ) will then travel back to the source and will generate another reﬂection if Z0 = Rs. Obviously, if Z0 = Rs, there will be no reﬂection at the source. If an impedance discontinuity exists on both sides of the transmission line, the signal will bounce back and forth between the driver and receiver until the reﬂections eventually reach steady state at the dc solution. For example, consider Figure 3-29, which shows one example for a time interval of a few τd , where τd is the time delay for the signal to propagate from one end of the transmission line to the other, which is derived from equation (3-31) and calculated by τd = l νp √ = l LC seconds (3-107) where l is the length of the transmission line and L and C are the inductance and capacitance values per unit length.† When the source transitions from 0 volts to vs, the initial voltage on the line at t = 0 is vi, determined by (3-106). At time t = τd , the incident voltage vi arrives at the load Rt . At this time a reﬂected component is generated with a magnitude of vi t , where t is the reﬂection coefﬁcient looking into the terminating load): t = Rt Rt − Z0 + Z0 The reﬂected component is added to the incident voltage vi, creating a total voltage at the load of vi t + vi. The reﬂected portion of the wave vi t then travels back to the source and at time t = 2τd generates a reﬂection off the source determined by vi t s, where s = Rs Rs − Z0 + Z0 is the reﬂection coefﬁcient looking into the source. At this time the voltage seen at the source will be the previous voltage (vi) plus the voltage from the reﬂection (vi t ) plus the reﬂected wave (vi t s). This reﬂecting and counter reﬂecting will continue until the line voltage has approached the steady-state dc value. As the reader can see, the reﬂections could take a long time to reach equilibrium if the termination and source resistors do not have values similar to the characteristic impedance of the transmission line. †Note that the phase velocity, υp, calculated in (3-31) is a frequency-domain variable that is valid only at a single frequency. The time delay will change with frequency because of non-TEM propagation (Section 3.3.5) and realistic dielectric behavior, as discussed in Chapter 6. However, for the purposes here, it is valid to assume frequency-invariant dielectric properties, which allows (3-107) to be applied over the broad range of frequencies contained in a trapezoidal digital waveform. 118 IDEAL TRANSMISSION-LINE FUNDAMENTALS It is apparent from the example above that hand calculation of multiple reﬂections can be rather tedious. An easier way to predict the effect of reﬂections on a signal is to use a lattice diagram. 3.5.4 Lattice Diagrams and Over- or Underdriven Transmission Lines A lattice diagram (sometimes called a bounce diagram) is a graphical technique used to solve the multiple reﬂections on a transmission line with linear loads. Figure 3-30 shows a sample lattice diagram for the transmission line shown in Figure 3-29. The left- and right-hand vertical lines represent the source end (z = 0) and load end (z = l) of the transmission line. The diagonal lines contained between the vertical lines represent the signal bouncing back and forth between the source and the load. The diagram progressing from top to bottom represents increasing time. Notice that the time increment is equal to the time delay τd of the transmission line as calculated with (3-107), and the reﬂection coefﬁcients looking into the source and into the load are labeled at the top of the vertical bars. The lowercase letters represent the magnitude of the reﬂected signal traveling on the line. The uppercase letters represent the voltages seen at the source, and the primed uppercase letters represent the voltage seen at the load end of the line. For example, referring to Figure 3-30, the near end of the line will be held at 0 l z Γs Γt vA 0 va td vA′ vb vB 2td vc 3td vB′ vd vC 4td t Figure 3-30 Structure of a lattice diagram for the transmission line in Figure 3-29. TRANSMISSION-LINE REFLECTIONS 119 a voltage of vA volts for a duration of 2τd . The voltage va is simply the initial voltage vi, which will remain constant until the reﬂection from the load reaches the source and is calculated with (3-106): vA = va = vs Zo Zo + Rs The voltage va will propagate down the transmission line toward the source. The voltage vA is the voltage va plus the voltage reﬂected off the load, vb, where vb = va t vA = va + vb The voltage vB is the sum of the incident voltage va plus the signal reﬂected from the load vb and the signal reﬂected off the source vc: vB = va + vb + vc where vc = vb s The reﬂections on the line eventually reach the steady-state voltage of the source, vs, if the line is open. However, if the line is terminated with a resistor Rt , the steady-state voltage is computed with the voltage divider between the source and the load (assuming that the transmission line is loss-free): vsteady state = vs Rt Rt + Rs Example 3-4 Multiple Reﬂections for Rs > Z0 Consider the transmission-line system and lattice diagram shown in Figure 3-31. The initial voltage launched onto the transmission line will be governed by the voltage divider between the source impedance Rs and the line impedance Z0: vi = vs Z0 Z0 + Rs =2· 50 50 + 75 = 0.8 V The initial signal, va = 0.8 V, will travel down the line until it reaches the load. In this particular case, the load is open and thus has a reﬂection coefﬁcient of 1. Subsequently, the entire signal is reﬂected back toward the source and is added to the incident signal of 0.8 V. So at time t = τd , or 250 ps in this speciﬁc example, the signal seen at the load is vA = va + vb = 0.8 + 0.8 = 1.6 V. The 0.8-V reﬂected signal vb will then propagate down the line toward the source. 120 IDEAL TRANSMISSION-LINE FUNDAMENTALS vs 75 Ω ts = 250 ps 50 Ω B 0-2 V A 0 Γs 0 250 ps 500 ps 1.76 V 750 ps 1000 ps 1.952 V (a) 0.8 V 0.8 V 0.16 V 0.16 V 0.032 V l z Γt 1.6 V 1.92 V Volts (b) 2.5 2 1.5 1 Waveform at A 0.5 Waveform at B 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time, ns (c) Figure 3-31 (a) Example of an underdriven transmission line; (b) lattice diagram; (c) digital waveform. When vb reaches the source, part of the signal will be reﬂected back toward the load, as determined with the reﬂection coefﬁcient looking into the source using (3-102): source = 75 75 − + 50 50 = 0.2 The value reﬂected back toward the load is vc = vb s = (0.2)(0.8) = 0.16 V. The reﬂected signal will be added to the signal already present on the line, which TRANSMISSION-LINE REFLECTIONS 121 will give a total magnitude at the source end vB = va + vb + vc = 0.8 + 0.8 + 0.16 = 1.76 V, with the reﬂected portion vc of 0.16 V traveling toward the load. This process is repeated until the voltage reaches a steady-state value of 2 V. If the same procedure is applied to the falling edge of a digital waveform, the signal integrity of a digital pulse propagating on this system can be calculated, as shown in Figure 3-31c. Notice how the reﬂections give the waveform a “stair-step” appearance at the receiver (node B), even though the unloaded output of the voltage source is a square wave. This effect occurs when the source impedance (Rs) is larger than the characteristic impedance (Z0) and is referred to as an underdriven transmission line. Example 3-5 Multiple Reﬂections for Rs < Z0 When the characteristic impedance of the transmission line is greater than the source impedance, as shown in Figure 3-32a, the reﬂection coefﬁcient looking into the source will be negative: source = 25 25 − + 50 50 = −1 3 When the lattice diagram is solved, as shown in Figure 3-32b, it is easy to show that a negative reﬂection at the source will produce a “ringing” effect. This is known as an overdriven transmission line. The resulting distorted digital waveform is shown in Figure 3-32c. Since the procedure for solving the lattice diagram is identical to Example 3-3, the exercise is left to the reader. 3.5.5 Lattice Diagrams for Nonideal Topologies Real bus designs rarely employ only a single transmission line. For example, even in point-to-point designs, the silicon driver is connected to the main bus through a package, which often employs transmission lines anywhere from 0.25 to 1.0 in. long. Also, many high-speed designs use add-in cards, where two separately manufactured printed circuit boards are interfaced through a connector. Furthermore, it is not uncommon to encounter designs where one driver is sending data to multiple receivers, such as a front-side bus of a multiprocessor system, necessitating the solution of multiple reﬂections in parallel. Consequently, it is important to explore techniques for understanding and solving systems with multiple transmission-line segments in a variety of topologies. Cascaded Topologies Consider the transmission-line structure depicted in Figure 3-33, which consists of two transmission-line segments cascaded in series. The ﬁrst section is of length l1 and has a characteristic impedance of Z01 ohms. The second section is of length l2 and has an impedance of Z02 ohms. Finally, the structure is terminated with a value of Rt . When the signal encounters the Z01: Z02 impedance junction, part of the signal (vc) will be 122 IDEAL TRANSMISSION-LINE FUNDAMENTALS vs 25 Ω td = 250 ps 50 Ω B 0-2 V A (a) 0 l z Γs Γt 0 250ps 500ps2.22 V 750ps 1000ps1.92 V 1.33 V 1.33 V −0.443 V −0.443 V 0.148 V t (b) 3 2.66 V 1.77 V 2.07 V 2 Volts 1 Waveform at A Waveform at B 0 1 2 3 4 5 −1 Time, ns (c) Figure 3-32 (a) Example of an overdriven transmission line; (b) lattice diagram; (c) digital waveform. reﬂected, as governed by the reﬂection coefﬁcient (3-102) looking into line 2 from line 1, 2 = Z02 Z02 − + Z01 Z01 and part of the signal (vb) will be transmitted, as governed by the transmission coefﬁcient, as deﬁned in equation (3-103): T2 = 2Z02 Z02 + Z01 = 1+ 2 TRANSMISSION-LINE REFLECTIONS 123 vs Rs Γ1 l1 l2 z01 z02 T2 T3 Γ2 Γ3 Rt Γ4 0 va td vc 2td vB vd 3td vf 4td vC vh vb ve vA′ vg vi vB′ Figure 3-33 Lattice diagram for two cascaded transmission lines with different impedance values and identical lengths (l1 = l2). Figure 3-33 also depicts how a lattice diagram can be used to solve for multiple reﬂections on a transmission-line system with a series of transmission lines with more than one characteristic impedance. Note that the transmission lines in this example are of equal length (l1 = l2), which simpliﬁes the problem because the reﬂections from each section will be in phase. For example, in Figure 3-33, the transmitted portion of ve adds directly to the reﬂection, vf . When the two transmission lines are of different lengths, the reﬂections from one section will not be in phase with the reﬂections from the other section, which complicates the diagram drastically. When the signal reaches the termination, the reﬂection is governed by the reﬂection coefﬁcient looking into the termination resistance at the load ( 4): 4 = Rt Rt − Z02 + Z02 The portion of the signal reﬂected off the termination resistor will travel back toward the source and experience another reﬂection when it reaches the junction between transmission lines, 3 = Z01 Z01 − Z02 + Z02 where 3 is the reﬂection looking into line 1 from line 2. Part of the signal will be reﬂected back toward the load as calculated by 3 and part transmitted toward the source, as dictated by the transmission coefﬁcient T3: T3 = 1 + 3 124 IDEAL TRANSMISSION-LINE FUNDAMENTALS The portion of the signal transmitted through the junction toward the source will experience another reﬂection when it arrives at Rs: 1 = Rs Rs − Z01 + Z01 The signal will subsequently bounce back and forth between the source and the termination load until equilibrium is reached. The voltage levels are calculated in the same manner as the single-line lattice diagram with a little more accounting. The initial voltage launched onto the line is va = vs Z01 Z01 + Rs and the voltage levels of the reﬂections on the line are vb = vaT2 vc = va 2 vd = vc 1 ve = vb 4 vf = vd 2 + veT3 vg = ve 3 + vd T2 vh = vf 1 vi = vg 4 giving source-side voltages of vB = va + vc + vd vC = va + vc + vd + vf + vh and load voltages of vA = vb + ve vB = vb + ve + vg + vi where the remaining reﬂections are left for the reader to calculate. Multireceiver Topologies So far in this book we have covered many issues that deal with an interconnect connecting two components. However, this is not always the case. Often, it is required that a single driver be connected to two or TRANSMISSION-LINE REFLECTIONS 125 vs Rs = Z0 l1 z01 0-2 V l2 z02 l2 = l3 z02 Receiver 1 Receiver 2 l3 Volts 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time, ns Figure 3-34 Signal integrity of a T-topology when the leg lengths and characteristic impedances are equal. more receivers. In these cases, the topology of the interconnect can affect system performance dramatically. For example, consider Figure 3-34, which is a case where one driver is connected to two receivers. In this example, the impedance of the base (Z01) is equal to the impedance of the two legs (Z02), and the legs are of equal length (l2 = l3). When the signal propagates to the junction, it will see an effective impedance of Z02/2, resulting in the waveform in the ﬁgure that steps up toward the ﬁnal value in the same manner as an underdriven transmission line as calculated in Example 3-4. When the impedance of the legs is twice that of the base (Z02 = 2Z01), the effective impedance the signal will see at the junction will be equal to the base, so that no reﬂections are generated. When the structure is unbalanced, as in the case where one leg is longer than the other, the signal integrity will deteriorate dramatically because the reﬂections will arrive at the junction at different times. To gain an intuitive understanding of how the multiple reﬂections from different legs interact, it is useful to solve a multiple-legged lattice diagram at least once, which is demonstrated in the following example. Example 3-6 Calculate the ﬁrst few reﬂections of the unbalanced T-topology shown in Figure 3-35 assuming that Z0 = Rs = 50 , l1 and l3 are lengths that corresponds to a propagation delay of 250 ps, and l2 has a delay of 125 ps. SOLUTION Referring to the lattice diagram in Figure 3-35, the ﬁrst and second vertical lines represents the electrical pathway between the driver and the junction, the third vertical line represents the pathway between the junction and the end of the short line (receiver 1), and the fourth vertical line represents the 126 IDEAL TRANSMISSION-LINE FUNDAMENTALS vs Rs = Z0 l1 z0 0-2 V Γ1 0 T3 T2 Γ2 Γ3 l2 z0 l3 z0 Receiver 1 Receiver 2 Γ4 Γ5 250ps 500ps 750ps 1000ps a a b a c bg A d ey B f Figure 3-35 Lattice diagram of a T-topology when the leg lengths are not equal. end of the long line (receiver 2). The initial voltage step launched onto line 1 is vi = vs Z0 Z0 + Rs =2· 50 50 + 50 =1 The reﬂection and transmission coefﬁcients looking from line 1 into the junction is 2 = (Z0/2) − Z0 (Z0/2) + Z0 = 25 − 50 25 + 50 = −1 3 T2 = 1 + 2 = 2 3 Consequently, the initial voltage launched into both legs (lines 2 and 3) is va = T2vi = 2 3 This voltage (va) travels down each leg and doubles when it arrives at the open circuit ( 4 = 5 = 1). Therefore, the voltage at receiver 1 (vα) occurs at t = 375 ps, which is the delay of line 1 plus line 2 (the short leg). vb = va 4 = 2 3 vα = va + vb = 4 3 TRANSMISSION-LINE REFLECTIONS 127 The voltage at receiver 2 (vA) occurs at t = 500 ps, which is the delay of line 1 plus line 2 (the long leg): vA = va + vg = va + va 5 = 4 3 For both legs, a reﬂection of vb = vg = 2 3 is reﬂected from the open circuit at the receivers; however, they will arrive at the junction at different times. As seen by the lattice diagram, at t = 500 ps, vb will arrive at the junction. Part of vb will be reﬂected back toward receiver 1, 3 = (Z0/2) − Z0 (Z0/2) + Z0 = 25 − 25 + 50 50 = −1 3 vc = vb 3 = 2 3 −1 3 = −2 9 and part will be transmitted onto line 1 toward the source and onto line 3 toward receiver 2: T3 = 1 + 3 = 2 3 The voltage at receiver 1 at t = 625 ps (vβ) is calculated: 2 vd = 4vc = − 9 vβ = vα + vc + vd = 4 3 − 2 9 − 2 9 = 8 9 To calculate the voltage at receiver 1 at t = 875 ps (vψ ), it is necessary to account for the portion of the signal reﬂected from receiver 2 at 500 ps and transmitted into the junction at 750 ps that travels toward receiver 1. This combination of reﬂections can be seen by observing the lattice diagram, where reﬂection g arrives at the junction simultaneously with reﬂection d. ve = 3vd = −2 9 −1 3 =2 27 vf = 4ve = 2 27 vg = 2 3 (from above) vψ = vβ + ve + vf + vgT3 + vgT3 4 = 8 + 2 + 2 + 2 2 + 2 2 (1) = 52 9 27 27 3 3 3 3 27 128 IDEAL TRANSMISSION-LINE FUNDAMENTALS By observing the lattice diagram, we see that the voltage at receiver 2 at t = 750 ps is calculated by accounting for the voltage reﬂected from receiver 1 at t = 375 ps and transmitted into the junction at t = 500 ps: vB = vA + T3vb + T3vb 5 = 4 3 + 2 3 2 3 +2 3 2 3 (1) = 20 9 This process can be continued until the waveform has reached steady state. The complete waveforms for this example are shown in Figure 3-36, with the ﬁrst few reﬂections (just calculated) labeled. Note that the complicated interactions between the reﬂections from each leg severely degrade the integrity of the signal. As more legs are added to the topology, it becomes more sensitive to differences in the electrical length of the legs. Furthermore, a mismatch between the source resistance and the characteristic impedance of line, differences between receiver loads, and impedance deltas between each leg will cause similar instabilities. vs Rs = Z0 l1 z0 0-2 V l2 z0 Receiver 1 l2 >l3 z0 Receiver 2 Volts 2.5 52 2 4 27 1.5 3 Receiver 1 1 0.5 8 9 0 −0.50 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time, ns Volts 2.5 20 9 2 4 1.5 3 Receiver 2 1 0.5 0 −0.50 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time, ns Figure 3-36 Signal integrity of a T-topology when the leg lengths are not equal. TRANSMISSION-LINE REFLECTIONS 129 So what can we learn form this? The answer is: symmetry. Whenever a topology is considered, the primary area of concern is symmetry. Make certain that the topology looks symmetrical from the point of view of any driving agent. This is usually accomplished by ensuring that the lengths, impedances, and loading are identical for each leg of the topology. The secondary concern is to minimize the impedance discontinuities at the topology junctions, although this may be impossible in some designs. 3.5.6 Effect of Rise and Fall Times on Reﬂections The rise and fall times of real digital waveforms begin to have a signiﬁcant effect on the wave shape when they become less than twice the delay (2τd) of the transmission line. Figures 3-37 and 3-38 show the effect that ﬁnite rise and fall times have on over- and underdriven transmission lines. Notice how signiﬁcantly the wave shape changes as the rise time exceeds twice the delay of the line. When the edge rate exceeds twice the line delay, the reﬂections are masked because the amount of time that it takes to transition from a low state to a high state (or vice versa) exceeds the period of the reﬂections. 3.5.7 Reﬂections from Reactive Loads In real systems, there are rarely cases where the loads are purely resistive. The input to a CMOS gate, for example, tends to be capacitive. Additionally, bond wires, vias, lead frames of the chip packages, chip sockets, and daughtercard vs 25 Ω 0-2 V ts = 250 ps 50 Ω A (a) tr = 10ps 3 Voltage at A 2 tr = 250ps 1 tr = 500ps tr = 750ps 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 −1 Time, ns (b) Figure 3-37 (a) Overdriven transmission line; (b) example of how increased rise and fall times mask the reﬂections. 130 IDEAL TRANSMISSION-LINE FUNDAMENTALS vs 75 Ω 0-2 V τd = 250 ps 50 Ω A (a) Voltage at A 3 tr = 10ps 2 tr = 250ps 1 tr = 500ps tr = 750ps 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 Time, ns (b) Figure 3-38 (a) Underdriven transmission line; (b) example of how increased rise and fall times mask the reﬂections. connectors tend to be inductive. This makes it necessary to understand how these reactive elements affect the reﬂections in a transmission-line system. In this section we brieﬂy introduce the effect that capacitors and inductors have on reﬂections. Reﬂection from a Capacitive Load When a transmission line is terminated in a reactive element such as a capacitor, the shape of the waveforms at the driver and the load will be dependent on the value of the capacitor, the characteristic impedance of the transmission line, and any resistive terminations that may be present. Essentially, a capacitor is a time-dependent load that will look initially like a short circuit when the signal reaches the capacitor and will then look like an open circuit after the capacitor is fully charged. Let’s consider the reﬂection coefﬁcient at time t = τd (the delay of the transmission line). At time t = τd, which is the time when the signal has propagated down the line and has reached the capacitive load, the capacitor will not be charged and will look like a short circuit. As described earlier in the chapter, a short circuit will have a reﬂection coefﬁcient of −1. This means that the initial wave of magnitude vi will be reﬂected off the capacitor with a magnitude of −vi, yielding an initial voltage of 0 V. The capacitor will then begin to charge at a rate dependent on τ , which is the time constant of an RC circuit, where C is the termination capacitor value and R is the characteristic impedance of the transmission line. Once the capacitor is fully charged, the reﬂection coefﬁcient will be 1 since the capacitor will resemble an open circuit. Equation (3-108), which is the step response of a simple network with a time constant τ , approximates the voltage at the end of a transmission TRANSMISSION-LINE REFLECTIONS 131 line terminated with a capacitor beginning at time t = τd , vcapacitor = vss 1 − e−(t−τd )/τ ] t > τd (3-108) where τ = CZ0 is the time constant, τd the time delay of the transmission line as given by (3-107), and vss the steady-state voltage determined by the voltage source vs and the voltage divider between the source resistance Rs and the termination resistance Rt . Note that (3-108) is an approximation because it assumes a step function with an inﬁnite edge rate (i.e ., the rise time is inﬁnitely fast). Figure 3-39 shows the response of a line terminated with a capacitive load. The waveform shape at node B follows equation (3-108). Notice that the waveform at the source (node A) dips toward zero at t = 500 ps (which is 2τd ) because the capacitor initially looks like a short circuit, so the reﬂection coefﬁcient is −1. Note that the shape of this wave at node A is also dictated by (3-108) when the exponent term is e−[(t−2τd )/τ], which simply shifts the time. The voltage reﬂected back toward the source is initially vi, where vi is the initial voltage launched onto the transmission line. After the capacitor is fully charged, it will look like an open and have a reﬂection coefﬁcient of +1. Consequently, the reﬂected waveform at the receiver will double. As seen in Figure 3-39b, the waveform at the receiver (B) reaches a steady-state value of 2 V after about three time constants [3τ = 3(50 )(2 pF = 300 ps] after arriving at the receiver, just as circuit theory predicts. If the line is terminated with a parallel resistor and capacitor, as depicted in Figure 3-40, the voltage at the capacitor will be dependent on the time constant vs Rs = 50 Ω A 0-2 V td = 250 ps Z0 = 50 Ω B CL = 2pF (a) 2.5 2 B 1.5 Volts 1 A 0.5 00 0.25 0.5 0.75 1 Time, ns (b) Figure 3-39 (a) Transmission line terminated with a capacitive load; (b) step response showing reﬂections from the capacitor. 132 vs Rs IDEAL TRANSMISSION-LINE FUNDAMENTALS td Rt Z0 CL Figure 3-40 Transmission line terminated in a parallel RC network. between CL and the parallel combination of Rt and Z0: τ1 = CLZ0Rt Rt + Z0 (3-109) Reﬂection from an Inductive Load In the real world, when a transmission line is terminated, there is usually a series inductance caused by the physical connection between the transmission line and the resistor. Some common examples of this inductive connection are bond wires, lead frames, and vias. When a series inductor appears in the electrical pathway of a transmission-line termination, as depicted in Figure 3-41a, it will also act as a time-dependent load. Initially, at vs Rs = 50 Ω A 0-2 V τd = 250 ps Z0 = 50 Ω L = 1 nH Rt = 50 Ω (a) 2 A 1.5 Volts 1 0.5 B 0 0.2 0.3 0.4 0.5 0.6 Time, ns (b) Figure 3-41 (a) Transmission line terminated with a series LR load; (b) step response showing reﬂections from the inductor. TRANSMISSION-LINE REFLECTIONS 133 time t = τd , the inductor will resemble on open circuit. When a voltage step is applied initially, no current ﬂows across the inductor. This produces a reﬂection coefﬁcient of 1, causing an inductive spike seen as a reﬂection at node A in Figure 3-41b. The value of the inductor will determine how long the reﬂection coefﬁcient will remain 1. If the inductor is large enough, the signal will double in magnitude. Eventually, the inductor will discharge its energy at a rate dependent on the time constant τ of an LR circuit. For the circuit depicted in Figure 3-41a, the wave shape of the rising edge at node B is calculated: vinductor = vss 1 − exp − (t − τd )(Z0 + Rt ) L t > τd (3-110a) Note that the wave shape calculated by (3-110a) will also be valid for the falling edge of the inductive spike, shown at node B in Figure 3-41b, if τd is adjusted to shift the waveform to the correct position in time (2τd ), the waveform is inverted, and dc is shifted to the appropriate level: vinductor = vss 1 + exp − (t − 2τd )(Z0 + Rt ) L (3-110b) Filtering Effects of Reactive Components Figures 3-39b and 3-41b show how the series inductor and the shunt capacitor affect the signal integrity. The series inductance will cause an inductive spike, which is seen as a positive reﬂection, the capacitance will cause a capacitive dip, which is seen as a negative reﬂection, and both will smooth the rising and falling edges seen at the receiver (node B). To understand why the edges are smoothed, we must explore how an inductor or a capacitor will ﬁlter the harmonics of a digital waveform. In Chapter 8 it will be shown that high-frequency harmonics are associated with the rising and falling edges of a digital waveform. Consequently, if the higher-frequency har- monics are ﬁltered out by the capacitive or inductive loads, the rising and falling times will be increased. Equation (3-111) shows that the impedance of the shunt capacitor will decrease with frequency, which means that the higher-order har- monics of the digital waveform will be shunted to ground, increasing the rise and fall times: Zcap = 1 j ωC (3-111) where ω = 2πf . Similarly, (3-112) shows that the series impedance of an inductor will increase with frequency, which will also tend to ﬁlter out the higher harmonics because they will experience larger impedances than the lower-frequency harmonics. Zind = j ωL (3-112) 134 IDEAL TRANSMISSION-LINE FUNDAMENTALS Consequently, for a digital pulse, reactive components such as inductors and capacitors will low-pass-ﬁlter the waveform, resulting in increased rise and fall times. The exception to this statement is when speciﬁc ﬁlters are constructed using reactive components to equalize a channel, which is described is Chapter 12. 3.6 TIME-DOMAIN REFLECTOMETRY In practice, it is often necessary to measure a system of transmission lines in a digital system to ensure compliance with design guidelines, to understand the equivalent circuit of a component, or simply to verify that the simulations are predicting the transient response correctly. One measurement method often used is called time-domain reﬂectrometry (TDR). A TDR measurement uses the premise of multiple reﬂections discussed in Section 3.5 to derive an impedance proﬁle of the device under test (DUT). Figure 3-42 depicts a generic TDR setup where a step function is driven onto the DUT through a 50- cable. A sampling oscilloscope is used to observe the waveform at A, which depicts the voltage proﬁle of the reﬂected waves. The voltage proﬁle is converted to an impedance proﬁle, which can be used to measure the characteristic impedance and propagation delay of a transmission line, estimate inductance and capacitance values of structures such as vias, bond wires, and lead frames, and deduce the form of an equivalent circuit for many applications. 3.6.1 Measuring the Characteristic Impedance and Delay of a Transmission Line One of the most common uses for TDR measurements is to determine the characteristic impedance of transmission lines. The impedance proﬁle is calculated Oscilloscope Sampler Voltage profile Step generator vs Rs = 50 Ω A Z0 = 50 Ω DUT Figure 3-42 Generic time-domain reﬂectometry measurement setup. TIME-DOMAIN REFLECTOMETRY 135 from the voltage by rearranging equation (3-102), ≡ vr vi = ZDUT − Z0 ZDUT + Z0 ZDUT = Z0 vi vi + − vr vr (3-113) where vi is the incident voltage transmitted and vr is the voltage reﬂected from the DUT. Figure 3-43 depicts the TDR waveforms of a 60- transmission line with a length that corresponds to a delay of 250 ps. The impedance of the transmission line under test is calculated from the ﬁrst voltage step that occurs at 500 ps using equation (3-113): ZDUT = 50 · 1 1 + − 0.091 0.091 = 60 We can also calculate the delay of the transmission line. Since a TDR is essentially a reﬂected voltage at the driver, the reﬂection of the DUT remains for the time it takes the signal to propagate to the end of the DUT and return to the source from the open circuit. Consequently, the duration of the reﬂection will correspond to twice the delay of the transmission line under test, as shown in Figure 3-43. vs Rs = 50 Ω A 0-2 V Zo = 50 Ω td = 250 ps ZDUT = 60 Ω DUT Voltage at A Voltage at A 2.5 2 1.5 2td = 500 ps 1 0.5 vi = 1.0 V 0 0 0.5 1 1.5 2 Time, ns 1.1 1.05 1 vi = 0.091 V 0.95 0.9 0.2 0.4 0.6 0.8 1 Time, ns Figure 3-43 TDR waveform measuring a 60- transmission line with a length that corresponds to a delay of 250 ps. 136 vs Rs = 50 Ω A 0-2 V IDEAL TRANSMISSION-LINE FUNDAMENTALS Zo = 50 Ω 2 inches ZDUT DUT Voltage at A 1.4 1.3 1.2 1.1 1 0.9 0 0.2 0.4 0.6 0.8 1 Time, ns 1.2 1.4 Figure 3-44 TDR proﬁle for Example 3-7. Example 3-7 Calculate the equivalent L and C per unit length of the 2-in. transmission line measured with TDR proﬁle shown in Figure 3-44. SOLUTION The impedance is calculated from the step voltage at 1.2 V, which corresponds to the reﬂections from the transmission line under test. vr = 1.2 − 1.0 = 0.2 V vi = 2Z0 Z0 + Rs =1 V ZDUT = Z0 vi vi + vr − vr = 50 · 1 + 0.2 1 − 0.2 = 75 The delay is calculated from the duration of the step: τd = 1 2 (1.1 ns − 0.5 ns) = 300 ps The distributed inductance and capacitance values are calculated by solving the impedance and delay equations simultaneously: Z0 = L = 75 C √ τd = l LC = 300 ps TIME-DOMAIN REFLECTOMETRY 137 The inductance per unit length is calculated by normalizing the delay to line length of the transmission line under test and multiplying by the impedance: τd √ = LC = 150 ps/in. 2 in. L= L√ LC = 75(150 × 10−12) = 11.25 nH/in. C The capacitance is calculated by dividing the normalized delay by the impedance: √ C = √ LC = 150 × 10−12 = 2.0 pF/in. L/C 75 3.6.2 Measuring Inductance and Capacitance of Reactive Structures In Section 3.5.7 we discussed reﬂections from reactive loads. In this section we show how to estimate the value of the capacitance or the inductance from a measured TDR proﬁle. The analysis presented here is somewhat idealized because it assumes an ideal step for a source. In reality, ideal steps are impossible to generate. However, modern TDR measurement equipment allows the generation of very fast rise times that vary anywhere between 9 and 35 ps, which are fast enough to approximate a step function for a variety of applications. Inductive Structures In a TDR measurement, narrow spikes such as those depicted in Figure 3-41 are indicative of an inductive component such as a bond wire, a connector pin, or a package lead frame. Assuming that the input step has a sufﬁciently fast rise time, the value of the inductor can be estimated by measuring the area under the inductive spike, as shown in Figure 3-45. The area is calculated by integrating equation (3-110b) after subtracting the dc offset and assuming that Rt = Z0: vss = vs Rt Rt + Rs = 50 50 + 50 = vs 2 Aind = ∞ vs 2τd 2 1 + e− (t−2τd )(Z0+Rt ) L = vs ∞ e−(2Zo/L)t dt = vsL 20 4Z0 L = 4Z0Aind vs − vs dt 2 (3-114) Therefore, through careful measurement of the area under the inductive spike, we can achieve a good approximation of the inductance. The accuracy is maximized when the rise time is very fast compared to the duration of the inductive spike. 138 IDEAL TRANSMISSION-LINE FUNDAMENTALS vs Rs = 50 Ω A Z0 = 50 Ω L Z0 = 50 Ω 0-2 V Rt = 50 Ω Voltage at A 2 1.8 1.6 Area = Aind 1.4 1.2 1 0.8 0.6 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 Time, ns Figure 3-45 The area under the reﬂection can be used to estimate the inductance. vs Rs = 50 Ω A 0-2 V Z0 = 50 Ω Z0 = 50 Ω C Rt = 50 Ω Voltage at A 1.2 1 0.8 0.6 Area = Acap 0.4 0.2 0 0.3 0.4 0.5 0.6 0.7 0.8 Time, ns Figure 3-46 The area under the reﬂection can be used to estimate the capacitance. For long rise times or very small inductance values, the area under the curve will be masked by a rise time similar to the reﬂections in Figure 3-37. Capacitive Structures In a TDR measurement, narrow dips such as that depicted in Figure 3-46 are indicative of an capacitive component such as a probe pad or a via pad. Assuming that the input step has a sufﬁciently fast rise time, the value TIME-DOMAIN REFLECTOMETRY 139 of the capacitance can be estimated by measuring the area under the capacitive dip. The area is calculated by integrating equation (3-108) after subtracting the dc offset, assuming that Rt = Z0 where τ = Z0C/2 vcapacitor = vs (1 − 2 e−[2(t−2τd )/ZoC]) Acap = ∞ vs − vs (1 − e−[2(t−2τd )/Z0C])dt 2τd 2 2 = vs ∞ e−2t /Z0C dt = vs CZ0 20 4 C = 4Acap vs Z0 (3-115) Example 3-8 Calculate the value of the shunt capacitance for the waveform in Figure 3-47 assuming the circuit shown in Figure 3-46. SOLUTION A ﬁne grid was overlaid on the waveform in Figure 3-47 so that the area under the curve can be estimated. The total number of squares under the capacitive dip is about 60. The area per square is Asquare = (0.04)(0.01 × 10−9) = 4 × 10−13 The total area is therefore Atot = 60Asquare = 2.4 × 10−11 Voltage at A 1.2 1 0.8 0.6 0.4 0.2 0 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 Time, ns Figure 3-47 TDR waveform for Example 3-8. 140 IDEAL TRANSMISSION-LINE FUNDAMENTALS vs Rs = 50 Ω A 0-2 V td = 250 ps Z0 = 50 Ω C = 1 pF L = 2 nH td = 250 ps Z0 = 30 Ω 2.00 Voltage at A 1.50 1.00 0.50 0.00 0 0.2 0.4 0.6 0.8 1 Time, ns 1.2 1.4 Figure 3-48 TDR proﬁle showing the relationship between the waveform and the individual circuit components. yielding an estimated capacitance of C = 4Acap = 4(2.4 × 10−11) = 0.96 × 10−12 F vs Z0 2(50) which is very reasonable because the waveform for this example was simulated using a capacitance value of 1.0 pF. More accuracy could be obtained by using a ﬁner grid. 3.6.3 Understanding the TDR Proﬁle Another very useful application of a TDR measurement is to deduce the form of the equivalent circuit. For example, consider the TDR waveform in Figure 3-48. The initial reﬂections from t = 0 to t = 0.5 ns correspond to the ﬁrst 50transmission line with a propagation delay τd of 250 ps. The capacitive dip at t = 0.5 ns corresponds to the capacitor, the peak at about t = 0.55 ns corresponds to the reﬂection from the inductor, and the ﬂat portion that begins at about t = 0.7 ns corresponds to the 30- transmission line with a propagation delay of 250 ps. Notice how the lumped elements (L and C) signiﬁcantly ﬁlter the edge and add an extra delay to the circuit, reducing the accuracy of the TDR measurement for structures following the reactive elements. REFERENCES Collins, Robert, 1992, Foundations for Microwave Engineering, McGraw-Hill, New York. PROBLEMS 141 Hall, S., G. Hall, and J. McCall, 2000, High-Speed Digital System Design, Wiley, New York. Hammerstad, E., and O. Jensen, 1980, Accurate models for microstrip computer-aided design, IEEE MTT-S International Microwave Symposium Digest, May, pp. 407–409. IPC, 1995, Design Guidelines for Electronic Packaging Utilizing High-Speed Techniques, IPC-D-317A, IPC, Chicago. Jackson, J. D., 1999, Classical Electrodynamics, 3rd ed., Wiley, New York. PROBLEMS 3-1 For the cross section in Figure 3-49, calculate the correct conductor width w to achieve 50- characteristic impedance, effective dielectric permittivity, propagation delay for a 10-in.-long line, and the equivalent inductance and capacitance per meter. er = 1 h = 3.9 er = 3.78 w t = 1.7 Figure 3-49 Cross section for Problem 3-1. 3-2 For the circuit shown in Figure 3-50 and the transmission line designed in Problem 3-1, use a lattice diagram to calculate the waveforms at nodes A and B when driven with a 1-V step. vs 35 Ω 0-1 V 10 inches 75 Ω Z0 = 50 Ω A B Figure 3-50 Circuit for Problem 3-2. 3-3 For the 50- cross section in Problem 3-1, use ﬁeld mapping techniques to calculate the effective dielectric permittivity, the propagation delay for a 10-in.-long line, and the equivalent inductance and capacitance per meter. Compare your answers to the results from Problem 3-1 to determine your accuracy. 3-4 Beginning with Laplace’s equation, derive a formula for the characteristic impedance of the stripline shown in Figure 3-51. Assume an inﬁnitely 142 IDEAL TRANSMISSION-LINE FUNDAMENTALS thin conductor with a uniform charge distribution. Compare the accuracy against equation (3-36c). h w εr h Figure 3-51 Cross section for Problem 3-4. 3-5 For the stripline in Problem 3-4, derive the impedance assuming realistic charge distribution. How does the charge distribution change the impedance? 3-6 For the stripline in Problem 3-4, calculate the characteristic impedance using ﬁeld mapping techniques. Compare your answer to those obtained in Problems 3-4 and 3-5. 3-7 For the 50- microstrip line designed in Problem 3-1, estimate the frequency where the dispersion due to the nonhomogeneous dielectric can no longer be ignored. 3-8 For the circuit shown in Figure 3-52, sketch the TDR waveform assuming a 2-V input step. Show all your calculations. vs Rs = 50 Ω td = 250 ps Z0 = 50 Ω 0-2 V TDR 2 nH td = 320 ps Z0 = 70 Ω td = 160 ps Z0 = 35 Ω 3 pF Figure 3-52 Circuit for Problem 3-8. 3-9 Draw the response of the circuit in Figure 3-53, where the device under test is (a) a 2-pF capacitor shunted to ground, (b) a series 2-pF capacitor, (c) a 4-nH series inductor, (d) a 4-nH inductor shunted to ground, and (e) a 1-in.-long 75- transmission line with εr = 4.0. PROBLEMS 143 vs 50 Ω td = 250 ps Z0 = 50 Ω 0-2 V TDR DUT 50 Ω Figure 3-53 Circuit for Problem 3-9. 3-10 For the microstrip transmission line designed in Problem 3-1, develop the equivalent circuit assuming that the input pulse is 100 ps wide with 25-ps rise and fall times. Perform a virtual TDR using the circuit in SPICE and conﬁrm that the impedance and delay match the values expected. 3-11 Calculate the ﬁrst few reﬂections of the stub topology shown in Figure 3-54 at nodes A, B, and C using a lattice diagram. Use the equivalent circuit in Problem 3-10 to simulate the structure in SPICE. Conﬁrm the results of your lattice diagram. vs 50 Ω 0-2 V td = 600 ps Z0 = 50 Ω A td = 250 ps Z0 = 50 Ω B td = 100 ps = 50 Ω Z0 C Figure 3-54 Topology for Problem 3-11. 3-12 For the topology in Figure 3-55, determine a set of design rules to ensure that the waveforms at node B, C, or D do not ring back below 0.7 V or above 0.3 V, and the rising and falling edges are linear between 0.3 and 0.7 V. Design the proper microstrip transmission-line cross sections to achieve proper impedance values for each segment, assuming a dielectric thickness of 4 mils and a relative dielectric permittivity εr = 4.2. Based on component placement requirements determined by a layout engineer, the length of line 1 (l1) can be as short as 1.5 in., l2 can be as short as 3 in., but it is impossible to route l3 shorter than 5.5 in. The design guidelines should produce limits on the lengths and impedance values of each leg as well as any other conditions needed to solve the problem. Do not design a point solution. Determine a solution space that guarantees proper signal integrity. Create equivalent circuits and verify the validity of your design 144 IDEAL TRANSMISSION-LINE FUNDAMENTALS guidelines through SPICE simulations. State all assumptions. Comment on the practicality of your design rules. D l1 vs 50 Ω A 0-2 V L = 3 inches Z0 l2 B l3 C Figure 3-55 Topology for Problem 3-12. 4 CROSSTALK 4.1 Mutual inductance and capacitance 146 4.1.1 Mutual inductance 147 4.1.2 Mutual capacitance 149 4.1.3 Field solvers 152 4.2 Coupled wave equations 153 4.2.1 Wave equation revisited 153 4.2.2 Coupled wave equations 155 4.3 Coupled line analysis 157 4.3.1 Impedance and velocity 157 4.3.2 Coupled noise 165 4.4 Modal analysis 177 4.4.1 Modal decomposition 178 4.4.2 Modal impedance and velocity 180 4.4.3 Reconstructing the signal 180 4.4.4 Modal analysis 181 4.4.5 Modal analysis of lossy lines 192 4.5 Crosstalk minimization 193 4.6 Summary 194 References 195 Problems 195 As described in Section 3.2, a signal propagates along an interconnect in the form of an electromagnetic wave established between two (or more) conductors. When neighboring transmission structures are in close proximity, the electric and magnetic ﬁelds from the signal will fringe and interact with adjacent conductors. The interaction of ﬁelds induces the coupling of energy from one transmission structure to another when a stimulus is applied. This is called crosstalk . Since Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 145 146 CROSSTALK most digital systems use signaling interfaces in which large numbers of transmission lines are routed in parallel through packages, connectors, and printed circuit boards, crosstalk can play an important role in determining the performance of the system. Trends toward smaller and faster systems will drive increased crosstalk levels in the future, resulting in two major impacts. First, crosstalk will affect signal integrity and timing by modifying the propagation characteristics of the lines (characteristic impedance and propagation velocity). Second, crosstalk couples noise onto lines, which harms signal integrity and reduces noise margins. We will see that signals on interconnects propagate in modes that are a function of the switching patterns and of the self- and mutual capacitances and inductances in the system. As was the case for isolated lines in Section 3.2.4, signals propagating in a coupled multiconductor transmission-line system are described by the wave equation. Whereas the solution to the isolated line system resulted in a pair of forward- and backward-traveling waves, for the coupled system with n signals there will be n forward- and n backward-traveling waves. Each forward–backward wave pair constitutes a mode, and we can analyze the behavior of the system in terms of the modes. In this chapter we introduce the mechanism that causes crosstalk, develop a mathematical formulation of the coupled wave equations, describe techniques for analyzing and modeling coupled systems, and discuss the impact of crosstalk on system performance. 4.1 MUTUAL INDUCTANCE AND CAPACITANCE From a circuit point of view, crosstalk is caused by mutual inductances and mutual capacitances between conductors. These two phenomena couple energy between lines via the magnetic (for mutual inductance) and electric (mutual capacitance) ﬁelds. We next examine each in more detail. Mutual inductance LM induces current from a driven line onto a quiet line by means of the magnetic ﬁeld, as shown in Figure 4-1. Conceptually, if the “victim” trace is in close enough proximity to a driven line such that its magnetic ﬂux lines intersect the victim trace, a current will be induced on that line. The mutual inductance creates a voltage noise on the victim in proportion to the rate of change of the current on the driven line according to vL = LM di dt (4-1) In equation (4-1), vL is the voltage coupled by the mutual inductance LM , in response to the transient current i. Mutual capacitance CM is the coupling of conductors via the electric ﬁeld. Conceptually, if the victim trace is in close enough proximity to a driven line such that its electric ﬁeld lines intersect the victim trace, a current is induced MUTUAL INDUCTANCE AND CAPACITANCE 147 H E Figure 4-1 Coupled PCB transmission lines. onto the victim line in proportion to the rate of change of voltage on the driven line: iC = CM dv dt (4-2) In equation (4-2), iC is the amount of current coupled through the mutual capacitance CM , when driven by the voltage signal v. I/O circuit rise and fall times decrease as data transfer rates increase, so (4-1) and (4-2) predict that both inductive and capacitive crosstalk will play a significant role in high-speed digital applications. Thus, we must account for mutual capacitance and inductance in our modeling and analysis of coupled systems, and we now proceed to examine each in more detail. 4.1.1 Mutual Inductance We start our discussion of mutual inductance by studying the simple inductive circuit shown in Figure 4-2. Given transient currents i1 and i2 injected into on lines 1 and 2, respectively, we can write expressions for v1 and v2 from Faraday’s law, v1 = L0 d i1 dt + LM d i2 dt v2 = L0 d i2 dt + LM d i1 dt (4-3) (4-4) where L0 is the self-inductance and LM is the mutual inductance between lines 1 and 2. From (4-3) and (4-4) we note that the potential differences v1 and v2 depend on both input currents and on the self- and mutual inductances. We can better understand the impact of mutual inductance by analyzing the situations where the input currents are equal (i1 = i2) and where they are opposite (i1 = −i2). In the equal-current case we assume that their transition times are equal, so that di1/t = di2/dt = di/dt. Applying these signals leads to di v1 = v2 = (L0 + LM ) dt (4-5) 148 CROSSTALK + v1 − i1 L0 LM i2 L0 + v2 − Figure 4-2 Coupled inductor circuit. Looking next at the opposing current ﬂow case, with the assumption that the transition time of current i1 is equal to that of i2, we have di1/dt = −di2/dt = di/dt, and derive v1 = −v2 = (L0 − di LM ) dt (4-6) When the two inputs are equal, we say that the system is being driven in even mode; in the case of opposite polarity inputs, we say that the system is driven in odd mode. Notice that the effective inductance of the system as seen by our input signals is changed by the mutual inductance and is a function of the switching pattern. In particular, the even-mode inductance is increased relative to the self-inductance by an amount equal to the mutual inductance. Correspondingly, the odd-mode inductance is decreased by the mutual inductance, giving us a relative comparison of self-, even-mode, and odd-mode inductances, Leven > L0 > Lodd.. As a ﬁrst step toward developing a general expression for the inductance, we write the equation for the voltage across the inductive elements as a function of inductances and input currents in matrix form: v1 v2 = L0 LM LM L0 d i1 /dt d i2 /dt (4-7) If we add a third line to our system, as shown in Figure 4-3, we can extend equation (4-7): v1 L11 L12 L13 di1/dt v2 = L21 L22 L23 di2/dt (4-8) v3 L31 L32 L33 di3/dt In equation (4-8), the diagonal elements L11, L22, and L33 represent the self-inductances on lines 1, 2, and 3, respectively. The mutual inductances are represented by Lij , where i and j correspond to the lines coupled by the mutual inductance. The inductance matrix is symmetric. In other words, the mutual inductances between lines do not depend on the direction, so that Lij = Lji. MUTUAL INDUCTANCE AND CAPACITANCE + i1 + i2 L11 v2 L22 v1 L12 − L13 149 − L33 L23 i3 + v3 − Figure 4-3 Circuit with three coupled inductors. As a ﬁnal step, we can generalize equation (4-8) to n inductively coupled lines, v1 L11 L12 · · · L1n di1/dt v...2 = L...n1 L22 ... L2n ... d i2/dt ... (4-9) vn Ln1 Ln2 · · · Lnn din/dt where Lii are the self-inductances and Lij are the mutual inductances. 4.1.2 Mutual Capacitance Our discussion of mutual capacitance follows a similar line of reasoning as for mutual inductance, beginning with the capacitive circuit shown in Figure 4-4. Given the input signals v1 and v2 on lines 1 and 2, we write expressions for the currents i1 and i2: i1 = Cg d v1 dt + CM i2 = Cg d v2 dt + CM dv1 − dv2 dt dt dv2 − dv1 dt dt = (Cg + CM ) d v1 dt − CM d v2 dt = (Cg + CM ) d v2 dt − CM d v1 dt (4-10) (4-11) In analyzing the behavior of the circuit, let us ﬁrst look at the case where line 1 has a transient signal dv /dt applied to it, whereas line 2 has no signal (dv2/dt = 0). Applying these inputs to (4-10) and (4-11) gives i1 = Cg + CM dv dt i2 = −CM d v1 dt (4-12) (4-13) 150 CROSSTALK i1 + v1 Cg CM − i2 + v2 Cg − Figure 4-4 Coupled capacitor circuit. Examination of equation (4-12) reveals that the effective capacitance of line 1 when it is driven in isolation is equal to the sum of the capacitance to ground and the mutual capacitance between lines 1 and 2. Conceptually, (4-12) says that the voltage signal applied to line 1 must charge up both the capacitance to ground and the capacitance between lines. Therefore, the sum of the capacitance to ground plus the mutual capacitance gives the total capacitance of line 1. Equation (4-13) indicates that signal in line 1 will impress an unwanted signal (i.e., crosstalk noise) on line 2 through the mutual capacitance. These effects are represented conceptually by the capacitance sketch for the three-conductor PCB (two signals plus a ground) shown in Figure 4-5. We can also further our insight into the impact of mutual capacitance by analyzing the even-mode (v1 = v2) and odd-mode (v1 = −v2) cases. For the even mode, since v1 = v2, we assume that the rise times (or fall times) are equal, and therefore that dv1/dt = dv2/dt = dv/dt. Substitution into (4-10) and (4-11) leads to i1 = i2 = Cg dv dt (4-14) For the odd mode we use dv1/dt = −dv2/dt = dv/dt, and ﬁnd that i1 = −i2 = Cg + 2CM dv dt (4-15) Cm 1 2 Cg Cg Figure 4-5 Coupled capacitances on a printed circuit board. MUTUAL INDUCTANCE AND CAPACITANCE 151 Equations (4-14) and (4-15) show that the effective capacitance of the system as seen by our input signals is changed by the mutual capacitance and is a function of the switching pattern. In particular, the even-mode capacitance is decreased relative to the total capacitance, by an amount equal to the mutual capacitance. Correspondingly, the odd-mode capacitance is increased by the mutual capacitance, giving us the relationship Ceven < Ctotal < Codd, where the total capacitance is Ctotal = C0 + CM . In developing a general expression for the capacitance, we write the equation for the current ﬂow in the capacitive elements as a function of capacitances and input voltages in matrix form: i1 i2 = Cg + CM −CM −CM Cg + CM d v1 /dt d v2 /dt (4-16) If we add a third line to our system, as shown in Figure 4-6, we can extend equation (4-16): i1 C11 i2 = −C21 i3 −C31 −C12 C22 −C32 −C13 dv1/dt −C23 dv2/dt C33 dv3/dt (4-17) In equation (4-17), the diagonal elements C11, C22, and C33 represent the total capacitances on lines 1, 2, and 3, respectively. The total capacitance is the sum of the capacitance to ground (e.g., C1g for line 1) plus the mutual capacitances between lines. The mutual capacitances are represented by Cij , where i and j correspond to the lines coupled by the mutual capacitance. In other words, Ci = Ci + Cij j =i As is the case with the inductance matrix, the capacitance matrix is symmetric since the capacitance does not depend on the polarity of the electric ﬁeld. We can generalize equation (4-17) to handle n capacitively coupled lines, i1 C11 −C12 · · · −C1n dv1/dt i...2 = −C... 21 C22 ... −C2n ... d v2/dt ... in −Cn1 −Cn2 · · · Cnn dvn/dt (4-18) where the Cii are the total capacitances and the Cij are the mutual capacitances, with Cij = Cji. Remember that total capacitance of line i is equal to sum of the capacitance to ground for line i and the mutual capacitances between line i and all of the other lines in the system. A feature of the capacitance matrix that immediately stands out is that the off-diagonal entries are negative. Although this may seem counterintuitive, it is 152 CROSSTALK +i1 v1 C1g − +i2 C12 C13 v2 C2g C23 − i3 + v3 C3g − Figure 4-6 Circuit with three coupled capacitors. simply a direct consequence of the fact that we deﬁned the diagonal elements to be the total capacitance of the individual traces. This is necessary to simplify the mechanics of our circuit calculations while guaranteeing correct results, as we illustrate with an example. Example 4-1 Equivalent Capacitance for a Coupled Pair We apply equation (4-18) to the two-line case, so that C= C11 −C21 −C12 C22 Driving the lines with dv1/dt = dv/dt and dv2/dt = 0 gives i1 = C11 (dv/dt) and i2 = −C21(dv/dt). Recalling that C11 = C1g + C12, we conﬁrm that our solution matches the result given in equations (4-12) and (4-13). Additional anal- ysis of the odd- and even-mode cases will also conﬁrm our earlier results, and is left as an exercise for the reader. 4.1.3 Field Solvers The capacitance and inductance matrices are typically obtained as the output from an electromagnetic ﬁeld solver. These tools model the electromagnetic ﬁelds between transmission lines in a multiconductor system, providing the basis for equivalent circuit models and the inputs to transmission-line simulators such as HSPICE. Field simulators fall into two general categories, two-dimensional (2D) quasistatic and three-dimensional (3D) full-wave solvers. Examples of commercially available tools include Linpar [Djordjevic et al., 1999] for the 2D quasistatic case, and HFSS for 3D full-wave solvers. Quasistatic tools use techniques similar to those outlined in Chapter 3, where Laplace’s equation is solved to calculate the capacitance for a given set of COUPLED WAVE EQUATIONS 153 boundary conditions. The inductance is then calculated from the capacitance using the speed of light as a conversion mechanism. Most 2D quasistatic tools will give the inductance and capacitance matrices (along with the resistance and conductance matrices) per unit of conductor length, under the assumption that the physical geometries and materials are uniform along the length of the transmission structures. They are typically easy to use and execute in a matter of seconds, but because of the length uniformity assumption, they cannot handle complex 3D transmission structures. Because they are static ﬁeld solvers, they typically do not calculate frequency-dependent effects such as internal inductance and skin effect resistance. This is not a signiﬁcant obstacle for printed circuit boards since the transmission structures are often uniform, and we can use other methods for including frequency dependence (refer to Chapters 5 and 6). The quasistatic assumption also requires that the signals propagate in transverse electromagnetic (TEM) mode, which means that the electric and magnetic ﬁelds are perpendicular and there is no ﬁeld component in the direction of wave propagation, as described in Section 2.3.2. The TEM assumption is geometry dependent, but for typical PCB traces used to design high-speed digital systems (50 , trace width about 5 mils), it is valid for frequencies well past 20 GHz. Full-wave 3D solvers, on the other hand, are capable of simulating complex physical structures and will predict frequency-dependent losses, internal inductance, dispersion, and most other electromagnetic phenomena, including radiation. These tools essentially solve Maxwell’s equations directly for an arbitrary geometry. Complex structures such as edge connectors and packages may require 3D tools to model their effects accurately at high data rates. The disadvantage of full-wave solvers is that they require more expertise to use, and simulations typically take hours or days rather than seconds. Additionally, the output from a full-wave simulator is often in the form of S-parameters, which typically require additional processing in order to use them for interconnect simulations for digital applications. As a result, design engineers typically employ 2D ﬁeld solvers whenever possible, making use of 3D full-wave tools only where necessary. 4.2 COUPLED WAVE EQUATIONS Before proceeding with the analysis of coupled systems, we ﬁrst derive the wave equations, to reinforce the notion of wave propagation and to allow us to study the effects of mutual inductance and capacitance. The wave equation is central to our analysis of transmission lines, and extension to coupled systems will lend insight into the effects of crosstalk on the propagation of signals in a coupled system. 4.2.1 Wave Equation Revisited We start our derivation of the transmission-line equations by focusing on the isolated line case shown in Figure 4-7. The circuit must satisfy Kirchhoff’s laws. 154 CROSSTALK L0 i(z) + v(z) − i(z + dz) + C0 v(z + dz) − dz Figure 4-7 Differential circuit subsection for a lossless transmission line. Application of Kirchhoff’s voltage law (KVL) gives the voltage drop across the incremental inductance: v(z) − v(z + dz) = −j ωL0i(z) dz (4-19) where j ωL0dz is the frequency-dependent impedance of the inductor. Equation (4-19) contains a frequency term, ω, implying a sinusoidal input which has the form v(t) = V0ejωt . A sinusoidal signal has the property that its derivative is a scaled version of the original signal, dv(t)/dt = d(V0ejωt )/dt = j ωV0ejωt = j ωv(t). Note that the current, i(z,t), is also a sinusoid, so that di(t)/dt = j ωi(t). Equation (4-19) is equivalent to Faraday’s law for the response of an inductor to a transient current [ v = L(di/dt)]. The analysis that follows is equally applicable for a digital input, since such a signal is composed of the superposition of multiple sinusoids, although the wave equation requires a separate solution for each frequency in the envelope of the driven signal. The next steps are to divide (4-19) through by dz , followed by differentiation with respect to z: dv dz = −j ωL0i d2v d z2 = −j ωL0 di dz (4-20) (4-21) Looking now at the incremental capacitance of the subsection, we apply Kirch- hoff’s current law to ﬁnd the change in current, as shown in equation (4-22), where (j ωC0 dz)−1 is the impedance of the capacitor: i(z + dz) − i(z) = −j ωC0v(z) dz (4-22) Equation (4-22) is equivalent to i = C0(dv/dt) in the time domain. Dividing through by dz gives di dz = −j ωC0v(z) (4-23) Substituting (4-23) into (4-21) to get an expression in terms of v yields the voltage wave equation: d2v d z2 + ω2L0C0v = 0 (4-24) COUPLED WAVE EQUATIONS 155 The same approach will give the wave equation for the current: d2i d z2 + ω2L0C0i = 0 (4-25) Equations (4-24) and (4-25) should be familiar as the wave equations for a uniform lossless transmission line. 4.2.2 Coupled Wave Equations Our goal now is to generalize equations (4-24) and (4-25) to the n-coupled-line case. Figure 4-8 describes the circuit for a subsection of a pair of coupled lines. Our derivation follows the same method that we used above for the isolated transmission line. We start by applying KVL to develop equation (4-26). Note that the voltage drop on line 1 depends on the mutual inductance LM and the current on the adjacent line i2(z), in addition to the line self-inductance L and driving current i1(z). We can also create a corresponding expression (4-27) for the voltage drop on line 2: v1(z) − v1(z + dz) = −j ωL0i1(z) dz − j ωLM i2(z) dz = −j ω[L0i1(z) + LM i2(z)] dz v2(z) − v2(z + dz) = −j ω[L0i2(z) + LM i1(z)] dz (4-26) (4-27) We can write the equations for the voltage drop across the coupled subcircuit in compact matrix form, where the boldface symbols represent the compact matrix: dv = −j ωLi dz (4-28) where dv = 1 dz dz v1(z) − v1(z + dz) v2(z) − v2(z + dz) L= L0 LM LM L0 i= i1(z, t) i2(z, t) taken in the limit as dz → 0 Applying the isolated line method to the current change caused by the capacitances of the coupled line yields i1(z) − i1(z + dz) = −j ω(Cg + CM )v1(z) dz + j ωCM v2(z) dz = −j ω[(Cg + CM )v1(z) − CM v2(z)] dz (4-29) 156 CROSSTALK i1(z) L0 + i2(z) L0 v1(z) + v2(z) −− LM CM Cg Cg dz i1(z + dz) + i2(z + dz) + v1(z + dz) v2(z + dz) −− Figure 4-8 Differential circuit subsection for two lossless coupled transmission lines. i2(z) − i2(z + dz) = −j ω[(Cg + CM )v2(z) − CM v1(z)] dz di = −j ωCv dz (4-30) (4-31) where di = 1 dz dz i1(z) − i1(z + dz) i2(z) − i2(z + dz) C= Cg + CM −CM −CM Cg + CM v= v1(z) v2(z) taken in the limit as dz → 0 Notice in equations (4-29) and (4-30) that the current change on line 1 caused by v1 is proportional to the sum of the capacitance to ground Cg and the mutual capacitance between lines CM . This is consistent with our earlier discussion of mutual capacitance. We can also reassure ourselves that this is correct by considering the response to a potential stimulus. Let us ﬁrst assume that a potential, v, is applied to line 1 (relative to ground), while no potential is applied to line 2. In this case, we must charge up both the capacitance to ground and the mutual capacitance between lines, a result that (4-29) predicts. In the second case we assume that the same potential is applied to both lines. In this situation, lines 1 and 2 remain at the same potential, so that no charge is stored in the electric ﬁeld between the lines. Therefore, line 1 need charge up only the capacitance to ground, a result that is also predicted by (4-29). Returning to our derivation, we differentiate (4-28) with respect to z to get d2v d z2 = −j ωL di dz (4-32) Substituting for di/dz from (4-31) results in the coupled voltage wave equation d2v d z2 = ω2LCv (4-33) COUPLED LINE ANALYSIS 157 Following our method, we can also derive the current wave equation for the coupled lines: d2i d z2 = ω2CLi (4-34) Equations (4-33) and (4-34) comprehend wave propagation on each line in the system due to the source on the line itself and to sources being coupled from other lines through the electromagnetic ﬁelds. Notice also that they bear a striking similarity to equations (4-24) and (4-25). In fact, (4-24) and (4-25) reduce to (4-33) and (4-34) for the case n = 1. A couple of additional observations about (4-33) and (4-34) are worth pointing out. First, in equation (4-34) the order of multiplication of the L and C matrices is reversed from that of (4-33). Since they are matrices, the product LC is not necessarily equal to CL. Second, the compact matrix equations are extensible to an arbitrary number of coupled transmission lines, providing the means for analyzing practical systems, as we demonstrate in the sections that follow. 4.3 COUPLED LINE ANALYSIS In Section 4.1 we demonstrated that the effective capacitance and inductance will change with different switching patterns in systems where there is signiﬁcant coupling between lines. This implies that the effective characteristic impedance and propagation delay will also be a function of the switching patterns. We also showed that coupling between lines can result in the appearance of noise on quiet lines. Therefore, both effects can have switching-dependent impacts on signal quality and timing characteristics of a coupled system, and we need to develop methods for quantifying them. 4.3.1 Impedance and Velocity In this section we explain the effects that crosstalk-induced impedance and velocity changes have on signal integrity and timing, and provide a simple method for analyzing multiconductor transmission lines to provide ﬁrst-order estimates of that impact. From Section 4.1, recall that the switching pattern will change the effective capacitance and inductance on a pair of couple lines, as summarized in Table 4-1. By inspection of the table we can easily construct equations for the effective impedances (Z0) and propagation velocities (νp) for each case: Z0,isolated = L0 Cg + CM Z0,even = L0 + LM Cg (4-35) (4-36) 158 CROSSTALK TABLE 4-1. Summary of Effective Capacitance and Inductance for a Couple Pair Mode Line 1 Line 2 Ceffective Leffective hi lo hi lo Even Cg hi lo hi lo L0 + LM hi lo hi lo Odd Cg + 2CM L0 − LM hi hi lo lo hi lo lo lo hi lo lo lo Quiet Cg + CM L0 lo hi hi hi hi hi lo Z0,odd = L0 − LM Cg + 2CM νp,isolated = 1 L0(Cg + CM ) νp,even = √ (L0 1 + LM )C0 νp,odd = 1 (L0 − LM )(Cg + 2CM ) (4-37) (4-38) (4-39) (4-40) Equations (4-35) through (4-40) are exact for the two-line case, and they give us a simple way to analyze a coupled pair via a lattice diagram or simulation of a single line using the effective characteristic impedance and effective propagation velocity. We call the models created using this method single-line equivalent models (SLEMs) [Hall et al., 2000]. It is interesting to note that the mutual inductance is always added or subtracted in the opposite manner as the mutual capacitance for odd- and even-mode propagation. The ﬁelds in Figure 4-9 help us understand why this is true. Considering odd-mode propagation as an example, the effect of mutual capacitance must be added because the conductors are at different potentials. Additionally, COUPLED LINE ANALYSIS 159 Electric field: Odd mode Electric field: Even mode Magnetic field: Odd mode Magnetic field: Even mode Figure 4-9 Odd- and even-mode electric and magnetic ﬁeld patterns for a simple two-conductor system. since the current in the two conductors ﬂows in opposite directions, the currents induced on each line due to the coupling of the magnetic ﬁelds always oppose each other and cancel the effect of the mutual inductance. Therefore, the mutual inductance must be subtracted and the mutual capacitance must be added to calculate the odd-mode characteristics. These characteristics of even- and odd-mode propagation are due to the assumption that the signals are propagating only in transverse electromagnetic (TEM) mode, so that the electric and magnetic ﬁelds are always orthogonal to each other. With homogeneous dielectrics, the product of L and C remains constant, since ﬁelds are conﬁned within the uniform dielectric: LC = 1 νp2 I (4-41) where I is the identity matrix. Thus, in a multiconductor homogeneous system such as a stripline array, if L is increased by the mutual inductance, C must be decreased by the mutual capacitance such that LC remains constant. Subsequently, a stripline, or buried microstrip, which is embedded in a homogeneous dielectric, should not exhibit velocity variations due to different switching modes. It will, however, exhibit pattern-dependent impedance variation. In a nonhomogeneous dielectric where the electric ﬁelds fringe through more than one dielectric material, such as an array of microstrip lines, LC is not constant for different propagation modes because the electromagnetic ﬁelds travel partially in air and partially in the board’s dielectric material. In a microstrip, the effective dielectric constant is a weighted average between air and the dielectric material of the board. Because the ﬁeld patterns change with different propagation modes, the effective dielectric constant will change depending on the ﬁeld densities contained within the board’s dielectric material and the air. Thus, the 160 0.005 in 0.005 in 0.005 in CROSSTALK 0.005 in 0.002 in εr = 4.0 Figure 4-10 Cross section of the PCB-based coupled transmission-line pair for Example 4-2. LC product will be mode dependent in a nonhomogeneous system. The LC product, however, will remain constant for a given mode. Therefore, a microstrip will exhibit both a velocity and an impedance change, due to different switching patterns. It should be noted that the description above holds for a single frequency. The product of LC varies with frequency but remains constant at each frequency point for a given mode. Example 4-2 The PCB transmission lines depicted in Figure 4-10 have the following inductances and capacitances: L= 3.592 × 10−7 3.218 × 10−8 3.218 × 10−8 3.592 × 10−7 H/m C= 8.501 × 10−11 −2.173 × 10−12 −2.173 × 10−12 8.501 × 10−11 F/m Assume that the waveform is driven into the line at t = 1 ns. We have designed the PCB traces to have a typical (isolated) characteristic impedance of approximately 65 with a length of 0.2794 m (11 in.). They are driven by a 1-V 65- source, and are terminated to ground in 65 at the far end. The rise and fall times are 0.1 ns. Compare the analytical results with those from a fully coupled simulation for even- and odd-mode propagation. SOLUTION Step 1: Calculate the impedances and velocities for all of the switching patterns. Z0,even = 3.592 × 10−7 + 3.218 × 10−8 H/m 8.501 × 10−11 − 2.173 × 10−12 F/m = 68.7 Z0,odd = 3.592 × 10−7 − 3.218 × 10−8 H/m 8.501 × 10−11 + 2.173 × 10−12) F/m = 61.2 COUPLED LINE ANALYSIS 161 Z0,isolated = 3.592 × 10−7 H/m 8.501 × 10−11 F/m = 65.0 νp,even = 1 [(35.92 + 3.218) × 10−8 H/m][(85.01 − 2.173) × 10−12 F/m] = 1.756 × 108 m/s 1 νp,odd = [(35.92 − 3.218) × 10−8 H/m][(85.01 + 2.173) × 10−12 F/m] = 1.873 × 108 m/s νp,isolated = 1 3.592 × 10−7 H/m 8.501 × 10−11 F/m = 1.810 × 108 m/s Step 2 : Calculate the even-mode waveform. Calculate the values for the initial voltage and current waves, reﬂection coefﬁcients, ﬁnal voltage and current levels, and propagation delay as preparation for a lattice diagram analysis. v(t = 0, z = 0) = RS Z0,even + Z0,even VS = 65 68.7 + 68.7 (1 V) = 0.514 V i(t = 0, z = 0) = v(t = 0, z = 0) = 0.514 V = 7.48 mA Z0,even 68.7 (z = 0) = RS RS − Z0,even + Z0,even = 65 65 − 68.7 − 68.7 = −0.028 (z = l) = RT RT − Z0,even + Z0,even = 65 65 − 68.7 + 68.7 = −0.028 v(t = ∞) = Rt RS + Rt VS = 65 65 + 65 (1 V) = 0.500 V i(t = ∞) = vS = 1.000 RS + Rt 65 + 65 = 7.69 mA td ,even = l νp,even = 11 in 1.756 × 108 m m/s 39.37 in. = 1.592 ns The corresponding lattice diagram is shown in Figure 4-11. Step 3 : Calculate the odd-mode waveform. Repeat the analysis for the odd-mode propagation, with the lattice diagrams shown in Figure 4-12. v(t = 0, z = 0) = RS Z0,odd + Z0,odd VS = 65 61.2 + 61.2 (1 V) = 0.485 V 162 CROSSTALK 0 Γ = −0.028 l z Γ = −0.028 1.000 ns 2.592 ns 4.182 ns 5.773 ns 7.364 ns v(z = 0) 0.000 V i(z = 0) 0.00 mA 0.514 V 7.14 mA 0.500 V 7.68 mA v (z = l) i(z = l) 70.4.581m4 AV 00.0.010m−00AV−.001.241VmA 00..00000mVA 0.000 V 0.00 mA 0.500 V 7.69 mA 0.500 V 7.69 mA 0.500 V 7.69 mA t Figure 4-11 Lattice diagram for even-mode propagation of the coupled line pair for Example 4-2. Note that the rising edge wave will be 0.485 V, starting from 0.000 V, while the falling edge wave will be −0.485 V, starting from 0.500 V. i(t = 0, z = 0) = v(t = 0, z = 0) = 0.485 V = 7.92 mA Zodd 61.2 (z = 0) = RS RS − Z0,odd + Z0,odd = 65 65 − 61.2 + 61.2 = 0.030 (z = l) = RT RT − Z0,odd + Z0,odd = 65 65 − 61.2 + 61.2 = 0.030 Rising edge v(t i(t = ∞) = = ∞) = Rt RS + Rt vS RS + Rt VS = = 65 65 65 + 1.000 + 65 65 = (1 V) = 0.500 7.69 mA V Falling edge v(t = ∞) = 0.000 V i(t = ∞) = 0.00 mA td ,odd = l νp,odd = 0.2794 m 1.873 × 108 m/s 109 ns s = 1.492 ns Step 4 : Compare the calculated results to simulated results. From Figure 4-13 we see that the results from SLEM analysis match results from fully coupled SPICE time-domain simulations using the L and C matrices. Having developed the SLEM approach for the two-line case, we want to generalize it to deal with an arbitrary number of coupled lines, since real systems COUPLED LINE ANALYSIS 163 0 Γ = 0.030 l z Γ = 0.030 1.000 ns 2.492 ns 3.984 ns 5.476 ns 6.976 ns v (z = 0) 0.000 V 0.485 V 0.500 V 0.500 V t i(z = 0) 0.00 mA 8.58 mA 7.69 mA 7.69 mA 7.09.248m5AV 0.00.100m010A..0V2144mVA 0.00.000m000A..0V0010mVA v (z = l ) i(z = l) 0.000 V 0.00 mA 0.499 V 7.68 mA 0.500 V 7.69 mA (a) 0 Γ = 0.030 l z Γ = 0.030 v (z = 0) i(z = 0) v (z = l ) i(z = l ) 1.000 ns 2.492 ns 3.984 ns 5.476 ns 6.976 ns 0.500 V 0.015 V 0.000 V 0.000 V t 7.69 mA −0.23 mA 0.00 mA 0.00 mA −7−.09.248m5AV −0−.00.100m−1−A00V..02144mVA 00.0.000m000.VA.00000mVA (b) 0.500 V 7.69 mA 0.001 V 0.01 mA 0.000 V 0.00 mA Figure 4-12 (a) Rising- and (b) falling-edge lattice diagrams for odd-mode propagation of the coupled line pair for Example 4-2. typically contain larger numbers of coupled lines. In our earlier discussion we noted that for even-mode switching, the effective inductance is increased by the mutual inductances between lines, while the effective capacitance is decreased by the mutual capacitance. We also know that for odd-mode switching, the effective inductance is decreased by the mutual inductances between lines, while the effective capacitance is increased by the mutual capacitances. Finally, the mutual inductances and capacitances for quiet lines do not change the effective 164 CROSSTALK Voltage [V] 0.6 v (z = 0) & v (z = 0) 0.5 v(z = l ) & v (z = l ) 0.4 0.3 0.2 0.1 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] (a) 0.6 v (z = 0) 0.5 v (z = l ) 0.4 0.3 0.2 0.1 0.0 v (z = 0) v (z = l ) −0.1 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] (b) Voltage [V] Voltage [V] 0.6 v (z = 0) & v (z = 0) 0.5 v(z = l) & v(z = l) 0.4 0.3 0.2 0.1 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] 0.6 v (z = 0) 0.5 v (z = l) 0.4 0.3 0.2 0.1 0.0 v (z = 0) v (z = l ) −0.1 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] Voltage [V] Figure 4-13 Comparison of (a) even- and (b) odd-mode calculated (left) and simulated (right) results for Example 4-2. inductance and capacitance. With this knowledge, we can write generalized approximations for the effective inductance and capacitance, adopting the matrix notation that we developed in Section 4.1 . Leff,n = Lnn + Ceff,n = Cnn − Lne − Lno Cne + Cno (4-42) (4-43) where Leff,n = effective inductance of line n Lnn = self-inductance of line n Lne = sum of the mutual inductances for the lines that switch in phase with line n (which we approximate as “even” mode) Lno = sum of the mutual inductances for the lines that switch out of phase with line n (which we approximate as “odd” mode) Ceff,n = effective capacitance of line n Cnn = total capacitance of line n Cne = sum of the absolute values of the mutual capacitances for the lines that switch in phase with line n Cno = sum of the absolute values of the mutual capacitances for the lines that switch out of phase with line n COUPLED LINE ANALYSIS 165 Recall from Section 4.1 that the elements on the matrix diagonals represent the self-inductance and total capacitance, respectively, while the off-diagonal elements represent the mutual terms. Once we obtain the effective inductance and capacitance, we can calculate the effective impedance and propagation velocity: Z0,eff,n = νp,eff,n = Leff,n Ceff,n 1 Leff,nCeff,n (4-44) (4-45) We can use equations (4-44) and (4-45) along with the physical length of the transmission line to analyze or simulate the behavior of any line as a simple noncoupled line while accounting for the coupling to the other lines in the system. This technique is best used during the early design phase of a bus, when I/O transceiver impedances and line-to-line spacing are being chosen. In addition, it is useful only for signals traveling in the same direction. For signals traveling in opposite directions, fully coupled simulations are required to comprehend the effects of crosstalk. At this point it is important to note that although the SLEM method gives correct results for systems with two coupled lines, it is an approximation that will not exactly match the actual modal impedances and velocities for three or more coupled lines. As such, its use should be restricted to early design exploration aimed at narrowing down the solution space. Final simulations should always be done with fully coupled models. The accuracy of the SLEM model (for three lines) is reasonable for cross sections in which the spacing/height ratio is greater than 1. When this ratio is less than 1, the SLEM approximation should not be used. In Section 4.4 we introduce a technique for producing exact solutions to systems of three or more coupled lines. 4.3.2 Coupled Noise Before describing the coupled noise mechanism and developing methods to quantify the noise, we need to provide some motivation for doing so. By now it should be clear that our goal is to transmit interchip digital data signals successfully. In doing so, we use an active transmitter circuit to drive the data onto the interconnect, where it propagates as an electromagnetic wave to a receiver circuit that senses the signal and restores it to the appropriate logic level. It is this signal restoration operation that can be affected by crosstalk noise (as well by any other sources of noise). In restoring the signals, the receivers have thresholds for distinguishing between logic levels. Crossing over a logic threshold will cause the receiver to switch the output state. Referring to Figure 4-14, we see a voltage transfer characteristic for an inverting receiver. When a rising input signal crosses vIH , it will cause the output of the receiver to switch from high to low. However, noise 166 vout vOVHC,mCin Slope = −1 CROSSTALK vOL,max VSS VSS Slope = −1 vIH,max vIH,min vin VCC Figure 4-14 Receiver transfer characteristic and switching thresholds. on the signal that causes an excursion back below vIL will cause erroneous switching (Figure 4-15). It is the job of the system designer to make sure that this does not happen. Refer to Chapter 11 for additional details on the operation of receivers for high-speed links. Qualitative Description Before developing a quantitative treatment of crosstalk-induced noise, we describe the behavior qualitatively by looking at the propagation of the aggressor and coupled noise wave. From our discussion of mutual inductance and capacitance it is apparent that energy is coupled from one line to another only during signal transitions (i.e., the rising and falling edges). Subsequently, we look at the propagation of a rising edge on the aggressor line. Figure 4-16 shows a pair of coupled lines terminated at both ends that represent a typical system. As the incident wave on the aggressor line is launched, it immediately begins coupling over to the victim line through the mutual capacitance and inductance. Current that couples through the mutual capacitance (iC) splits into forward-traveling (if ) and backward-traveling (ib) components in the victim line, as shown in Figure 4-17. Current that couples through the mutual inductance (iL) travels back toward the source end. As a result, we have a forward-coupled wave that is a function of the difference between the capacitively coupled current and the inductively coupled current. Since it is based on the difference between capacitive and inductive coupling, the amplitude may have the same polarity as COUPLED LINE ANALYSIS Clean signal vIH vIL 167 Threshold violation Voltage (V) Time (ns) Figure 4-15 Receiver threshold violation. z=I z=0 t=0 Z0 Z0 Aggressor: Z0, td, I Z0 Near-end ×talk pulse at t = 0 Z0 Victim: Z0, td, I Far-end ×talk pulse at t = 0 Figure 4-16 Initial launch of aggressor signal and coupling of noise. v1 i1 Z0 Cg dz Line 1 L0 dz ∆iC LM dz ∆ib CM dz ∆if Z0 ∆vb Z0 Cg dz L0 dz Line 2 iL Z0 ∆vf Figure 4-17 Coupled circuit subsection for crosstalk noise analysis. 168 CROSSTALK t = ½ td ⋅I Aggressor: Z0, td, I Near-end ×talk pulse at t = ½ td ⋅I Victim: Z0, td, I Far-end ×talk pulse at t = ½ td ⋅I Figure 4-18 Propagation of incident aggressor signal and coupled noise pulses. the aggressor signal, or it may have opposite polarity. The forward coupled wave then begins to propagate toward the far end of the victim line (z = l). We also have a backward coupled wave that is a function of the sum of the capacitively and inductively coupled currents. The backward traveling noise will have the same polarity as the aggressor signal, which is a direct result of the summing of the inductive and capacitive coupling. The backward crosstalk noise propagates toward the near end of the victim line, where it is immediately detectable. As the incident wave on the aggressor propagates toward the far end, it continues to couple energy over to the victim line. As Figure 4-18 shows, the forward crosstalk pulse on the victim line propagates alongside the aggressor signal. Since coupling continues along the length of the line, the amplitude of the far-end noise pulse grows as it propagates along the length of the coupled pair. The backward crosstalk pulse propagates back toward the near end (z = 0). The coupling of energy to the victim line continues as the aggressor propagates along the line until it reaches the far end at time t = τd l, where τd is the signal propagation delay per unit length and l is the line length. Alternatively, we could deﬁne the time as t = l/νp, where νp is the propagation velocity of the signal. At that point, assuming that we have a matched termination, the coupling ceases since the aggressor does not generate reﬂected waves. As Figure 4-19 demonstrates, arrival of the far-end crosstalk noise occurs simultaneously with arrival of the aggressor signal. Since the far-end crosstalk travels along with the aggressor signal, the crosstalk noise pulse grows in amplitude but does not grow in width. Coupling occurs only during the signal transition, so the width of the forward coupled pulse will be approximately equal to the rise (or fall) time of t = td ⋅I Z0, td, I Near-end ×talk pulse at t = td ⋅I Z0, td, I Far-end ×talk pulse at t = td ⋅I Figure 4-19 Propagation of coupled noise pulses as the aggressor reaches the far end (z = l). COUPLED LINE ANALYSIS 169 z=0 z=l t = 2td ⋅I Z0 Z0, td, l Near-end ×talk pulse at t = 2 td ⋅I Z0 Z0 Z0, td, l Figure 4-20 Completion of the noise pulse at the near end (z = 0). the aggressor signal, as the ﬁgure illustrates. Even though additional energy is no longer being coupled at this point, the backward crosstalk wave must still travel back to the near end. This takes a full propagation delay (td = τd l) to complete and is shown in Figure 4-20. The shape of the near-end pulse depends on the electrical length of the coupled lines relative to the transition time of the aggressor signal. Consider the case for which the coupled length is less than one-half of the signal rise time. The beginning of the signal transition will reach the far end before the rising edge at the driving end has completed one-half of the transition. As the coupled noise propagates back to the near end, the signal at the far end continues to change, and therefore to couple more energy to the victim. Not until the rising-edge transition has ﬁnished propagating all the way to the far end does coupling stop. In that case, the near-end crosstalk pulse will have a similar shape (but different amplitude) to the far-end pulse, as shown in Figure 4-21a. On the other hand, if the coupled length is greater than one-half the signal rise time, the near-end crosstalk pulse will reach maximum amplitude and will then begin to spread in time, a phenomenon known as saturation, which is shown in Figure 4-21b. To understand this effect, consider a case where the coupled electrical length is equal to several rise times. For this situation, we can visualize the signal edge as a traveling wave on the aggressor line. Recalling that the lines couple only during the transition, we can imagine the backward crosstalk as a train of pulses of equal magnitude with a width equal to the rise time that propagate back to the near end. As a result, the backward noise does not grow in magnitude, but instead, spreads out in time. Figure 4-22 illustrates the coupled pulse propagation as just described. As a ﬁnal note, we must realize that the crosstalk pulse magnitudes and shapes that we just described, and for which we will develop quantitative models in the next section, are speciﬁc to the matched termination case. Terminating the lines simpliﬁes the analysis by eliminating the need to deal with reﬂected crosstalk and crosstalk from reﬂected aggressor signals. In general, the characteristics of the crosstalk noise are heavily dependent on the amount of coupling and the termination. For cases with imperfect termination and/or complex topologies, we recommend using a simulator to analyze the behavior of the system. Hall et al. [2000] describe crosstalk shapes for some general nonperfect termination schemes. 170 CROSSTALK tr Driven signal td ⋅I Near-end crosstalk tr Far-end crosstalk tr (a) tr Driven signal td ⋅I Near-end crosstalk Far-end crosstalk tr 2 td ⋅I (b) Figure 4-21 Forward- and backward-coupled noise pulses: (a) nonsaturated; (b) saturated. Quantitative Development Having developed an intuitive understanding of crosstalk and the characteristics of crosstalk noise, we now derive the equations for noise at each end of a quiet transmission line (line 2 in Figure 4-17) that is induced by coupling from an adjacent coupled line (line 1) that is actively driven. Noise will be coupled from line 1 to line 2 through the mutual inductance LM dz and mutual capacitance CM dz. Line 1, whose characteristic impedance is Z0, has an incident pulse of magnitude v1 (voltage) and i1 (current), and is terminated in its characteristic impedance at both ends. We want to come up with expressions for the backward ( vb) and forward ( vf ) noise pulses on line 2, which also has a characteristic impedance of Z0 with termination at both ends. Our derivation follows the method presented by Seraphim et al. [1989]. COUPLED LINE ANALYSIS 171 z=I z=0 t=0 Z0 Z0 Z0 Aggressor: Z0, td, I Near-end ×talk pulse at t = 0 Z0 Victim: Z0, td, I Far-end ×talk pulse at t = 0 (a) z=I z=0 t = ½td ⋅I Z0 Aggressor: Z0, td, I Near-end ×talk pulse at t = ½td ⋅I Z0 Z0 Victim: Z0, td, I t = td ⋅I Far-end ×talk pulse at t = ½td ⋅I (b) Z0, td, I Near-end ×talk pulse at t = td ⋅I Z0, td, I Far-end ×talk pulse at t = td ⋅I (c) z=0 z=I t = 2 td ⋅I Z0 Z0, td, I Near-end ×talk pulse at t = 2 td ⋅I Z0 Z0 Z0, td, I (d) Figure 4-22 Summary of propagation of forward- and backward-coupled noise: (a) initial wave launch; (b) halfway down the line; (c) one full trip down the line; (d) round trip. 172 CROSSTALK We begin the development by applying Ohm’s law at each end of line 2 to get the amplitudes of the noise pulses at the near end ( vb) and far end ( vf ): vb = ibZ0 vf = if Z0 (4-46) (4-47) Current is coupled from line to line through the mutual capacitance: iC = CM dz dv1 dt (4-48) The coupled current splits into separate branches on line 2 that ﬂow in both directions: iC = ib + if (4-49) Combining equations (4-46) through (4-49) gives an expression for the voltage pulses created by the coupling through the mutual capacitance, vb + vf = Z0CM dz dv1 dt (4-50) Next we deﬁne the capacitive coupling coefﬁcient as the ratio of the mutual capacitance between lines to the total capacitance of the line: KC ≡ CM Cg + CM (4-51) Along with the capacitive coupling coefﬁcient deﬁnition, we apply expressions for the characteristic impedance of the line, Z0 = L0/Cg + CM , to equation (4-50), resulting in vb + vf = L0CM2 dz dv1 Cg + CM dt (4-52) By substituting νp = 1/ L0(Cg + CM ) and performing some algebra, we arrive at the following expression for the sum of forward and backward crosstalk induced by the mutual capacitance: vb + vf = 1 νp KC dz dv1 dt (4-53) Turning now to the inductance, we note that the mutual inductance acts as a coupling transformer. Current on line 1 induces a voltage on line 2 that travels in the direction opposite that of the incident signal on line 1. As a result, it creates a COUPLED LINE ANALYSIS 173 voltage difference across the differential segment (dz ) which travels back toward the source: vb − vf = LM dz di1 dt (4-54) In the same way that we deﬁned a capacitive coupling coefﬁcient, we deﬁne the inductive coupling coefﬁcient as the ratio of the mutual inductance between lines to the self-inductance of the line: KL ≡ LM L0 (4-55) Application of Ohm’s law at the driven end of line 1 (i1 = v1/Z0) yields vb − vf = dz LM Z0 d v1 dt = dz L2M (Cg + CM ) L0 dv1 L0 L0 dt = dz L0(Cg + CM ) L2M L20 d v1 dt which reduces to another expression relating the forward and backward coupled noise to a coupling coefﬁcient, in this case the inductive coupling coefﬁcient: vb − vf = dz KL νp d v1 dt (4-56) Since (4-53) and (4-56) give us two expressions with two unknowns, we can solve them for vb and vf and take dz → 0 in the limit, to get dvf = KC − KL dv1 dz 2νp dt dvb = KC + KL dv1 dz 2νp dt (4-57) (4-58) Forward Crosstalk Integrating (4-57) from z = 0 to z = l gives an expression for forward crosstalk: vf = 1 2 (KC l − KL) νp d v1 dt (4-59) By approximating dv 1/dt as the ratio of the voltage swing v and a 10 to 90% rise time tr , we have our ﬁnal expression for the forward crosstalk: vf = 1 2 (KC l − KL) νp v tr (4-60) 174 CROSSTALK Noting again that the forward crosstalk is a function of the difference between capacitive and inductive coupling, we see that the coupled pulse will have the same polarity as the aggressor signal if there is more capacitive coupling than inductive coupling in the system, and vice versa for the case where inductive coupling dominates. Equation (4-60) also suggests that it is possible to have no forward crosstalk if we can ﬁnd a case where KC = KL. In fact, this is always true for coupled lines in a homogeneous dielectric (proof is left for the reader as Problem 4-9). On the other hand, for typical transmission lines in inhomogeneous dielectrics, such as microstrip PCB traces, the inductive coupling is generally greater than the capacitive coupling, so that the forward crosstalk pulse has the opposite magnitude from that of the aggressor signal. As we described earlier, the width of the forward crosstalk pulse is tpw,f ∼= tr (4-61) where tr is the rise time of the signal. Note that (4-60) and (4-61) are equally applicable for a falling-edge transition. Reverse Crosstalk To get an expression for the reverse (near-end) crosstalk, we must take into account the fact that the coupling region travels in the direction opposite to the coupled waved. The output wave at the left in Figure 4-17 is a superposition of the waves coupled at earlier times that propagate and sum at the near end. This requires that we integrate from z = 0 to z = l while accounting for the travel time of the wave: vb = KC + KL 2νp l dv(t − 2z/νp) dz z=0 dt (4-62) After integration we have an expression for the coupled noise at the near end. vb (t ) = KC + 4 KL v1(t) − v1 t − 2l νp (4-63) The apparent reduction of the effect of the coupling coefﬁcient after the integration of (4-62) is caused by the fact that the energy coupling of the backward crosstalk is spread out over a pulse width of 2l/vp. The width of the backward coupled pulse is calculated with tpw,b = 2τd l (4-64) where τd is the propagation delay per unit length and l is the coupled length. Equations (4-63) and (4-64) assume that the backward crosstalk has saturated, which is realistic for multi-Gb/s links. Finally, we note that the equations that we derived in this section apply to situation in which both lines are terminated at each end. Other conﬁgurations, such as when the near end is not terminated, will have different equations to COUPLED LINE ANALYSIS 175 describe the amplitudes and shapes of the crosstalk pulses. The modiﬁed crosstalk equations can be derived by considering the effect of reﬂections as described in Section 3.5. Example 4-3 We now analyze the coupling from an active line to a quiet line for the PCB transmission lines from Example 4-2. Recall that the lines had the following inductances and capacitances: L= 3.592 × 10−7 3218 × 10−8 3218 × 10−8 3.592 × 10−7 H/m C= 8.501 × 10−11 −2.173 × 10−12 −2.173 × 10−12 8.533 × 10−11 F/m The 0.2794-m-long traces have a typical (isolated) characteristic impedance of approximately 65 and are terminated to ground in 65 at the far end. They are driven by a 1-V 65- source with a 100-ps rise time. Compare the analytical results with those from a fully coupled simulation. SOLUTION Step 1 : We start by calculating the impedance and propagation velocity: Z0,isolated = 3.592 × 10−7 H/m 8.501 × 10−11 F/m = 65.0 νp,isolated = 1 3.592×10−7 H/m 8.501 × 10−11 F/m = 1.810 × 108 m/s Step 2 : Since we plan to analyze the coupled noise, we need the coupling coefﬁcients. KC = 2.173 8.501 × × 10−11 10−11 F/m F/m = 0.0256 KL = 3.218 3.593 × × 10−8 10−7 H/m H/m = 0.0896 Step 3 : Analysis for a rising edge: v(t = 0, z = 0) = Z0 RS + Z0 VS = 65 65.0 + 65.0 (1 V) = 0.500 V i(t = 0, z = 0) = v(t = 0, z = 0) = 0.500 V = 7.69 mA Z0 65.0 (z = 0) = RS RS − + Z0 Z0 = 65 65 − 65.0 + 65.0 = 0.000 176 CROSSTALK (z = 0.2794 m) = RT RT − Z0 + Z0 = 65 65 − 65.0 + 65.0 l td = νp = 0.2794 m 1.810 × 108 m/s = 0.000 109 ns = 1.544 ns s vf = 1 2 (KC − KL) l νp v tr = 1 (0.0256 − 0.0896) 2 0.2794 m 1.820 × 108 m/s × 0.500 V 100 ps 1012 ps s = −0.247 V vb = KC + 4 KL v(t) − v t−2 l νp = 0.0256 + 0.0896 (0.500 V) = 0.014 V 4 tpw,f = tr = 100 ps tpw,b = 2τd l = 2 11 in. 1.810 × 108 m/s m 39.37 in. 109 ns s = 3.088 ns Since the reﬂection coefﬁcients are zero, a lattice diagram is not necessary, as we can construct the waveform directly from our calculations. Step 4 : Comparison to simulation. Figure 4-23 compares our calculated results with those from SPICE time-domain simulations. We see that although waveforms nearly match, they are not identical. In particular, the rising edge of the active signal in the SPICE simulation has grown to approximately 200 ps at the receiver end of the transmission line. The degradation of the rising edge of the active signal can be attributed to the crosstalk mechanism using an energy conservation argument. To conserve energy, the active line must give up an amount of energy that is equal to the amount coupled to the quiet line. Recall that the reactive nature of the coupling mechanism means that energy is coupled to the quiet line only during signal transition, so that the rising edge is degraded as a direct result of the coupling. In addition, the increase in rise time at the receiver causes the width of the crosstalk pulse at the receiver end to be approximately 200 ps rather than the 100 ps predicted by our calculation. As the example demonstrates, the crosstalk model presented here is an approximation. This model begins to break down when the coupled length is long enough such that the difference in propagation delay between the even and odd modes MODAL ANALYSIS 177 Voltage (V) Voltage (V) 0.6 0.5 vAggressor(z = 0) 0.4 0.3 0.2 vAggressor(z = I ) 0.1 vvictim(z = 0) 0.0 −0.1 −0.2 vvictim(z = I ) −0.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time (ns) 0.6 0.5 vAggressor(z = 0) 0.4 0.3 0.2 vAggressor(z = I ) 0.1 vvictim(z = 0) 0.0 −0.1 −0.2 vvictim(z = I ) −0.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time (ns) Figure 4-23 Comparison of (a) calculated and (b) simulated results for Example 4-3. exceeds the rise time: td,even − td,odd > tr When this occurs, the forward crosstalk saturates. At that point the amplitude ceases to grow; instead, the pulse width increases. In the next section we show how modal analysis allows us to explain the noise coupling mechanism in terms of modal propagation velocities and gives us the means to calculate accurately the behavior of the system. 4.4 MODAL ANALYSIS Earlier we introduced the notion of propagation modes in the even- and odd-mode cases for a two-line system. When we have more than two lines, the notion of even- and odd-mode signals does not apply directly. A system with n coupled lines can propagate n distinct modes, each of which is a function of the electric and magnetic ﬁeld strengths (or the capacitances and inductances) and of the driven signals on each of the lines. Each mode can have a different effective impedance and propagation velocity. We can calculate the modal impedances and velocities, although the mathematics is somewhat complicated. Modal decomposition and analysis provides us with the means to model the behavior of a coupled system using multiple single line simulations, which dramatically simpliﬁes the analysis without sacriﬁcing accuracy. To do so, we must ﬁrst express the behavior of the lines in terms of the propagation modes. The technique of modal decomposition transforms the n × n inductance and capacitance matrices into sets of n vectors, known as eigenvectors, that are each weighted by a constant value called an eigenvalue. What makes this technique interesting is that the eigenvectors are orthogonal, which allows us to diagonalize the L and C matrices so that the off-diagonal matrix entries are zero. Once we decompose the inductance and capacitances into diagonal matrices, we can analyze the behavior of the coupled line system as a set of n isolated transmission lines, each propagating modal voltage and modal current waves. In this approach we use the modal inductance and capacitance matrices Lm and Cm to transform the line voltages and currents to modal voltages and currents. The modal voltages and currents 178 CROSSTALK are then used to drive the isolated lines in a simulation or hand analysis such as a lattice diagram. The simulated or calculated response is then converted back from modal voltages and currents to line voltages and currents. The material that follows describes the modal decomposition and analysis technique. 4.4.1 Modal Decomposition If the matrices LC and CL can be diagonalized, the diagonal components are orthogonal to each other. This allows us to represent the line voltages as a linear combination of modal voltages. To do this it is ﬁrst necessary to determine the relationship between line voltages and modal voltages. We can use the coupled voltage wave equation (4-33) to derive a voltage transform matrix Tv; v = Tvvm (4-65) where Tv is a matrix that contains the eigenvectors of LC, which translate normal line voltages in v to modal voltages in vm. Once we have done so, we can simulate the system as n isolated lines using vm as input rather than as n coupled lines with v as input. Similarly, equation (4-34) for the coupled current wave is used to derive Ti: i = Tiim (4-66) where Ti is a matrix that contains the eigenvectors of CL, which translate normal line currents in i to modal currents in im. From (4-28) and (4-31), the matrix equations for an n × n system become d dz v i = 0 −j ωC −j ωL 0 v i (4-67) To derive the modal inductances, we start with equation (4-28) and substitute the modal voltages, currents, and transform matrices for the line voltage and current: d dz (Tvvm) = −j ωLTiim Multiplying both sides by T−v 1 gives d dz (T−v 1Tvvm) = −j ωT−v 1LTiim dvm dz = −j ωT−v 1LTiim The modal inductance matrix is deﬁned as (4-68) (4-69) Lm = T−v 1LTi (4-70) MODAL ANALYSIS 179 By following a similar approach, we can also derive the equation for modal capacitance: d dz (Ti im) = −j ωCTv vm dim dz = −j ωT−i 1CTvvm Cm = T−i 1CTv (4-71) (4-72) (4-73) Lm and Cm are diagonal, Lm11 0 0 0 0 Lm = 0 0 0 Lm22 0 0 0 ··· 0 0 0 Lm(n−1)(n−1) 0 0 0 0 00 0 Lmnn Cm11 0 0 0 0 Cm = 0 0 0 Cm22 0 0 0 ··· 0 0 0 Cm(n−1)(n−1) 0 0 0 0 00 0 Cmnn (4-74) (4-75) where the diagonal elements are deﬁned by (4-70) and (4-71). Another useful relationship that we will ﬁnd useful is [Paul, 1994] Ti = (T−v 1)T (4-76) To carry out the modal transmission line analysis, we need to put the modal quantities into the wave equation, starting with (4-33): d 2 Tv vm d z2 = ω2TvLmT−i 1TiCmT−v 1Tvvm = ω2TvLmCmvm Multiplying through by T−v 1 gives the wave equation expressed in terms of the modal voltage, inductance, and capacitance: d 2 vm d z2 = ω2LmCmvm (4-77) Following the same process with (4-15) gives d 2 im d z2 = ω2CmLmim (4-78) Again, since the modal quantities are orthogonal, we can use them to simulate the system as n isolated lines rather than n coupled lines. 180 CROSSTALK 4.4.2 Modal Impedance and Velocity Transmission-line analysis requires that we determine the modal impedance and propagation velocity of the lines. Recall that for an isolated line, νp = ω β (4-79) with the phase shift per unit length for the lossless case: √ β = I m (R + j ωL)(G + j ωC) = ω LC (4-80) leading to νp = √1 LC (4-81) By analogy, we can write an expression for the modal velocity matrix in terms of the modal inductance and capacitance matrices: νpm = L−m1C−m1 (4-82) Because Lm and Cm are square, diagonal matrices, their inverses will be, too, as will be the product of their inverses. Since the product is diagonal, the square root in (4-82) can be evaluated to yield the result νpm,ii = 1 Lm,ii Cm,ii (4-83) The subscript i represents the row and column in each matrix. By similar means, we ﬁnd an equation to express the modal impedances: Zm,ii = Lm,ii Cm,ii (4-84) 4.4.3 Reconstructing the Signal The observable line voltage and currents that compose the signal propagating on the interconnect are linear combinations of the modal values. After the system is decomposed into its orthogonal values and a set of n single lines that correspond to each mode is analyzed to determine the modal voltage and currents, the observable line voltage and currents must be reconstructed using MODAL ANALYSIS 181 equations (4-65) and (4-66). For example, expanding (4-65) for an n = 2 signal conductor system yields v1 v2 = Tv1 Tv3 Tv2 Tv4 vm1 vm2 v1 = Tv1Vm1 + Tv2vm2 v2 = Tv3Vm1 + Tv4vm2 where v1 is the line voltage observed on line 1, v2 is the line voltage observed on line 2, and vm1 and vm2 are the modal voltages calculated from the analysis of the single transmission lines with modal impedance and velocities. Notice that the line voltages are reconstructed from linear combinations of scaled modal voltages. 4.4.4 Modal Analysis In this section we outline the method for using the modal quantities. 1. Find, Tv, the eigenvectors of LC. 2. Find, Ti, the eigenvectors of CL. Tv and Ti are related by Ti = (T−v 1)T. 3. Use Tv and Ti to calculate the modal inductances, capacitances, and volt- ages and/or currents. 4. Use the modal inductance and capacitance to calculate the modal impedance and propagation velocity. 5. Carry out a conventional transmission-line analysis, such as a lattice dia- gram, or a single-line simulation for each mode. 6. Convert the modal quantities back into observable line quantities. The following examples illustrate the application of modal analysis to a coupled transmission-line pair. Example 4-4 Using the inductance and capacitance matrices and the interconnect design from Example 4-3, calculate the line voltages for signals switching in the odd mode using modal analysis. Compare the results to those of a fully coupled simulation. SOLUTION Our analysis follows the method described above. Step 1 : Start by calculating the voltage eigenvector: LC = 3.0466 × 10−17 1.9551 × 10−18 1.9551 × 10−18 3.0466 × 10−17 s2/m2 In ﬁnding the eigenvector of LC, we will make use of its symmetry, which has the form LC = a b b a 182 CROSSTALK Following the method of O’Neil [1983], we subtract λI: LC − λI = a b b a −λ 1 0 0 1 = a−λ b b a−λ The eigenvalue Tv of LC satisﬁes the following relationship: (LC − λI)Tv = a−λ b b a−λ Tv1 Tv2 = 0 0 Carrying out the multiplication yields (a − λ)Tv1 + bTv2 = 0 bTv1 + (a − λ)Tv2 = 0 From inspection of the equations above, we observe that Tv will satisfy the equation if Tv1 = −Tv2, or Tv1 = Tv2. As a result, we have a pair of eigenvectors Tv1 = 1 −1 and Tv2 = 1 1 Combining them gives the Tv matrix: Tv = Tv1 Tv2 = 1 −1 1 1 As a ﬁnal step, we normalize the voltage eigenvector, Tv: 11 −1 1 Tv = √ 12 + 12 = 0.707 −0.707 0.707 0.707 Step 2: The current eigenvector is equal to the transpose of the inverse of the voltage eigenvector [O’Neil, 1983]: Ti = (T−v 1)T = 0.707 −0.707 0.707 0.707 Note that because we derived Tv and T i from a general symmetric 2 × 2 matrix, they are applicable to any other 2 × 2 matrix, so we do not need to recalculate them for any other system of two coupled lines. This is true only for the 2 × 2 case. Whereas the matrices will be symmetrical for larger numbers of coupled lines, the eigenvectors will change as a function of the values of matrix entries. MODAL ANALYSIS 183 Step 3: The modal inductances and capacitances are calculated from equations (4-70) and (4-73). Lm = T−v 1LTi = 0.707 −0.707 0.707 −1 3.592 × 10−7 0.707 3.218 × 10−8 3.218 × 10−8 3.592 × 10−7 H/m × 0.707 −0.707 0.707 0.707 = 3.270 × 10−7 0 0 3.914 × 10−7 H/m Cm = T−i 1CTv = 0.707 −0.707 0.707 −1 8.501 × 10−11 0.707 −2.173 × 10−11 −2.173 × 10−11 8.501 × 10−11 F/m × 0.707 −0.707 0.707 0.707 = 8.718 × 10−11 0 0 8.284 × 10−11 F/m To calculate the modal voltage, we must ﬁrst have the voltage inputs. For the odd mode, the input on line 1 is 1 V and line 2 is −1 V. Therefore, the input voltage matrix is vin = 1 −1 and solving (4-65) for the modal voltages yields vm = 0.707 −0.707 0.707 −1 0.707 −1 1 = vodd veven = 1.414 0 Note that there is no voltage propagating in the even mode, which is expected since we drove the system in the odd mode. Step 4: The modal impedance and propagation velocity are calculated from (4-83) and (4-84). Zm = Lm11 Cm11 Lm22 = Cm22 3.270 × 10−7 H/m/8.718 × 10−11 F/m 3.914 × 10−7 H/m/8.284 × 10−11 F/m = 61.25 68.74 ohms 184 CROSSTALK z=0 z=I 1.414 V 65 Ω 61.25 Ω, 1.873 × 108 m/s, 10 in 65 Ω Figure 4-24 Equivalent modal circuit for the transmission-line pair in Example 4-4. νpm = √1 Lm11 1 Cm11 = √ Lm22Cm22 1 (3.270 × 10−7 H/m)(8.718 1 × 10−11 F/m) (3.914 × 10−7 H/m)(8.284 × 10−11 F/m) = 1.873 × 108 1.756 × 108 m/s This gives us the information that we require to construct the modal circuit for the odd mode, as shown in Figure 4-24. Step 5: Transmission-line analysis. Beginning with the low-to-high transition at the voltage source, we use the odd-mode equivalent circuit in Figure 4-24 to calculate the voltage and current waves that are launched, assuming that the rising edge occurs at time t = 1 ns. v(t = 1, z = 0) = (1.414 61.25 V) 61.25 + 65 = 0.686 V i(t = 1, z = 0) = v(t = 0, z = 0) = 0.686 V A mA Zm 61.25 V/ A = 11.20 mA The propagation delay to the far end (z = l = 0.254 m) is td ,odd = length νpm,odd = 0.254 m 1.873 × 108 m/s 109 ns s = 1.356 ns The voltage reﬂection coefﬁcient at the far end is = 65 65 − 61.25 + 61.25 = 0.030 The voltage reﬂection coefﬁcient at the source end is = 65 65 − 61.25 + 61.25 = 0.030 MODAL ANALYSIS 185 0 Γ = −0.030 l z Γ = −0.030 1 ns 2.356 ns 3.712 ns 5.069 ns 6.421 ns v(z = 0) i(z = 0) 0.000 V 0.00 mA 0.625 V 12.14 mA 0.707 V 10.88 mA 11.02.068m6AV 00.0.000m10000A..V.300.0302000mmVVAA v (z = l ) i(z = l ) 0.000 V 0.00 mA 0.706 V 10.87 mA 0.707 V 10.88 mA 0.707 V 10.88 mA t Figure 4-25 Odd-mode lattice diagram for Example 4-4. The ﬁnal steady-state voltage and current for the odd mode are v(t = ∞) = (1.414 V) 65 65 + 65 = 0.707 V i(t = ∞) = 65 1.414 + 65 = 10.88 mA We now have enough information to construct the odd-mode lattice diagram analysis shown in Figure 4-25. Step 6: We now convert the modal voltages and currents back to line voltages and currents. In converting back to line voltages and currents, we use (4-65) and (4-66) along with the values calculated in the odd-mode lattice diagram analysis. v = Tvvm i = Tiim with vm = from lattice 0 with im = from lattice 0 For example, the ﬁrst voltage level on the odd-mode lattice diagram is 0.686 V. The equivalent observable line voltages are calculated as v1 = Tv1Vm1 + Tv2vm2 = 0.707(0.686) + 0.707(0) = 0.485 V v2 = Tv3Vm1 + Tv4vm2 = −0.707(0.686) + 0.707(0) = −0.485 V 186 CROSSTALK Translating each point on the odd-mode lattice diagram to line voltage and currents allows us to construct the lattice diagrams for lines 1 and line 2, as shown in Figure 4-26a and b. As a ﬁnal check, we compare waveforms that we calculated from lattice diagrams created using modal analysis against a fully couple simulation, using a 100-ps rise time, as shown in Figure 4-27a and b. Note that the results are identical. Example 4-5 For the transmission-line pair used in Example 4-4, calculate the crosstalk noise on line 2 when line 1 is driven with a IV, 100 ps rise time step. SOLUTION The solution for the problem of coupling from an active line to a quiet line requires that we carry out the model analysis for both modes (even and odd). To calculate the modal voltage, we must ﬁrst have the voltage inputs. For this example, the input on line 1 is 1 V and on line 2 is 0 V. Therefore, the input voltage matrix is vin = 1 0 and solving (4-65) for the modal voltages using the matrix Tv from Example 4-4 yields vm = 0.707 −0.707 0.707 −1 0.707 −1 1 = vodd veven = 0.707 0.707 Note that there is voltage propagating in both the even and odd modes. Table 4-2 summarizes the relationship between the modal and line voltages for the odd, even, and quiet cases. Since the modal voltages for the case where line 1 is switching and line 2 is held low are vm = 0.707 0.707 TABLE 4-2. Modal Voltages for the Coupled PCB Transmission-Line Pair of Example 4-5 Case Line 1 Line 2 vin (V) vm (V) Even mode Low → high Low → high 1 1 0.707 −0.707 0.707 −1 0.707 1 1 = 0 1.414 Odd mode Low → high High → low 1 −1 0.707 −0.707 0.707 −1 0.707 −1 1 = 1.414 0 Quiet Low → high Low → low 1 0 0.707 −0.707 0.707 −1 0.707 1 0 = 0.707 0.707 MODAL ANALYSIS 187 0 l z 1 ns 2.356 ns 3.712 ns 5.069 ns 6.421 ns v(z = 0) i(z = 0) 0.000 V 0.00 mA 0.485 V 7.92 mA 0.500 V 7.69 mA 7.902.4m85AV 0.015 00.0.010m00A.V24 V mA 0.000 0.00 V mA v (z = l) i(z = l) 0.000 V 0.00 mA 0.500 V 7.68 mA 0.500 V 7.69 mA 0.500 V 7.69 mA t (a) 0 l z 1.000 ns 2.356 ns 3.712 ns 5.069 ns 6.421 ns v(z = 0) i(z = 0) 0.500 V 7.69 mA 0.015 V −0.23 mA 0.000 V 0.00 mA −7−.902.4m85A V −−00.0.010−−m000.A.V02145mVA 0.000 0.00 V mA v (z = l) i(z = l) 0.500 V 7.69 mA 0.000 V 0.01 mA 0.000 V 0.00 mA 0.000 V 0.00 mA t (b) Figure 4-26 Line voltage–current lattice diagram for Example 4-4: (a) line 1 (rising edge); (b) line 2 (falling edge). we start by calculating the propagation delays for both the odd and even modes: td ,even = l νpm,even = 0.2794 m 1.756 × 108 m/s = 1.446 ns td ,odd = l νpm,odd = 0.2794 m 1.873 × 108 m/s = 1.356 ns 188 CROSSTALK Voltage [V] 0.6 0.5 0.4 v(z = 0) v(z = l) 0.3 0.2 0.1 0.0 −0.1 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] (a) Voltage [V] 0.6 0.5 0.4 v(z = 0) v(z = l) 0.3 0.2 0.1 0.0 −0.1 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] 0.6 0.5 0.4 v(z = 0) v(z = l) 0.3 0.2 0.1 0.0 −0.1 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] (b) Voltage [V] 0.6 0.5 0.4 v(z = 0) v(z = l) 0.3 0.2 0.1 0.0 −0.1 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time [ns] Voltage [V] Figure 4-27 Comparison of Odd Mode Transient Waveforms for Example 4-4: (a) line 1 (low-to-high transition); (b) line 2 (low-to-high transition). Left, model analysis; right, coupled simulation. Next we calculate the launch voltage and current for each mode: veven(t = 1, z = 0) = vm,even Zm,even Zm,even + Rs = (0.707) 68.74 68.74 + 65 = 0.363 V ieven(t = 1, z = 0) = veven(t = 1, z Zm,even = 0) = 0.363 68.74 V A V/ mA = 5.29 mA A vodd(t = 1, z = 0) = vm,odd Zm,odd Zm,odd + Rs = (0.707) 61.25 61.25 + 65 = 0.343 V iodd(t = 1, z = 0) = vodd(t = 1, z Zm,odd = 0) = 0.343 61.25 V A V/ mA = 5.60 mA A We also want to know the ﬁnal voltages for each mode: veven(t = ∞) = vm,even RT RT + RS = (0.707) 65 65 + 65 vodd(t = ∞) = vm,odd RT RT + RS = (0.707) 65 65 + 65 = 0.354 V = 0.354 V MODAL ANALYSIS 189 0 Γ = −0.0279 l z Γ = −0.0279 1.000 ns 2.492 ns 3.984 ns 5.476 ns 6.967 ns v (z = 0) 0.000 V 0.343 V 0.354 V 0.354 V t i(z = 0) 0.00 mA 6.07 mA 5.44 mA 5.44 mA v(z = l) 0500..0.6.300040m3m400VAVA..01170mVA 00.0.000m00A0V.0.0000mVA 0.000 V 0.352 V 0.354 V (a) 0 Γ = −0.0279 l z Γ = −0.0279 i(z = l ) 0.00 mA 5.43 mA 5.44 mA 1.000 ns 2.592 ns 4.182 ns 5.773 ns 7.364 ns v (z = 0) 0.000 V i(z = 0) 0.00 mA 0.363 V 5.29 mA 0.354 V 5.44 mA v(z = l) i(z = l ) 50..23963mVA 00.0.000−m00AV−.001.105VmA 0.00.0000 V mA 0.000 V 0.353 V 0.354 V 0.00 mA 5.44 mA 5.44 mA 0.354 V 5.44 mA t (b) Figure 4-28 Lattice diagrams for coupling onto a quiet line in Example 4-5: (a) odd mode; (b) even mode. ieven(t = ∞) = iodd(t = ∞) = 0.354 65 V A V/ mA = 5.45 mA A We now perform lattice diagram analysis for each of the modes, as shown in Figure 4-28a and b. 190 CROSSTALK Voltage (V) Voltage (V) 0.6 0.5 vAggressor(z = 0) 0.4 0.3 vAggressor(z = l ) 0.2 0.1 0.0 −0.1 vVictim(z = 0) vVictim(z = l) −0.2 −0.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time (ns) (a) 00..56 0.4 0.3 0.2 0.1 0.0 −0.1 −0.2 −0.3 0.0 vAggressor(z = 0) vAggressor(z = l ) vVictim(z = 0) vVictim(z = l) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time (ns) (b) Figure 4-29 Comparison of coupling to a quiet line for the pair in Example 4-5: (a) model analysis calculation; (b) HSPICE W-element simulation. Finally, we convert the modal voltages from the odd- and even-mode lattice diagrams back to line voltages while taking into account the difference in propagation delays for the two modes. In converting back to line voltages and currents, we use (4-65) and (4-66) along with the values calculated in the odd-mode lattice diagram analysis. For example, the ﬁrst voltage level on the odd-mode lattice diagram is 0.343 V, and the ﬁrst voltage level on the even-mode lattice diagram is 0.363 V. The equivalent observable line voltages are calculated as v1 = Tv1vm1 + Tv2vm2 = (0.707)(0.343) + (0.707)(0.363) = 0.499 V v2 = Tv3vm1 + Tv4vm2 = (−0.707)(0.343) + (0.707)(0.363) = 0.014 V In matrix form, this is represented as vline = v1 v2 = 0.499 0.014 This amounts to a merging of the even- and odd-mode lattice diagrams in Figure 4-28a and b, as shown in the fourth column of Table 4-3. The modal voltages are converted to the transmission-line signals, vline, contained in the ﬁfth column of the table using equation (4-65). The table contains the necessary information for constructing the waveforms on each line. Figure 4-29 shows that the waveforms calculated with modal analysis match results from simulation, giving a much more accurate result than did analysis using equations (4-60) through (4-64). Modal analysis predicts correctly not only the magnitude for the far-end crosstalk, but also the rise-time degradation of the signal and the shapes and duration of both forward and backward coupled noise pulses. Study of Figure 4-30 reveals some additional insight into the nature of the forward coupled noise. In particular, the far-end noise pulse exists over the time interval between the arrival of the odd-mode signal and the arrival of the even-mode signal. In effect, far-end crosstalk noise is a function of the difference in propagation velocity between odd mode and even mode. The effect is illustrated MODAL ANALYSIS 191 TABLE 4-3. Modal and Line Voltages for the Coupled PCB Transmission-Line Pair of Example 4-5 Event Location t (ns) vm (V) vline = TVvm(V) Notes Waveform launch z=0 1.000 0.000 0.000 0.000 0.000 t =0 z=0 1.100 0.343 0.363 0.499 0.014 t = tr Odd-mode incident wave z=l 2.492 0.000 0.000 0.000 0.000 t = td,odd z=l 2.592 0.352 0.000 0.249 −0.249 t = td,odd + tr Even-mode incident wave z=l 2.592 0.352 0.000 0.249 −0.249 t = td,even z=l 2.692 0.352 0.353 0.499 −0.0007 t = td,even + tr Odd-mode ﬁrst reﬂection z=0 3.984 0.343 0.363 0.499 0.014 t = 2td,odd z=0 4.084 0.354 0.363 0.507 0.006 t = 2td,odd + tr Even-mode ﬁrst reﬂection z=0 4.182 0.354 0.363 0.507 0.006 t = 2td,even z=0 4.282 0.354 0.354 0.499 0.000 t = 2td,even + tr Odd-mode second reﬂection z=l 5.476 0.352 0.353 0.499 −0.0007 t = 3td,odd z=l 5.576 0.354 0.353 0.500 −0.0007 t = 3td,odd + tr Even-mode second reﬂection z=l 5.773 0.354 0.353 0.500 −0.0007 t = 3td,even z=l 5.873 0.354 0.354 0.501 0.000 t = 3td,even + tr for this example in Figure 4-30. In the ﬁgure, the line voltage contributions from each mode are calculated at the far end of line 2 (the quiet line). Odd-mode component: vodd = Tv3vm1 = −0.707vm1 Even-mode component: veven = Tv4vm2 = 0.707vm2 192 CROSSTALK Voltage (V) 0.30 0.20 0.10 0.00 −0.10 −0.20 −0.30 2.0 0.30 0.20 0.10 0.00 −0.10 −0.20 −0.30 2.0 Odd mode signal (arrives early) 2.5 3.0 3.5 4.0 Time (ns) Even mode signal (arrives late) 2.5 3.0 3.5 4.0 Time (ns) Voltage (V) 0.30 0.20 Noise pulse 0.10 (Sum of odd and even mode signals) 0.00 −0.10 −0.20 −0.30 2.0 2.5 3.0 3.5 4.0 Time (ns) Voltage (V) Figure 4-30 Far-end crosstalk generation due to even- and odd-mode propagation velocity differences. Voltage on line 2: v2 = Tv3vm1 + Tv4vm2 = −0.707vm1 + 0.707vm2 = veven + vodd When the even- and odd-mode components are summed to produce the observable line voltage, a pulse occurs at the far end when the modal velocities differ because each component arrives at a different time. Note that this same phenomenon will induce a ledge in the received waveform on the driving line that can be seen when the rise time is short compared to the difference between evenand odd-mode propagation delays. We also note that the description of crosstalk as a function of the difference in even and odd propagation velocities correctly predicts zero far-end crosstalk for coupled lines in homogeneous dielectrics, since the even and odd velocities will be equal. As a ﬁnal note on modal analysis, we acknowledge the difﬁculty of solving for the eigenvectors by hand. We encourage the use of computer tools in ﬁnding the eigenvectors wherever possible. Commercially available math packages such as Mathcad and Mathematica contain routines for calculating eigenvectors and eigenvalues. For readers who are interested in developing their own code, a book by Press et al. [1989] provides fully documented and widely used routines that are easily capable of handling the relatively small matrices that are part of coupled transmission-line analysis. 4.4.5 Modal Analysis of Lossy Lines Although our discussion has focused on the application of modal analysis to lossless coupled lines, it is equally valid for analyzing lossy lines. When including CROSSTALK MINIMIZATION 193 the loss terms, equations (4-65) and (4-66) still apply, but we start from the lossy coupled line equations [Paul, 1994]: d dx v(z) i(z) = 0 −Y −Z 0 v(z) i(z) Z = R + jωL Y = G + jωC (4-85) (4-86) (4-87) In essence, we calculate the eigenvectors Tv and Ti from the product (R + j ωL)(G + j ωC) for the lossy case. Recalling from our example that we calcu- lated Tv and Ti from the product LC, we realize that it is equivalent to using (R + j ωL)(G + j ωC) with R = G = 0. 4.5 CROSSTALK MINIMIZATION Since all of the major components in an interconnect system (i.e., PCB, packages, connectors) can have enough crosstalk to harm system performance, we present some crosstalk reduction guidelines in this section. Because it is often not possible to reduce crosstalk without affecting system cost, we include discussion of trade-offs along with reduction techniques in Table 4-4. In particular, we note that in cost-sensitive applications such as desktop personal computers, adding layers in the printed circuit board represents signiﬁcant added cost to the system. Another technique that sometimes ﬁnds use is the placement of guard traces between signals. These are connected to the ground return layers using plated via holes in the board. This technique requires careful attention to the design to provide the desired crosstalk beneﬁt. Inductance of the traces will tend to create a potential difference at points that are a signiﬁcant distance from the ground vias. When this occurs, the guard traces can radiate the coupled energy, thereby s w M1: Signal h Dielectric (er) M2: Ground M3: Signal M4: Signal M5: VCC M6: Signal Figure 4-31 PCB layer stackup for Example 4-6. 194 CROSSTALK TABLE 4-4. Crosstalk Reduction Techniques and Trade-offs Approach Trade-off Increase the spacing, s, between the PCB and package traces. Decrease the thickness of the dielectric, h, between the ground (return) layer and the signal layer in PCBs and packages to couple the transmission lines more tightly to the return layer and reduce the coupling to adjacent signals. Use differential signaling. Route traces on adjacent PCB signal layers orthogonally to each other. Route the signals in PCBs and packages using striplines or embedded microstrips to eliminate velocity variations. Minimize parallel run lengths between signals in PCBs and packages. Reduce signal edge rates. Insert power/ground pins between signal I/O pins in connectors, sockets, and packages. Can add cost by requiring additional layers in PCBs and/or packages to route all traces. Operating limits for characteristic impedance may limit the minimum acceptable thickness (recall that Z0 decreases as h decreases). The minimum manufacturable dielectric thickness may also limit the effectiveness of this approach. Can add cost by requiring additional layers in PCBs and/or packages to route all traces, since each signal requires two traces. Can also add cost to packages, sockets, and connectors, due to increased pin counts. Can add cost by requiring additional layers in PCBs and/or packages to route all traces, since restricting signal routing direction may decrease routing efﬁciency. Striplines require at least six layers in the PCB. Refer to Figure 4-31. Can add cost by requiring additional layers in PCBs and/or packages to route all traces, since restricting signal routing direction may decrease routing efﬁciency. May limit the maximum performance, since rise and fall times typically must scale with data rate. Cost increases with pin count. defeating their purpose. As a result, the guard traces must be connected to the ground layer at multiple points. The distance allowed between vias is inversely proportional to the frequency content of the signals, making it impractical for multigigabit per second data rates. 4.6 SUMMARY In this chapter we described the coupling mechanisms that cause crosstalk in digital systems. The SLEM modeling technique and the equations for coupled PROBLEMS 195 noise provide the means to make ﬁrst-order engineering estimates of the impacts of crosstalk on high-speed systems, while modal decomposition provides a more accurate analytical technique that does not require the use of coupled simulations. Guidelines for crosstalk reduction provide designers with a toolbox for managing coupled noise. REFERENCES In addition to the references cited in the text, the reader may ﬁnd that others provide useful background on the topic of crosstalk. In particular, a book by Paul [1994] is the classic text on the analysis of coupled transmission lines; books by Bakoglu [1990] and Poon [2002] also contain general discussions of crosstalk. For an alternative treatment of modal decomposition, refer to Young’s [2001] book. Bakoglu, H. Brian, 1990, Circuits, Interconnections, and Packaging for VLSI , Addison-Wesley, Reading, MA. Djordjevic, Antonije, Miodrag Bazdar, Tapan Sarkar, and Roger Harrington, 1999, LINPAR for Windows: Matrix Parameters for Multiconductor Transmission Lines, Software and User’s Manual, Version 2.0 , Artech House, Norwood, MA. Hall, Stephen, Garrett Hall, and James McCall, 2000, High Speed Digital System Design, Wiley-Interscience, New York. O’Neil, P., 1983, Advanced Engineering Mathematics, Wadsworth, Belmont, CA. Paul, Clayton, 1994, Analysis of Multiconductor Transmission Lines, Wiley-Interscience, New York. Poon, Ron, 1995, Computer Circuits Electrical Design, Prentice Hall, Upper Saddle River, NJ. Press, William, Saul Teukolsky, William Vetterling, and Brian Flannery, 2002, Numerical Recipes in C++: The Art of Scientiﬁc Computing, 2nd ed., Cambridge University Press, Cambridge, UK. Seraphim, Don, Ron Lasky, and Che-Yu Li, eds., 1989, Principles of Electronic Packaging, McGraw-Hill, New York. Young, Brian, 2001, Digital Signal Integrity, Prentice Hall, Upper Saddle River, NJ. PROBLEMS 4-1 Use the SLEM method to calculate the effective even- and odd-mode impedances and propagation velocities for the coupled striplines whose capacitance and inductance matrices are shown below. Estimate the impacts of crosstalk on the propagation delay for a 0.5-m coupled length. L= 3.480 × 10−7 1.951 × 10−8 1.951 × 10−8 3.480 × 10−7 H/m C= 1.271 × 10−10 −7.213 × 10−12 −7.213 × 10−12 1.271 × 10−10 F/m 196 CROSSTALK 4-2 For coupled striplines from Problem 4-1, calculate the near- and far-end noise for the isolated switching case, and compare your results against simulation with each line terminated at both ends in the characteristic impedance calculated. 4-3 Use the SLEM method to calculate the effective impedance and propagation velocity for the coupled striplines shown whose capacitance and inductance matrices are shown below when all three lines switch in the same direction. Compare the propagation delay for a 0.5-m coupled length when only the middle line switches. 3.480 × 10−7 5.268 × 10−8 1.687 × 10−8 L = 5.268 × 10−8 3.461 × 10−7 5.268 × 10−8 H/m 1.687 × 10−8 5.268 × 10−8 3.480 × 10−7 1.087 × 10−10 −1.172 × 10−11 −7.918 × 10−11 C = −1.172 × 10−11 1.105 × 10−10 −1.172 × 10−11 F/m −7.918 × 10−11 −1.172 × 10−11 1.087 × 10−10 4-4 Estimate crosstalk pulse amplitudes and pulse widths on the middle lines of the three-line system from Problem 4-3 when the two outer lines are switching from low to high. Assume that each line is terminated at both ends in its characteristic impedance. Compare your results against a fully coupled simulation. 4-5 Sketch the far-end crosstalk pulse for a two-line case with no termination at the near end, and matched termination at the far end. 4-6 Use the inductance and capacitance matrices below to determine the switching activity on lines 1 and 2, given the waveform on line 3 in Figure 4-32. 3.544 × 10−7 1.914 × 10−8 5.161 × 10−9 L = 1.914 × 10−8 3.826 × 10−7 1.914 × 10−8 H/m 5.161 × 10−9 1.914 × 10−8 3.544 × 10−7 8.266 × 10−11 −1.108 × 10−11 −2.354 × 10−11 C = −1.108 × 10−11 1.001 × 10−10 −1.108 × 10−11 F/m −2.354 × 10−11 −1.108 × 10−11 8.266 × 10−11 4-7 Use the inductance and capacitance matrices from Problem 4-6 to determine the switching activity on lines 1 and 3 given the waveform on line 2 in Figure 4-33. PROBLEMS Voltage (V) 0.02 v(l = 0) 0.00 −0.02 −0.04 −0.06 −0.08 v(l = 0.254m) −0.10 −0.12 −0.14 −0.16 −1.0 0.0 1.0 2.0 3.0 4.0 5.0 Time (ns) Figure 4-32 Transient response for Problem 4-6. 197 6.0 Voltage (V) 0.05 v (l = 0) 0.00 −0.05 −0.10 −0.15 v(l = 0.254m) −0.20 −0.25 −0.30 0.0 1.0 2.0 3.0 4.0 5.0 6.0 Time (ns) Figure 4-33 Transient response for Problem 4-7. 4-8 The circuit in Figure 4-34 uses transmission lines that have the induc- tance and capacitance matrices given in Problem 4-1. In this problem, line 1 undergoes a rising-edge transition at z = 0; simultaneously, line 2 undergoes a falling-edge transition at the other end of the pair (z = l). Calculate the waveform on line 1 at the receiving end (z = l) for a 0.254-m 198 CROSSTALK coupled length, and compare your results to those from a fully coupled simulation. z=0 z=I Vs1 = 1 volt tr 1 = 100 ps 63 Ω 63 Ω Line 1 Line 2 63 Ω 63 Ω Vs 2 = 1 volt tf 2 = 100 ps Figure 4-34 Coupled line system for Problem 4-8. 4-9 Prove that a coupled pair in a homogeneous medium has no forward crosstalk noise. 4-10 Use modal analysis to determine the resistor values required to terminate both even and odd modes using a pi network for a two-transmission-line system. 4-11 Calculate modal reﬂection coefﬁcients for the two lines in Problem 4-1 when they are terminated in 50 at each end. 4-12 The PCB transmission lines depicted in Figure 4-35 have the following inductances and capacitances: z=0 z=I Vs1 = 1 volt tr 1 = 100 ps 65 Ω 65 Ω Line 1 Line 2 65 Ω 65 Ω (a) 0.005 in 0.005 in 0.005 in 0.002 in 0.005 in er = 4.0 (b) Figure 4-35 PCB-based coupled transmission-line pair for Problem 4-12: (a) schematic; (b) cross section. PROBLEMS 199 L= 3.537 × 10−7 9.559 × 10−8 9.559 × 10−8 3.537 × 10−7 H/m C= 8.533 × 10−11 −1.205 × 10−11 −1.205 × 10−11 8.533 × 10−11 F/m The traces have a length of 0.254 m (10 in.) and are terminated to ground with 65 . They are driven by a 1-V 65- source, with 100-ps rise and fall times. Analyze the crosstalk noise using the approximate model from Section 4.3.2 and modal analysis, and compare your results with simulation. Do all of the results agree? If not, why not? 4-13 Use modal analysis to: (a) Calculate the modal impedance and velocity for the three-coupled-line case in Problem 4-3. (b) Simulate a 0.5-m coupled length and compare your results to those obtained using SLEM analysis. 4-14 Develop expressions for the forward and backward crosstalk amplitudes for the circuit shown in Figure 4-17, in which the victim line is open circuited at the near end. 4-15 Use modal analysis to derive equations for the near- and far-end crosstalk amplitudes for the terminated coupled pair shown in Figure 4-22. 4-16 Project: Develop a tool (Matlab, C++, etc.) to perform modal analysis for an arbitrary number of coupled lines, using the inductance and capacitance matrices as input. 5 NONIDEAL CONDUCTOR MODELS 5.1 Signals propagating in unbounded conductive media 202 5.1.1 Propagation constant for conductive media 202 5.1.2 Skin depth 204 5.2 Classic conductor model for transmission lines 205 5.2.1 Dc losses in conductors 206 5.2.2 Frequency-dependent resistance in conductors 207 5.2.3 Frequency-dependent inductance 213 5.2.4 Power loss in a smooth conductor 218 5.3 Surface roughness 222 5.3.1 Hammerstad model 223 5.3.2 Hemispherical model 228 5.3.3 Huray model 237 5.3.4 Conclusions 243 5.4 Transmission-line parameters for nonideal conductors 244 5.4.1 Equivalent circuit, impedance, and propagation constant 244 5.4.2 Telegrapher’s equations for a real conductor and a perfect dielectric 246 References 246 Problems 247 As digital systems evolve and technology pushes for smaller and faster designs, the geometric dimensions of the physical platform are shrinking. Smaller dimensions and higher data transmission rates necessitate the use of proper techniques to model both the frequency dependent resistive losses and inductance. Without proper models that accurately predict these quantities, simulation-based bus design for multigigabit data rates is not possible. Frequency-dependent resistive losses, for example, will affect bus performance by decreasing the signal amplitude and slowing edge rates, which in turn affects voltage and timing margins, respectively. In addition, frequency dependent inductance models are required to Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 201 202 NONIDEAL CONDUCTOR MODELS preserve causality , which is discussed in Chapter 8 and Appendix E. In prior days it was possible to utilize simpler conductor models for digital designs because bandwidth demands were much lower. However, as digital data rates increase, the assumptions and approximations of traditional conductor models begin to break down. Consequently, the signal integrity engineer is now required to learn new techniques that compensate for variables that were insigniﬁcant in past designs. So far in this book we have covered the topics of electromagnetic theory for signal integrity engineers, transmission-line fundamentals, and crosstalk. Up to this point, the conductors were assumed to be inﬁnitely conductive and the dielectric was assumed to be a perfect insulator. In this chapter we develop modeling techniques to predict properly the electrical behavior of conductors used to design transmission lines on printed circuit boards, multichip modules, and chip packages. First, classic electromagnetic theory will be used to derive the frequency dependence of resistance and inductance for smooth conductors with ﬁnite conductivity. Next, three different methodologies for modeling the effects of rough copper on the electrical parameters of the transmission lines will be introduced. Detailed analysis of how currents ﬂow on a rough surface will give physical insights into the mechanisms of surface roughness losses. Finally, a new circuit model for a transmission line that accounts for realistic conductors will be introduced along with a modiﬁed version of the telegrapher’s equations that account for realistic conductor losses. 5.1 SIGNALS PROPAGATING IN UNBOUNDED CONDUCTIVE MEDIA The topic of uniform plane waves propagating in a lossless media was discussed in Section 2.3, where the inﬂuences of the material properties µ and ε were observed. In Chapter 3 we described how the waves propagated when conﬁned to the physical dimensions of a transmission line, yet the problem was still idealized because it was assumed that the dielectric was a perfect insulator and the conductor was inﬁnitely conductive. To derive the equations that govern the propagation of waves on realistic transmission lines, it is ﬁrst necessary to comprehend how an electromagnetic wave propagates in unbounded, conductive media. 5.1.1 Propagation Constant for Conductive Media To derive the equations that govern electromagnetic waves propagating in conductive or lossy media, we begin with the loss-free forms of Maxwell’s differential equations presented in Chapter 2 and modify them appropriately to obtain a wave equation that accounts for loss. To begin, the time-harmonic forms of Maxwell’s equations are repeated here: ∇ × E + jωB = 0 ∇ × H = J + jωD (2-33) (2-34) SIGNALS PROPAGATING IN UNBOUNDED CONDUCTIVE MEDIA 203 ∇ ·D = ρ ∇·B =0 (2-35) (2-36) where J = σ E, D = εE, and B = µH , as derived from equations (2-6) through (2-9). Simpliﬁcation of Ampe`re’s law (2-34) by replacing the current density term with σ E and D = εE yields ∇ × H = σ E + jωεE = jω σ + ε E = jω ε − j σ E jω ω (5-1) Comparison to the solution of Ampe`re’s law in a loss-free medium (∇ × H = j ωεE) allows us to deﬁne the complex permittivity for a conductive or lossy media by analogy: ε = ε − j σ = ε − jε ω (5-2) where the real component is the dielectric permittivity discussed in Chapters 2 and 3 (ε = ε0εr ) and the imaginary component accounts for the losses in the medium where the wave is propagating. The term σ can be thought of as the conductivity of the material, which will be quite high for a metal and quite low for a dielectric. If (5-2) is inserted into the time-harmonic solution of the electric ﬁeld derived in Section 2.3.4, Ex (z, t) = Ex+e−γ z + Ex−eγ z (2-54) then the complex propagation constant for a lossy media can be derived. The complex propagation constant for a plane wave was derived in Section 2.3.4: γ = α + jβ (2-42) As discussed extensively in Chapter 2, if the wave is propagating in a loss-free medium (where α = 0 ), (2-42) reduces to jβ = √ ω µε rad/m (5-3) where ε = εr ε0 and µ = µr µ0. Substitution of the complex permittivity (5-2) into (5-3) provides the form of the complex propagation constant of an electromagnetic wave traveling in a conductive medium: γ =ω µ ε −jσ ω (5-4) Setting (5-4) equal to (2-42) and separating into real and imaginary components yields the general form of the attenuation constant α and the phase constant β for 204 NONIDEAL CONDUCTOR MODELS an electromagnetic wave propagating in a conductive media with a conductivity of σ : √ α = ω √µε 2 1/2 1+ σ 2 −1 εω Np/m (5-5) √ β = ω √µε 2 1/2 σ2 1+ +1 εω rad/m (5-6) Note that µ = µ0 for virtually every practical digital design because the conductors are almost always constructed from copper, which is not magnetic. Both α and β have units of 1/m; however, the dimensionless terms neper (Np) and radian (rad) are used to communicate the attenuation and phase meanings in the wave equation. 5.1.2 Skin Depth As described in Section 2.3.4, α is the attenuation constant, which will modify a wave propagating in the z-direction as described by E(z, t ) = Re Ex+e−γ zejωt = Re Ex+e−αze−jβzejωt = e−αzEx+ cos(ωt − βz) (2-43) The factor e−αz is known as the wave decay for a wave propagating in the +z-direction. The wave attenuation in a conductive region is governed by the term 1 + (σ/ε ω)2, as shown in equation (5-5). As the conductivity of the medium is increased, the attenuation constant α becomes larger and the wave decay increases with distance and time. Consequently, for a good conductor such as copper, the wave will decay very rapidly. The decay of an electromagnetic wave propagating into a conductor is measured in terms of the skin depth. The skin depth, denoted δ, is simply the distance of penetration where the settling exponent −αz of the wave decay factor is −1 (e−αz = e−1). The skin depth is therefore given in units of meters: δ= 1 α (5-7) Since the term J = σ E in Ampe`re’s law (2-34) is no longer neglected, a current density J must accompany the electric ﬁeld E in the conductive region: Jx+(z, t) = e−αzσ Ex+ cos(ωt − βz) (5-8) Therefore, at a penetration distance of one skin depth (1/α), the ﬁeld intensity and the current density have been attenuated by a factor of e−1, or approximately 36.7%, meaning that approximately 63.6% of the current density exists within a distance of δ from the conductor surface. Note that for a perfect conductor, the CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 205 Amplitude Attenuation 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 0 JX = sEX e−1 Wave motion into the conductive media (z) d (z) 1.0 2.0 3.0 4.0 5. 0 Skin depth Figure 5-1 Penetration depth δ associated with the amplitude attenuation of the current density for a plane wave propagating into a conductive region. conductivity σ is inﬁnite and therefore α is also inﬁnite. If α is inﬁnite, equation (5-7) says that δ must be inﬁnitely small. Therefore, for a perfect conductor, the current only ﬂows on the surface and the wave cannot penetrate the conductor. Figure 5-1 shows how the current density decays as the wave propagates into a conductive medium. Note that although the magnitude of the current density is oscillatory, it remains within the envelope deﬁned by the exponential decay of the skin depth. If a good conductor is deﬁned such that σ/εω 1, (5-5) reduces to α = ωµσ 2 (5-9) Therefore, the skin depth for a metal with conductivity σ at a frequency (ω = 2πf ) is given by (5-10) in units of meters: δ= 1 = 2 α ωµσ (5-10) Figure 5-2 shows the magnitude of the skin depth in copper as a function of frequency. Note that even at 1 GHz, the skin depth is only 2 µm, meaning that most of the current is ﬂowing in a very small area. 5.2 CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES Classic conductor models are derived on the assumption of perfectly smooth surfaces. Although realistic conductors used to construct printed circuit boards (PCBs), packages, and multichip modules for high-speed digital designs rarely employ smooth conductors, a study of classical transmission-line losses will 206 NONIDEAL CONDUCTOR MODELS Skin depth, µm 7 6 5 4 3 2 1 0 0 2 4 6 8 10 12 14 16 18 20 Frequency, GHz Figure 5-2 Skin depth δ in copper as a function of frequency. provide the theoretical basis needed to derive frequency-dependent physically consistent models of real-world conductors, which are often purposely roughened to promote adhesion to dielectric layers during the manufacturing process. The resistive loss induced by general transmission-line conductors can be broken down into two components: low frequency, or dc, and high frequency, or ac. First, the dc losses will be derived and the formulas will be modiﬁed to include the frequency-dependent effects of ac resistance at high frequencies. 5.2.1 DC Losses in Conductors Dc losses are of particular concern in small-geometry conductors, very long lines, and multiload (also known as multidrop) buses. Long copper telecommunication lines, for example, must have repeaters every few miles to receive and retransmit the data because of signal degradation. Additionally, designs of multiprocessor computer systems with long buses experience resistive drops that can encroach on the logic threshold levels and reduce the noise margins. The dc loss of a transmission line depends primarily on two factors: the conductivity of the metal and the cross-sectional area of the conductor where the current is ﬂowing. Figure 5-3 shows the current distribution in a transmission line at dc. Traditionally, dc resistance is deﬁned to be the value at 0 Hz. However, for the purposes of this chapter, dc will be assumed to be valid for all frequencies where the skin depth is larger than the conductor thickness t, which w t At dc, current flows through entire area of the cross section where Area = A = wt Figure 5-3 Current distribution in a microstrip at dc. CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 207 ensures almost uniform current density through the cross section of the signal conductor. At dc the current will spread out as much as possible and ﬂow through the entire cross section of the conductor, and the resistive loss can be found with Rdc = l σ Across section = l σ wt ohms (5-11) where l is the length, w the width, t the thickness of the signal conductor, and σ the conductivity of the metal. Note that (5-11) has neglected the dc losses of the current return path in the reference plane. This is an adequate approximation because at dc, the current will spread out and ﬂow through the entire plane, which is several orders of magnitude larger than the signal conductor. Consequently, the cross-sectional area where the current ﬂows in the return path will be much larger and the associated resistance will be much smaller. 5.2.2 Frequency-Dependent Resistance in Conductors By extending the dc equation (5-11), the frequency dependence of the resistance in a transmission line can be approximated. Frequency-dependent resistance will be referred to as ac resistance or skin effect resistance in the remainder of the book. At low frequencies, the ac resistance will be identical to the dc resistance because the skin depth will be much greater than the thickness of the conductor. The ac resistance will remain equal to the dc resistance until the frequency increases to a point where the skin depth is smaller than the conductor thickness. Microstrip Conductor Losses (Smooth Conductors) Figure 5-4 depicts the current distribution on a microstrip line at high frequencies. Notice that the current distribution is concentrated on the bottom edge of the transmission line. This is because the ﬁelds between the signal line and the ground plane pull the charge to the bottom edge, and the skin depth is much smaller than the conductor thickness. Also notice that the current density is greater near the corners of the conductor. This is because the charge density increases signiﬁcantly in the proximity of a sharp edge, as described in Sections 3.4.4 and 3.4.5, and the current density along the conductor will vary in the same way. Furthermore, there is still signiﬁcant ﬁeld concentration along the thickness (the t dimension in Figure 5-4) of the conductor. w t h d (Skin depth) Reference plane Figure 5-4 Current distribution in a microstrip with an ideal reference plane at high frequencies where the skin depth δ is small compared to the thickness t. 208 NONIDEAL CONDUCTOR MODELS The skin effect will cause the cross-sectional area where the current is ﬂowing to decrease as the frequency increases. Consequently, the frequency-dependent losses in the conductor can be approximated using the dc resistance formula by setting t = δ: Rac = l σ wδ = √l σ w 2/ωµσ = l w π µf σ ohms (5-12) Note that the approximation is valid only when the skin depth is smaller than the conductor thickness. Notice that the ac resistance is directly proportional to the square root of the frequency f and inversely proportional to the conductivity σ . Equation (5-12) assumes that all the current is ﬂowing in the ﬁrst skin depth, which is not correct. Section 5.1.2 deﬁnes the skin depth such that only about 63% of the current density is contained in this depth. To test the validity of equation (5-12), the effective area of an exponential decay can be calculated by integrating e−αz from z = 0δ to z = ∞δ and comparing it to the case where all the current is conﬁned to one skin depth. To visualize the differences, refer to Figure 5-5, which plots penetration depth into a conductive medium in terms of skin depths versus the total current density. If 100% of the current is assumed to ﬂow within one skin depth, the area under the curve is J δ = 1, where J is the current density and δ is the skin depth. Integrating the wave decay term e−αz from z = 0δ to z = ∞δ, the area under the curve also yields J δ: J z=∞δ e−αz dz = J = Jδ z=0δ α Since the effective areas under each curve are identical, it is a valid approximation to assume that all the current is ﬂowing in an area conﬁned by the conductor width Current density (J ), A/m2 1 0.9 Jd = 1 0.8 0.7 0.6 0.5 J z = ∞d e−az dz = J = Jd 0.4 z = 0d a 0.3 0.2 0.1 0 0 1 2 3 4 5 6 7 8 9 10 Skin depth, d Figure 5-5 If all the current is approximated to be in one skin depth, the total area under the curve is identical to the realistic behavior, where the current density decays exponentially with increasing skin depths. CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 209 Approximate AC Resistance, Ohms 4.0 w = 127 mm t = 50 µm 3.5 Reference plane Resistance changes with f when d < t d (Skin depth) 3.0 (l = 10 inches (0.254 m)) 2.5 2.0 Rac_conductor ≈ l swd = l w pmf s = 521.5 × 10−6 f 1.5 Rdc = l swt = 0.254 m = 0.68 Ω (5.8 × 107)(127µm)(50µm) 1.0 0.5 Resistance is ~constant at the dc value when d > t 0.0 0 5 10 15 20 25 30 35 40 45 50 Frequency, MHz Figure 5-6 Frequency-dependent resistance of a microstrip transmission line with an ideal reference plane. and a single skin depth. As this section progresses, more accurate methods of calculating the losses will be presented. Figure 5-6 plots the total resistance of a transmission line over a wide fre- quency range. Note that the resistance will remain at approximately the dc value until the frequency where the skin depth is smaller than the thickness of the conductor, after√which the ac resistance begins to take effect and increases proportional to f . Note that the discontinuity shown in Figure 5-6 is artiﬁ- cial and drawn for instructional purposes. In reality, current is not completely conﬁned to a single skin depth, so the transition from Rdc to Rac is more gradual and not discontinuous. The discontinuity generally will not have any ill effects on simulated waveforms; however, if it is desirable to smooth the curve to provide more realistic behavior, a root-sum-square function can be used: Rtotal ≈ Rd2c + Ra2c (5-13) The resistance of the signal conductor, however, is only one part of the total ac resistance. The portion that is not included in equation (5-12) is the resistance of the return current on the reference plane. The return current will ﬂow underneath the signal line in the reference plane, will be largely concentrated in one skin depth, and will spread out perpendicular to the trace direction with the highest amount of current concentrated directly beneath the signal conductor. Equation (5-14) was derived by Collins [1992] using conformal mapping 210 NONIDEAL CONDUCTOR MODELS J(d) ∝ J0 2 1+ d h w h J(d) Reference plane −3h −2h −1h 0 1h 2h 3h Distance from center, d Figure 5-7 Current distribution in the reference plane of a microstrip. techniques and shows how the current density varies with increasing distance from the signal conductor center: J (d) ∝ 1 + J0 (d/ h)2 (5-14) where d is the distance from the conductor center and h is the height above the ground plane and J0 is the total current density. Figure 5-7 is a graphical representation of this current density distribution. An approximation of the ground plane resistance can be derived using a technique similar to that used to ﬁnd the ac resistance of the signal conductor. First, assume that all the current will be conﬁned to one skin depth δ. Next, an effective width weff where the current will ﬂow must be determined. Integrating (5-14) from −∞ to +∞ with h = 1, ∞ −∞ 1 + 1 (d /1)2 dd = tan−1 (d )|∞ −∞ = π allows us to determine that (5-14) must be normalized by π so that the total current density is unity for an inﬁnite plane. If the effective width is chosen somewhat arbitrarily at ±3h from the center of the conductor and the normalized current density function is integrated, J0 π 3 −3 1 + 1 (d /1)2 dd = J0 2 tan−1(3) π = 0.795J0 (5-15) it can be shown that about 80% of the total current density is contained within a distance of ±3h of the signal conductors center. Using this approximation, CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 211 weff = 6h and an approximate formula can be derived for the ground-plane resistance in a microstrip transmission line in units of ohms: Rac, ground ≈ l σ weffδ = l 6h π µf σ The total resistance is the sum of (5-12) and (5-16) in ohms: (5-16) Rac, micro = l w πµf + l σ 6h πµf = σ π µf σ 1+ 1 w 6h (5-17) Equation (5-17) should be considered a good “back of the envelope” estimation of the ac resistance for a microstrip transmission line [Hall et al., 2000] A more exact formula for the ac resistance of a microstrip was derived using conformal mapping techniques by Collins [1992] and is shown in equation set (5-18). This formula is signiﬁcantly more cumbersome than (5-17), but should yield the most accurate results. Rtrace = LR 1 π + 1 π2 ln 4π w t Rs w where LR is given by 1 LR = 0.94 + 0.132 w − 0.0062 w 2 h h when w ≤ 1 h2 when 1 < w ≤ 10 2h Rground = w/ h (w/ h) + 5.8 + 0.03(h/w) Rs w when 1 ≤ w ≤ 10 (5-18) 10 h where Rs = ωµ 2σ Rac, micro = Rtrace + Rground For practical micristrip lines, formulas based on smooth conductors should simply be used as an approximation because realistic conductor surfaces are generally rough, which will increase the conductor losses signiﬁcantly at frequencies where the skin depth begins to approach the magnitude of the roughness proﬁle. The extra losses caused by surface roughness are calculated in Section 5.3. Example 5-1 Calculate the approximate frequency where ac resistance must be used to calculate the ohmic losses of a microstrip transmission line constructed with a copper conductor with a conductivity of σ = 5.8 × 107( · m)−1 and the following cross-sectional dimensions: w = 5 mils, h = 3 mils, t = 2.1 mils. 212 NONIDEAL CONDUCTOR MODELS SOLUTION The ac resistance will exist when the frequency gets high enough so that the skin depth is smaller than the conductor thickness. Above this frequency, dc resistance ceases to exist; only ac (skin effect) resistance is present. The frequency can be calculated by setting a skin depth equal to the conductor thickness using (5-10) and solving for the frequency: δ = 2.1 mils = 2 ωµσ Since ω = 2πf, µ = µ0 = 12.56 × 10−7 H/m and 2.1 mils = 55.3 × 10−6 m: f = 2 2π σ µ0(53.3 × 2 10−6) = 1.53 × 106 Hz Therefore, at 1√.53 MHz, dc resistance does not exist and ac resistance begins to increase with f . Stripline Losses (Smooth Conductors) In a stripline transmission line, the currents of a high-frequency signal are concentrated in the upper and lower edges of the conductor. The current density will be dependent on the proximity of the local reference planes. If the stripline is referenced equidistant from both planes, the current will be divided equally in the upper and lower portions of the conductor as depicted in Figure 5-8. In an offset transmission line, the current densities on the upper and lower edges of the transmission line will be dependent on the relative distances between the ground planes and the conductor (h1 and h2 in Figure 5-8). The current density distributions in each stripline reference plane will be governed by an equation similar to equation J(d )∝ 1 2 1+ d hx h1 w t h2 Reference plane d (Skin depth) Reference plane d Figure 5-8 Current distribution in the signal; conductor and reference planes of a stripline. CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 213 (5-14) and will differ only in the magnitude, which is a function of the respective distances between the reference planes and the strip (h1 and h2). Thus, the resistance of a stripline can be approximated by the parallel combination of the resistance in the top and bottom portions of the conductor. The resistance equations for the upper and lower sections of the stripline may be obtained by applying equation (5-17) or (5-18) for the appropriate value of h. These two resistance values must then be put in parallel to obtain the total resistance for a stripline: Rac, strip = (R(h1)ac, micro)(R(h2)ac, micro) R(h1)ac, mico + R(h2)ac, micro (5-19) 5.2.3 Frequency-Dependent Inductance Section 5.1 describes how the skin effect will force high-frequency current to ﬂow primarily in a small layer near the periphery of the conductor, and Section 5.2 describes how this translates into frequency-dependent resistance. Another consequence of the skin effect is a frequency-dependent inductance. To conceptualize where this frequency dependence comes from, consider two ﬁlaments of current that form a loop with the return plane immediately below the conductor over a differential length of transmission line z, as depicted in Figure 5-9. Loop (a) passes through the center of the signal conductor, and loop (b) exists only on the conductor surfaces. As described in Section 2.5.2, the inductance is proportional to the loop area, L11 ≡ ψ1 I1 (2-97) where ψ1 is the magnetic ﬂux, which depends on the loop area. Therefore, loop (a) will have a higher inductance than loop (b) simply because the loop is larger. At low frequencies, the skin depth will be large compared to the conductor thickness and there will be signiﬁcant current ﬂowing in the interior of the conductor, length = ∆z length = ∆z Signal Conductor ∆Iinternal ∆Iexternal Reference Conductor (a) (b) Figure 5-9 Loops for a ﬁlament of current: (a) in the center of the conductor; (b) on the surface of the conductor. 214 NONIDEAL CONDUCTOR MODELS similar to loop (a). As frequency increases, the skin effect will begin to force the current to ﬂow on the periphery of the conductor, so internal currents will decrease. At very high frequencies, there will be almost no current ﬂowing in loop (a) because the skin effect has forced all the current to ﬂow on the sur- face, causing the inductance of loop (b) to dominate. The inductance associated with loop (a), where the currents are ﬂowing internal to the conductor, is com- monly known as the internal inductance. The inductance associated with loop (b), where all currents are ﬂowing on the surface of the conductor, is commonly known as the external inductance. The total inductance is the sum of the external and internal parts: Ltotal = Linternal + Lexternal (5-20) As frequency increases, the skin effect causes the current to ﬂow very close to the surface, with minimal current ﬂowing interior to the conductor. At inﬁnite frequencies, all the current is ﬂowing on the surface and the internal contribution is zero. Consequently, as frequency increases, the effect of Linternal will decrease and Ltotal will asymptote to Lexternal. The external inductance is calculated with quasistatic techniques assuming that all the charge is on the surface of the conductor and solving Laplace’s equation as shown in Sections 3.4.2, 3.4.3, and 3.4.6. The internal inductance is derived by observing Ampe`re’s (2-34) and Faraday’s (2-33) laws for a good conductor. For a good conductor, the conduction current J will be much larger than the displacement current j ωD, and therefore Ampe`re’s law reduces to ∇ ×H ≈ J = σE (5-21) Faraday’s law is repeated here: ∇ × E + j ωµH = 0 (2-33) Taking the curl of (2-33) yields the following equation: ∇ × (∇ × E) = −j ωµ∇ × H (5-22) If it is assumed that the free charge is negligible and the source of the electric ﬁeld is the time-varying magnetic ﬁeld, Gauss’s law reduces to ∇ · εE = 0, allowing the use of the vector identity (see Appendix A) ∇ × (∇ × ψ) = ∇(∇ · ψ) − ∇2ψ ∇ × (∇ × E) = 0 − ∇2E = −j ωµ∇ × H to simplify the equation, which reduces to ∇2E = j ωµ∇ × H (5-23) CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 215 Substituting Ampe`re’s law (5-21) into (5-23) and multiplying each side by σ allows us to write (5-23) in terms of the current density (since J = σ E): ∇2σ E = j ωµσ σ E → ∇2J = j ωµσ J (5-24) Equation (5-24) is known as the diffusion equation for the current density. Similar equations govern the electric and magnetic ﬁelds. Assuming a z-directed current, the solution to (5-24) is found using standard techniques. d 2 Jz dx2 = j ωµσ Jz d2 dx2 − j ωµσ Jz = 0 The roots of the characteristic equation yield √ ± −4(−j ωµσ ) = ± 2 ωµσ + j ωµσ = ±(1 + j ) ωµσ 2 2 2 √ Comparing to equation (5-10), note that ωµσ/2 = 1/δ, giving a solution to the diffusion equation for current density of Jz = J0e−(1+j )x/δ (5-25) where only the negative exponent is a valid solution because the current density decreases as the electromagnetic wave penetrates into the conductor. Equation (5-25) says that the current density will decrease very rapidly into the surface of the conductor and is essentially conﬁned to a layer at the surface equal to a few skin depths, as depicted in Figure 5-10. The total current can be found by integrating (5-25): JT = ∞ 0 J0e−(1+j )x/δdx = − 1 2 J0 δ(j − 1) (5-26) z J0 x J0 J0 d m, s Figure 5-10 Diffusion of currents into a semi-inﬁnite conductive material. 216 NONIDEAL CONDUCTOR MODELS The surface impedance can be deﬁned as the ratio of the electric ﬁeld at the surface and the total current density: Zs = E0 JT (5-27) Substituting (5-26) into (5-27) and expressing the current density in terms of the conductivity and the electric ﬁeld (J0 = σ E0) yields Zs = E0 − 1 2 σ E0δ(j − 1) = (1 + 1 j) σδ ohms/square (5-28) The surface impedance is expressed in terms of an area of unit width and unit length, so the term ohms/square is used. Note the similarity of the real part of (5-28) to equation (5-12), which was the series resistance caused by the skin effect of a transmission line. For the case where l = w, the real part of equations (5-28) and (5-12) are identical. If the real part of (5-28) is resistance, the imaginary part must therefore be reactance (the impedance of inductance). Since the impedance of inductance is j ωL, (5-28) can be expressed in terms of a series resistance (due to the skin effect) and a series inductance (the internal inductance): Zs = Rac + j ωLinternal (5-29) Therefore, the internal inductance can be calculated directly from the ac resis- tance: Linternal = Rac ω (5-30) Equation (5-30) highlights an important relationship between the skin effect resistance (ac resistance) and the internal inductance. As the skin effect forces the current to the periphery of the conductor, the resistance increases; however, since current ceases to ﬂow in the interior of the conductor, the inductance must decrease. Figure 5-11 plots both the internal inductance and the ac resistance. Note that when the ac resistance becomes signiﬁcant, the internal inductance is almost negligible. Example 5-2 Calculate the total inductance and the resistance at 2 GHz of a microstrip transmission line constructed with copper of conductivity σ = 5.8 × 107( · m)−1, a dielectric constant of εr = 4.0 and the following cross-sectional dimensions: t = 0.5 mil, h = 2 mils, w = 3 mils. SOLUTION Step 1: Determine if the ac or dc resistance should be used. The methodology presented in Example 5-1 could be used, however, it is easier simply to calculate CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 217 Skin effect resistance, Ω/m Internal inductance, nH/m 300 Linternal 250 200 Rac 150 100 50 0 1.00E+07 1.00E+08 1.00E+09 1.00E+10 Frequency, Hz 100 90 80 70 60 50 40 30 20 10 0 1.00E+11 Figure 5-11 Example of how the skin effect changes resistance and internal inductance with frequency for a copper microstrip as shown in Figure 5-6. the skin depth in copper at 2 GHz using (5-10) and compare it to the conductor thickness t: δ= 2 ≈ 1.41 µm 2πf µ0σ Since t = 0.5 mil = 12.7 µm, δ < t, so the ac resistance must be used. Step 2: The impedance and effective dielectric permittivity must be calculated using equation (3-36b), which assumes a perfect conductor. If it is unknown whether a particular impedance formula includes the effect of a realistic conductor, the lack of a metal conductivity or a magnetic permeability variable indicates the assumption of inﬁnite conductivity. From (3-36b), Z0 ≈ 55 and εeff ≈ 2.95. Step 3: Calculate the phase velocity with (2-52): νp = c √ µr εeff = 3 × 108 m/s √ (1)(2.95) = 1.75 × 108 m/s Step 4: The external inductance is solved with (3-31) and (3-33). Since (3-36b) assumes a perfect conductor, the inductance is the external value. νp = √1 = 1.75 × 108 → √ LC = 5.73 × 10−9 s/m LC ≈ 146 × 10−12 s/in. L Z0 = = 55 C Lexternal = L = L · √ LC = 55(146 × 10−12) = 8.03 nH/in. C 218 NONIDEAL CONDUCTOR MODELS Step 5: Calculate the ac resistance at 2 GHz using (5-17). Note that (5-18) could be used as well. Rac, micro = π µ0f σ 1+ 1 w 6h = 0.0117 1 (3 mils)(25.4 × 10−6 m/mil) + 6(2 1 mils)(25.4 × 10−6 m/mil) = 191.3 /m = 4.86 /in. Step 6: Calculate the internal inductance using (5-30). Linternal = Rac ω = 4.86 2π(2 × /in. 109 Hz) = 0.387 nH/in. Step 7: Calculate the total inductance using (5-20). Ltotal = Linternal + Lexternal = 8.03 + 0.387 = 8.42 nH/in. 5.2.4 Power Loss in a Smooth Conductor In high-speed digital design, surface treatment of the copper foils used to construct printed circuit boards (PCBs) signiﬁcantly affects the power losses experienced by a signal propagating on a transmission line. In this section, the power losses of an electromagnetic wave impinging on a ﬂat, smooth plane are examined. In later sections we explore the consequences of rough conductor surfaces. First, we assume that the ﬁelds in the vicinity of a good but not perfect conductor will behave approximately the same as for a perfect conductor. In Section 3.2.1 it was shown that the electric ﬁelds terminate normal to a perfect conducting surface and the magnetic ﬁelds are tangential to the surface. Furthermore, in Section 5.1.2 it was shown that ﬁelds inside a conductor will attenuate exponentially and are measured in terms of the skin depth δ. At high frequencies, the boundary condition shown in (3-3) is true for a good conductor, except for a thin transitional layer. To derive an equation to predict the loss for a smooth plane, we ﬁrst assume that just outside the conductor there exists only a normal component of the electric ﬁeld (E⊥) and a tangential component of the magnetic ﬁeld (H ), which are the identical boundary conditions used for perfect conductors. Following the approach outlined by Jackson [1999], Maxwell’s equations are then used to calculate the ﬁelds within the transition layer. If a tangential H exists just outside the surface, the same H must exist just inside the conductor surface. With the neglect of the displacement current, (2-33) and (2-34) become CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 219 ∇ × Ec + j ωµHc = 0 ∇ × Hc = J = σ Ec (5-31) (5-32) where Ec and Hc denote the ﬁeld values inside the conductor. If it is assumed that n is the normal vector pointing outward from the conductor surface and z is the normal coordinate inward into the conductor, the gradient operator is ∇ ≈ −n(∂/∂z), reducing Maxwell’s equations to Hc = j n × ∂ Ec /∂ z µω (5-33) Ec = −n × ∂ Hc /∂ z σ (5-34) These equations can be solved to yield the ﬁelds inside the conductor. The ﬁrst step is to take the partial derivative of (5-34): ∂ Ec ∂z = −1n× σ ∂ 2 Hc ∂ z2 Next is to take the cross product of (5-33) with the unit vector, (5-35) j n × Hc = n µω × n × ∂Ec ∂z so the vector identity from Appendix A can be used to simplify the math, a × (b × c) = (a · c)b − (a · b)c n × Hc = j µω n · ∂Ec n − (n · n) ∂Ec ∂z ∂z where n · ∂Ec/∂z = 0 because when (5-35) is substituted, the form becomes proportional to n · n × ∂2Hc/∂z2 , which is zero. Furthermore, n · n = 1, yielding n × Hc = −j µω ∂ Ec ∂z Next, equation (5-35) is substituted for ∂Ec/∂z, yielding n × Hc = j µω 1 σ n × ∂ 2 Hc ∂ z2 = j µωσ ∂2 ∂ z2 n × Hc 220 NONIDEAL CONDUCTOR MODELS After rearranging the equation, we get a more manageable form: −j µωσ n × Hc = ∂2 ∂ z2 n × Hc Since µωσ = 2/(2/µωσ ) = 2/δ2, the equation can be written in terms of the skin depth, −j 2 δ2 n × Hc = ∂2 ∂ z2 n × Hc which yields the differential equation ∂2 2j ∂z2 n × Hc + δ2 n × Hc = 0 (5-36) To solve (5-36), we reasonably assume that the ﬁeld inside the conductor is a function of the external applied ﬁeld, Hc = H||f (z) (5-37) where f (z = 0) = 1, which says that at the surface of the conductor, Hc = H||. This allows us to substitute (5-37) into (5-36) and solve the equation for f (z). ∂2 ∂z2 f (z) + 2j δ2 f (z) = 0 The solution to (5-38) is f (z) = Ae−z/δe−jz/δ (5-38) where f (z = 0) = 1 = A, yielding Hc = H||e−z/δe−j z/δ (5-39) where H|| is the tangential magnetic ﬁeld applied to the surface of the conductor. The electric ﬁeld inside the conductor is calculated from (5-34): Ec = −n × ∂ Hc /∂ z σ = −1n σ × − 1 δ (1 + j )H|| e−z/δ e−j z/δ = 1+j σδ n × H|| e−z/δ e−j z/δ = (1 + j ) ωµ 2σ n × H|| e−z/δ e−j z/δ (5-40) CLASSIC CONDUCTOR MODEL FOR TRANSMISSION LINES 221 Since the tangential component of the electric ﬁeld must remain continuous, as shown in equation (3-8), the electric ﬁeld just outside the conductor surface can be calculated with (5-40) evaluated at z = 0: E|| = (1 + j ) ωµ 2σ n × H|| (5-41) Section 3.2.1 says the electric ﬁeld must terminate normal to a perfectly conducting surface, however, equation (5-41) shows that for a good conductor, a tangential component of E must exist just outside the conductor. Since (5-40) describes the electric ﬁeld decaying with increasing depth z into the surface, there must be power ﬂow into the conductor. The time-averaged value of the Poynting vector described in Section 2.6.1 is used to calculate the power absorbed per unit area: Save = az (E+)2 2η (2-121) The intrinsic impedance (2-53) at the surface of the conductor, ηs, is calculated with (5-37) and (5-41): ηs(z = 0) = Ec = (1 + j ) Hc ωµ 2σ (5-42) √ It is interesting to note that since µω/2σ = 1/σ δ, (5-42) reduces to equation (5-28), which is the series impedance of a transmission line: (1 + j ) ωµ = (1 + j ) 1 2σ σδ Finally, the power ﬂow per unit area into the conductor is calculated using equation (2-121): 2 Ec Save = 2η = ωµ 4σ 2 H|| 2σ = ωµ ωµ 4σ 2 H|| 2σ 2 ωµσ √ From (5-10), δ = 2/ωµσ , yielding (5-43), which is the time-averaged power absorbed by a ﬂat conducting plane per unit area: 2 ωµδ H|| Save = Pplane = 4 W/m2 (5-43) Equation (5-43) will allow for the solution of the resistive power losses of a good conducting plane provided that the applied magnetic ﬁeld H|| has been solved for the idealized case of a perfect conducting plane. 222 NONIDEAL CONDUCTOR MODELS The power loss from a plane can also be put in more intuitive form. Equation (2-7), which is simply Ohm’s law, can be used to calculate the current density from (5-40): J = σ Ec = σ µω (1 + j ) 2σ n × H|| e−z/δ e−j z/δ √ Again, since µω/2σ = 1/σ δ, the equation can be rewritten as 1 J = σ Ec = (1 + j ) δ n × H|| e−z/δe−j z/δ (5-44) Equation (5-44) simply says that most of the current density will be conﬁned to a very small thickness, as described in Section 5.1.2. Note that the cross product in (5-44) implies that the current will ﬂow perpendicular to the magnetic ﬁeld. To derive a more intuitive form of the power dissipated by a ﬂat plane, it is ﬁrst necessary to deﬁne an effective surface current. If the current density (5-44) is integrated to get the total current, an equivalent surface current can be calculated for use in the classic time-averaged power equation P = 1 2 RI 2: ∞ Jeff = J dz 0 amperes (5-45) Equation (5-45) simply calculates the total current that is decaying exponentially into the conductor surface. To calculate the power per unit area, we make the approximation that all the current exists on the surface and use the real part of the surface impedance in (5-28): Pplane = 1 RI 2 2 = 1 2σ δ 2 Jeff W/square (5-46) 5.3 SURFACE ROUGHNESS To account properly for the frequency variation of Linternal and Rac, the nonideal effects of the copper surface must be considered. The problem is that most (if not all) commercial 2D ﬁeld solvers calculate the resistance and inductance assuming smooth conductors. Real copper surfaces, however, are purposely roughened to promote adhesion to the dielectric when manufacturing printed circuit boards. The resulting copper surfaces have a “tooth structure” as depicted in Figure 5-12. When the tooth height is comparable to the skin depth, the smooth copper assumptions break down. The root-mean-square (RMS) tooth height of common copper foils used to manufacture printed circuit boards range from approximately 0.3 to 5.8 µm, with peak heights exceeding 11 µm [Brist et al., 2005]. The skin depth in copper at 1 GHz is about 2 µm, indicating that for many copper foils, most of the current will be ﬂowing in the tooth structure for multigigabit designs SURFACE ROUGHNESS 223 Trace Skin depth Ground Plane Figure 5-12 Realistic conductors used to manufacture transmission lines exhibit a rough surface often called the “tooth structure.” When the skin depth is similar to the tooth size, power dissipation is increased. [Hall et al., 2000]. Since the rough copper surface affects current ﬂow, it will also affect power dissipation, and thus insertion loss. Insertion loss, which is described extensively in Chapter 9, is a common way to measure the frequency-dependent losses in the form of a transfer function by injecting a sinusoidal waveform at port 1 (such as the input to a transmission line) and measuring at port 2 (such as the output). Expressed in terms of power, the insertion loss is S21(f ) = 20 log P2(f ) P1(f ) dB where S21 is the insertion loss in decibels, P2 is the power measured at the output of the transmission line, and P1 is the power injected into the input of the transmission line. Insertion loss is a convenient method to evaluate the power loss of a transmission line. Note that the ratio of powers reduce to a ratio of voltages when the port impedances are identical, as shown in equation (9-21), which is why the form 20 log is used instead of 10 log to calculate the magnitude in decibels [Hall et al., 2000]. Figure 5-13 shows the measured results of two identical transmission lines built with rough and relatively smooth copper. The copper foil used to construct the test boards was characterized with an optical proﬁlometer prior to lamination yielding an RMS tooth height of hRMS = 1.2 µm for the smoother copper and hRMS = 5.8 µm for the rough copper. Note the signiﬁcant increase in insertion losses, and therefore power losses, due to increased roughness proﬁle. At high frequencies, surface roughness will increase the ohmic losses of a transmission-line conductor signiﬁcantly . 5.3.1 Hammerstad Model The traditional way to account for surface roughness losses in a transmission-line model is to use the Hammerstad equation: Rac = KHRs f (5-47) 224 0 −5 −10 NONIDEAL CONDUCTOR MODELS hRMS = 1.2 µm hRMS = 5.8 µm S21(dB) −15 −20 −25 0 5 10 15 20 25 30 Frequency, GHz Figure 5-13 Measured results of identical 7-in. transmission lines with relatively smooth (hRMS = 1.2 µm) and rough (hRMS = 5.8 µm) copper showing how surface roughness affects losses. √ where Rs f is the classic skin resistance for a smooth conductor as calculated in (5-17) and (5-18) and KH is the Hammerstad coefﬁcient: KH = 1 + 2 π arctan 1.4 hRMS δ 2 (5-48) where hRMS is the root-mean-square value of the surface roughness height and δ is the skin depth [Hammerstad and Jensen, 1980; Brist et al., 2005]. The Hammerstad coefﬁcient is used to model the extra losses caused by the copper surfaces on a transmission line that are often purposely roughened to promote adhesion to the dielectric. The frequency dependence of the skin effect resistance and total inductance using the Hammerstad correction for surface roughness is implemented with† √ RH(f ) = KHRs f Rdc when δ < t when δ ≥ t LH(f ) = Lexternal + Lexternal + RH(f ) 2πf RH(fδ=t ) 2πfδ=t when δ < t when δ ≥ t (5-49a) (5-49b) †Note that the method shown here for calculating the internal portion of the inductance (Linternal = Rac/ω) for a rough conductor is an approximation based on the result for a smooth conductor. The approximation will induce causality errors that tend to be small enough to ignore, so this method is generally acceptable. For the interested reader, Appendix E derives the internal inductance using a more rigorous approach based on the discussion in Chapter 8. SURFACE ROUGHNESS 225 where KH is calculated with (5-48), t is the conductor thickness, δ the skin depth, and fδ=t the frequency where the skin depth equals the thickness of the conductor. When the skin depth (δ) is larger than the conductor thickness, the dc value of the resistance and low-frequency inductance where the skin depth is equal to the conductor thickness should be used. This approach has been shown to be accurate by comparison to vector network analyzer measurements of insertion loss for copper surface roughness proﬁles less than approximately 2 µm (RMS). Figure 5-14 shows the accuracy of the Hammerstad model by comparing measured transmission-line structures constructed with relatively smooth and very rough copper to simulations using (5-49). Notice that the Hammerstad model is considerably less accurate for the rough copper case for frequencies greater than about 5 GHz. To understand why the accuracy breaks down for some copper types, it is useful to explore the assumptions behind (5-48), which assumes a 2D corrugated surface similar to that shown in Figure 5-15 [Pytel, 2007]. The ﬁrst published work providing a theoretical investigation into the power loss from copper roughness was in 1948 by Samuel Morgan (Bell Laboratories), who studied the effects up to 10 GHz providing loss equations for current ﬂow transverse and parallel to corrugated structures similar to that depicted in Figure 5-15. He concluded S21(dB) S21(dB) 0 −5 −10 −15 −20 −25 0 0 −5 −10 −15 −20 5 10 15 20 25 30 Frequency, GHz −25 0 5 10 15 20 25 30 Frequency, GHz (a) (b) Measurement Hammerstad Model Figure 5-14 Example of the Hammerstad model accuracy for (a) relatively smooth (hRMS = 1.2 µm) and (b) rough (hRMS = 5.8 µm) copper foils; 7-in. microstrip line. I hRMS Figure 5-15 Two-dimensional corrugated surface assumption behind the Hammerstad equation. 226 NONIDEAL CONDUCTOR MODELS that power loss is proportional to the surface area of the roughness structure, current ﬂow transverse to the corrugated surface could increase the power losses by 100%, and current ﬂow parallel to the grooves increases losses by about 30%. In 1975, a Norwegian scientist named Erik Hammerstad used Morgan’s work to ﬁt the data to an arctan function, producing the Hammerstad equation shown in (5-48), which became the standard equation in industry to account for the effects of surface roughness [Pytel, 2007]. The model assumes that at high frequencies, when the skin depth becomes small compared to the tooth height, the current will begin to follow the contour of the corrugated surface, which will increase the losses. To understand why the Hammerstad equation breaks down for some roughness proﬁles, the surfaces of relatively smooth and rough copper surfaces were measured using an optical proﬁlometer. The surface shown in Figure 5-16a [Hall et al., 2007] can be described as corrugated with sparse protrusions on the surface, suggesting that the Hammerstad equation (5-48) might be adequate for approximating the surface roughness losses for a transmission line manufactured with this copper foil. Figure 5-14a depicts a measurement of a transmission line that was constructed with the copper foil depicted in Figure 5-16a compared to a model created using (5-47) and (5-48). Note that the model and measurement (a) (b) Figure 5-16 Surface proﬁle measurement of (a) relatively smooth and (b) rough copper foil used to construct PCBs. SURFACE ROUGHNESS 227 correlate very nicely until about 15 GHz, indicating that the Hammerstad model is adequate to model the losses for this case. Conversely, Figure 5-16b depicts an optical proﬁlometer measurement of a very rough copper foil. Note the signiﬁcant difference between the corrugated surface assumed by the Hammerstad equation and the surface depicted in Figure 5-16b, suggesting that (5-48) will not work well for this type of 3D surface proﬁle. Figure 5-14b depicts the measurement of a transmission line constructed with the rough copper sample shown in Figure 5-16b compared to a model created with (5-47) and (5-48). Note that the accuracy breaks down after about 5 GHz, indicating that an alternative modeling methodology is needed for losses induced by this surface proﬁle. Example 5-3 Assuming the transmission line in Example 5-2, calculate the frequency where surface roughness begins to affect the losses for an RMS tooth height of 1.8 µm, calculate the ac resistance and total inductance at an operating frequency of 2 GHz, and determine how surface roughness changes the resistance and inductance compared to the smooth case. SOLUTION Step 1: To determine the approximate frequency where the surface roughness will begin to affect the ac losses, the skin depth must be set equal to the RMS roughness height, and the frequency is calculated using (5-10): fr = 2 2π σ µ0(1.8 × 10−6)2 = 1.34 × 109 Hz Since this is close to the operating frequency of 2 GHz, the surface roughness will be signiﬁcant and cannot be ignored. Note that the roughness will inﬂuence the losses at frequencies below fr ; however, if fr f , the surface roughness will have no signiﬁcant effect, and it can be ignored. Step 2: Calculate the ac resistance at 2 GHz using (5-47) and (5-48). The ac resistance at 2 GHz for a smooth surface was calculated in Example 5-2. R = KHRs f = KH π µ0f σ 1+ 1 w 6h = KH · 4.86 /in. Since the RMS roughness height is less than 2 µm, (5-48) is suitable to correct the smooth formula to estimate the extra losses due to the rough surface. δ(2 GHz) = 2 ≈ 1.41 µm 2πf µ0σ 228 NONIDEAL CONDUCTOR MODELS 2 1.8 × 10−6 2 KH = 1 + π arctan 1.4 1.41 × 10−6 = 1.73 Therefore, the ac resistance including the surface roughness is R = KH · 4.86 /in. = (1.73)(4.86) = 8.4 /in. Step 3: Calculate the internal inductance using (5-30). Linternal = Rac ω = 8.4 2π(2 × /in. 109 Hz) = 0.669 nH/in. Step 4: Calculate the total inductance using (5-20), where Lexternal is as calculated in Example 5-2. Ltotal = Linternal + Lexternal = 8.03 + 0.669 = 8.7 nH/in. Step 5: Compare the values to the smooth case in Example 5-2. Note that the surface roughness increases the resistance and internal portion of the inductance signiﬁcantly compared to the identical values for a smooth conductor from Example 5-2. Smooth conductor Ltotal = 8.42 nH/in. Rac = 4.86 /in. Rough conductor (hRMS = 1.8 µm) Ltotal = 8.7 nH/in. Rac = 8.4 /in. 5.3.2 Hemispherical Model The rough surface depicted in Figure 5-16b can be characterized as random protrusions sitting on a ﬂat plane, which precludes rigorous derivation of an analytical formula to calculate the extra losses due to current ﬂowing in the tooth structure. Subsequently, for the rough copper, an approximation of the tooth structure is required so that an analytical solution can be derived. As a ﬁrst approximation, a hemispherical boss sitting on a plane can be used to represent the individual surface protrusions as shown in Figure 5-17 [Hall et al., 2007]. The complete surface is modeled using N hemispheres randomly distributed on a ﬂat plane. A TEM (transverse electromagnetic) wave is assumed incident on the hemisphere at a grazing angle of 90◦ with respect to the ﬂat plane and with the H ﬁeld (the magnetic ﬁeld intensity) tangential to the surrounding plane as shown in Figure 5-17. To ﬁnd the power dissipated by the structure, the absorption and scattering of the incident TEM wave on the hemisphere must be calculated. The problem of scattering of a plane wave from the hemispherical protrusion on the ﬂat surface can be approximated using superposition. First the power losses of a sphere are calculated and then divided in half because the SURFACE ROUGHNESS 229 l Atile = lw E H I Abase qm w E I H Figure 5-17 Simpliﬁed hemispherical shape approximating a single surface protrusion: top and side views; current direction and applied TEM ﬁeld orientations shown. structure is a hemisphere, and then the power loss of the ﬂat plane surrounding the hemisphere is calculated. The total power dissipated is then calculated simply by adding the results. For a good conducting protrusion (as opposed to a perfect electrical conductor protrusion) a plane electromagnetic wave incident on a conducting sphere will be partly scattered and partly absorbed. The total power scattered and absorbed from a sphere divided by the incident ﬂux is known as the total cross section and is calculated by Jackson [1999] (with units of square meters) as σtot = − π k2 (2m + 1) Re[α(m) + β(m)] m (5-50) √ where k = 2π/λ, λ = c/f ε , c is the speed of light, and the scattering coefﬁcients are approximated assuming that kr 1, where r is the sphere radius, and are given by [Jackson, 1999] α(1) = − 2j 3 (kr )3 1 − (δ/r)(1 + j ) 1 + (δ/2r)(1 + j ) β (1) = − 2j 3 (kr )3 1 1 − + (4j (2j / / k2 k2 r r δ)[1/(1 δ)[1/(1 − − j j )] )] (5-51a) (5-51b) 230 NONIDEAL CONDUCTOR MODELS The Poynting vector, which was described in Section 2.6, gives the power of an electromagnetic wave in units of watts per square meter: |S| = 1 2 Re(E0 × H0∗) = 1 2 η|H0 |2 (5-52) Subsequently, the total power absorbed or scattered is calculated by multiplying (5-50) and (5-52) and dividing by 2 so the results are for that of a hemisphere: Phemisphere = 1 2 1 2 η|H0 |2σtot = −Re 1 4 η|H0|2 3π k2 (α(1) + β (1)) (5-53) √ where η = µ0/ε0ε and H0 is the magnitude of the applied magnetic ﬁeld. Note that reasonable accuracy (at least up to 30 GHz) can be obtained when only the ﬁrst term (m = 1) in (5-50) is considered when calculating (5-53). Equation (5-53) calculates the power loss of a hemisphere. Now, the losses of the ﬂat plane surrounding the protrusion must be accounted for. The time-averaged power absorbed by a ﬂat conducting plane per unit area is calculated from equation (5-43): d Pplane da = µ0 ωδ 4 |H0|2 (5-54) To approximate the losses of a single hemispherical boss sitting on a ﬂat plane of ﬁnite conductivity, the power loss of the hemisphere is added to the loss of the plane less the base area of the hemisphere: Ptot = −Re 1 4 η|H0 |2 3π 2k2 (α(1) + β (1)) + µ0 ωδ 4 |H0 |2(Atile − Abase) (5-55) where Atile is the tile area of the plane surrounding the protrusion (see Figure 5-17) and Abase is the base area of the hemisphere. Note that (5-55) is an approximation because it assumes that the magnetic ﬁeld (H0) on the tile is not affected by the presence of the hemisphere and the loss of the surrounding plane is simply a function of the area. To gain an intuitive understanding of how a propagating electromagnetic wave behaves in the presence of a protrusion, it is useful to observe the ﬁelds and solve for the surface currents on a PEC (perfect electrical conducting) sphere, which is a good approximation of how the current will ﬂow at very high frequencies when the skin depth is small compared to the sphere. In Section 3.2.1 the boundary conditions for a PEC were described; the electric ﬁeld must emanate from and terminate normal to a perfectly conducting surface, and the magnetic ﬁeld must be tangential to the conductor surface. First consider Figure 5-18a, which depicts the front cross-sectional view of a hemispherical protrusion sitting on a conducting plane where the current ﬂow is out of the page. Note that the electric ﬁelds are drawn perpendicular to the conductor surface and the magnetic ﬁelds are tangent SURFACE ROUGHNESS I H E 231 E I H Flat copper plane Electric field Magnetic field Surface current flowing out of the page (a) Current streamlines (b) Magnetic field at the plane surface Figure 5-18 (a) Front view of ﬁelds and current showing the normal electric ﬁeld and tangential magnetic ﬁeld; (b) current streamlines ﬂowing over the top of a single surface protrusion. to the surface. Figure 5-18b shows the top view of the same protrusion with the current ﬂowing from the left to the right. If TEM is assumed, the magnetic ﬁeld will be perpendicular to the electric ﬁeld as described is Section 2.3.2. The magnetic ﬁeld intensity lines depicted in Figure 5-18b are those near the surface of the plane and not on the sphere. Note that these ﬁelds bend around the protrusion because they must satisfy the boundary conditions and remain tangent to the PEC hemisphere. Furthermore, equation (5-44) says that lines of constant magnetic ﬁeld intensity are orthogonal to the lines of surface current ﬂow, indicating that if the magnetic ﬁeld bends around the hemisphere, an area of low current density perpendicular to the current ﬂow will be induced. The surface current density on the hemisphere (Jeff) (in A/m2) can be derived by deﬁning the magnetic ﬁeld in terms of a magnetic scalar potential [Orlando and Delin, 1991; Huray et al., 2007; Huray, 2008]: Jeff = −Re 3 2 H0 sin θmej ωt (5-56) where θm is the angle between the applied magnetic ﬁeld and the current ﬂow (see Figure 5-17). If the uniform current streamlines on the plane are matched with those calculated with (5-56), the inﬂuence of a hemisphere sitting on a plane can be observed. Figure 5-18b depicts how current will ﬂow in the presence of a spherical protrusion. Note that the current on the ﬂat portion of the plane is drawn toward the protrusion with the highest density on the top and minimal current density on the side perpendicular to the current ﬂow. The current crowding at the top of the protrusion effectively decreases the area where current ﬂows, increases the path length, and thus helps explain the physical mechanisms that cause extra 232 NONIDEAL CONDUCTOR MODELS losses on a rough surface. It also appears to support Morgan’s claim that the surface area is the key factor to surface roughness losses. To calculate a new correction factor for use in (5-47), the ratio of the power absorbed with and without a good conducting protrusion present must be derived. This is accomplished simply by dividing (5-55) by (5-54): Ptot Pplane = Ks = Re 1 4 η|H0|2(3π/ k2)(α(1) + β (1)) + (µ0 ωδ/4)|H0|2(Atile − Abase) µ0 ωδ 4 |H0 |2 Atile This equation can be simpliﬁed to eliminate the variable of the magnetic ﬁeld, yielding Ks = Re η(3π/4k2)(α(1) + β(1)) + (µ0 ωδ/4)(Atile − Abase) (µ0ωδ/4)Atile (5-57) Note that (5-57) becomes invalid when the skin depth is greater than the surface protrusion height. At these frequencies the power dissipated by a ﬂat plane with an area equal to the base of the protrusion will be greater than the power dissipated by the protrusion. Subsequently, a knee frequency can be deﬁned when Ks = 1, where the roughness begins to affect the losses signiﬁcantly. Below the knee frequency, the correction factor, Ks, is unity. Subsequently, implementation of this correction factor is shown as Khemi = 1 Ks when Ks ≤ 1 when Ks > 1 (5-58) To implement (5-58) accurately, the surface shown in Figure 5-16b must somehow be represented by equivalent hemispheres. The additional surface area of simple hemisphere bosses are insufﬁcient to account for the measured surface roughness losses. This is not surprising when one compares the additional hemisphere model area to the 3D surface of the rough copper sample shown in Figure 5-16b. To account for additional surface area the root mean square (RMS) volume of the rough surface must be calculated, and volume equivalent hemispheres are created to determine Abase. The RMS distance between peaks in the roughness proﬁle is used to calculate the tile area, Atile. To obtain these input parameters, the surface is measured using a proﬁlometer, as shown in Figure 5-19. To facilitate the volume equivalent model, the tooth shape is approximated as one-half of a prolate spheroid instead of a hemisphere, because it more closely resembles the shape of the protrusion. A prolate spheroid is a surface of revolution obtained by rotating an ellipse about its major axis. A symmetrical egg (i.e., with the same shape at both ends) would approximate a prolate spheroid SURFACE ROUGHNESS 233 Height, µm 10 dpeaks 8 6 4 2 htooth bbase 0 20 40 60 80 100 Position, µm Figure 5-19 Example of a surface proﬁlometer measurement of a rough copper sample showing peak heights that range from 0.7 to 8.5 µm. The ﬂat surface is assumed to be at 0.5 µm. [Mathworld, n.d.]. The spheroid volume is in turn equal to one-half the volume of a sphere to calculate the radius of a hemisphere with the same volume as the hemispheroid-shaped surface protrusion: re = 3 htooth bbase 2 2 (5-59) where bbase is the tooth base width, htooth the tooth height, and re the radius of a hemisphere with equivalent tooth volume. The base area of the hemisphere, Abase, is then calculated: Abase = π bbase 2 2 (5-60) The square tile area of the surrounding ﬂat plane is calculated based on the distance between peaks: Atile = dp2eaks (5-61) If the RMS values of dpeaks, htooth, and bbase values are calculated, the surface shown in Figure 5-16b and measured in Figure 5-19 can be represented as the equivalent surface in Figure 5-20. A comparison between the correction factor calculated from the Hammerstad model (5-48) and the hemispherical model (5-58) with the modiﬁed equivalent volume is shown in Figure 5-21. Note that the hemisphere model saturates at a much higher value than Hammerstad, which will always saturate at a value of 2. The implementation shown in (5-58) causes a nonphysical discontinuity at the frequency where the model transitions from dc to ac behavior, which is the point where the surface of the ﬂat plane with an area equal to the base of the hemisphere has more loss than the hemisphere. Some engineers may be concerned that the discontinuity may induce nonphysical glitches into the time-domain responses. However, simulations of pulses as fast as 30 Gb/s were 234 NONIDEAL CONDUCTOR MODELS dpeaks, RMS re,RMS Figure 5-20 Equivalent surface represented by hemispheres with the same RMS volume as that of the measured surface proﬁle. Surface roughness correction factor 2.8 Hemisphere model (5-58) 2.4 2 Hammerstad model (5-48) 1.6 1.2 0.8 0 10 20 30 40 50 Frequency, GHz Figure 5-21 Hammerstad correction factor (5-48) compared to the hemisphere model (5-58). RMS roughness: hRMS = 5.8 µm, dpeaks,RMS = 9.4 µm. performed using this method in HSPICE and Nexxim with no apparent nonphysical aberrations observed. The frequency dependence of the skin effect resistance and total inductance using the hemisphere correction for surface roughness is implemented with (5-62a) and (5-62b).† When the skin depth δ is larger than the conductor thickness (which includes the roughness proﬁle), the dc value of the resistance and low-frequency inductance where the skin depth is equal to the total conductor thickness should be used: †Just as with equation (5.49b), the method shown here for calculating the internal portion of the inductance (Linternal = Rac/ω) for a rough conductor is an approximation based on the result for a smooth conductor. The approximation will induce causality errors that tend to be small enough to ignore, so this method is generally acceptable. Appendix E derives the internal inductance using a more rigorous approach based on the discussion in Chapter 8. SURFACE ROUGHNESS 235 0 Hemisphere model −5 Measurement −10 S21, dB −15 −20 −25 0 5 10 15 20 25 30 Frequency,GHz Figure 5-22 Accuracy of the hemisphere surface roughness model (5-58); 7 in. microstrip; εr / tan δ = 3.9/0.0073 at 1 GHz; RMS roughness of copper foil hRMS = 5.8 µm, dpeaks,RMS = 9.4 µm. (Adapted from Hall et al. [2007].) √ Rhemi(f ) = KhemiRs f Rdc when δ < t when δ ≥ t Lhemi(f ) = Lexternal + Rhemi(f 2πf ) Lexternal + Rhemi(fδ=t ) 2πfδ=t when δ < t when δ ≥ t (5-62a) (5-62b) where Khemi is calculated with (5-58), t is the conductor thickness, δ is the skin depth, and fδ=t is the frequency where the skin depth equals the thickness of the conductor. Figure 5-22 depicts the accuracy of the hemisphere model for very rough copper. The small deviation that occurs between the model and the measurement in Figure 5-22 is because (5-57) essentially uses superposition to combine the losses of the tile and protrusion and does not account for the interaction between the hemisphere and the plane. Consequently, the formula is very accurate (1) at low frequencies where the skin depth is large compared to the protrusion and the loss is due primarily to the plane, and (2) at high frequencies where the skin depth is small compared to the protrusion and the loss is due primarily to the roughness. At intermediate frequencies, where the skin depth is on the same order as the roughness height, an error is introduced. Additionally, the roughness shape was approximated with a spheroid and the interactions between spheres were neglected. Nonetheless, for very rough copper surfaces, Figure 5-22 shows that the methodology produces very reasonable accuracy over a wide bandwidth with a simple, easy-to-use formula. More important, it provides valuable intuition into the mechanisms that cause surface roughness losses. Note that the method of calculating the surface roughness correction factor K should be chosen carefully. If relatively smooth copper is being used, with 236 NONIDEAL CONDUCTOR MODELS a RMS value of the surface roughness less than about 2 µm, Hammerstad’s formula (5-48) has been shown to adequately approximate the surface roughness losses. However, for very rough copper (which is often preferred by PCB vendors due to its decreased tendency of delaminating), equation (5-58) using equivalent volume hemispheres will approximate the surface roughness losses with more accuracy. Example 5-4 Calculate the surface roughness correction factor of the microstrip transmission line in Example 5-2 at 5 GHz. The RMS value of the tooth height htooth = 5.8 µm, the RMS base width of the tooth structures is bbase = 9.4 µm, and the distance between peaks dpeaks = 9.4 µm. SOLUTION Step 1: Use equations (5-59) through (5-61) to calculate a sphere with the same volume as a spheroid-shaped surface protrusion: re = 3 htooth bbase 2 = 5 µm 2 Abase = π bbase 2 = 69.4 µm2 2 Atile = dp2eaks = 88.4 µm2 Note that the diameter of the equivalent volume hemisphere (10 µm) is actually larger than the edge length of the tile (9.4 µm), meaning that the hemispheres will overlap. This is not a concern since it is understood that the surface roughness shape is not hemispherical in nature and the shape of the protrusion is assumed to be a spheroid. However, the equivalent volume hemisphere allows for a much simpler solution to the electromagnetic ﬁelds without sacriﬁcing much accuracy. Step 2: Use (5-57) to calc√ulate the correction factor at 5 GHz. To calculate the intrinsic impedance η = µ0/ε0ε , the value of the dielectric permittivity directly under the rough surface should be used and not the effective value for the microstrip as calculated in Example 5-2. η = µ0 = √377 = 3√77 = 188.5 ε0ε ε 4 δ(5 GHz) = 2 ≈ 0.935 µm 2πf µ0σ SURFACE ROUGHNESS 237 λ = √c = 3 × 10√8 = 0.03 m f ε 5 × 109 4 k = 2π = 209.3 m−1 λ α(1) = − 2j 3 (k re )3 1 − (δ/re)(1 + j ) 1 + (δ/2re)(1 + j ) = − 2j 3 [209.3(5 × 10−6)]3 1 1 − + 0.935 × 10−6 5 × 10−6 (1 0.935 × 10−6 2 · 5 × 10−6 (1 + + j ) j) = −j 0.764 × 10−9 1 − 0.187(1 + j ) 1 + 0.0935(1 + j ) = −1.78 × 10−10 − j 5.53 × 10−10 For this example, it can be shown that β(1) is negligible, so it is ignored. Ks = Re η(3π/4k2)α(1) + (µ0ωδ/4)(Atile − Abase) (µ0ωδ/4)Atile = 1.8 × 10−12 + 1.75 × 10−13 8.15 × 10−13 = 2.42 In this case, the series resistance of a transmission line manufactured with a copper conductor with this roughness proﬁle would be approximately 2.42 times as high at 5 GHz as the same transmission line constructed with a smooth conductor. 5.3.3 Huray Model In 2006 at the University of South Carolina, Paul G. Huray was researching new wideband modeling techniques for surface roughness that would provide better accuracy than both the Hammerstad and hemisphere models. Upon observation of scanning electron microscope (SEM) photographs of copper foil samples used to manufacture printed circuit boards (PCBs), he observed that the structures appeared to be constructed of conducting “snowballs,” as shown in Figure 5-23. Subsequently, he formed a material and physical basis of a theoretical model that is composed of a distribution of spherical shapes [Olufemi, 2007; Hurray, 2009]. Printed circuit boards are a “stackup” of layers of copper conductors and intervening layers of an insulating propagating medium such as FR4 joined under heat and pressure. To assure that the copper sheets do not delaminate from the dielectric layers, manufacturers typically electrodeposit an additional surface layer of copper on a relatively smooth copper foil that creates irregular features as large as 11 µm, to promote good adhesion. The electroplated copper produces a surface 238 NONIDEAL CONDUCTOR MODELS Figure 5-23 SEM photograph of rough copper at 5000× magniﬁcation at a 30◦ angle. roughness proﬁle that can be described as spherical particles, joined together in a network to form a distorted surface, as shown in Figure 5-24. An individual snowball is located a distance xi below a ﬂat copper surface, has radius ai, and, when it experiences an external electromagnetic ﬁeld intensity, as it would in the region below the copper trace when a signal is propagating on a transmission line, it will dissipate power similar to the hemisphere described in equation Electrodeposited copper spheres ai xi Flat Copper Plane Figure 5-24 Cross section of a distribution of copper spheres that create a 3D rough surface in the form of copper “pyramids” on a ﬂat conductor. SURFACE ROUGHNESS 239 (5-53) with twice the magnitude since it is a whole sphere: Psphere = 1 2 η|H0|2σtot = −Re 1 2 η|H0 |2 3π k2 (α(1) + β (1)) (5-63) For a distribution of copper spheres such as the pyramidal conﬁguration that creates 3D copper roughness proﬁles shown in Figures 5-23 and 5-24, superposition of the sphere losses can be used to calculate the total losses of the structure [Olufemi, 2007; Huray, 2009]. Since roughness losses are proportional to the surface area of the tooth structure, the number and size of the spheres can be chosen to approximate the surface roughness correction factor very accurately. To do this, the general tooth shape must be approximated, the average snowball size must be measured, and the number of spheres that correspond to the surface area of the protrusion must be calculated. If SEM photographs are analyzed extensively, it may be possible to build a model with multiple snowball sizes that very closely replicated the actual tooth shape. However, since electron microscopes are typically not available, proﬁlometers will provide approximate tooth shapes, and a reasonable model can be constructed. The total power dissipated by the tooth structure is simply the sum of the power dissipated from the total number of spheres (N ) required to replicate the surface area of the tooth structure: N PN,spheres = − Re n=1 1 2 η|H0|2 3π k2 (α(1) + β (1)) n (5-64) √ where α(1) and β(1) are calculated with (5-51a) and (5-51b),η = µ0/ε0ε , and H0 is the magnitude of the applied magnetic ﬁeld. To calculate a new surface roughness correction factor for use in (5-47), the ratio of the power absorbed with and without surface roughness must be derived using (5-65) and (5-54): KHuray = Pﬂat + PN,spheres Pﬂat = (µ0ωδ/4)Atile + N n=1 Re 1 2 η(3π/ k2)(α(1) + β (1)) n (5-65) (µ0ωδ/4)Atile To calculate the total number of spheres (N ) needed to represent the surface area of the roughness proﬁle, it is useful to choose a geometric shape that resembles a typical tooth. If one-half of a spheroid (a hemispheroid) is chosen as the approximate tooth shape, laboratory data indicate that reasonable accuracy can be obtained. However, the SEM photograph in Figure 5-23 clearly shows that a spheroid is only an approximation because the real tooth shape has more surface area. Consequently, this assumption is expected to slightly underpredict the surface roughness losses at high frequencies. A detailed statistical analysis of the 240 NONIDEAL CONDUCTOR MODELS SEM photograph could provide a more accurate representation of the speciﬁc size and number of spheres; however, the spheroid assumption produces reasonable results compared to measured surfaces. To construct a Huray surface roughness model, the surface is measured as in Figure 5-19 and the RMS values of bbase and htooth are calculated. Next, the surface area of a hemispheroid is calculated with a base bbase and a height htooth. An appropriate snowball radius is chosen (usually between 0.5 and 1 µm), and the total number of copper spheres (N ) are calculated so that the total surface area of all N spheres is equal to that of the hemispheroid constructed using the RMS values measured with the proﬁlometer. The procedure is demonstrated with the following example. Example 5-5 Calculate the surface roughness correction factor using the Huray equation at 5 GHz. The RMS value of the tooth height htooth = 5.8 µm, the RMS base width of the tooth structures bbase = 9.4 µm, and the distance between peaks dpeaks = 9.4 µm. SOLUTION Step 1: Calculate the number of spheres required. If proﬁliometer measurements indicate that the RMS value of the tooth height htooth = 5.8 µm with an RMS spacing between protrusions of bbase = 9.4 µm and the tooth shape is approximated as a hemispheroid, the number of spheres can be calculated. The lateral surface area of a hemispheroid is given by Alat = π htooth bb2asehtooth arcsin 1 − (bbase/2)2/(htooth)2 1 − (bbase /2)2 (htooth )2 + bbase 2 Plugging in the values for bbase and htooth yields the surface area: Alat = 161 µm2 Assuming a 0.8-µm sphere radius, the surface area of a single sphere is calcu- lated: Asphere = 4π a2 = 8 µm2 The number of spheres that would have the same surface area as the hemi- spheroid is Alat = 161 = 20.125 Asphere 8 Step 2: Calculate the correction factor at 5 GHz. Since the peak–peak distance is 9.4 µm, Atile = (9.4 µm)2 is chosen for the tile area. However, any tile size could be used as long as the total number of spheres that equals the surface area of SURFACE ROUGHNESS 241 the proﬁle within the tile Example 5-4, k =√209.3 size is m−1, η calc√ulated. = µ0/ε0 The ε= spher√e radius a =√0.8 µm. From 377/ ε = 377/ 4 = 188.5 , and δ(5 GHz) = 2/2πf µ0σ ≈ 0.935 µm. The scattering coefﬁcient is calculated from equation (5-51a). For this example, it can be shown that β(1) is negligible, so it is ignored. α(1) = − 2j 3 (ka)3 1 − (δ/a)(1 + j ) 1 + (δ/2a)(1 + j ) = − 2j 3 [209.3(0.8 × 10−6)]3 1 1 − + 0.935 × 10−6 0.8 × 10−6 (1 0.935 × 10−6 2 · 0.8 × 10−6 (1 + + j) j) = −1.93 × 10−12 − j 1.05 × 10−12 The Huray surface roughness correction factor is calculated with (5-66) at 5 GHz: KHuray = Pﬂat + PN Pﬂat spheres = (µ0ωδ/4)Atile + N n=1 Re 1 2 η(3π / k2)(α (1) + β (1)) n (µ0ωδ/4)Atile = (µ0ωδ/4)Atile + 20 Re 1 2 η(3π/ k2 )α (1) (µ0ωδ/4)Atile = 8.15 × 10−13 + 7.8 × 10−13 8.15 × 10−13 = 1.95 Therefore, the Huray model predicts that the series resistance of a transmission line manufactured according to the roughness proﬁle deﬁned above would be approximately 1.95 times higher than the same transmission line constructed with a smooth conductor at a frequency of 5 GHz. Figure 5-25 shows a comparison of the insertion losses of a transmission line constructed with rough copper modeled using the Huray equation compared to measured results. Note that the Huray model correctly predicts the shape of the insertion loss curve with less than 1.5 dB of error at 30 GHz. Simulations show that if 23 spheres are used in this example, the model ﬁts the measured results almost exactly [Olufemi, 2007]. However, aside from detailed statistical analysis of the SEM photographs, there is no known deterministic methodology to arrive at the perfect ﬁt. The methodology presented allows the use of a proﬁlometer and produces reasonable wideband results. Figure 5-26 shows the Huray equation (5-65) constructed with twenty 0.8-µm spheres as calculated in Example 5-5, compared to the Hammerstad and hemisphere models for the roughness proﬁle assumed in Examples 5-4 and 5-5. Note that the Hammerstad equation saturates at 2, which does not provide enough loss 242 0 −5 −10 NONIDEAL CONDUCTOR MODELS Huray roughness model Measurement S21, dB −15 −20 −25 0 5 10 15 20 25 30 Frequency, GHz Figure 5-25 Accuracy of the Huray surface roughness model (5-65) constructed with N = 20 spheres with a radius of a = 0.8 µm, assuming a hemispheroid tooth shape; 7-in. microstrip; εr / tan δ = 3.9/0.0073 at 1 GHz; RMS roughness of copper foil hRMS = 5.8 µm, dpeaks,RMS = 9.4 µm. Surface roughness correction factor 2.8 Hemisphere model (5-58) 2.4 Huray model (5-65) 2 Hammerstad model (5-48) 1.6 1.2 0.8 0 10 20 30 40 50 Frequency, GHz Figure 5-26 Surface roughness correction factors for rough copper; Huray model (5-65), hemisphere model (5-58), and Hammerstad model (5-48); RMS roughness: hRMS = 5.8 µm, dpeaks,RMS = 9.4 µm. for rough copper. The hemisphere model overpredicts at middle frequencies and slightly underpredicts at high frequencies. As demonstrated by Figure 5-25, a properly constructed Huray model predicts a realistic correction curve. The frequency dependence of the skin effect resistance and total inductance using the Huray equation for surface roughness is implemented with (5-66a) and SURFACE ROUGHNESS 243 (5-66b).∗ When the skin depth (δ) is larger than the conductor thickness, the dc value of the resistance and low-frequency inductance where the skin depth is equal to the conductor thickness should be used: √ RHuray(f ) = KHurayRs Rdc f when δ < t when δ ≥ t LHuray(f ) = Lexternal + RHuray(f ) 2πf Lexternal + RHuray(fδ=t 2πfδ=t ) when δ < t when δ ≥ t (5-66a) (5-66b) where KHuray is calculated with (5-65), t is the conductor thickness, δ is the skin depth, and fδ=t is the frequency where the skin depth equals the thickness of the conductor. 5.3.4 Conclusions For rough copper, Figure 5-14b shows that the Hammerstad model slightly overpredicts surface roughness losses at low frequencies and signiﬁcantly underpredicts losses at high frequencies. However, Figure 5-14a shows that the Hammerstad method works well for copper proﬁles that are relatively smooth with a corrugated surface proﬁle. The Hammerstad model should be used only for RMS copper roughness proﬁles less than about 2 µm. The hemisphere model is an improvement over Hammerstad for rough copper but still overpredicts the losses at middle frequencies and underpredicts the losses at high frequencies, which causes the “belly” in the simulated results in Figure 5-22. The major beneﬁt gained by studying the hemisphere model is a physical understanding of how the ﬁelds and surface currents behave in the presence of a protrusion. The hemisphere is the simplest geometry that can be solved analytically to predict the impact of a rough surface compared to a smooth surface. Furthermore, the hemisphere model is not appropriate for smoother copper proﬁles that are corrugated in nature. The hemisphere model is valid only for copper proﬁles that can be characterized by a distribution of distinct protrusions, such as the surfaces shown in Figures 5-16b and 5-23. The Huray approach is the most diverse and accurate modeling methodology for surface roughness. It can be used for any type of copper as long as a detailed image of the surface proﬁle can be obtained with a proﬁlometer or a scanning electron microscope. The surface area of the roughness proﬁle prior to lamination must be determined, and the number of appropriately sized spheres less than 1µm ∗Just as with equations (5.49b) and (5.62b), the method shown here for calculating the internal portion of the inductance (Linternal = Rac/ω) for a rough conductor is an approximation based on the result for a smooth conductor. The approximation will induce causality errors that tend to be small enough to ignore, so this method is generally acceptable. Appendix E derives the internal inductance using a more rigorous approach based on the discussion in Chapter 8. 244 NONIDEAL CONDUCTOR MODELS in radius must be calculated so that the total surface area is equal to that of the roughness proﬁle. 5.4 TRANSMISSION-LINE PARAMETERS FOR NONIDEAL CONDUCTORS As bus data rates increase and physical implementations of high-speed digital designs shrink, the transmission-line losses become more important. Consequently, the engineer must have the ability to calculate the response of a transmission line successfully and account for realistic conductor behavior. The next two sections will (1) describe how to include ac resistance and internal inductance in an equivalent circuit, and (2) modify telegrapher’s equations to comprehend realistic conductors. 5.4.1 Equivalent Circuit, Impedance, and Propagation Constant In deriving the equivalent circuit for a transmission line in Section 3.2.3, which is shown in Figure 3-9, the conductor was considered to be inﬁnitely conductive, meaning that all the current ﬂows only on the surface because the skin depth δ = 0, as shown by taking the limit of (5-10): limσ →∞[δ] = limσ →∞ 2 =0 ωµσ Furthermore, the assumption of perfect conductors did not allow the calculation of a resistive term or an internal inductance term because there was no ﬁeld penetration into the conductor. As described in Section 5.1.2, physical conductors manufactured with metals of ﬁnite (although good) conductivity behave very similar to perfect conductors except for a small transition region where internal currents exist that are mostly conﬁned to a few skin depths. As described in Sections 5.2.2 and 5.2.3, the skin effect leads directly to frequency-dependent resistance and internal inductance terms that must be comprehended in the equivalent circuit. Fortunately, the form of the equivalent circuit derived in Section 3.3 is also applicable to a line whose conductors have ﬁnite conductivity. To begin this derivation, the series impedance of an ideal transmission line with inﬁnite conductivity is calculated in units of ohms. Zexternal = j ωLexternal (5-67) The idealized parameters must be modiﬁed to include the surface impedance (also known as the internal impedance): Zs = Rac + j ωLinternal (5-29) TRANSMISSION-LINE PARAMETERS FOR NONIDEAL CONDUCTORS 245 The total inductance was calculated in (5-20): Ltotal = Linternal + Lexternal (5-20) indicating that the series impedance contribution from the inductance is simply j ω(Linternal + Lexternal). To calculate the total series impedance of a transmission-line segment, the resistive component is added: Zseries = Rac + j ω(Linternal + Lexternal) = Rac + j ωLtotal ohms (5-68) leading to the equivalent circuit for a transmission line with a conductor of ﬁnite conductivity, as shown in Figure 5-27, where Ns is the number of segments, C z = z C and L z = z Ltotal, as calculated in Section 3.2.3, and R z = z Rac, where z is the length of the differential section of transmission line and C, Ltotal, and Rac are the capacitance, inductance, and resistance per unit length. The characteristic impedance, which was deﬁned in equation (3-33), can be calculated by dividing the series impedance as deﬁned by (5-68) by the parallel admittance of the capacitance, Yshunt = j ωC, for a short section of transmission line of length z. Z0 = Zseries = Yshunt RAC + j ωLtotal j ωC √ Note that the units in (5-69) are ohms/(1/ohms) = ohms (5-69) (ohms)2 = ohms. L∆z R∆z C∆z 1 L∆z R∆z C∆z (a) 2 L∆z R∆z C∆z Ns L∆z R∆z C∆z (b) Figure 5-27 (a) Model for a differential element of a transmission line; (b) full model. 246 NONIDEAL CONDUCTOR MODELS The propagation constant can be derived by inserting the complex values of the series impedance and shunt admittance into the loss-free formula that was derived in Section 3.2.4, which takes the form √ γ = α + jβ = 0 + j ω LC = (j ωL)(j ωC) = ZlosslessYlossless Substitution of Zseries and Yshunt in place of the loss-free values of the series impedance and the parallel admittance yields the propagation constant for a transmission line with an ideal dielectric and a conductor with a ﬁnite conductivity, as shown in γ = α + jβ = ZseriesYshunt = (RAC + j ωLtotal)j ωC (5-70) Equations (5-69) and (5-70) account for realistic conductor effects, such as skin effect resistance, internal inductance, and surface roughness, as outlined throughout this chapter. 5.4.2 Telegrapher’s Equations for a Real Conductor and a Perfect Dielectric The time-harmonic forms of the telegrapher’s equations for a transmission line with a perfect dielectric and perfect conductor were shown in equations (3-25) and (3-26): dv(z) = −j ωLi(z) dz di(z) = −j ωCv(z) dz (3-25) (3-26) To adjust these formulas to account for a realistic conductor with a ﬁnite con- ductivity, the equivalent circuit must be observed. Notice that the right-hand side of equation (3-25) is simply the impedance of an inductor. Therefore, the series impedance of a differential slice of an ideal transmission line of length dz is based on external inductance since a perfect conductor has inﬁnite conductivity. To account for the realistic conductor effects, the series impedance of a realistic conductor as deﬁned in equation (5-68) is simply substituted into the right hand of (3-25): dv(z) dz = −[Rac + j ω(Linternal + Lexternal)]i(z) (5-71) Consequently, the classic form of the telegrapher’s equations for a perfectly insulating dielectric and a realistic conductor are ∂v(z, t) = − ∂z Rac + Ltotal ∂ ∂t i(z, t) (5-72a) ∂i(z, t) = −C ∂v(z, t) ∂z ∂t (5-72b) PROBLEMS 247 REFERENCES Brist, G., S. Hall, S. Clouser, and T. Liang, 2005, Non-classical conductor losses due to copper foil roughness and treatment, CircuiTree, May. Collins, Robert, 1992, Foundations for Microwave Engineering, McGraw-Hill, New York. Hall, S., G. Hall, and J. McCall, 2000, High-Speed Digital System Design, Wiley, New York. Hall, Stephen, Steven G. Pytel, Paul G. Huray, Daniel Hua, Anusha Moonshiram, Gary A. Brist, and Edin Sijercic, 2007, Multi-GHz causal transmission line modeling using a 3-D hemispherical surface roughness approach, IEEE Transactions on Microwave Theory and Techniques, vol. 55, no. 12, Dec. Hammerstad, E., and O. Jensen, 1980, Accurate models for microstrip computer-aided design, IEEE MTT-S International Microwave Symposium Digest, May, pp. 407–409. Huray, Paul, 2008, Foundations of Signal Integrity, Wiley, Hoboken, NJ. Huray, P. G., S. Hall, S. G. Pytel, F. Oluwafemi, R. Mellitz, D. Hua, and P. Ye, 2007, Fundamentals of a 3D “snowball” model for surface roughness power losses, Proceedings of the IEEE Conference on Signals and Propagation on Interconnects, Genoa, Italy, May 14. Jackson, J. D., 1999, Classical Electrodynamics, 3rd ed., Wiley, New York. Mathworld, n.d., http://mathworld.wolfram.com/ProlateSpheroid.html. Olufemi, Oluwafemi, 2007, Surface Roughness and Its Impact on System Power Losses, ProQuest, Ann Arbor, MI. Orlando, Terry P., and Kevin A. Delin, 1991, Foundations of Applied Superconductivity, Addison-Wesley, Reading, MA. Pytel, Steven G., 2007, Multi-gigabit data signaling rates for PWBs including dielectric losses and effects of surface roughness, Ph.D. dissertation, Department of Electrical Engineering, University of South Carolina. PROBLEMS 5-1 Draw the resistance, inductance, and impedance curves from 0 to 10 GHz for a microstrip transmission line constructed with copper that has the parameters w = 5 mils, h = 3 mils, t = 2.1 mils, εr = 4.1, bbase = 10.2 µm, htooth = 4.5 µm, dpeaks = 12 µm, and σ = 5.8 × 108 S/m. Assume a general tooth shape that is hemispheroid in nature. Create the model with the hemisphere model and the Huray model. Discuss the physical mechanisms that cause the differences between the curves. 5-2 Create an equivalent-circuit model valid at 15 GHz of the transmission lines deﬁned in Problem 5-1. 5-3 For the transmission-line description in Problem 5-1, under what conditions is the Hammerstad approximation valid? 5-4 For the transmission line deﬁned in Problem 5-1, when do ac losses overtake dc losses? 248 NONIDEAL CONDUCTOR MODELS 5-5 What causes internal inductance? Is internal inductance larger or smaller for a thick conductor line versus a thin conductor? What is the relationship between inductance and resistance? In physical terms, explain why the relationship must hold. 5-6 For a stripline transmission line with the parameters w = 5 mils, h1 = h2 = 6 mils (refer to Figure 5-8), t = 0.7 mil, εr = 4.1, bbase = 10.2 µm, htooth = 0.9 µm, and dpeaks = 15 µm, calculate the resistance and inductance at 1 GHz. 5-7 For the transmission-line models deﬁned in Problem 5-1, when do the surface roughness losses become signiﬁcant? 5-8 For a periodic roughness pattern such as that shown in Figure 5-20, how will the current ﬂow between and on top of the surface protrusions? Assuming TEM ﬁelds, draw the magnetic and electric ﬁelds and the associated current. What is signiﬁcant about the periodic pattern? What assumptions break down for a periodic pattern? 5-9 For a 10-in. transmission line with the parameters w = 5 mils, h = 3 mils, t = 2.1 mils, εr = 4.1, and σ = 5.8 × 108S/m, sourced with 5 mA of current, what is the total power per unit area dissipated by the reference plane? 5-10 For the transmission line described in Problem 5-1, how much error would be introduced by neglecting surface roughness at 10 GHz? 5-11 For the transmission line described in Problem 5-1, how much surface roughness would be required to double the power loss of the transmission line? 5-12 For a plane wave impinging onto an inﬁnitely thick slab of a goodconducting medium, draw the electric ﬁeld outside the conductor and the magnetic ﬁelds inside the conductor. Compare and contrast this to the boundary conditions for a perfect conductor. 6 ELECTRICAL PROPERTIES OF DIELECTRICS 6.1 Polarization of dielectrics 250 6.1.1 Electronic polarization 250 6.1.2 Orientational (dipole) polarization 253 6.1.3 Ionic (molecular) polarization 253 6.1.4 Relative permittivity 254 6.2 Classiﬁcation of dielectric materials 256 6.3 Frequency-dependent dielectric behavior 256 6.3.1 Dc dielectric losses 257 6.3.2 Frequency-dependent dielectric model: single pole 257 6.3.3 Anomalous dispersion 261 6.3.4 Frequency-dependent dielectric model: multipole 262 6.3.5 Inﬁnite-pole model 266 6.4 Properties of a physical dielectric model 269 6.4.1 Relationship between ε and ε 269 6.4.2 Mathematical limits 271 6.5 Fiber-weave effect 274 6.5.1 Physical structure of an FR4 dielectric and dielectric constant variation 275 6.5.2 Mitigation 276 6.5.3 Modeling the fiber-weave effect 277 6.6 Environmental variation in dielectric behavior 279 6.6.1 Environmental effects on transmission-line performance 281 6.6.2 Mitigation 283 6.6.3 Modeling the effect of relative humidity on an FR4 dielectric 284 6.7 Transmission-line parameters for lossy dielectrics and realistic conductors 285 6.7.1 Equivalent circuit, impedance, and propagation constant 285 6.7.2 Telegrapher’s equations for realistic conductors and lossy dielectrics 291 References 292 Problems 292 Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 249 250 ELECTRICAL PROPERTIES OF DIELECTRICS As the speed of digital systems continues to increase with Moore’s law, the electrical performance of the dielectric layers of the printed circuit board, package, or multichip modules becomes signiﬁcantly more important. Dielectric materials that worked perfectly well for slower designs become increasingly difﬁcult to design with because new phenomena, such as frequency-dependent dielectric permittivity and loss tangents, environmental factors, and localized interactions between the electromagnetic signal and the ﬁber weave reinforcement of the board, become signiﬁcant and can no longer be ignored. Without properly accounting for the high-speed dielectric phenomena, it becomes impossible to properly predict phase delay and signal losses, leading to nonphysical behavior of the transmission line model. In short, simulation-based digital bus designs exceeding 3 to 5 GHz are not possible without accounting for the effects covered in this chapter. Dielectrics, more commonly called insulators, are substances whose charges in the molecules and atoms are bound and therefore cannot move over macroscopic distances under the inﬂuence of an applied ﬁeld. Ideal dielectrics do not contain any free charge (such as in conductors), and chemical structure is macroscopically neutral. When a ﬁeld is applied to a dielectric, the bound charges do not move to the surface of the material as they would in a conductor, but the electron clouds associated with the atomic and molecular structures of the dielectric can be distorted, reoriented, or displaced, inducing electric dipoles. When this happens, the dielectric is said to be polarized . The polarizability of a dielectric leads directly to the deﬁnition of the relative permittivity, dielectric losses, and the relationship between energy propagation and losses. 6.1 POLARIZATION OF DIELECTRICS For a metal, conductivity is caused by the redistribution of free charges over a macroscopic distance. For example, Figure 5-4 shows how the current, and therefore the charge are largely contained within one skin depth on the bottom of the conductor. For dielectrics, the applied ﬁeld only displaces a few electrons per atom over very small, subatomic distances. In a dielectric, the electrons are tightly bound to the atoms, and only a negligible number of electrons are available for conduction of electric current. The difference in electrical behavior of a conductor versus a dielectric is essentially the difference between free and bound charges. 6.1.1 Electronic Polarization When a dielectric material constructed of nonpolar molecules is exposed to an external electric ﬁeld, the electrons react by shifting with respect to the nucleus opposite the applied ﬁeld. This establishes numerous small electric dipoles that align with the electric ﬁeld. When the external electric ﬁeld is removed, the electric dipoles return to the neutral position, as shown in Figure 6-1a. Essentially, POLARIZATION OF DIELECTRICS 251 − −+ + − −+ + − + − + E (a) (b) Figure 6-1 (a) Atoms with no external electric ﬁeld applied; (b) electric dipoles induced as the externally applied electric ﬁeld distorts the electron cloud. the electron clouds of the atoms are distorted by the applied electric ﬁeld, similar to stretching a spring as depicted in Figure 6-1b. This gives the material the ability to store electric energy (as potential energy). To demonstrate the relationship between the applied electric ﬁeld and the distortion of the electron cloud, a very simpliﬁed model of an atom can be employed by assuming a point positive charge of qe+ representing the nucleus encompassed by a sphere, with a charge of qe− representing the electron cloud. When an electric ﬁeld is applied, the electron cloud is displaced until the attractive force between the positively charged nucleus and the negatively charged sphere is equal to the force of the applied electric ﬁeld. By equating these two forces, the displacement of the electron cloud can be estimated. Note that this model assumes that the electron cloud remains spherical when it is displaced. In reality, when the electron cloud is moved by the external electric ﬁeld, it is elongated and stretched. However, this simple derivation will provide valuable insight into the mechanisms that cause electronic polarization. Consider the spherical cloud of uniform charge density as in Figure 6-2a. The charge density per unit volume is QQ ρ= V = 4 3 π re3 where Q is the total charge distributed over the volume of the sphere V and re is the radius of the electron cloud. The electric ﬁeld is solved using the integral form of Gauss’s law (2-59): εE · ds = ρ dV = Qenclosed S V In this case, only the electric ﬁeld inside the sphere is relevant because the force required to move the electron cloud away from the positive nucleus must be calculated. Assuming that the electron cloud encompasses free space, ε = ε0, Gauss’s law reduces to ε0Er 4π r2 = 4 3 π r 3ρ 252 ELECTRICAL PROPERTIES OF DIELECTRICS qe − re qe + qe − x r E qe + Original position of electron cloud (a) (b) Figure 6-2 (a) Atom with no external electric ﬁeld applied; (b) the applied electric ﬁeld displaces the electron cloud a distance r, inducing an electric dipole. and for surfaces with a radius of less than re, ε0Er 4π r2 = Q r re 3 Qr → Er = 4π ε0re3 Assuming the total charge Q = qe, the force required to move the electron cloud a distance of r is calculated with equation (2-57): F = qeEr = qe2r 4π ε0re3 (6-1) Consequently, if an external electric ﬁeld with a magnitude equal but opposite to Er is applied to the atom, it will move the electron cloud center a distance of r with respect to the nucleus, polarizing the atom, as shown in Figure 6-2b. The electric dipole moment (or electric dipole for short) is a measure of the polarity of a system of electric charges. In the simple case of two point charges, one with charge +qe and one with charge −qe, the electric dipole moment is p = qer (6-2) where r is the displacement vector pointing from the negative charge to the positive charge. This implies that the electric dipole moment vector points in the same direction as the electric ﬁeld. Solving (6-1) for qe and multiplying by r produces the dipole moment of a single atom in units of coulomb-meters per atom: p = qer = 4π ε0re3Er (6-3) Note that (6-3) indicates that the polarization is directly proportional to the electric ﬁeld strength. For a volume of dielectric that has N atoms, the dipole moment per unit volume is P = Np (6-4) POLARIZATION OF DIELECTRICS 253 where P is the polarization vector. To simplify the notation, (6-3) is often denoted as p = αeEr (6-5) where αe = 4π ε0re3 and is called the electronic polarizability of the material. 6.1.2 Orientational (Dipole) Polarization If the material contains polar molecules, such as liquid water, which possess a permanent dipole moment, the orientation of the molecules is usually randomized by thermal agitation so that the dipoles are oriented in random directions, as shown in Figure 6-3a. When the material is exposed to an applied electric ﬁeld, the polar molecules will tend to orient themselves with the electric ﬁeld, as shown in Figure 6-3b. The orientational polarizability of the material is denoted αo. An analysis similar to that shown in section 6.1.1 can be performed to derive αo, but it is more involved and generally not practical because a detailed analysis of the intermolecular distances must be known. A more detailed discussion may be found in a book by Elliott [1993]. 6.1.3 Ionic (Molecular) Polarization Polarization of a dissolved ionic substance may be considered to be a special case, because the solution, unlike the ideal dielectric described above, consists of mobile, dissociated positive and negative charged particles, or ions. For example, NaCl dissolved in water exists principally as individual Na+ and Cl− ions rather than as neutral ion pairs. Each ion is associated with several of the neutral but polar water molecules, where the latter are oriented in a “sphere of hydration” surrounding each ion, as illustrated in Figure 6-4a for a positive sodium ion. Note that the partial negative charge on the oxygen of the water molecule is + −+ + +− − ++ − ++ − ++ − ++ − E ++ − ++ +− + −+ + (a) − ++ − ++ (b) Figure 6-3 (a) Polar molecules with no applied electric ﬁeld, showing somewhat randomized orientation due to thermal agitation; (b) the applied electric ﬁeld tends to align the polar molecules due to orientational polarization. 254 ELECTRICAL PROPERTIES OF DIELECTRICS + −+ − +− + + ++ − + + + Na+ − + +− ++ +− + +− + − +− + Cl− Cl− Na+ Na+ Cl− E Na+ Sphere of hydration Water Molecule (a) (b) Figure 6-4 (a) In the absence of an external electric ﬁeld, polar molecules and ions will tend to arrange themselves so that the total polarity is neutral. Example depicts the positive sodium ion associated with and surrounded by the polar water molecule. (b) When an external electric ﬁeld is applied, the positive and negative ions will be displaced relative to each other, inducing ionic or molecular polarization. oriented toward the positive sodium and that the hydrated ion maintains a net positive charge. When this solution is exposed to an external electric ﬁeld, the positive ions are displaced with respect to the negative ones, resulting in physical migration—and separation—of the oppositely charged hydrated ions, as shown in Figure 6-4b. If the poles of the applied ﬁeld represent an electron sink and source, and if the applied potential is sufﬁciently high, oxidation and reduction (redox) reactions occur at the poles and there is net current in the circuit. In the absence of the physical requirements for electron transfer (i.e., redox reactions), only displacement occurs in response to the applied ﬁeld. In a solid material (such as NaC1), the individual ions are not able to dissociate. However, when an electric ﬁeld is applied, the electrons associated with the molecule will be displaced with respect to the Na and C1 nuclei over subatomic distances creating a dipole moment. This is called ionic polarizability, is denoted as αi, and is also known as molecular polarizability. For a detailed discussion, see the books by Elliott [1993] and Pauling [1948]. 6.1.4 Relative Permittivity The discussion above was primarily for instructional purposes. In general, the polarizability of a material is not calculated directly, because it would require precise knowledge of the atomic dimensions for each atom or molecule in the material. For practical applications, the phenomenon is measured indirectly by looking at the behavior of the dielectric permittivity versus frequency. POLARIZATION OF DIELECTRICS 255 To derive the relationship between the polarizability of a material and the dielectric permittivity, the total polarization per unit volume is observed: P = N (αe + αi + αo)Etot (6-6) where N is the number of dipoles per unit volume. The electric ﬁeld Etot in (6-6) consists of the externally applied ﬁeld plus the molecular ﬁeld created by the alignment of the electric dipoles. If the dielectric material were removed, the value of the applied electric ﬁeld E0 could be measured and the electric ﬂux density would be given by D0 = ε0E0 (6-7) When the dielectric material is included, the effect of the polarization on the electric ﬁeld is added and the electric ﬂux density becomes D = ε0E0 + P (6-8) Because P and therefore Etot in (6-6) is proportional to the applied electric ﬁeld, E0, as implied by (6-3), a constant of proportionality can be chosen to relate D and E0. This value, the dielectric permittivity ε, has been used throughout the book: D = εE0 (6-9) Comparing (6-8) and (6-9) allows us to choose another constant to relate P to the applied electric ﬁeld, E0: D = ε0E0 + P = ε0(1 + χ )E0 (6-10) where χ= P ε0E0 (6-11) where χ is known as the electric susceptibility and is dimensionless. Comparing (6-9) and (6-10) shows that the dielectric permittivity can be expressed in terms of χ: ε = ε0(1 + χ ) (6-12) Furthermore, in Section 2.1 the dielectric permittivity is deﬁned as ε = εr ε0 (2-10b) indicating that the relative dielectric permittivity εr can be deﬁned in terms of χ and therefore the polarizability of the material: εr = 1 + χ (6-13) This means that the relative dielectric permittivity is a direct result of how the material reacts to an applied electric ﬁeld. 256 ELECTRICAL PROPERTIES OF DIELECTRICS 6.2 CLASSIFICATION OF DIELECTRIC MATERIALS In Section 5.1.1 it was shown that when a plane wave propagates in a conductive medium such as a metal, Ampe`re’s law reduces to ∇ × H = jω ε − j σ E ω (6-14) In the case of a conductor, which was the topic of Chapter 5, the conductivity term σ represented the metal losses. Conductivity implies the movement of free charges within a material. In the case of a good dielectric, the charges are bound. However, as discussed in Section 6.1, the interaction between the molecular or atomic structure of the dielectric and the applied electric ﬁeld changes the orientation of the bound charges within the material. As an example, equation (6-1) calculated the force required to displace the electron cloud of an atom. Consequently, as the electric dipoles within a material attempt to remain aligned with the time-varying electric ﬁeld, energy is consumed, which manifests itself as dielectric losses. Subsequently, the term σ in (6-14) can be thought of as the equivalent conductivity of the dielectric, which represents the losses due to the polarization of the material. This allows the deﬁnition of the complex permittivity for a lossy dielectric, as was done for conductors in Chapter 5: ε = ε − j σdielectric = ε − j ε ω (6-15) The imaginary portion of the complex dielectric permittivity represents the dielectric losses, and the real portion represents the dielectric permittivity ε = εr ε0, discussed throughout this book. For most practical purposes, materials are clas- siﬁed by the real part of (6-15) divided by the permittivity of free space: εr = ε ε0 (6-16a) which is known as the relative dielectric permittivity, and the loss tangent, tan |δ| = ε ε (6-16b) which is simply the ratio of the imaginary and real components of (6-15). 6.3 FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR In Section 6.1 we showed that when a dielectric is subjected to an external electric ﬁeld, the positive and negative charges bound to the atoms and molecules are displaced relative to their average positions, causing electric dipoles to be formed which are quantiﬁed using the polarization vector P . Furthermore, the FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR 257 relative dielectric permittivity εr was introduced to account for the presence of P in a dielectric. When the applied ﬁelds begin to alternate in polarity, the polarization vector P , and subsequently the dielectric permittivity, is affected. Since the electric dipoles in a material will not align instantaneously with the applied time-varying ﬁelds, the polarization and relative permittivity are a function of the frequency of the alternating ﬁeld. In this section we explore the frequency dependence of dielectric materials and derive useful models that can be used to simulate these effects. 6.3.1 DC Dielectric Losses Dc losses imply a dielectric with conduction electrons that are free to move according to Ohm’s law: J = σd E (6-17) where J is the current density and σd is the conductivity of the dielectric. Assuming a time-harmonic ﬁeld, substitution of (6-17) into Ampe`re’s equation gives ∇ × H = J (x, t) + ∂D(x, t) ∂t = σd E(x) + j ω[ε − j ε ]E(x) = j ωε0 εr − j εr − j σd ε0ω E(x) (6-18) Rearranging (6-18) yields the dependence of Ampe`re’s equation on permittivity by adding the term σd /ε0ω to (6-15) to account for the conduction electrons. Do not get σd in (6-18) confused with the term σdielectric in (6-15), which is the effective conductivity of the dielectric due to the energy it takes to polarize the electric dipoles in the material. The term σd used in (6-18) is true conductivity, similar to that of a metal where electrons are not bound and are free to move. Note that a pole is created at ω = 0 based on the rearrangement of (6-18), rendering the equation invalid at dc. Since the dielectric conductivity σd is very small in practical dielectrics (σd /ε0ω 1), the dc term is almost always neglected [Huray, 2009]. 6.3.2 Frequency-Dependent Dielectric Model: Single Pole Following the derivation in Section 6.1.1, assume that an atom in the absence of an applied electric ﬁeld is represented by a positive nucleus and a negative electric cloud with centers that coincide. Since protons and neutrons are much heavier than electrons, we assume that when the external electric ﬁeld is applied, the nucleus remains stationary and the electron cloud moves. Therefore, when the electric ﬁeld is applied, the electron cloud is displaced, and when the ﬁeld is removed, the electron cloud returns to its original position, similar to the 258 ELECTRICAL PROPERTIES OF DIELECTRICS k m b x m F Figure 6-5 Analogy to an electron cloud in the presence of an electric ﬁeld: When the mass (electron) is pulled downward by a force (electric ﬁeld), it stretches the spring (displaces the electron cloud), displacing the mass (electron) by x through a medium with damping factor b. The spring constant is k . spring and mass system shown in Figure 6-5. When the electric ﬁeld is applied (the mass is pulled downward, stretching the spring by F ), the electron (the mass) is displaced by a distance of x and an electric dipole is created. When the electric ﬁeld is removed (the mass is no longer being pulled downward), the electron cloud returns to its neutral position (the spring is no longer stretched). This analogy suggests that a simple mechanical spring model can be used to model the polarization and therefore the dielectric permittivity as a function of frequency [Huray, 2009]. The classic differential equation that governs the solution of a damped driven harmonic oscillator (and therefore a damped spring, as shown in Figure 6-5) is d2x m dt2 + dx b dt + kx = F ejωt where m is the mass, b the damping factor, k the spring constant, and Fejωt the driving force. The spring equation can be rewritten in terms of the applied electric ﬁeld and the electron mass. From equation (2-57), the force can be written in terms of the electric ﬁeld and the charge, F = qE, where q is the charge of the dipole, m the mass of the negative charge relative to the stationary nucleus, and x the displacement distance: d2x dt2 + b m dx dt + k x m = qE0 ejωt m (6-19) The homogeneous solution to (6-19) is uninteresting because it is heavily damped and will not contribute to the steady-state solution. The particular solution is FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR 259 solved assuming that x takes the form x = Aejωt . Therefore, dx = j Aωejωt dt d2x dx2 = −ω2Aejωt which are substituted back into (6-19), so the coefﬁcient A can be solved for: − ω2Aejωt + j b Aωejωt + k Aejωt = qE0 ejωt m m m A −ω2 + j b ω + k = qE0 mm m A = −ω2 + q E0 / m j (b/m)ω + k/m Therefore, x = Aej ωt = −ω2 (q E0 / m)ej ωt + j (b/m)ω + k/m Dropping the time-harmonic function and rearranging yields x = (k/m − q E0 / m ω2) + j (b/m)ω (6-20) The natural (resonant) frequency of the oscillator is deﬁned when, ω2 = k /m, which allows the deﬁnition of ω0: ω02 = k m (6-21) Therefore, (6-20) can be simpliﬁed: x = ω02 − q E0 / m ω2 + j (b/m)ω From equation (6-2), the electric dipole moment is (6-22) p = qx = ω02 − q 2 E0 / m ω2 + j (b/m)ω and subsequently, the polarization vector is (6-23) P = Np = ω02 N (q2E0/m) − ω2 + j (b/m)ω 260 ELECTRICAL PROPERTIES OF DIELECTRICS and the electric susceptibility is deﬁned from (6-11): χ = P ε0E0 = ω02 N (q2/ε0m) − ω2 + j (b/m)ω leading to an expression for the relative dielectric constant from (6-13) [Huray, 2009]: εr = 1 + χ = 1 + ω02 N (q2/ε0m) − ω2 + j (b/m)ω (6-24) The real and imaginary parts are shown as [Balanis, 1989] εr = 1 + N (q2/ε0m)(ω02 − ω2) (ω02 − ω2)2 + (ω(b/m))2 εr = N q2 ω (b/ m) ε0m (ω02 − ω2)2 + ω b m 2 (6-25a) (6-25b) Equations (6-24) and (6-25) calculate the frequency response for a material that exhibits an atomic or molecular structure with only one natural or resonant frequency, an example of which is shown in Figure 6-6. At the natural (i.e., resonant) frequency of the harmonic oscillator, the imaginary portion of the complex permittivity will peak, which in turn dramatically increases the dielectric losses, which are quantiﬁed with the loss tangent (6-16). Also note that the real portion of the dielectric permittivity is almost constant until the operating frequency approaches the resonant frequency of the oscillator, which in this case is an atomic structure. In the vicinity of the natural frequency, the real portion begins 1.5 er ′ Relative Dielectric Permitivity 1.0 anomalous dispersion 0.5 er ′′ 0.5 1.0 5.0 10.0 Frequency, GHz 50.0 100.0 Figure 6-6 Frequency response for a pure material that exhibits an atomic or molecular structure with only one natural or resonant frequency. FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR 261 to drop and the imaginary portion begins to increase, implying a relationship between the real and imaginary parts of the complex permittivity. This relationship is discussed in detail in Section 6.4. Finally, the value of the permittivity settles out at a new level after the operating frequency ω has surpassed the natural frequency ω0. 6.3.3 Anomalous Dispersion Perhaps the most interesting part of the curve shown in Figure 6-6 is the area where the relative dielectric permittivity drops below 1. This area is often referred to as anomalous dispersion. When this is ﬁrst encountered, it often causes great confusion, as demonstrated by observing the deﬁnition of the speed of light in a vacuum, which was introduced in Section 2.3.4 in units of meters per second: c≡√ 1 µ0εr ε0 when εr = 1 This implies that if εr < 1, the velocity will exceed the speed of light in a vacuum (c ≈ 3 × 108 m/s). One of the consequences of Einstein’s theory of special relativity is that speeds greater than c are not attainable, yet the phase velocity of a wave traveling at frequencies where εr < 1 in Figure 6-6 appears to break this fundamental law of physics. The apparent conﬂict with the laws of special relativity comes from a widely mistaken assumption that all quantities with units of velocity must obey this rule. In fact, special relativity only places an upper value on the speed of material bodies that include signals, or information. Since a single-frequency harmonic plane wave is not a material body and is not a signal, it cannot be used to transmit information by itself. To understand this, the deﬁnition of phase velocity must be examined from Chapter 2. To determine how fast the wave is propagating, it is necessary to observe the cosine term for a small duration of time, t. Since the wave is propagating, a small change in time will be proportional to a small change in distance z, which means that an observer moving with the wave will experience no phase change because she is moving at the phase velocity (νp). However, the only way to measure the velocity of a signal is to turn on the transmitter, time how long it takes for a response to arrive at the receiver, and divide by the distance. When velocity is measured in this way, it cannot exceed c. To understand why, consider the propagation of information, such as a digital pulse. If the pulse is decomposed into its Fourier components, each will propagate with its own velocity, some slower than c and some faster than c; however, when all components are combined, the total velocity of the pulse cannot exceed the speed of light. A question that often arises when this subject is discussed is the possibility of modulating a narrowband signal with a frequency that coincides with an area of anomalous dispersion to sidestep the laws of special relativity and transmit information faster than c. However, a single-frequency plane wave still cannot carry information unless another signal is combined with it. For example, 262 ELECTRICAL PROPERTIES OF DIELECTRICS if a narrowband signal was simply turned on and off as a simplistic form of modulation, this is equivalent to convolving the narrowband signal with a step response, which contains an inﬁnite number of Fourier components. The proof that the velocity of a signal used to transmit information cannot exceed the speed of light is too complicated to address here. However, Brillouin [1960] showed that a signal propagating in a realistic dielectric medium cannot exceed c. 6.3.4 Frequency-Dependent Dielectric Model: Multipole As discussed in Section 6.1, the polarization of a realistic dielectric may consist of any number of ionic, orientational, and electronic mechanisms, depending on the molecular and atomic structure of the material. In general, ε and ε depend on frequency in complicated ways, exhibiting several resonances over a wide range. To account for this, a separate harmonic oscillator model is constructed for each molecular or atomic resonance, and the results are combined using superposition. Thus, for a material with n natural frequencies, the relative dielectric permittivity is represented by summing the responses from n independent harmonic oscillator models: εr = 1 + n i=1 ωi2 Ni (q2/ε0m) − ω2 + j (bi /m)ω (6-26) An example is shown in Figure 6-7, where three harmonic oscillator models were chosen arbitrarily with natural frequencies ω1 = 20 GHz, ω2 = 100 GHz, and ω3 = 400 GHz. Note how the permittivity tends to “stair-step” down to smaller values as the frequency increases, separated by areas of anomalous dispersion, producing a change in the steady-state value ( εr ) at frequencies higher than 2.5 Relative Dielectric Permitivity, εr′ 2.0 1.5 1.0 1 10 100 Frequency, GHz 1000 Figure 6-7 Example of a frequency response for a material that exhibits three atomic or molecular resonances at 20, 100, and 400 GHz. FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR 263 the resonance. The summation indicates that even low-frequency values of the permittivity depend on the high-frequency resonances. Although (6-26) is quite useful for developing an understanding of the physical mechanisms that cause the dielectric to vary with frequency, it does not lend itself well to simulating dielectrics for digital designs because minuscule details of the molecular structure are required. A more pragmatic approach is to represent (6-26) in terms of measured permittivity. If (6-26) is multiplied by ε0 and the right-hand term is simpliﬁed by dividing the top and bottom by 1/ω02, the top part, (1/ω02)[N (q2/m)], has units of farads per meter: 1 F = s4 · A2 m2 · kg 1 C=A·s 1 ω02 q2 N m → s2 1 A2 · s2 m3 kg = F/m which are the same units as ε , allowing (6-26) to be rewritten as ε = ε∞ + n i=1 εi 1 − ω2/ω12i + j ω/ω2i − j σd ε0ω (6-27) where ω1i and ω2i[(1/ωi2)(b/m)ω = ω/ω2i] are the frequencies where the dielectric variations are occurring, ω = 2πf is the operating frequency, ε the variation of dielectric permittivity over the frequency of interest, σd the true conductivity of the dielectric material, ε∞ the dielectric permittivity value at very high frequencies in the area of interest, and ε0 the permittivity of free space. The term j (ω/ω2i ) accounts for the damping of the molecular dipoles (orientational polarization) in the mid-frequency ranges, the term ω2/ω12i accounts for resonance of induced atomic and molecular dipoles (ionic and electronic polarization), and the ﬁnal term, j (σd /ε0ω) (derived in Section 6.3.1), accounts for the low-frequency loss of the dielectric, which is usually ignored. Figure 6-8 shows a conceptualized plot of the real and imaginary dielectric permittivity as a function of frequency. It should be noted that the ﬁgure does not represent any particular dielectric material; rather, it is simply a guide to help the reader conceptualize when the different forms of polarization begin to become signiﬁcant. For most dielectric materials used in digital design, labora- tory measurements show that the permittivity is dominated by the damping factor in (6-27)(the j ω/ω2i term) and not resonances (the ω2/ω12i term). For orientational polarization (where polar molecules attempt to remain aligned with the time varying electric ﬁeld), the damping factor tends to be high. Consequently, the classic model is derived from the concept of orientational polarization even though other forms may affect the damping. When the high frequency resonance term is ignored, the dielectric equation reduces to (6-28), which is applicable for 264 Orientatiional or dipole e′ ELECTRICAL PROPERTIES OF DIELECTRICS Ionic or molecular Electronic e′′ ~109 ~1010 −1014 Frequency ~1016 Figure 6-8 Conceptual complex dielectric permittivity variations as a function of frequency, showing approximate regions where each polarization mechanism exists. most practical high-speed digital platforms built on commonly available dielectric materials. ε = ε∞ + n i=1 εi 1 + j (ω/ω2i ) (6-28) Note that equation (6-28) is identical to the well-known Debye equation and is often curve-ﬁt to measured data to build a very accurate dielectric model library, used for designing high-speed digital systems. There is evidence based on laboratory measurements that dielectric resonances can exist as low as 30 GHz in FR4 dielectrics, which may affect some very high frequency designs. When designing digital systems with signiﬁcant harmonics past about 20 GHz, care should be taken to examine carefully the measured phase delay and loss characteristics of transmission lines built on a representative dielectric so that any resonance can be accounted for. In Chapter 9 we describe methodologies for extracting the loss characteristics and the phase velocity from S-parameter measurements. A dramatic narrowband increase in the loss accompanied by a simultaneous increase in the phase velocity is a telltale sign that a dielectric resonance exists in the frequency of interest. Equations (6-27) and (6-28) are much more useful than (6-26) for one very important reason: They can be ﬁt empirically. Since equation (6-26) requires intimate knowledge of the atomic substructure, it is not very useful for practical applications. However, the dielectric permittivity can be measured using the phase delay (see Chapter 9 for details), the results of which can be ﬁt to (6-27) by choosing the appropriate poles (ω1i and ω2i), ε∞ and ε , to essentially curve-ﬁt the dielectric behavior to a physically consistent model. The particular implementation of (6-27) is dependent on the characteristics of the material. FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR 265 4.0 0.04 εr ′ εr ′′ 0.035 3.8 0.03 3.6 0.025 3.4 Measured εr′′ 0.020 Modeled εr′′ 0.015 3.2 Measured εr′′ Modeled εr′′ 0.010 10 15 20 25 30 35 40 Frequency, GHz Figure 6-9 Measured dielectric response curve ﬁt to equation (6-27). Subsequently, for a generic material, the most straightforward usage requires a measured response of εr and tan δ so the damping poles and resonant peaks can be identiﬁed. Example 6-1 Use equation (6-27) to create an empirical model for the dielectric measured in Figure 6-9. SOLUTION Figure 6-9 depicts the measured value of the real and imaginary permittivity of a dielectric sample between 15 and 35 GHz. Note that the measured data shows small ε r peaks in the vicinity of 19 and 32 GHz, indicating the presence of two resonant induced dipoles, which are probably ionic or molecular in nature. Consequently, ignoring the dc term, the implementation of (6-27) for this material over the frequency range 15 to 35 GHz is ε = ε∞ + 1 + ε1 j (ω/ω1) − ω2/ω22 + 1 + ε2 j (ω/ω3) − ω2/ω42 where ω1 = 2π (20 GHz), ω2 = 2π (19 GHz), ω3 = 2π (63 GHz), ω4 = 2π (32 GHz), ε1 = 0.0163, ε2 = 0.012, and ε∞ = 3.8. The resonant terms ω2 and ω4 were chosen at the peaks, ε1 and ε2 are the variations in the real part of the dielectric constant in the vicinity of the peaks, and ω1 and ω3 are damping terms that were varied until the width and the height of the peaks match the data measured. In this case, the dielectric model is valid only between 15 and 35 GHz, which are the limits of the measured data [Hall et al., 2007]. 266 ELECTRICAL PROPERTIES OF DIELECTRICS 6.3.5 Inﬁnite-Pole Model Most high-speed digital platform designers do not have the luxury of wideband complex permittivity measurements that allow a reasonable ﬁt to (6-27). Consequently, a methodology is required that will allow the frequency dependence of the complex dielectric permittivity to be calculated from a single data point obtained from the vendor-provided material data sheet, which usually lists the dielectric constant (εr ) and loss tangent (tan δ) at 1 GHz and/or 10 GHz. To do this, it is ﬁrst necessary to assume that the primary source of frequency dependent dielectric behavior is from damping, as in equation (6-28), which is a reasonable approximation for frequencies below 20 GHz, and is often valid to much higher frequencies. Equation (6-28) can be simpliﬁed using the method described by Djordjevic et al. [2001] by increasing the number of terms and assuming that ω2i decreases linearly on the logarithmic scale. The variation ε is assumed to be the total variation of ε between the lower frequency of interest, ω1 = 10m1, and the upper frequency of interest, ω2 = 10m2, which is assumed to be distributed uniformly over the logarithm of frequency so that ε /(m2 − m1) is the variation per decade, which is a linear decay of ε on a logarithmic scale. By taking an inﬁnite number of terms, the sum in (6-28) becomes (6-29), which is referred to as the inﬁnite pole model : n ε →ε m2 dx i=1 1 + j (ω/ω2i ) m2 − m1 x=m1 1 + j (ω/10x ) = ε ln[(ω2 + j ω)/(ω1 + j ω)] m2 − m1 ln(10) ε − jε = ε∞ + m2 ε − m1 ln[(ω2 + j ω)/(ω1 ln(10) + j ω)] (6-29) By comparing the results to laboratory measurements, it has been shown that (6-29) is reasonably accurate for typical dielectrics used to design modern digital platforms under dry and nominal environmental conditions (0 to about 50% relative humidity). However, for dielectric materials that have an afﬁnity to absorb water (such as FR4) when exposed to very humid environmental conditions (i.e., 95% relative humidity), the model breaks down for frequencies greater than 10 GHz and (6-28) must be curve ﬁt to measured data to achieve realistic dielectric behavior. More details on how dielectric properties change with the environment is discussed in Section 6.6. To calculate the wideband dielectric permittivity and the loss tangent from a single-frequency data point, it is necessary to know the value of εr and tan δ at that frequency, which allows the calculation of ε and ε using ε = ε − j σdielectric = ε − j ε ω tan |δ| = ε ε (6-15) (6-16b) FREQUENCY-DEPENDENT DIELECTRIC BEHAVIOR 267 Equating the real and imaginary terms of the permittivity to the real and imaginary components of (6-29) allows the calculation of ε and ε∞, which can then be inserted into (6-29) so the response at all other frequencies can be calc√ulated. Recall that for a complex number, z = a + j b = rejθ , where r = a2 + b2 and ln(z) = ln(r) + j (θ + 2πk), where k is an integer. This allows the log term in (6-29) to be simpliﬁed assuming that ω1 ω ω2: Re ln ω2 ω1 + + j j ω ω = ln ω2 ω2 + ω22 + ω12 ≈ ln ω2 ω allowing the real part of (6-29) to be approximated as ε ≈ ε∞ + ε ln(ω2/ω) m2 − m1 ln(10) (6-30a) The imaginary part is simpliﬁed by noting that for the assumption ω1 ω ω2, the logarithm function in (6-29) is almost constant at a value of −π/2, allowing the imaginary part of (6-29) to be approximated as [Pytel, 2007] ε≈ ε −π/2 m2 − m1 ln(10) (6-30b) Figure 6-10 depicts an example of the modeled and measured relationship between the loss tangent and the dielectric permittivity as calculated with (6-29) and measured with a Fabry–Perot open-cavity resonator. The modeled response was calculated from a known data point at 1 GHz, where εr / tan δ = 3.9/0.0073, which was measured using a split post resonator. Note the deviation between tan d er, real 0.02 0.018 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0 Model (er) Measurement (er) Model (tan d) Measurement (tan d) 10 20 30 40 Frequency, GHz 4 3.95 3.9 3.85 3.8 3.75 3.7 3.65 3.6 3.55 3.5 50 Figure 6-10 Measured dielectric response compared to the response calculated from a single data point at 1 GHz using the inﬁnite-pole model (6-29). 268 ELECTRICAL PROPERTIES OF DIELECTRICS the measured tan δ data points and that the model has a maximum error of only 0.0009 at 30 GHz with a maximum error less than 0.01 for εr at 20 GHz [Hall et al., 2007]. Example 6-2 Calculate the frequency dependence of the dielectric permittivity using the inﬁnite-pole Debye model (6-29) assuming a nominal relative humidity of 30% and a measured data point at 1 GHz where tan δ = 0.0073 and εr = 3.9 to 10 GHz. SOLUTION Step 1: Calculate the inﬁnite-pole variables ε∞ and ε . Choose a lower and an upper frequency limit that are well beyond the frequency of interest. We choose 10 rad/s (1.6 Hz) for the lower limit and 100 Grads/s (16 GHz) for the upper limit: ω1 = 101 → m1 = 1 and ω2 = 1011 → m2 = 11 The real part of the complex permittivity at 1 GHz is ε = εr ε0 = 3.9ε0. The imaginary part of the complex permittivity is calculated with (6-16): tan |δ| = ε ε ε = ε tan |δ| = (3.9)(0.0073) = 0.028 The imaginary part of the permittivity is substituted into (6-30b) and ε is calculated: 0.028 = 11 ε − 1 −π/2 ln(10) → ε = −0.417 where the negative sign is ignored because it is already included in (6-15). Next, ε∞ is calculated using (6-30a): 3.9 ≈ ε∞ + 0.417 11 − 1 ln(1011 /2π (109 )) ln(10) → ε∞ = 3.85 Step 2: Calculate the complex frequency-dependent dielectric properties using (6-30a) and (6-30b): ε (f ) ≈ 3.85 + 0.0178 ln 1011 2πf tan |δ| = ε ε = 3.85 + 0.028 0.0178 ln(1011/2πf ) PROPERTIES OF A PHYSICAL DIELECTRIC MODEL 269 er, real tan d 3.92 tan d 3.91 3.90 3.89 3.88 er 3.87 0.0073 0.0072 0.0071 0.0070 0.0069 2.0 4.0 6.0 8.0 10.0 Frequency, GHz Figure 6-11 Dielectric properties versus frequency from Example 6-2. Figure 6-11 shows a plot of the dielectric properties versus frequency. As a double check, it can be conﬁrmed that at 1 GHz, the relative permittivity is 3.9 and the loss tangent is 0.0073. 6.4 PROPERTIES OF A PHYSICAL DIELECTRIC MODEL For a dielectric model to remain physical, it must predict correctly the frequencydependent behavior observed in real life. A common mistake in digital design is to use transmission-line models that have frequency-invariant dielectric properties. Although this approximation works ﬁne for low frequencies, as system data rates increase past about 2 Gb/s, such methods will introduce large errors into the simulations, rendering the analysis almost useless. In this section we describe the properties that must be satisﬁed to ensure a physically realizable model. Although the focus here is on dielectric models, any model is required to pass these tests to ensure behavior consistent with nature. First, the relationship between the dielectric permittivity and the loss tangent is discussed, and then speciﬁc mathematical tests are introduced to ensure a model that obeys the laws of physics. 6.4.1 Relationship Between ε and ε In Section 5.2.3, speciﬁc relationships were shown to exist between the real and imaginary parts of the series impedance of a transmission line in the form of ac resistance and internal inductance. There is a similar relationship between the real and imaginary portions of the complex permittivity, which in turn implies a relationship between the relative dielectric permittivity εr and the loss tangent tan δ. To conceptualize the relationship, assume that the frequency range is below 20 GHz, where measurements have shown that damping is the dominant dielectric response. In this region, large polar molecules are attempting to 270 ELECTRICAL PROPERTIES OF DIELECTRICS align themselves with the time-varying electric ﬁeld (e.g., orientational polarization), which consumes energy and therefore dissipates power. As the frequency increases, it becomes more difﬁcult for the polar molecules to follow the electric ﬁeld, meaning that more energy is consumed at higher frequencies. Think of pulling a paddle through the water when rowing a boat. If you increase the rate of paddling, it becomes more difﬁcult to move the oar though the water, increasing the amount of energy used. Following this analogy, the dielectric loss, and therefore the loss tangent, should tend to increase with frequency. If the deﬁnition of the volume energy density from Section 2.4.2 is observed, it shows that the energy stored in a charge distribution is proportional to ε: we = ε E2 2 J/m3 (2-75) If energy (and therefore power) is being dissipated by the polarization of the molecules, the total energy stored (power transmitted through the dielectric) must decrease proportionally. Since ε dictates how much energy can be stored in an electric ﬁeld, the permittivity must therefore decrease with a corresponding increase in the loss tangent. This general trend is demonstrated in Figures 6-10 and 6-11. Note that as the frequency of operation approaches the frequency of a pole or a resonance in the dielectric models derived earlier, this analogy breaks down. However, for the majority of high-speed digital designs that are built on common PCB materials, this analogy holds true. The analogy above hints toward a speciﬁc relationship between the real and imaginary parts of the dielectric permittivity that must be accounted for properly during the design of high-speed buses. To demonstrate this, the single-pole Debye model (6-31) can be separated into real and imaginary parts: ε − jε = ε∞ + 1 + ε j (ω/ω0) ε = ε∞ + ε 1 + (ω/ω0)2 ε = 1 ε (ω/ω0) + (ω/ω0)2 Next, (6-31b) and (6-31c) are combined to yield (6-31a) (6-31b) (6-31c) ε = ε∞ + ε ω0 ω ε ε = (ε − ε∞) ε − ε∞ − 1 (6-32a) (6-32b) Note that the real part of the permittivity (ε ) is a function of the imaginary part (ε ), and vice versa. Equations (6-32) show that a speciﬁc relationship governs PROPERTIES OF A PHYSICAL DIELECTRIC MODEL 271 0.025 0.020 0.015 0.010 0.005 tan d 0 10 20 30 40 w0 Frequency, GHz Figure 6-12 When a harmonic oscillator is being driven at a frequency much larger than its natural or resonant frequency (ω0 < ω), the losses vanish. This example plots the loss generated from the Debye equation (6-28) with a pole at ω0 = ω2i = 5 GHz. the real and imaginary parts of the complex permittivity. Furthermore, if ω0 ω, the loss tangent reduces to ω tan |δ| ≈ ω0 1 − ε∞ ε∞ + ε (6-33) which indicates an increase in the loss tangent with frequency. So for dielectrics that can be described by the Debye equation, equation (6-32) shows that ε (and subsequently εr ) will decrease with frequency, and (6-33) shows that the loss tangent will increase with frequency. A question arises regarding the very high frequency response of the Debye model. If the model is subjected to a very high frequency oscillatory forcing function, where ω0 < ω, there will be no time for the system to respond before the forcing function has switched direction, so the losses will vanish as ω becomes large, as is demonstrated by Figure 6-12, which has a pole at ω0 = 5 GHz. 6.4.2 Mathematical Limits For a transmission-line model to be physically consistent with the laws of nature, certain mathematical limits must be obeyed. Presently, the vast majority of engineers utilize models to design high-speed digital systems that employ frequency-invariant values of the dielectric permittivity and loss tangent. This assumption, although perfectly valid at low frequencies, induces amplitude errors and phase miscalculations for digital data rates faster than 1 to 2 Gb/s propagating on transmission lines because realistic dielectric materials have frequency-dependent properties that must be modeled correctly. As data rates increase, the spectral content of the digital pulse trains also increase and errors 272 ELECTRICAL PROPERTIES OF DIELECTRICS induced by the frequency-invariant approximation are ampliﬁed, especially for longer line lengths. When models are used that do not obey these rules, incorrect solutions are determined, lab correlation becomes difﬁcult or impossible, and the design time is increased signiﬁcantly. In this section we introduce speciﬁc limitations that dielectric models must obey to remain physically consistent with nature. Speciﬁcally, the conditions of causality, analytic functions, reality, and passivity are explored. Causality An in-depth analysis of the relationship between the real and imaginary parts of complex permittivity was done by Ralph Kronig and Hendrik Anthony Kramers in the early twentieth century [Balanis, 1989]. They developed the Kramers–Kronig relations, which describe the relationships between the real and imaginary parts of any complex function that is analytic in the upper half-plane: ε (ω) = 1 + 2 π ∞ 0 ω (ω εr )2 (ω ) − ω2 dω ε (ω) = 2ω π ∞ 0 1 − εr (ω ) (ω )2 − ω2 dω (6-34a) (6-34b) The Kramers–Kronig formalism is applied to response functions. In physics, a response function χ(t − t ) describes how a property P (t) of a physical system responds to an applied force F (t ). For example, P (t) could be the angle of a pendulum and F (t ) the applied force of a motor driving the pendulum motion. The response χ(t − t ) must be zero for t < t since a system cannot respond to a force before it is applied. Such a function is said to be causal . From a commonsense point of view, it makes sense that an effect cannot precede its cause, which is the fundamental principle of causality that every physical model must respect, as expressed mathematically by h(t) = 0 when t < 0 (6-35) The causality requirement is described mathematically in Chapter 8. Analytic Functions The response of a dielectric to an applied electric ﬁeld is quantiﬁed in terms of ε, and therefore a causal dielectric model cannot respond prior to the electric ﬁeld being applied, which makes sense for a physical system. It can be shown that this causality condition implies that the Fourier transform of the complex dielectric permittivity ε(ω) is analytic [Jackson, 1998], which in turn implies a speciﬁc relationship between the real ( ε ) and imaginary ( ε ) parts. For a complex function f (x + jy) = u(x, y) + j v(x, y), if u and v have continuous ﬁrst partial derivatives and satisfy the Cauchy–Riemann equations, f is said to be analytic [LePage, 1980]. The Cauchy–Riemann equations dictate PROPERTIES OF A PHYSICAL DIELECTRIC MODEL 273 speciﬁc relationships between the real and imaginary parts of such a complex function: ∂u = ∂v ∂x ∂y ∂u = −∂v ∂y ∂x (6-36a) (6-36b) It can be shown that the dielectric models presented earlier in the book satisfy the Cauchy–Riemann equations assuming that the dielectric response is a function of complex frequency. For example, if ω = ωR + j ωI is inserted into the Debye model in (6-31), ε(ωR + j ωI ) = ε∞ + 1 + j [(ωR ε + j ωI )/ω0] = ω0 (ω0 ε (ω0 − ωI ) − ωI )2 + ωR2 − j (ω0 ω0 ε ωR − ωI )2 + ωR2 it can be shown that ∂Re(ε) = ∂Im(ε) ∂ ωR ∂ ωI ∂Re(ε) = − ∂Im(ε) ∂ ωI ∂ ωR Therefore, the real and imaginary parts of a realistic dielectric model are interrelated, meaning that the imaginary response can be calculated from the real response, and vice versa. Reality Another constraint is that a physical model must produce a response that is real in the time domain, as discussed brieﬂy in Section 2.3.3. If a function is real in the time domain, its Fourier transform must satisfy the complex-conjugate rule [LePage, 1980]: F (−ω) = F (ω)∗ (6-37a) For example, consider Gauss’s law: ∇ · D = ∇ · εE = ρ If D and E are real functions in the time domain, they must satisfy the complex conjugate rule in (6-37a) and therefore so must the permittivity [Jackson, 1998]: ε(−ω) = ε(ω)∗ (6-37b) 274 ELECTRICAL PROPERTIES OF DIELECTRICS Passivity A physical system is denoted as passive when it is unable to generate energy. For example, an n-port network is said to be passive if t vT(τ ) · i(τ ) dτ ≥ 0 −∞ (6-38) where vT(τ ) is the transpose of a matrix containing the port voltages and i(τ ) a matrix containing the currents. Integral (6-38) represents the cumulative net energy (power) absorbed by the system up to time t. In a passive system, this energy must be positive for all t. This requirement is satisﬁed if (1) the system absorbs more energy than it generates, and (2) any generation of energy happens after the absorption. A noncausal system that generates energy before it absorbs it would be considered to be nonpassive. Summary It has been shown that a realistic dielectric model must be: 1. Causal , meaning that the model cannot respond to a stimulus until some time after it has been applied. 2. Analytic, which says that the function must satisfy the Cauchy–Riemann equations, meaning that the real and imaginary parts of the permittivity are related. 3. Real in the time domain, which means that its Fourier transform must satisfy the complex-conjugate rule (6-37). 4. Passive, which means that the model cannot generate energy. The dielectric models derived in Section 6.3 satisfy these mathematical constraints. 6.5 FIBER-WEAVE EFFECT As described brieﬂy in Section 3.1, printed circuit boards (PCBs) are most commonly constructed from an FR4-type material. Many dielectric materials used in the electronics industry, including FR4, have been treated historically as a homogeneous dielectric medium for the propagation of digital signals in PCBs. As data rates for system buses push into the multi-Gb/s range, the composition of FR4, in which woven ﬁberglass bundles are embedded in epoxy resin, causes this assumption to break down. At high frequencies, local variations in dielectric constants create spatially dependent values of the time delay τd and the characteristic impedance Z0. If not controlled properly, the spatially dependent transmission-line parameters can severely degrade voltage and timing margins, especially in differential signal-based buses, described in detail in Chapter 7. In this section we demonstrate how PCB physical structures create local, systematic dielectric constant variation, and describe material and design options for mitigating the effect. FIBER-WEAVE EFFECT High er; Low Z0 Low er; High Z0 275 trace a Top view trace b Glass Bundles Epoxy Pool Cross-sectional view Figure 6-13 Physical structure of FR4 dielectric; note how the physical position of the transmission line with respect to the ﬁberglass weave dictates the impedance. 6.5.1 Physical Structure of an FR4 Dielectric and Dielectric Constant Variation FR4 dielectric is a composite material made from a matrix of woven bundles of ﬁberglass embedded in an epoxy resin. The physical structure of FR4 is illustrated in Figure 6-13, which demonstrates how local material variation makes the homogeneous dielectric assumption inaccurate when propagating signals with multi-GHz frequency content. The reinforcing ﬁberglass bundles have a dielectric constant (εr ) of approximately 6, whereas εr is close to 3 for the epoxy resin in which the bundles are embedded. The bulk dielectric constant is dependent on the glass/resin volume ratio: εr = εrsnVrsn + εglsVgls (6-39) where εrsn and εgls are the dielectric permittivities and Vrsn and Vgls are the volume ratios of the epoxy resin and glass. From Figure 6-13 we see that there is a signiﬁcant gap between bundles over which a signal trace may be routed. This creates the opportunity for a differential pair to be routed such that one trace (a) is located over a ﬁberglass bundle, while the other (b) is routed above the gap between bundles. The result is that a signal on trace a will see a higher effective dielectric permittivity (εr,eff) and a lower impedance then trace b. Furthermore, it should be noted that this is a statistically likely scenario, since the ﬁber weave is usually manufactured so that the bundles are parallel to the edge of the board. The most common layout methodology is to route the transmission lines of the system buses at either 0◦ or 90◦ with respect to the board edges, which maximizes the chance that spatially dependent effective dielectric permittivity (εr,eff) will be observed. The effective dielectric permittivity variability can be shown experimentally by using a test board structure as shown in Figure 6-14. If the structures are designed with a center–center trace distance (xt ) slightly larger than the expected 276 high er, low Z0 xt ELECTRICAL PROPERTIES OF DIELECTRICS low er, high Z0 xt xt xt Glass bundle xt xw >xt xw xw xw xw Figure 6-14 Depiction of transmission lines in relation to the ﬁber bundles in a printed circuit board. distance between adjacent ﬁberglass bundles (xw), routing several traces parallel to the ﬁberglass bundles across a large section of a PCB, the relative signal trace location with respect to the ﬁberglass bundles will varied in a systematic fashion to allow determination of the worst-case difference in εr,eff between traces. Figure 6-15 shows effective dielectric permittivity extracted from time-domain reﬂectometry (TDR) measurements on test structures similar to Figure 6-14. The effective dielectric permittivity is calculated from the measured time delay τd of the transmission line, which is determined from the TDR proﬁle similar to Example 3-7. Next, equation (2-52) is used to calculate the effective permittivity where µr = 1 and εr = εr,eff: νp = c √ µr εr,eff → 1 νp = τd = √εr,eff c → εr,eff = (cτd )2 (6-40) Note that the effective dielectric constant is a composite function of the glass cloth, epoxy resin, and surrounding air for microstrip transmission-line structures. Since the relative location of the traces is slowly varied across the PCB, the worst-case difference in effective dielectric constant is equal to the difference between minimum and maximum measured εr,eff. The example in Figure 6-15 shows a worst-case difference in the effective dielectric constant to be εr,eff ∼ 0.23 for microstrip structures using 2116-type ﬁberglass cloth for the board, which corresponds to an actual dielectric permittivity difference for the material ( εr ) of approximately 1.5 to 2.25 εr,eff, based on empirical observations of measured data. Empirical data from numerous FR4-based test boards indicate that the differences in the measured effective dielectric permittivity for a microstrip structure can be as high as εr,eff = 0.4, leading to a material value of εr ∼ 0.8 to 0.9. Although these spatial variations might seem small, the analysis in Chapter 7 will demonstrate that it can have a severe impact for relatively short differential transmission lengths at data rates of 5 to 10 Gb/s. 6.5.2 Mitigation Routing pairs at angles relative to the direction of weave will average out the ﬁber-weave effect by constantly changing the relative position of the traces FIBER-WEAVE EFFECT 277 Trace over bundle 3.75 3.70 3.65 er,eff 3.60 3.55 3.50 3.45 1 Trace between bundles 6 11 16 21 26 31 36 41 46 51 56 61 Trace Figure 6-15 Example of εr, eff variations due to the ﬁber-weave effect for 64 parallel transmission lines routed parallel to a board edge constructed with FR4 dielectric material. 3.75 3.70 3.65 εr,eff 3.60 3.55 3.50 3.45 1 6 11 16 21 26 31 36 41 46 51 56 61 Trace Figure 6-16 Example of the reduction in εr, eff variations from the ﬁber-weave effect for 64 parallel transmission lines routed 45◦ with respect to a board edge constructed with FR4 dielectric material. relative to the bundles. As Figure 6-16 shows, routing at 45◦ angles minimizes the impact. It has been shown that offsets as small as 5 to 10◦ are adequate to mitigate most of the spatial effects. 6.5.3 Modeling the Fiber-Weave Effect The preceding discussion indicates that the traditional approach of modeling a transmission line using a uniform dielectric constant throughout the dielectric 278 ELECTRICAL PROPERTIES OF DIELECTRICS w h er 1 er 2 between bundles over a bundle Figure 6-17 Two-dimensional ﬁber-weave modeling structure. layer is no longer sufﬁcient. We need to modify our approach to comprehend the localized dielectric variation. Figure 6-17 shows the proﬁle of the crosssectional geometry description used as 2D ﬁeld solver input for calculating the transmission-line parameters. Simple in principle, this new approach is complicated by the fact that the effective dielectric constant measured results from the combined effect of multiple dielectrics: resin, ﬁberglass, solder mask, and air. As a starting point, the effective dielectric constant of the FR4 material in each region can be estimated from measured data using the method described in equation 6-39 for multicomponent dielectrics. To determine the actual values of εr1 and εr2 (see Figure 6-17), they should be varied until the delay of the 2D simulation is equal to the measured response from the test board. Alternatively, accurate representation of the cross section, including proper glass/resin height ratios and solder mask thickness using a 2D electromagnetic ﬁeld solver will yield correct values of the effective dielectric permittivity. Example 6-3 From the spatial εr,eff variations shown in Figure 6-15, calculate the actual (noneffective) dielectric permittivity for a trace over a bundle and between bundles for a microstrip transmission line with a w/ h ratio of 2 and t h. SOLUTION Step 1: Use equation (3-35) to estimate the relationship between εr and εr,eff. a = 1 + 1 49 ln 24 + (2/54)2 24 + 0.432 + 1 18.7 ln 1+ 2 18.1 3 b = 0.564 εr − 0.9 0.053 εr + 3 εeff(u, εr ) = εr + 2 1 + εr − 2 1 1 + 10 2 −ab The relationships between εr and εr,eff are plotted in Figure 6-18a. The value of εr that corresponds to the minimum and maximum values of εr,eff in Figure 6-15 are chosen to create equivalent cross sections to represent the case where the trace is over a bundle and between bundles. ENVIRONMENTAL VARIATION IN DIELECTRIC BEHAVIOR 279 er,eff 4 ∆er,eff = 0.23 3.5 3 2.5 2 1.5 ∆er = 0.35 1 1 1.5 2 2.5 3 3.5 4 4.5 5 er (a) w w h er = 4.6 h er = 4.95 between bundles (b) over a bundle Figure 6-18 (a) Relationship between εr and εr, eff; (b) equivalent cross sections for a trace directly over a bundle and between bundles for Example 6-2. Trace between bundles: εr,eff = 3.5 (from Figure 6-15) and εr ∼ 4.6 (from Figure 6-18a) Trace over a bundle: εr,eff = 3.72 and εr ∼ 4.95 Step 2: Create the equivalent cross sections, as shown in Figure 6-18b. 6.6 ENVIRONMENTAL VARIATION IN DIELECTRIC BEHAVIOR One problem often overlooked in high-speed digital design is the impact that the relative humidity (RH) of the environment has on the electrical performance of a dielectric material. The electrical properties of a dielectric are partially a function of the amount of moisture in the material. The extent to which a material absorbs moisture is characterized by its moisture diffusivity and saturated moisture concentration. Moisture diffusivity describes the rate of change of a material’s moisture concentration. The saturated moisture concentration provides an expression for the limit to the amount of moisture that a material can contain. It is important to note that both are a function of temperature and relative humidity. Another measure of a material’s susceptibility to moisture is its maximum moisture uptake. This is typically reported on material data sheets as % weight and is related to the saturated moisture concentration. If a printed circuit 280 ELECTRICAL PROPERTIES OF DIELECTRICS board (PCB) constructed with a dielectric that tends to absorb moisture (such as FR4) is exposed to a humid environment (such as Malaysia) for a signiﬁcant amount of time, both the loss tangent and the dielectric permittivity will increase. To demonstrate this, Figure 6-19 shows an example of the insertion losses of an identical structure measured in a dry environment such as an Arizona winter (15% RH and 60◦F) and a humid environment such as Malaysia (95% RH and 95◦F). As explained in Chapter 9, this dramatic increase in insertion losses corresponds to an increase in dissipated power. In linear terms, the example in Figure 6-19 represents approximately 50% more losses due to the dielectric at 10 GHz. Figure 6-20 depicts measured values of the same dielectric material for both low- and high-humidity environments. Note that at about 7.5 GHz, the increase in the loss tangent is greater than 50%. Conversely, Figure 6-21 shows 0.0 Insertion loss (S21), dB −4.0 Arizona Winter Low RH −8.0 −12.0 −16.0 Malaysia Summer High RH −20.0 0 5 10 15 20 Frequency, GHz Figure 6-19 Example of the impact of the environment on PCB losses for a microstrip structure constructed on an FR4 dielectric. 0.022 0.02 95% Relative humidity and 36°C 0.018 0.016 1.57 × increase tan d 0.014 10% Relative humidity and 20°C 0.012 0.01 0 2 4 6 8 10 12 Frequency, GHz Figure 6-20 Example of the measured variation in tanδ for 7628-FR4 dielectric for different environmental extremes. ENVIRONMENTAL VARIATION IN DIELECTRIC BEHAVIOR 281 4.8 95% Relative humidity and 36°C 4.7 Dielectric permitivity (er) 4.6 1.04 × increase 4.5 10% Relative humidity and 20°C 4.4 4.3 4.2 0 2 4 6 8 10 12 Frequency, GHz Figure 6-21 Example of the measured variation in εr for 7628-FR4 dielectric for different environmental extremes. a relatively small increase of less than 5% in the dielectric permittivity for the environmental extremes [Hamilton et al., 2007]. The mechanism that explains this dramatic increase in losses and the minor increase in permittivity is probably a combination of orientational and ionic polarization. The water molecule is relatively heavy, and it therefore takes a signiﬁcant amount of energy for the molecule to follow the external time-varying electric ﬁeld, as shown in Figure 6-3. Furthermore, the relative dielectric permittivity of distilled water is εr ∼ 81. Since the dielectric permitivity of the FR4 dielectric does not change much (∼5%) between dry and humid conditions whereas the loss tangent changes dramatically (∼50%), it indicates that (1) a relatively small amount of water is absorbed by the dielectric, and (2) a small amount of absorbed water (per unit volume of dielectric) has a large impact on the dielectric losses. 6.6.1 Environmental Effects on Transmission-Line Performance The absorption of water in an FR4 dielectric can be observed with vector network analyzer (VNA) measurements. Using linear values for S21 and S11, the power loss of a transmission line as derived in Chapter 9 is αloss = 1 − S221 − S121 where αloss is a measure of the power dissipated by the transmission line from conductor, dielectric, and radiation losses. The humidity-induced change in αloss is deﬁned as the difference between the measured loss of a transmission line in a dry state (αloss,dry) and the measured loss of a transmission line that has been exposed in a humid environment for a signiﬁcant amount of time (αloss,humid). This humidity-induced change in αloss from the dry state is termed αhumid: αhumid = αloss,humid − αloss,dry (6-41) 282 ELECTRICAL PROPERTIES OF DIELECTRICS aloss Dahumid 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0 10 20 30 Frequency, GHz aloss, dry Dahumid aloss, humid (55 days @ 38°C / 95%RH) 0.16 0.12 0.08 0.04 0.00 40 Figure 6-22 Example of how a humid environment will change the losses of 7628-FR4 dielectric. (Adapted from Hamilton et al. [2007].) Hamilton et al. [2007] detail several experiments where the relative humidity was controlled with an environmental chamber and measurements were taken with a vector network analyzer (VNA) to determine how dry and wet environments inﬂuence dielectric losses. Figure 6-22 depicts an example of a measurement showing how humidity changes the loss characteristics of a test board built on an FR4 dielectric. The dry case (αloss,dry) was preconditioned in an environmental chamber to bake out the moisture already absorbed by the dielectric and then stored in a dry box at 20◦C and 10% RH. The wet case (αloss,humid) was conditioned in an environment of 37◦C and 95% RH for 55 days. Note that αhumid peaks at about 5 GHz and then decreases to almost zero at high frequencies. The peak is caused by the increase in power dissipation caused by the afﬁnity of the dielectric material to absorb water from a humid environment as demonstrated by the increased slope of αhumid prior to the peak. After 5 GHz, the insertion loss (S21) for both humid and dry conditions will increase (meaning that the transmitted power decreases), causing both αloss,humid and αloss,dry to approach unity. Consequently, αhumid will initially peak because αloss,humid will increase faster than αloss,dry, then approach zero. This measurement highlights how a transmission line can exhibit a signiﬁcant increase in dissipated power losses if the dielectric material has an afﬁnity to absorb water from a humid environment. For design purposes, the question arises of how long it takes for moisture to diffuse into the dielectric. Obviously, each dielectric material will exhibit its own moisture absorption characteristics. For example, Teﬂon (PTFE) generally exhibits very little or no tendency to absorb moisture, whereas common solder-mask materials exhibit a very high tendency to absorb moisture. For the purposes of this discussion, we concentrate on an FR4 dielectric sample. Figure 6-23 shows the measured change in insertion loss (S21) at 10 GHz for a microstrip transmission line built on 7628-FR4 dielectric that was preconditioned to a dry state by baking the moisture out of the sample and then exposed to humid ENVIRONMENTAL VARIATION IN DIELECTRIC BEHAVIOR 283 −6 −6.5 Day 0, sample is preconditioned to be dry −7 S21 at 10 GHz (dB) −7.5 −8 Sample stored in an environmental chamber at −8.5 38°C and 95% relative humidity for 55 days −9 −9.5 0 10 20 30 40 50 60 Days Figure 6-23 Insertion loss at 10 GHz of a transmission line built on 7628-FR4 dielectric, showing how losses change with time exposed to a humid environment. (Adapted from Hamilton et al. [2007].) conditions in an environmental chamber and measured at speciﬁc intervals for 55 days. In this example, the dielectric transmission-line losses at 10 GHz were −6.3 dB under dry conditions at t = 0. When the structure was exposed to a humid environment, the dielectric became almost fully saturated within the ﬁrst 7 days, increasing the loss to −8.9 dB. Finally, the transmission-line loss stabilized to a value of −9.3 dB of loss at t = 48 days. Stripline transmission lines tend to absorb moisture at a signiﬁcantly slower rate than microstrip lines because there is much less dielectric surface area exposed to the surrounding environment. Hamilton et al. [2007] report experimental data for striplines built on 7628-FR4 dielectric with 90% coverage of the dielectric material by copper planes, showing that it takes about 5 months for the stripline to reach the same loss value as a microstrip does in 7 days under identical conditions. This means that striplines are not immune to environmental conditions; however, they absorb moisture at a signiﬁcantly slower rate. 6.6.2 Mitigation Unfortunately, there are very few practical methods for mitigating the effect of the environment on transmission-line performance. In consumer products such as laptop or desktop mother boards, the full range of environmental conditions must be accounted for in the design to ensure that performance speciﬁcations are maintained in both dry and humid environments (such as a dry Arizona winter or a humid Florida summer). This means that an FR4-based transmission-line model used in the design must comprehend as much as a 50% or more variation in the loss tangent, which can signiﬁcantly reduce solution spaces, particularly for long bus lengths and high data rates. Generally, the use of stripline transmission 284 ELECTRICAL PROPERTIES OF DIELECTRICS lines does not cure the effect of the environment because products are generally designed to operate under all reasonable conditions, and it must be assumed that there will be periods where the digital design is powered down long enough for the dielectric to absorb enough water to increase the loss tangent. However, striplines can help mitigate the problem under certain conditions because it takes a very long time for the dielectric contained between copper layers to absorb water. In some cases, the product usage model may allow mitigation of the environmental effects by not allowing large changes in relative humidity for normal operation. For example, computer servers are often operated in a controlled temperature/humidity room, allowing engineers to design for minimum environmental variation, thus increasing the performance of the platform. Additionally, some computers are often powered on continuously. In this case, the heat generated by the CPU and the movement of air by the fan will produce a low-humidity environment inside the computer chassis. However, in this case, the engineer must understand that if the computer is powered down for any signiﬁcant amount of time, the board dielectric material may begin to absorb moisture. The absorbed moisture will increase the loss of the transmission lines and decrease the performance of the high-speed buses until the absorbed moisture diffuses out, which could take several hours to several days, depending on the dielectric material and the environmental conditions inside the chasses during operation and transmission-line structure. 6.6.3 Modeling the Effect of Relative Humidity on an FR4 Dielectric The modeling method employed to simulate the effect of moisture absorption in a dielectric material is entirely dependent on the dielectrics afﬁnity to absorb water, how the absorbed moisture interacts with the chemical structure of the dielectric, and the frequency range of interest. The dielectric most commonly used in the electronics industry is FR4, which readily absorbs moisture. To understand how to account properly for environmental variations in the relative humidity, we consider how to model FR4 dielectric for two cases: 1. Low frequencies (below 2 GHz). Below 2 GHz, the effect of moisture absorption has very little effect on total transmission-line losses (at least for FR4 dielectrics). This behavior is demonstrated in Figure 6-19, where the difference in insertion loss is minimal until approximately 2 GHz. This is because at low frequencies, the total loss is usually dominated by the skin effect resistance of the signal conductor, as described in Section 5.2. For typical transmission lines, the losses of the dielectric begin to dominate above about 1 to 2 GHz. 2. High frequencies (2 to 50 GHz). For this frequency range, the dry and nominal environmental corners (up to approximately 50% relative humidity for FR4) can be modeled with the inﬁnite-pole Debye model, as described in Section 6.3.5 and shown in equation (6-29). However, for TRANSMISSION-LINE PARAMETERS 285 high-humidity conditions, the assumptions used to derive the inﬁnite-pole model break down at frequencies greater than about 10 GHz. Speciﬁcally, the approximation that ω2i in equation (6-28) decreases linearly on the logarithmic scale is no longer valid for a dielectric that has absorbed a signiﬁcant amount of moisture. Therefore, for humid conditions (above 50% relative humidity) and frequencies above approximately 10 GHz, FR4 dielectric must be ﬁt empirically to a multipole model as described in Section 6.3.4 with equation (6-27) and demonstrated in Example 6-1. It is sometimes necessary to account for induced dipole resonances with the squared term in equation (6-27) (ω2/ω12i). 6.7 TRANSMISSION-LINE PARAMETERS FOR LOSSY DIELECTRICS AND REALISTIC CONDUCTORS As bus data rates increase and physical implementations of high-speed digital designs shrink, transmission-line losses become more important. Consequently, the engineer must have the ability to calculate the response of a transmission line successfully and account for realistic conductor and dielectric behavior. In the next two sections we build on the equivalent circuit derived in Section 5.4, which included nonideal conductor effects, and describe how to include dielectric losses in the equivalent circuit and modify the telegrapher’s equations to comprehend dielectric losses. 6.7.1 Equivalent Circuit, Impedance, and Propagation Constant Building directly on Section 5.4.1, where we developed the equivalent circuit, impedance, and propagation constant for a transmission line with a realistic conductor, we can complete the transmission-line model by adding the effects of dielectric losses. The assumption of a perfect dielectric did not allow calculation of a loss term. Fortunately, the form of the equivalent circuit derived in Section 3.2 is also applicable to a line whose dielectrics have an imaginary term as in equation (6-15), and therefore losses. To begin this derivation, the shunt admittance of an ideal transmission line with a loss-free dielectric is calculated as Yshunt = j ωC (6-41) which has units of siemens. The equivalent circuit is easily modiﬁed to include a shunt resistor with a conductance G with units of siemens, in parallel with the capacitance to account for the dielectric losses, as shown in Figure 6-24, where Ns is the number of segments, C z = z C and L z = z Ltotal, as calculated from Section 3.23 and R z = z Rac from Section 5.4.1 and G z = z G, where z is the length of the differential section of transmission line and C, Ltotal, Rac, and G are the capacitance, inductance, resistance, and conductance per unit length. 286 ELECTRICAL PROPERTIES OF DIELECTRICS L∆z R∆z C∆z G∆z 1 R∆z L∆z C∆z (a) 2 R∆z L∆z C∆z Ns R∆z L∆z C∆z G∆z G∆z G∆z (b) Figure 6-24 (a) Model for a differential element of a transmission line; (b) full model. For this equivalent circuit, equation (6-42) calculates the shunt admittance by adding a conductance term G: Yshunt = G + j ωC (6-42) The formula for G can be derived by ε = ε − j σdielectric = ε − j ε ω tan |δ| = ε ε (6-15) (6-16b) which show clearly that the dielectric losses will be proportional to ε and tan |δ|, and therefore G. Furthermore, we know from Section 6.4.1 that the real and imaginary parts of the dielectric permittivity must be related. Consequently, there must also be a relationship between the conductance G and the capacitance C. If the dielectric losses are treated as an equivalent conductivity, we can say that the dielectric carries a current of J = σdielectricE [equation (2-7)]. Equation (3-1) says that the voltage between the signal conductor and the reference plane is v = − b a E · d l and that the total current is calculated from (2-20) as i = S J · ds. Therefore, in circuit terms, the conductance G can be written G = i v = S J · ds − b a E · dl = σdielectric S E · ds − b a E · dl (6-43a) TRANSMISSION-LINE PARAMETERS 287 Similarly, from equations (2-76), (2-59) and (3-1), we can write the capacitance in the form C = Q v = ε − S E · ds b a E · dl (6-43b) Dividing (6-43a) by (6-43b) yields the following ratio: G = σdielectric C ε (6-44) From (6-15) and (6-16b), σdielectric = tan |δ|ε ω which is substituted into (6-44) to yield G(ω) = tan |δ|ωC (6-45) where G(ω) is the frequency-dependent conductance in units of siemens per unit length, ω = 2πf , and C is the capacitance per unit length for the transmission line. The characteristic impedance, which was deﬁned in equation (3-33), can be calculated by dividing the series impedance as deﬁned by (5-68) by the parallel admittance deﬁned by (6-42) for a short section of transmission line of length z: Z0 = Zseries = Yshunt Rac + j ωLtotal G + jωC ohms (6-46) √ Note that the units in (6-46) are ohms/(1/ohms) = (ohms)2 = ohms. The propagation constant can be derived by inserting the complex values of the series impedance and shunt admittance into the loss-free formula derived in Section 3.2.4 in (3-30), which takes the form √ γ = α + jβ = 0 + j ω LC = (j ωL)(j ωC) = ZlosslessYlossless Substitution of Zseries and Yshunt in place of the loss-free values of the series impedance and the parallel admittance yields the propagation constant for a transmission line with a lossy dielectric and a conductor with a ﬁnite conductivity: γ = α + jβ = ZseriesYshunt = (R + j ωL)(G + j ωC) (6-47) 288 ELECTRICAL PROPERTIES OF DIELECTRICS If the loss parameters of equation (6-47) are small compared to j ωLtotal and j ωC, which is the case for most practical transmission lines, (6-47) can be approximated [Johnk, 1988]: γ ≈ α + jβ = 1 R C +G L √ + j ω LC 2 L C (6-48a) α≈ 1 R C +G L 2 L C √ β ≈ ω LC (6-48b) (6-48c) As described in Section 3.2.4, the voltage propagating on a loss-free transmission (α = 0) line can be written √ √ v(z) = v(z)+e−jzw LC + v(z)−ejzw LC (3-29) However, for a lossy transmission line, the voltage is described by multiplying equation (3-29) by the decay factor e−αz: √ √ v(z) = v(z)+e− α+jw LC z + v(z)−e α+jw LC z = v(z)+e−γ z + v(z)−eγ z (6-49) Equation (6-49) describes the voltage propagating on a lossy transmission line. Example 6-4 Calculate the frequency-dependent voltage at the output of a 0.5-m transmission line that is perfectly terminated in its characteristic impedance with the following properties: Lext = 2.5 × 10−7 H/m, Cquasistatic = 1.5 × 10−10 F/m, σ = 5.8 × 107 S/m, µ = 4π × 10−7 H/m, l = 0.5 m, and w = 100 × 10−6 m, εr,eff = 3.32, and tan δ = 0.0205 at 1 GHz. SOLUTION Step 1: Calculate the frequency-dependent parameters of the conductor. The resistance can be calculated approximately using equation (5-12), and is plotted in Figure 6-25a. Rac = l w π µf σ The inductance is calculated with equations (5-20) and (5-30) and plotted in Figure 6-25b. Ltotal = Linternal + Lexternal Linternal = Rac ω (5-20) (5-30) TRANSMISSION-LINE PARAMETERS 289 Rac, ohms/meter 150 290 285 L, nH/meter 100 280 275 50 270 265 1 2 3 4 5 Frequency, GHz (a) 1 2 3 4 5 Frequency, GHz (b) 170 165 160 155 150 145 1 2 3 4 5 Frequency, GHz (c) G, ohm−1/meter 0.08 0.06 0.04 0.02 0.00 1 2 3 4 5 Frequency, GHz (d) Figure 6-25 Frequency-dependent transmission-line parameters for Example 6-4. C, pF/mter Step 2: Calculate the frequency-dependent parameters of the dielectric. First, the real and imaginary parts of the dielectric constant must be found: ε = εr = 3.32 tan δ = ε = 0.0205 ε ε = 0.068 ε = ε − j ε = 3.32 − j 0.068 Equations (6-30a) and (6-30b) are solved simultaneously to get ε and ε∞: 3.32 = ε∞ + m2 ε − m1 ln(ω2/ω) ln(10) 0.068 = m2 ε − m1 −π/2 ln(10) where m1 = 1 and m2 = 11 are chosen to correspond to a frequency range of 10 rad/s to 100 Grad/s, ω is 2π (1 GHz), and ω2 = 1011. ε = 0.997 ε∞ = 3.2 290 ELECTRICAL PROPERTIES OF DIELECTRICS The frequency-dependent capacitance is calculated by dividing Cquasistatic by the quasistatic value of the effective dielectric permittivity and multiplying by the frequency-dependent dielectric permittivity calculated by equation (6-30a). The frequency-dependent capacitance is plotted in Figure 6-25c: C(ω) = Cquasistatic εr,eff ε∞ + m2 ε − m1 ln(ω2/ω) ln(10) where m1 = 1 and m2 = 11 are chosen to correspond to a frequency range of 10 rad/s to 100 Grad/s. The conductance, G(ω), is calculated using equation (6-45) and plotted in Figure 6-25d: G(ω) = tan |δ|ωC(ω) (6-45) where the loss tangent is calculated with (6-30a) and (6-30b): ε −π/2 tan |δ = | ε ≈ m2 − m1 ln(10) ε ε∞ + m2 ε − m1 ln(ω2/ω) ln(10) Step 3: Calculate the frequency-dependent voltage. Since the transmission line is perfectly terminated, no reﬂections will be generated. Therefore, the loss-free voltage wave will behave as described by equation (6-49) with v(z)− = 0. vout = vine−αze−jβz = vine−γ z Therefore, using equation (2-31) to simplify, vout = vine−αz[cos(−βz) + j sin(−βz)] where γ is deﬁned by equation (6-47) and z is the line length: γ (ω) = α + jβ = (R + j ωL)(G + j ωC) (6-47) Next, the magnitude of the voltage is calculated: vout,mag = Re(vout)2 + Im(vout)2 and is plotted in Figure 6-26 assuming that vin = 1 V. TRANSMISSION-LINE PARAMETERS 291 1.0 0.8 vout, volts 0.6 0.4 0.2 5 10 15 20 Frequency, GHz Figure 6-26 Magnitude of voltage at the output of a 0.5-m transmission line of Example 6-4. 6.7.2 Telegrapher’s Equations for Realistic Conductors and Lossy Dielectrics The time-harmonic forms of the telegrapher’s equations for a transmission line with a perfect dielectric and perfect conductor were shown in Section 3.2.4: dv(z) = −j ωLi(z) dz di(z) = −j ωCv(z) dz (3-25) (3-26) These formulas were adjusted in Section 5.4.2 to account for a realistic conductor with a ﬁnite conductivity. The classic form of the telegrapher’s equations for a perfectly insulating dielectric and a realistic conductor are ∂v(z, t) = − ∂z Rac + Ltotal ∂ ∂t ∂i(z, t) = −C ∂v(z, t) ∂z ∂t i(z, t) (5-72a) (5-72b) To include the effects of a lossy dielectric, equation (5-72) must be modiﬁed to include the conductance G, as deﬁned in equation (6-45). Notice that the right-hand side of equation (3-26) is simply the inverse impedance of a capacitor (i.e., the conductance). To account for the dielectric losses, G is simply added into the right-handside of (3-26): di(z) = −[G + j ω(C)]v(z) dz (6-50) 292 ELECTRICAL PROPERTIES OF DIELECTRICS Consequently, the classic form of the telegrapher’s equations for a lossy dielectric and a realistic conductor are ∂v(z, t) = − ∂z Rac + Ltotal ∂ ∂t i(z, t) (5-72a) ∂i(z, t) = − G + C ∂ v(z, t) ∂z ∂t (6-51) REFERENCES Balanis, Constantine, 1989, Advanced Engineering Electromagnetics, Wiley, New York. Brillouin, L., 1960, Wave Propagation and Group Velocity, Academic Press, New York. Djordjevic, A. R., R. M. Biljic, V. D. Likar-Smiljanic, and T. K. Sarkar, 2001, Wide- band frequency domain characterization of FR-4 and time domain causality, IEEE Transactions on Electromagnetic Compatability, vol. 43, no. 4, Nov., pp. 662–667. Elliot, R., 1993. Electromagnetics, IEEE Press, Piscataway, NJ. Hall, Stephen, Steven G. Pytel, Paul G. Huray, Daniel Hua, Anusha Moonshiram, Gary A. Brist, and Edin Sijercic, 2007, Multi-GHz causal transmission line modeling using a 3-D hemispherical surface roughness approach, IEEE Transactions on Microwave Theory and Techniques, vol. 55, no. 12, Dec. Hamilton, Paul, Gary Brist, Guy Barnes, Jr., and Jason Schrader, 2007, Humidity dependent losses in FR4 dielectric, presented at the IPC Apex Conference. Huray, Paul, 2009, Foundations of Signal Integrity, Wiley, Hoboken, NJ. Jackson, J. D., 1998, Classical Electrodynamics, 3rd ed., Wiley, New York. Johnk, Carl T. A., 1988, Engineering Electromagnetic Fields and Waves, Wiley, New York. LePage, Wilbur P., 1980, Complex Variables and the Laplace Transform, Dover Publications, New York. Pauling, Linus, 1948, The Nature of the Chemical Bond , Cornell University Press, Ithaca, NY. Pytel, Steven G., 2007, Multi-gigabit data signaling rates for PWBs including dielectric losses and effects of surface roughness, Ph.D. dissertation, Department of Electrical Engineering, University of South Carolina. PROBLEMS 6-1 If the dielectric permittivity of a dielectric is known to be εr = 4.2 and tan δ = 0.021 at 1 GHz, plot the frequency dependent values εr (f ) and tan δ(f ) from 1 MHz to 20 GHz. 6-2 Given the plot of the dielectric permittivity in Figure 6-27a and the loss tangent plot in Figure 6-27b, derive an equation for a frequency-dependent dielectric model. 6-3 Given the real part of the dielectric permittivity in Figure 6-28, calculate the loss tangent. PROBLEMS 293 3.95 3.90 er 3.85 0.025 0.020 0.015 0.010 0.005 10 20 30 40 50 Frequency, GHz (a) tan d 10 20 30 40 50 Frequency, GHz (b) Figure 6-27 (a) Permittivity data and (b) loss tangent data for Problem 6-2. er 4.10 4.05 4.00 3.95 3.90 10 20 30 40 Frequency, GHz Figure 6-28 Dielectric permittivity for Problem 6-3. 294 ELECTRICAL PROPERTIES OF DIELECTRICS 6-4 Show that the Debye dielectric model [equation (6-28)] satisﬁes the Cauchy–Riemann equations for i = 1: ε = ε∞ + n i=1 1 εi + j (ω/ω2i ) (6-28) 6-5 Given the cross section in Figure 6-29, estimate the bulk value of the dielectric permittivity. Assume that εr,glass = 6.1 and εr,epoxy = 3.2. 3.8 mils 2.1 mils 6.2 mils Epoxy Glass bundle 13 mils Figure 6-29 Cross section for Problem 6-5. 6-6 Assume that a series of 1-in.-long parallel transmission lines routed adjacent to each other are measured. The TDR waveforms are shown in Figure 6-30. Estimate the variation in the dielectric permittivity due to the ﬁber-weave effect. Volts 1.10 tr = 25 ps 1.05 0-1V step 1.00 50 ohms 50 ohms τd,Z0 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 Time, ns Figure 6-30 TDR for Problem 6-6. 6-7 Assuming a conductor width of 5 mils and a loss-free dielectric thickness of 4 mils, create an equivalent-circuit model of the corner cases in PROBLEMS 295 Problem 6-6. Simulate them to show that the model matches the TDR measurements. 6-8 What is the best way to minimize losses for a given cross section of a transmission line with constant impedance? What is the best way to minimize losses if constant impedance is not required? Plot the voltage magnitude versus frequency for a 10-in.-long transmission line to prove your theory for each case. 6-9 Given the relationship between the ﬁber-weave and transmission-line dimensions shown in Figure 6-31, determine the minimum angle needed to mitigate the impedance and velocity swings caused by the ﬁber-weave effect. Assume that the maximum and minimum delay cases of the TDR measurements of Figure 6-30 correspond to the cross sections of Figure 6-31. 10 mils 5 mils Epoxy Glass 10 mils 8 mils 10 mils 8 mils Figure 6-31 Cross sections for Problem 6-9. 6-10 For the cross section shown in Figure 6-32, plot the frequency-dependent R, L, C, G, Z0, and τd from 100 MHz to 20 GHz. Assume that the surface of the copper is speciﬁed to have a RMS tooth height of 0.8µm. 5 t = 1.2 4 er = 3.78, tan d = 0.022 at 7 GHz Figure 6-32 Cross section for Problem 6-10. 7 DIFFERENTIAL SIGNALING 7.1 Removal of common-mode noise 299 7.2 Differential crosstalk 300 7.3 Virtual reference plane 302 7.4 Propagation of modal voltages 303 7.5 Common terminology 304 7.6 Drawbacks of differential signaling 305 7.6.1 Mode conversion 305 7.6.2 Fiber-weave effect 310 References 313 Problems 313 When the interconnections between drivers and receivers on a bus are implemented with a dedicated transmission line for each bit, the signaling scheme is said to be single ended . Buses designed with single-ended signaling generally work well up to approximately 1 to 2 Gb/s. As data rates increase, it becomes increasingly difﬁcult to maintain adequate signal integrity because digital systems are notoriously noisy. For example, large arrays of I/O circuits used to drive digital information onto the bus induce noise on the power and ground planes called simultaneous switching noise (for a complete description, see a book by Hall et al. [2000]). There are many other sources of noise that can severely distort the integrity of the digital waveform such as crosstalk (as discussed in Chapter 4) and nonideal current return paths (as discussed in Chapter 10). With single-ended signaling, each data bit is transmitted on a single transmission line and latched into the receiver with the bus clock. The decision of whether the bit is a 0 or a 1 is determined by comparing the received waveform to a reference voltage vref. If the received waveform has a voltage greater than vref, the signal is latched in as a 1, and if it is below vref, it is latched in as a logic 0. Noise coupled Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 297 298 DIFFERENTIAL SIGNALING onto the driver, receiver, transmission lines, reference planes, or clock circuits will degrade the ideal relationship between the transmitted waveform and vref. If the magnitude of the noise is large enough, the incorrect digital states will be latched into the receiver, and bit errors will occur. Figure 7-1 depicts how noise can make the determination of a logic 0 or 1 uncertain. Ground noise vref Power noise Driver latch Data from D Q CPU Clock Zo = 50 Ω Zo = 50 Ω Crosstalk noise Zo = 50 Ω Receiver latch Comparator D Q Clock A Ideal received bit (State = 1) Received bit with noise (State is uncertain) Volts at A Volts at A vref vref Time Time Figure 7-1 How system noise can severely degrade signal integrity on single-ended buses. The ideal versus noisy receiver voltages compared to the reference voltage. Driver latch Data Q from D CPU Q Clock Zo = 50 Ω Zo = 50 Ω D+ Receiver latch D+ Diff Amp D Q D− Clock Volts at D+, D− VD+− VD− D− Time Time Figure 7-2 Differential signaling where each bit is transmitted from the driver to a receiver using a pair of transmission lines driven in the odd mode. The signal is recovered at the receiver with a differential ampliﬁer. REMOVAL OF COMMON-MODE NOISE 299 A strategy that reduces the effect of the system noise dramatically is to dedicate a pair of transmission lines for each bit on the bus. The two transmission lines are driven 180◦ out of phase (in the odd mode), and the differences between the voltages are used to recover the signal at the receiver using a differential ampliﬁer. This technique is called differential signaling and is depicted in Figure 7-2. 7.1 REMOVAL OF COMMON-MODE NOISE Differential signaling is very effective for removing common-mode noise, which is deﬁned as noise present on both legs of the differential pair. If the bus is designed properly so that the legs of the differential pairs are in close proximity to each other, the noise on D+ will be approximately equal to the noise on D−. Therefore, assuming that the receiver’s differential ampliﬁer has a reasonable common-mode rejection ratio, the noise will be eliminated. For example, assume that there is noise with a magnitude of vnoise coupled equally onto both legs of a differential pair. The output of a differential ampliﬁer with unity gain is vdiff = (vD+ + vnoise) − (vD− + vnoise) = vD+ − vD− (7-1) which removes the common-mode noise. Figure 7-3 shows an example of a differential interconnect with noise present on the ground plane (vnoise). In this case, the noise is common to both legs of the driver. Figure 7-4a shows a simulated stream of data bits at the input to a differential ampliﬁer with signiﬁcant common-mode noise. Note that the digital state of the single-ended waveforms vD+ and vD− is indeterminable because the noise is so large. However, since the noise is common mode, the bit stream can be recovered when the signals are subtracted by a differential ampliﬁer (vD+ − vD−), as shown in Figure 7-4b. idrive vD+ D D Zdiff R R R + − vnoise R vD− Figure 7-3 Differential driver and interconnect with common-mode noise present on the ground. Noise on both the power and ground planes is very common in digital designs. 300 DIFFERENTIAL SIGNALING volts volts 1.0 0.8 0.6 0.4 0.2 0.0 −0.2 vD+ vD− −0.4 0.0 1.0 2.0 3.0 4.0 5.0 Time, ns (a) vdiff = vD+ − vD− 0.6 0.4 0.2 0.0 0.2 −0.4 −0.6 0.0 1.0 2.0 3.0 4.0 5.0 Time, ns (b) Figure 7-4 Example of how common-mode noise is eliminated with differential signaling: (a) single-ended waveforms at each leg of a differential receiver showing commonmode noise; (b) differential waveform. 7.2 DIFFERENTIAL CROSSTALK Although crosstalk noise on a differential pair has a signiﬁcant common-mode component, it also has a differential component because the distance between an aggressor and each side of the pair differs (in some cases, if the aggressor is on another layer, the crosstalk could be 100% common mode, but this is rare). Consequently, each leg will experience slightly different crosstalk, which will not be rejected by the differential ampliﬁer. Nonetheless, under certain conditions, differential signaling can reduce crosstalk signiﬁcantly. If the crosstalk between single-ended lines is compared to the crosstalk between differential pairs, the differential crosstalk will be less if the spacing is similar. The drawback is that the differential pair will occupy signiﬁcantly more board area. Additionally, if the single-ended pairs are spaced far enough apart so that the same board area as the differential pairs is occupied, the single-ended crosstalk is typically lower simply because the signals are spaced so far apart. For example, consider Figure 7-5, which shows three cases: 1. Two single-ended coupled transmission lines spaced 10 mils apart (Figure 7-5a). 2. Two differential pairs with an interpair spacing of 10 mils (Figure 7-5b). Note that the spacing between pairs is identical to the spacing between single-ended signals in case 1. 3. Two single-ended coupled transmission lines spaced 28 mils apart (Figure 7-5c), which occupies the same board area as the differential pairs of case 2. Figure 7-6 shows the far end crosstalk for each case when the aggressors are driven with 100-ps-wide pulses and line voltages of 0.5 V. The differential crosstalk is calculated as the difference between V + and V −: vdifferential = (V +) − (V −) DIFFERENTIAL CROSSTALK 301 0V 0.5 V 0.5 V A, A+ A− 0V 100 ps 4 5 5 A S = 10 V 1.3 4 A = aggressor V = victim (a) εr = 4.5, tan δ = 0.017 5 5 5 5 A+ 4 A− S = 10 V+ 4 V− (b) 5 5 A S = 28 V 4 (c) Figure 7-5 Cross sections used to compare differential versus single-ended crosstalk: (a) two single-ended signals; (b) two differential pairs with the same pair spacing as that of the single-ended signals in (a); (c) two widely spaced single-ended signals that occupy the same board area as the differential pairs (dimensions in mils). 0.25 0.2 0.15 0.1 0.05 vdifferential = (V+) − (V−) Differential S = 10 mils (Figure 7- 5b) Single-ended S = 28 mils (Figure 7- 5c) Far end crosstalk, volts 0 −0.05 −0.1 −0.15 Single-ended S = 10 mils (Figure 7- 5a) −0.2 −0.25 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 Time, ns Figure 7-6 Crosstalk for the cross sections in Figure 7-5 demonstrating that differential signaling does not reduce crosstalk compared to single-ended signaling if the same board area is used. However, if the spacing between differential pairs is similar to the spacing between single-ended signals, crosstalk in reduced (5-in. microstrip). 302 DIFFERENTIAL SIGNALING Note that the differential crosstalk (case 2) is less than the identically spaced single-ended crosstalk (case 1), but greater than a single-ended case that occupies the same board space as two differential pairs (case 3). This scenario is typical for modern printed circuit boards, but is not necessarily always true. The beneﬁt of differential signaling for crosstalk is dependent on the speciﬁc geometries and dielectric properties of the design. 7.3 VIRTUAL REFERENCE PLANE Another advantage of differential signaling is that the complementary nature of the electric and magnetic ﬁelds creates a virtual reference plane that provides a continuous return path for the current. Figure 7-7 shows the ﬁeld patterns for a differential pair driven in the odd mode. Note that halfway between the conductors there exists a plane that is normal to the electric ﬁelds and tangent to the magnetic ﬁelds. These conditions are consistent with Section 3.2.1, which describes the boundary conditions of a perfectly conducting plane. Therefore, for a differential pair, there exists a virtual reference plane between the conductors. The existence of the virtual reference plane is extremely helpful for cases when a nonideal reference exists that helps preserve signal integrity. Some common examples of a nonideal reference include connector transitions, via ﬁelds, layer transitions, and routing over a slot in the reference plane. In Chapter 10 we describe in detail the Electric field is perpendicular to the virtual plane Magnetic field is tangent to the virtual plane Electric field Magnetic field Virtual reference plane Figure 7-7 For an odd-mode (or differential) signal, the ﬁelds orient so that an ideal virtual reference plane existed between the conductors. When the physical plane is interrupted, the virtual plane provides a continuous reference for a differential signal and helps preserve signal integrity. PROPAGATION OF MODAL VOLTAGES 303 ill effects of nonideal reference planes. Additionally, when the ﬁelds are conﬁned between the conductors (i.e., strongly coupled to the virtual reference plane), they are less apt to fringe out to other signals, which helps reduce crosstalk. 7.4 PROPAGATION OF MODAL VOLTAGES For a multiconductor system, in Chapter 4 we discussed how all digital signal states are composed of linear combinations of the modal voltages. For a single differential pair, the two modes are the even and odd modes. Ideally, the differential pair is driven with lines 1 and 2 180◦ out of phase so that all of the energy is contained solely in the odd mode. However, if common-mode noise is cou- pled onto the differential pair, some energy will exist simultaneously in the even mode. This is easy to show mathematically using modal analysis, as discussed in Section 4.4. For example, consider a differential pair with line voltages vD+ and vD−, which are composed of linear combinations of the odd- and even-mode voltages: vD+ vD− = [TV ] vodd veven (7-2) where [TV ] is a matrix containing the eigenvectors of the product LC as developed in Section 4.4.1. If the differential pair is driven exactly 180◦ out of phase and there is no noise present, the odd- and even-mode voltages are calculated, where [TV ] = 0.707 −0.707 0.707 0.707 from Example 4-4: 1 −1 = 0.707 −0.707 0.707 0.707 vodd veven (7-3) resulting in vodd = 1.41443 V veven = 0 V which proves that all the energy is contained on the odd mode. However, if common-mode noise is present, energy is introduced into the even mode. If vnoise is the voltage noise introduced to each leg of the pair, the even and odd modes are calculated: 1 + vnoise −1 + vnoise = 0.707 −0.707 0.707 0.707 vodd = 1.41443 V veven = 1.41443vnoise V vodd veven (7-4a) (7-4b) (7-4c) 304 DIFFERENTIAL SIGNALING Therefore, the common-mode noise propagates in the even mode. The voltage present on each leg of the differential pair can be calculated from the modal voltages by solving (7-2): vD+ = 0.707vodd + 0.707veven = 1 + 0.707(1.41443vcm) = 1 + vcm vD− = −0.707vodd + 0.707veven = −1 + 0.707(1.41443vcm) = −1 + vcm In a differential bus, the line voltages are subtracted with a differential ampliﬁer at the receiver which eliminates the noise. vD+ − vD− = (1 + vcm) − (−1 + vcm) = 2 To summarize, a differential signaling scheme eliminates energy in the even mode. 7.5 COMMON TERMINOLOGY For systems with more than two signal conductors, the terms even mode and odd mode are no longer applicable. When analyzing differential buses, it is common to refer to a pair being driven with two signals 180◦ out of phase as the differential mode and two signals driven in phase as the common mode. The differential- and common-mode terms are simply naming conventions and are not technically modes at all. Analysis of the modal voltages propagating on multiconductor systems as described in Section 4.4 will show that the digital states do not correspond directly to modal voltages for systems with more than two signal conductors. The differential-mode impedance is deﬁned as twice the odd mode, and the common-mode impedance is one-half the even mode. The odd- and even-mode impedance values are described in Section 4.3.1. Zdifferential = 2Zodd Zcommon = Zeven 2 (7-5a) (7-5b) It should be noted that equations (7-5a) and (7-5b) are typically used for the purpose of specifying design guidelines. The actual impedance of a differential pair may not correspond directly to Zdifferential if there is signiﬁcant coupling to adjacent pairs. Equations (7-5a) and (7-5b) are representative of the true pair impedance values only if the interpair coupling is weak. Remember: In a multiconductor system with N signal conductors, there will be N unique modal impedance values. It is also true that for a system with N signal conductors, there will be N modal propagation velocities. If the transmission lines are routed in a homogeneous dielectric (such as a stripline), all the modal velocities will be identical. However, for a nonhomogeneous dielectric (such as a microstrip), the differential- and DRAWBACKS OF DIFFERENTIAL SIGNALING 305 common-mode propagation velocities are approximately equal to the odd- and even-mode velocities if the interpair coupling is weak. Odd- and even-mode velocities are deﬁned in Section 4.3.1: νdifferential ≈ νodd νcommon ≈ νeven (7-6a) (7-6b) To quantify the voltage propagating in the differential and common modes, the following deﬁnitions are often used: Vdm = V − V Vcm = V + 2 V (7-7a) (7-7b) where Vdm is the voltage propagating in the differential mode, Vcm is the voltage propagating in the common mode, V represents the voltage propagating on line 1 (the same as vD+ used in the two-signal-conductor example), and V represents the complementary signal propagating on line 2 (vD− used in the two-signal-conductor example). 7.6 DRAWBACKS OF DIFFERENTIAL SIGNALING Differential signaling is a powerful tool used to design high-speed digital buses that dramatically reduces the amount of common-mode noise seen at the receiver, which allows higher data rates to be realized. However, differential signaling is not a panacea. Most obviously, a differential bus consumes signiﬁcantly more area on the printed circuit board than does a single-ended bus. Sometimes, differential buses are so large that they force designers to use printed circuit boards with more layers, which increases cost. The increased number of signals drives package, socket, and connector pin counts to higher levels, which complicates designs and further increases system costs. Also, the impedance tolerance of differential pairs tends to be lower than single ended because the variations in etching and plating proﬁles of the conductors affect mutual inductance and capacitance and, consequently, differential impedance. Furthermore, small asymmetries in the differential pair can have a large impact on signal integrity if not controlled properly. 7.6.1 Mode Conversion In Section 7.4 we showed how common-mode noise coupled onto a differential pair causes voltage to propagate in the even mode, which is rejected by the receiver assuming that the common-mode rejection ratio is high enough. Another mechanism that causes even-mode voltage to propagate are phase errors 306 DIFFERENTIAL SIGNALING between V and V . Ideally, the phase difference between the waveforms propagating on V and V is 180◦, which keeps the energy in the odd mode. However, if the phase relationship between V and V deviates from 180◦ as the signals propagate toward the receiver, some of the energy will be converted from odd mode to even mode. This phenomenon has many names, including mode conversion, differential-to-common mode conversion, and ac common-mode conversion (ACCM conversion). In this text we use the term ACCM conversion. ACCM conversion is caused by asymmetry between V and V in the differential pair. The asymmetry can be caused by length differences, coupling differences, etching differences, proximity effects, termination differences, bends, or anything else that would make one leg of the pair look electrically different from the other. Some of these examples are shown in Figure 7-8. To demonstrate how intrapair asymmetry affects voltage and timing noise, consider the simple (but very common) example where lengths of lines 1 and 2 in the pair are not equal, as shown in Figure 7-8a. Assuming that the signal is launched differentially, the difference in propagation delay between lines 1 and 2 will change the phase relationship at the receiver because the voltage propagating on one leg of the pair will arrive early, converting part (or all) of the differential signal to common mode. Figure 7-9 shows how a perfect differential signal at the transmitter (Tx) is distorted at the receiver (Rx) when asymmetry exists in the pair. The distortion is proportional to the amount of voltage that exists in the common mode. The amount of signal that is converted to common mode depends on the length of the pair, the difference in propagation delays, and the frequency. This line 1 line 2 (a) V V ∆l Z01 V Z02 V (b) S1 V V S2 l2 l1 VV (c) (d) Figure 7-8 Examples of asymmetry in a differential pair that can cause differential to common-mode conversion: (a) routing-length differences; (b) impedance differences due to etching variation; (c) crosstalk differences; (d) length differences due to bends. DRAWBACKS OF DIFFERENTIAL SIGNALING V Tx V l1 l2 Signal at Tx V V Differential Signal Vdm 307 Rx Signal at Rx Common Mode Signal Vcm Figure 7-9 When asymmetry exists in the differential pair, part of the signal gets converted to common mode at the receiver. can be shown by calculating the voltage on each leg of the transmission line at the receiver when the signals are launched 180◦ (π) out of phase: V (ω, l1) = v1+e−αl1 ej (ωt−βl1) V (ω, l2) = v2+e−αl2 ej (ωt+π−βl2) (7-8a) (7-8b) where β is the propagation constant as deﬁned in equation (6-48c), α the atten- uation constant as deﬁned in equation (6-48b), l1 the length of line 1, and l2 the length of line 2. Note that since there is no backward-propagating component (v−), all reﬂections are perfectly terminated in this example. The differential-to-common mode conversion (ACCM) is calculated from (7-8a) and (7-8b) with α = 0: ACCM = V (z = l1) + V (z V (z = 0) − V (z = = l2) 0) = v1+ej (ωt−βl1) v1+ej (ωt) + v2+ej (ωt+π−βl2) − v2+ej (ωt+π) = v1+ e−j β l1 v1+ + v2+e−jβl2 − v2+ (7-9) where V (z = l1) and V (z = l2) are the voltages at the receiver, and V (z = 0) and V (z = 0) are the voltages at the driver. At low frequencies where the wavelength is large, the phase delay difference between lines 1 and 2 is small, so the numerator of (7-9) is approximately zero. However, as the frequency increases, the phase difference becomes large. When 308 DIFFERENTIAL SIGNALING the phase difference reaches 180◦(π), the differential signal launched at the driver is converted completely to a common-mode signal at the receiver, and equation (7-9) is unity. Example 7-1 For the differential pair shown in Figure 7-10, calculate the frequency where the differential signal injected at the driver is 100% converted to a common-mode signal at the receiver. SOLUTION Step 1: Calculate the propagation constant of the transmission lines. Equation (2-46) deﬁnes the propagation constant in terms of the wavelength: β = 2π λ rad/m equation (2-45) deﬁnes wavelength in the terms of the frequency where the speed of light in a vacuum has been replaced with the speed of light in the media (νp): f = νp Hz λ and equation (2-52) calculates the speed of light in the media (assuming that µr = 1): νp = c √ εr m/s Therefore, the propagation constant is calculated as a function of frequency: √ β = 2πf εr = (41.866 × 10−9)f c rad/s Step 2: Use equation (7-9) to plot the differential-to-common mode conversion. Since V (z = 0) = 1 and V (z = 0) = −1, the terms v1+ = 1 and v2+ = −1. The plot is shown in Figure 7-11. When ACCM = 1, the phase error due to the length mismatch equals 180◦ and the differential signal launched at the driver shows Driver V(z = 0) = 1 V(z = 0) = −1 l1 = 0.254 meters εr = 4.0 l2 = 0.260 meters Receiver V(z = l1) V(z = l2) Figure 7-10 Figure for Example 7.1. DRAWBACKS OF DIFFERENTIAL SIGNALING 309 1.0 0.8 ACCM = 100% 0.6 ACCM 0.4 ACCM = 50% 0.2 5 10 15 20 Frequency, GHz Figure 7-11 Differential common-mode conversion plotted for Example 7-1 showing that a differential signal launched at the driver becomes common mode at the receiver when the frequency is about 12.5 GHz. up as a common-mode signal at the receiver. Therefore, the frequency where the differential-to-common mode conversion is 100% is 12.5 GHz. In a differential system, only the difference between V (z = l1) and V (z = l2) is recovered. Consequently, the percentage of differential energy converted to common mode looks like loss to the differential bus in the frequency domain. In Example 7-1 the differential receiver would see no signal at 12.5 GHz. At 4 GHz, approximately 50% of the energy has been converted to the common mode and only one-half the signal would be sensed by the differential receiver. Note that the differential-to-common mode conversion in Example 7-1 decreases after 12.5 GHz. Do not be tempted to operate the bus in this frequency range. The phase difference between signals on the differential pairs will continue to increase until they are 360◦ out of phase, where (7-9) predicts zero mode conversion. If a digital bus were operated so that the phase difference was 360◦ at the receiver, bit 1 on line 1 would align with bit 2 on line 2 and the incorrect data would be latched into the receiver. This can easily be demonstrated by observing the real portion of V (z = l1) and V (z = l2) as calculated with (7-8a) and (7-8b). Figure 7-12a shows the differential-to-common mode conversion (ACCM) for the differential pair in Example 7-1 when l1 = 0.254 m and l2 = 0.340 m. Figure 7-12b shows the real part of V (z = l1) and V (z = l2) at the receiver. Note that at low frequencies, the waveforms are 180◦ out of phase and the differential-to-common mode conversion is zero. At about 880 MHz, the differential-to-common mode conversion is 100%, and the waveforms in Figure 7-12b are in phase. At ∼1.75 GHz, the ninth peak of V (z = l1) is 180◦ out of phase with the seventh peak of V (z = l2), and equation (7-9) predicts that ACCM = 0. Although the magnitude of the differential-to-common mode conversion begins to decrease after it peaks, the phase error is large, so the bus will not function properly. 310 DIFFERENTIAL SIGNALING ACCM 1.0 0.8 0.6 0.4 0.2 0.5 1.0 1.5 2.0 Frequency, GHz (a) 1.0 9th peak of V Real Voltage 0.5 V(z = l1) V(z = l2) 0.5 1.0 1.5 2.0 −0.5 −1.0 Frequency, GHz (b) 7th peak of V Figure 7-12 The differential-to-common mode conversion peaks when the difference in phase delay between lines 1 and 2 in a differential pair is 180◦. As the phase delay difference increases with frequency, the ACCM becomes zero again, but the phase error is so large that the bus will not operate properly. 7.6.2 Fiber-Weave Effect In Section 6.5.1 we described how FR4 and similar dielectrics are composite materials made from a matrix of woven bundles of ﬁberglass embedded in an epoxy resin. Figure 7-13 depicts how it is possible for one leg of the differential signal to be routed between glass bundles and the other to be routed over a glass bundle that effectively gives each trace a unique propagation velocity. The reinforcing ﬁberglass bundles have a dielectric permittivity εr of approximately 6, whereas εr is close to 3 for the epoxy resin in which the bundles are embedded. When a differential pair is aligned with the reinforcing ﬁber matrix of the dielectric in this manner, it causes an imbalance in the differential pair that causes differential-to-common mode (ACCM) conversion. Even if the lines are routed symmetrically, the difference in the dielectric permittivity will make the electrical delay shorter in one line versus the other. Figure 7-13 depicts a cross section of a differential pair that is asymmetric due to the ﬁber-weave effect. DRAWBACKS OF DIFFERENTIAL SIGNALING 311 10 mils V V Glass Epoxy Glass 16.7 mils Figure 7-13 The ﬁber-weave matrix used to construct composite dielectrics such as FR4 can cause differential signals to be asymmetric due to the differences in permittivity of the glass weave (εr ∼ 6) and the epoxy (εr ∼ 3). The glass pitch shown is typical for many glass weaves used in the printed circuit board industry. Unless speciﬁc measures are used to eliminate the alignment of the differential pair with the ﬁber weave, the traditional approach of modeling the transmission line using a uniform dielectric permittivity throughout the dielectric layer is no longer sufﬁcient. The approach must be modiﬁed to comprehend the localized dielectric variation. Figure 7-14 shows the proﬁle of the cross-section geometry description used as 2D ﬁeld solver input for calculating the transmission-line parameters. Simple in principle, this approach is complicated by the fact that the measured effective dielectric permittivity results from the combined effect of multiple dielectrics: resin, ﬁberglass, solder mask, and air. As a starting point, the effective dielectric permittivity of the FR4 material in each region can be calculated using the measured value of the propagation delay as in Figure 6-15, and the actual dielectric differences can be estimated as described in Section 6.5.1. Alternatively, the results from a 2D ﬁeld simulation with a cross section that describes the composite nature of the dielectric can be used if sufﬁcient information is available. Many commercial ﬁeld solvers have this capability, including Ansoft’s Q2D. Nonetheless, once reasonable values for εr1 and εr2 are estimated, the ﬁber-weave effect can be approximated by generating the appropriate transmission-line parameters for each leg of the pair using a cross section similar to that shown in Figure 7-14. The differential-to-common mode (ACCM) conversion due to dielectric permittivity variations can be calculated using the same procedure that was used to derive equation (7-9). The only difference is that the line length (l) stays constant w S Spair w S h er 1 er 2 er 1 er 2 over a bundle between bundles over a bundle between bundles Figure 7-14 Cross section for modeling the ﬁber-weave effect in differential pairs. The dielectric permittivity is adjusted to account for a trace routed over a glass bundle or over an epoxy pool. 312 DIFFERENTIAL SIGNALING ACCM ACCM l = 5 inches 1.0 0.8 0.6 0.4 0.2 l =10 inches 1.0 0.8 0.6 0.4 0.2 5 10 15 20 Frequency, GHz (a) 5 10 15 20 Frequency, GHz (b) Figure 7-15 Waveforms for Example 7-2: (a) length = 5 in. (b) length = 10 in. and the propagation constant (β) changes: ACCM = V (z = l) V (z = 0) + − V (z V (z = = l) 0) = v1+e−jβ1l v1+ + − v2+e−jβ2l v2+ (7-10) where β1 = 2πf √εr,eff1/c, β2 = 2πf √εr,eff2/c, l the differential pair length, and v1+ and v1− are the driving voltages. Example 7-2 Determine the frequency where the differential-to-common mode conversion is 100% for a 10-in. (0.254-m) and a 5-in. (0.178-m) differential pair routed where one leg is over a bundle and one leg is between glass bundles. Use the measured data shown in Figure 6-15. SOLUTION Step 1: Determine the maximum spread in the effective dielectric permittivity. From Figure 6-15 the spread is 0.23: εeff ≈ 3.73 − 3.5 = 0.23 Step 2: Calculate β1 and β2: β1 = 2πf √εr,eff1 c = √ 2πf 3.73 3 × 108 = f · 40.429 × 10−9 rad/s β2 = 2πf √ εr,eff2 c = √ 2πf 3.5 3 × 108 = f · 39.163 × 10−9 rad/s Step 3: Plot the differential-to-common mode conversion using (7-10). The plots are shown in Figure 7-15a and b. When the length is 10 in., the differential-to-common mode conversion is 100% at about 10 and 20 GHz for 5 in. Mitigation of the ﬁber-weave effect was discussed brieﬂy in Section 6.5.2. PROBLEMS 313 REFERENCE Hall, Stephen, Garrett Hall, and James McCall, 2000, High Speed Digital System Design, Wiley-Interscience, New York. PROBLEMS 7-1 Assume that a differential pair is routed so that the ﬁber-weave effect occurs similar to Figure 7-16. Assume that the line is 7 in. long and the metal is copper with a surface roughness of 0.5 µm. If the differential-to-common mode conversion is calculated with losses, how does the answer differ from the case where it is calculated with no losses? Which answer better predicts the percentage of the signal converted to the common mode? w = 5 mils h = 4 mils 10 mils V V Glass Epoxy Glass 16.7 mils Figure 7-16 The permittivity/loss tangent for the glass weave is εr = 6/tan δ = 0.0002 and the epoxy εr = 3/tan δ = 0.025. 7-2 Derive metrics based on line lengths that will help predict the magnitude of the voltage noise on a digital signal when the ﬁber-weave effect is present. Use the results of Problem 7-1 for the derivation. 7-3 How would the differential pair of Problem 7-1 affect the timing of a digital signal? (Hint: Consider the phase noise.) 7-4 Describe three different methods for mitigating differential-to-common mode conversion. 7-5 Derive a formula to predict differential-to-common mode conversion for the case where the impedance of each leg does not match. 7-6 Use modal analysis described in Section 4.4 to calculate the differential waveform expected for a 10-in. transmission line with a cross section of Figure 7-16. 314 DIFFERENTIAL SIGNALING 7-7 Show that the modal voltages correspond to speciﬁc digital states for two signal conductors. Use the following L and C matrices: L= 3.77 × 10−7 5.18 × 10−8 5.18 × 10−8 3.77 × 10−7 H/m C= 8.21 × 10−11 −4.26 × 10−12 −4.26 × 10−12 8.21 × 10−11 F/m 7-8 Show that the modal voltages do not correspond to speciﬁc digital states for three or more signal conductors. Use the following L and C matrices: 3.77 × 10−7 5.17 × 10−8 1.08 × 10−8 L = 5.17 × 10−8 3.77 × 10−7 3.21 × 10−8 H/m 1.08 × 10−8 3.21 × 10−8 3.77 × 10−7 8.21 × 10−11 −4.25 × 10−12 −3.74 × 10−12 C = −4.25 × 10−12 8.21 × 10−11 −4.25 × 10−12 F/m −3.74 × 10−12 −2.04 × 10−12 8.19 × 10−11 7-9 Use modal analysis to calculate the far-end (forward) crosstalk between two single-ended lines using the matrices in Problem 7-7, and compare it to the crosstalk from a differential pair to a single-ended line using the matrices from Problem 7-8. 7-10 Create a SPICE model that proves how routing guidelines that force signal lines to be routed at 10◦ or greater angles with respect to the board edge mitigate the ﬁber-weave effect. (Hint: Review Chapter 6.) 8 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS 8.1 Frequency-domain effects in time-domain simulations 316 8.1.1 Linear and time invariance 316 8.1.2 Time- and frequency-domain equivalencies 317 8.1.3 Frequency spectrum of a digital pulse 321 8.1.4 System response 324 8.1.5 Single-bit (pulse) response 327 8.2 Requirements for a physical channel 331 8.2.1 Causality 331 8.2.2 Passivity 340 8.2.3 Stability 343 References 345 Problems 345 Modern high-speed digital design requires extensive signal integrity simulations to assess the electrical performance of the system prior to fabrication of prototypes. To ensure accurate results from the simulations, careful attention must be given to system component models such as transmission lines, vias, connectors, and packages. For a model to be physically consistent with the laws of nature, certain mathematical constraints must be obeyed to ensure proper balance between signal propagation, energy storage, and losses. For example, the vast majority of engineers designing high-speed digital systems today utilize simpliﬁed modeling techniques that employ frequency-invariant values of dielectric permittivity, loss tangent, and inductance for transmission-line models. A review of Chapters 5 and 6 will remind the reader that transmission-line models have frequency-dependent properties that must be modeled correctly if a Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 315 316 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS realistic response is expected. Such assumptions, although valid at low frequencies or for very short electrical structures, induce amplitude and phase errors for digital data rates faster than 1 to 2 Gb/s. In fact, the computer currently being used to write this chapter was designed using traditional modeling techniques, which assume frequency-invariant electrical properties of dielectrics and conductors. Such assumptions, however, cease to produce valid results for higher data rates. As data rates increase, bandwidth demands skyrocket, form factors shrink, and new phenomenan that were insigniﬁcant in past designs become signiﬁcant. When the correct model assumptions are not used, incorrect solution spaces are determined, lab correlation becomes difﬁcult or impossible, and the design time is increased signiﬁcantly. In this chapter we outline some of the most important techniques used to determine if a physical channel model is adequate for the design at hand. First, the fundamentals of calculating the channel response and methods of implementing frequency-domain phenomena in time-domain simulations are explored. Next, the mathematical requirements for a channel that is consistent with nature are explained, and methodologies for testing these requirements are deﬁned. It is not always necessary for a model to obey these physical rules if the error is small enough; however, it is an important concept for the modern-day digital designer to understand to ensure full comprehension of the modeling assumptions. 8.1 FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS Although high-speed digital design is focused largely on the signal integrity of time-domain digital waveforms, many of the phenomena that heavily inﬂuence the propagation of signals on interconnects are best described in the frequency domain. Examples covered in previous chapters include skin effect resistance, surface roughness, internal inductance, and frequency-dependent dielectric properties. Consequently, it is important for the digital engineer to understand the relationship between a time-domain waveform and its equivalent frequency-domain representation. In fact, many modern buses have component speciﬁcations in terms of frequency-domain parameters because it is the most convenient way to describe the wideband behavior. In this section we outline some of the fundamental principles for linear time-invariant systems that will allow the engineer to translate between the frequency and time domains. 8.1.1 Linear and Time Invariance A system is linear if the relationship between the input and output of the system satisﬁes the superposition property. For example, if the input to the system is the FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 317 sum of two component signals, x(t) = c1x1(t) + c2x2(t) where c1 and c2 are constants, the output of the system will be y(t) = c1y1(t) + c2y2(t) where yn(t) is the output resulting from the input xn(t) (n = 1 and 2). To generalize, a linear system with input will have output x(t) = cnxn(t) n y(t) = cnyn(t) n for any constants cn and where the output yn(t) results from the input xn(t). Time invariance means that whether an input to the system is applied at t = 0 or t = τ , the output will be identical except for a time delay of τ . For example, if the output due to input x(t) is y(t), the output due to input x(t − τ ) is y(t − τ ). Simply put, a time delay at the input should produce a corresponding time delay at the output. 8.1.2 Time- and Frequency-Domain Equivalencies Any time-domain waveform in an LTI system has an equivalent spectrum in the frequency domain. This means that any time-domain signal, such as a digital waveform, can also be described fully in terms of its frequency-domain parameters. This concept is important because it allows the frequency-dependent nature of electromagnetic models to be integrated into time-domain waveforms so that the signal integrity of digital bits propagating on a bus can be analyzed. The relationship between a time-domain signal and its frequency-domain equivalent is described with the Fourier transform: F (ω) = |b| (2π )1−a ∞ f (t)ejbωt dt −∞ f (t) = |b| (2π )1+a ∞ F (ω)e−jbωt dω −∞ (8-1a) (8-1b) 318 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS where (8-1a) translates a time-domain signal into a frequency-domain representation and (8-1b) translates the frequency-domain representation into a time-domain waveform. When using the Fourier transform, it is important to ensure that the correct conventions are used. Different technical and scientiﬁc ﬁelds have different conventions that can confuse the analysis if consistency is not maintained. The conventions are based on the choices for a and b. Some popular conventions are (a = 0, b = −1) for modern physics, (a = 1, b = −1) for systems engineering, (a = −1, b = 1) for classical physics, and (a = 0, b = −2π) for signal processing [Wolfram, 2007]. In this chapter we utilize both the systems engineering and signal processing conventions, depending on the application. The Fourier transform simply relates a time-domain waveform to a frequency response by superimposing an inﬁnite number of sinusoidal functions to reconstruct the original waveform. For example, if the Fourier expansion of a 50% duty cycle square wave is calculated, f (t) = A + 2A 1 sin(n2πf t) 2 π n=1,3,5. . . n (8-2) the wave shape can be approximated by superimposing several individual harmonics, as demonstrated in Figure 8-1. As the harmonic n is increased, the reconstructed waveform better approximates the original. The Fourier transform shown in equation (8-1a) can be used to calculate the real and imaginary parts of the individual sinusoids needed to reconstruct the time-domain waveform, which is known as the spectrum of the waveform. For example, the frequency spectrum of a perfect square pulse, with a width of 2 time units as shown in Figure 8-2a, is calculated using the systems engineering convention (a = 1, b = −1). ∞ F (ω) = f (t)e−jωt dt −∞ n = 1 (frequency of first harmonic) n = 1, 3 n = 1, 3, 5 n = 1, 3, 5, 7 Figure 8-1 Fourier expansion for a 50% duty cycle square wave. FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 319 f (t) 1.0 0.8 0.6 0.4 0.2 F (w) 2.0 1.5 1.0 0.5 −2 −1 1 Time (a) −40 −20 2 20 40 Frequency, w (b) Figure 8-2 (a) Square wave f (t), which is the most primitive approximation of a digital bit; (b) frequency-domain equivalent spectrum of the square wave. 1 = e−jωt dt −1 = j e−jωt t=1 = − j (ejω − e−jω) ω t=−1 ω However, since (ejω − e−jω)/2j = sin ω, the frequency spectrum can be simpliﬁed: F (ω) = − j sin ω 2j ω = 2 sin ω = 2 sinc ω ω (8-3) Equation (8-3) is the frequency spectrum of a square-wave pulse and is plotted in Figure 8-2b. This means that the frequency-domain representation of a square wave takes the form of a sinc function, and vice versa. Note that for a perfect step or pulse, an inﬁnite number of harmonics is required to reproduce the waveform. The centering of the square wave around t = 0 in Figure 8-2a simpliﬁes the mathematics by eliminating the imaginary portion of the transform. For a realistic digital pulse with only positive time values, the spectrum becomes complex. For example, the Fourier transform of a square pulse with a width of 2 time units is shown in Figure 8-3 calculated with system engineering conventions (a = 1, b = −1). Note that the spectrum contains both real and imaginary components. The examples shown in Figures 8-2 and 8-3 demonstrate three very important concepts that link the time-domain waveform to its frequency-dependent equivalent. 1. The spectrum of the time-domain waveform has both positive and negative frequency components. 320 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS f (t ) 1.0 0.8 0.6 0.4 0.2 time −1 1 2 3 (a) F (ω) 2.0 1.5 Re[F (ω)] 1.0 Im[F (ω)] 0.5 −20 −10 −0.5 Frequency, ω 10 20 −1.0 −1.5 (b) Figure 8-3 (a) Square wave f (t) with only positive time values; (b) frequency-domain equivalent spectrum of the square wave. Note that the spectrum has both real and imaginary values. 2. The spectrum of a time-domain waveform with only positive time values is complex. 3. Since time-domain signals that are physically observable have no imaginary component, they must be real. Reality in the time domain is guaranteed when the positive frequencies of the Fourier transform are the complex conjugate of the negative frequencies [LePage, 1980]: F (−ω) = F (ω)∗ (8-4) The behavior of equation (8-4) can be observed in Figure 8-3b. For example, at the frequency ω = −2 rad/s, the spectrum has a value of F (−2) = −0.4 + j 0.74 and at ω = 2 rad/s, the spectrum has a value of F (2) = −0.4 − j 0.74 Since F (−2) = F (2)∗, the spectrum is for a real time-domain waveform. FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 321 f (t ) 1.0 0.8 0.6 0.4 0.2 square trapezoid F (w) 2.0 1.5 1.0 0.5 −2 −1 1 2 −20 −10 10 20 Time Frequency, ω Figure 8-4 Frequency response of a trapezoidal wave compared to an ideal square wave. 8.1.3 Frequency Spectrum of a Digital Pulse Although a square pulse is a ﬁrst-order approximation of a digital bit, a bet- ter approximation is a trapezoid. Figure 8-4 shows the frequency response of a trapezoidal wave compared to an ideal square wave. Notice the shape of the trapezoid’s frequency response is very similar to that of a square wave, except that the magnitude of the harmonics are smaller, especially at high frequencies. This demonstrates another important concept: The rise and fall times of a dig- ital waveform determine the magnitude of the high-frequency harmonics in the frequency-domain spectrum. Figure 8-4 also indicates that the spectrum of a trapezoid can be derived by applying a low-pass ﬁltering function to the spectrum of a square wave. This allows a relationship to be deﬁned between the spectral content of a digital waveform and the rise and fall times. To begin the derivation of this relation- ship, some unique properties of the square-wave spectrum need to be observed. Equation (8-3) shows that the harmonics of a square wave are a sinc func- tion. Taking the limits of the sinc function allows the frequency spectrum to be generalized: sin ω ≈ 1 ω 1/ω when ω is small when ω is large (8-5) The quantity 2/ω is compared to the spectrum calculated with equation (8-3) in Figure 8-5. Note that the quantity 1/ω is equivalent to a slope of 20 dB/decade on a log-log plot: 20 log 1/10ω = −20 dB/decade 1/ω meaning that the spectral content of a square wave will fall off at a rate of −20 dB/decade. One way to approximate the spectrum of a trapezoidal wave is to apply a low-pass ﬁltering function to the harmonics of a square wave until the desired 322 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS F(w) 2.0 1.5 2 1.0 w 0.5 −20 −10 10 20 Frequency, ω Figure 8-5 The spectrum of a square wave ﬁts within the envelope of 2/ω at high frequencies. rise and fall times are realized. The easiest way to apply this ﬁltering is to use a simple low-pass one-pole ﬁltering response, such as an RC network. The step response of a simple one-pole ﬁlter is vout = 1 − e−t/τ vinput (8-6) where vinput is the input voltage to the ﬁlter, vout the output voltage, and τ the time constant. If the rise times are deﬁned with the 10% and 90% voltage magnitude points, the time constant required to degrade a step to a speciﬁc t10–90% can be calculated. The rise time of a unit step after it passes though a one-pole ﬁlter with a time constant of τ is calculated as t10–90% = t90% − t10% = 2.3τ − 0.105τ = 2.195τ (8-7) Note that t10% and t90% are calculated from 0.1 = 1 − e−t10%/τ and 0.9 = 1 − e−t90%/τ . An example is shown in Figure 8-6, where the time constant was calculated assuming an RC network with R = 50 and C = 5 pf. The 3-dB bandwidth of a one-pole ﬁlter is f3dB = 1 2π τ →τ = 1 2πf3dB Solving for τ and substituting into (8-7) produces the well-known relationship between the spectral content of an edge and the rise time: t10–90% = 2.195τ = 2.195 2πf3dB ≈ 0.35 f3dB (8-8) FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 323 input step 1.0 0.8 1 − e−t/t vout/vin 0.6 t = RC = (50 ohms) (5 × 10−12F) = 250 ps 0.4 0.2 t10 − 90% = 2.195t 500 1000 1500 2000 2500 Time, ps Figure 8-6 If a unit step is driven into a single-pole network with time constant τ , the resulting 10 to 90% rise time can be calculated. Equation (8-8) is a good “back of the envelope” calculation that estimates the spectral bandwidth of a digital signal with a rise/fall time of t10−90%. Equations (8-5) and (8-8) can be used to estimate the spectral envelope of a trapezoidal digital signal. The spectrum of a square wave will fall off at −20 dB/decade, as described by (8-5). When the frequency (f3dB) that corresponds to the bandwidth of an edge with rise and fall times of t10−90% is reached, the low-pass ﬁltering function becomes signiﬁcant, and also falls off at a rate of −20 dB/decade. This allows us to draw the approximate spectral envelope of a digital pulse, as shown in Figure 8-7. wRaovlle-osffpoefctsrquumare 20 dB/dec t10 − 90% T Harmonic amplitude waRveolal-nodff eodf gsequsapreectrum 40 dB/dec 1/T 0.35/t10 − 90% Frequency Figure 8-7 Approximation of the spectral envelope of a digital pulse. 324 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS 8.1.4 System Response When a system is linear and time invariant, the output and input can be represented with a convolution, where bold face type indicates a matrix: ∞ y(t) = h(t) ∗ x(t) = h(t − τ )x(τ ) dτ −∞ (8-9a) where h(t) is the system impulse response matrix, and each element hij (t) is the response at port i when Dirac’s delta function (an ideal impulse) is applied at port j with all other inputs set to zero. This is an important concept, because if the impulse response of a system is known, the response of the system with any input, x(τ ), can be determined. When the system impulse response is represented in the frequency domain, it is referred to as the transfer function H (ω): Y(ω) = H(ω)X(ω) (8-9b) where Y(ω) is the frequency response matrix of the of the system output and X(ω) is the frequency response of the system input. It is usually much more convenient to analyze systems using transfer functions in the frequency domain rather than impulse responses in the time domain. However, due to the equivalency between a time-domain waveform and a frequency-domain spectrum, the inverse Fourier transform of the transfer function is the impulse response: h(t) = F−1{H (ω)} (8-10) The relationship of (8-10) is convenient because it is generally much easier to measure the transfer function in the laboratory using a vector network analyzer (VNA) than to measure the impulse response because an ideal impulse is impossible to produce. In fact, in Chapter 9 we describe methods to obtain the impulse response from S-parameters which can be measured in the laboratory using a vector network analyzer. It should be noted that since real laboratory instruments do not provide measured values for negative frequencies, the relationship of equation (8-4) must be used to construct the negative frequency response from the complex conjugate of the positive (measured) frequency response. Another useful property is that convolution in the time domain is equivalent to multiplication in the frequency domain. Generally, it is much simpler to handle the convolution of an input and an impulse response in the frequency domain by multiplying the spectrums and performing an inverse Fourier transform to convert back to the time domain. Example 8-1 Determine the wave shape of an ideal, 2-ns-wide square wave propagating through the low-pass RC ﬁlter shown in Figure 8-8a. Assume that R = 50 , C = 5 pF, and the amplitude of the square wave is 1 V. FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 325 x(t ), volts 1.0 0.8 0.6 0.4 0.2 −2 −1 1 time, ns F{x(t )} 2 (a) R vin vout C H(w) = vout vin X(w) × 10−9 2.0 1.5 1.0 0.5 −10.0 −5.0 5.0 10.0 frequency, ω × 109 H(w) 1.0 0.8 0.6 0.4 0.2 −10.0 −5.0 Re{H(w)} Im{H(w)} 5.0 10.0 frequency, ω × 109 (b) Figure 8-8 (a) Spectrum of a 2-ns-wide square wave for example 8-1; (b) transfer function of an RC ﬁlter with R = 50 and C = 5 pF. SOLUTION Step 1: Calculate the spectrum of the square wave using equation (8-1a). For this example, system engineering conventions (a = 1, b = −1) were chosen. X(ω) = = ∞ x(t)e−jωt dt −∞ ∞ us (t + 1 × 10−9)us (−t + 1 × 10−9)e−jωt dt −∞ 1×10−9 = e−jωt dt −1×10−9 e−j ωt t=1×10−9 =j ω t=−1×10−9 2 sin[ω 1 × 10−9 ] = ω X(ω) is plotted in Figure 8-8a. 326 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS Step 2: Calculate the spectrum of the ﬁlter. The response of the RC ﬁlter can be calculated using basic circuit techniques where the impedance of a capacitor is Zc = 1/j ωC: H (ω) = R 1/j ωC + 1/j ωC = 1 j ωRC +1 The ﬁlter response is plotted in Figure 8-8b. H (ω) represents the transfer function in the frequency domain and the impulse response if converted to the time domain. Step 3: Calculate the response of the output using equation (8-9b), which is plotted in Figure 8-9a: 2 sin ω 1 × 10−9 Y (ω) = X(ω)H (ω) = j ω2RC + ω Y (ω) is the frequency-domain equivalent of the pulse after it has passed though the ﬁlter. This operation is identical to convolving the ﬁlter’s impulse response with the square pulse in the time domain. Step 4: Convert Y (ω) back to the time domain using equation (8-1b) and system engineering conventions: y(t) = F−1{Y (ω)} The inverse Fourier transform was evaluated using Mathematica and plotted in Figure 8-9b. An alternative to the impulse response of a system is the step response. Often, the step response is more conducive to laboratory measurements because time-domain reﬂection (TDR) instruments are widely available and are capable Y(w) × 10−9 2.0 1.5 1.0 0.5 Re{Y(ω)} Im{Y(ω)} F−1{Y(ω)} y (t), volts 1.0 0.8 0.6 0.4 0.2 −10.0 −5.0 5.0 10.0 frequency, ω × 109 (a) −2 −1 1 2 time, ns (b) Figure 8-9 (a) Spectrum of the driving function for example 8-1 convolved with the ﬁlter response; (b) output of the ﬁlter when driven with a square wave. FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 327 of injecting rise times as fast as 9-25 ps into a test structure, which can approximate a step function. The unit step function is the integral of the impulse function (Dirac delta function), t us(τ ) = δ(t) dt −∞ dus(t) = δ(t) dt (8-11a) (8-11b) where δ(t) is the impulse function and us(t) is the unit step function. 8.1.5 Single-Bit (Pulse) Response Perhaps the most useful means to characterize a channel in the time domain is to use the single-bit response, also known as the pulse response. The utility of the pulse response, as opposed to the impulse response, is that it is directly measurable in the laboratory. The pulse response is obtained by driving a system with a waveform that corresponds to a single digital bit of information for the system being designed. The data rate (DR) is deﬁned as the maximum number of bits per seconds the system will support. This means that the maximum data rate is determined by the width of a single bit: DR = 1 tbit (8-12a) where tbit is the width of a single data bit, as shown in Figure 8-10a. Sometimes, tbit is referred to as a unit interval (UI). This means that the maximum fundamental frequency of the digital pulse train, where alternating bits of 1 and 0 are transmitted sequentially, is half the data rate, as shown in Figure 8-10b: ffundamental = 2 1 tbit (8-12b) The pulse response is more practical than the impulse response for two reasons. First, the pulse response will give the engineer an intuitive feeling of how the bus will operate because it represents how the system will respond to an actual waveform that will be propagating on the bus. Second, it can be used to calculate the worst-case eye and the worst-case bit pattern using the peak distortion analysis, which is described in Chapter 13. Analytically, the pulse response is calculated by convolving the input waveform with the system impulse response in the time domain. It is usually more convenient to perform the convolution in the frequency domain, Y (ω) = F{xpulse(t)} · H (ω) (8-13a) 328 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS 1 ∆t bit (a) 1 0 1 2∆t bit (b) Figure 8-10 Deﬁnitions used to calculate (a) the maximum data rate and (b) the fundamental frequency of a bit stream. where Y (ω) is the response in the frequency domain, F{xpulse(t)} is the Fourier transform of the input pulse and H (ω) is the transfer function of the system. The pulse response is then computed by taking the inverse Fourier transform of Y (ω): y(t) = F−1{Y (ω)} (8-13b) Example 8-2 Calculate the 10-Gb/s pulse response of the 0.5-m transmission line calculated in Example 6-4. Assume that the input pulse has rise and fall times of 33 ps and a magnitude of 1 V. SOLUTION Step 1: Calculate the real and imaginary parts of the transmission-line transfer function in the frequency domain. This is done with the equation derived in Example 6-4: vout = vine−αz[cos(−βz) + j sin(−βz)] where γ is deﬁned by equation (6-47) and z is the line length, γ (ω) = α + jβ = (R + j ωL)(G + j ωC) and the frequency-dependent values of R, L, C, and G are as calculated in Example 6-4. Next the transfer function is calculated: H (ω) = vout = e−αz[cos(−βz) + j sin(−βz)] vin FREQUENCY-DOMAIN EFFECTS IN TIME-DOMAIN SIMULATIONS 329 H(ω) 1.0 Re{H (ω)} Im{H(ω)} −1 −0.5 frequency, GHz 0.5 1 −1.0 (a) 2.0 1.5 1.0 0.5 h (t) 2.5 5.0 7.5 10 time, ns (b) Figure 8-11 (a) Transfer function of the transmission line for example 8-2 (b) impulse response calculated with the inverse Fourier transform of the transfer function. Using the values derived in Example 6-4, the real and imaginary parts of the transfer function are plotted in Figure 8-11a. Step 2: Calculate the impulse response by taking the inverse Fourier transform of H (ω) using equation (8-10): h(t) = F−1{H (ω)} The impulse response is plotted in Figure 8-11b. Due to the complexity of the integration, for this example the fast Fourier transform (FFT) was used instead of the Fourier integral to evaluate the spectral response of the waveforms. The FFT is a numerical method used to calculate the Fourier transform of an arbitrary waveform. The FFT was equated by sampling the positive and negative frequency response 2000 times in steps of 100 MHz, which gives a time-domain resolution of t= 1 Nf = 1 2000(100 × 106) = 5 × 10−12 s where f = 100 × 106 Hz and N = 2000. The FFT is described by Press et al. [2005] and is built into many commercial tools, such as Mathematica, Matlab, Mathcad, and even Microsoft Excel. 330 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS 1.0 x(t), volts X(w) 0.8 0.4 0.6 0.2 0.4 0.2 −0.2 50 100 150 200 time, ps −0.4 (a) Re{X(w)} Im{X(w)} 5 10 15 20 25 frequency, GHz (b) Figure 8-12 (a) Waveform driving the transmission line for example 8-2; (b) Fourier transform of the input waveform. Step 3: Calculate the Fourier transform of the input waveform shown in Figure 8-12a: X(ω) = F{xpulse(t)} The Fourier transform of the input waveform is plotted for positive frequencies in Figure 8-12b. Note that it takes the familiar form of a sinc function. Again, the FFT was used in this example. Step 4: Calculate the pulse response. First, the spectrum of the pulse response Y (ω) is calculated using equation (8-9b): Y (ω) = X(ω)H (ω) Finally, the pulse response is calculated by taking the inverse Fourier transform of Y (ω): y(t) = F−1{Y (ω)} The pulse response, which depicts the input waveform after it has propagated from one end of the transmission line to the other, is plotted in Figure 8-13. Note that the pulse arrives at approximately 3.1 ns. This result can be veriﬁed by estimating the total delay from the quasistatic values of the inductance given in Example 6-4 and capacitance using equation (3-107): √ τd = l LC = 0.5 (2.5 × 10−7)(1.5 × 10−10) = 3.06 × 10−9 s The quasistatic approximation will not be identical to the delay calculated with frequency-dependent parameters, but it should be close. Therefore, the delay of the waveform in Figure 8-13 passes the sanity check. It is interesting to note how the transmission line distorts the pulse as it propagates down the transmission line. For example, the amplitude of the waveform REQUIREMENTS FOR A PHYSICAL CHANNEL 331 1.0 x(t ) y(t ) 0.8 y(t), volts 0.6 0.4 0.2 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time, ns Figure 8-13 Pulse response of the transmission line for Example 8-2. in Example 8-2 (Figure 8-13) is degraded signiﬁcantly due to the conductor and dielectric losses. Furthermore, the output pulse is wider than the input pulse because the frequency components of the digital waveform (as calculated with a Fourier transform) will travel at different speeds due to the frequency-dependent dielectric permittivity used to calculate the capacitance. The velocity differences between each harmonic will distort the waveform by spreading it out in time, which is known as dispersion. 8.2 REQUIREMENTS FOR A PHYSICAL CHANNEL In this section we introduce speciﬁc limitations that channel models of a LTI system must obey to remain physically consistent with nature. Speciﬁcally, the conditions of causality, passivity, and stability are described and deﬁned by the appropriate mathematical conditions that must be met to ensure physical behavior. The analysis is restricted to linear and time-invariant electrical networks, which is appropriate for all passive components used in modern bus design, such as transmission lines, vias, packages, connectors, and so on. 8.2.1 Causality One seemingly obvious requirement of a model that obeys the laws of nature is that an output cannot precede its input. In other words, in the real world we live in, an effect cannot precede its cause. This fundamental principle is called causality. Mathematically, a linear time-invariant system is causal only if for every input all the elements of its impulse response hij vanish for t < 0: h(t) = 0 when t < 0 (8-14a) 332 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS 0.5 0.4 Noncausal waveform 0.3 Causal waveform 0.2 volts 0.1 0.0 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Time, ns Figure 8-14 Simulated pulse response for a 20-in. transmission line, showing the outputs of a causal and a noncausal model. Note how the noncausal waveform arrives early. But more generally, if a system has known delay [τ ], the system is causal if h(t) = 0 when t < [τ ] (8-14b) An example of a causal and a noncausal pulse response is shown in Figure 8-14, where a pulse response was simulated by driving a 100-ps-wide digital pulse into a causal and a noncausal 20-in. transmission-line model.† Note that the noncausal pulse response has a component that arrives early, which is a telltale sign of a nonphysical model. In this example, the noncausal transmission-line model was created using frequency-invariant values of the loss tangent and the dielectric permittivity. The causal model was created using the inﬁnite-pole dielectric model presented in Chapter 6. Unfortunately, it is not always easy to ascertain whether a model is causal by observing the pulse response. Furthermore, it is not always obvious if it matters. For example, a very short transmission line may be noncausal, but the error could be so small that it would not affect the ﬁnal result signiﬁcantly. However, if numerous noncausal models are cascaded together to simulate a bus, the causality errors could accumulate and signiﬁcant waveform miscalculations could be realized. Fortunately, it is not very difﬁcult to create a causal model by ensuring that the dielectric and conductor properties outlined in Chapters 5 and 6 are followed. The discussion above prompts an inevitable question: What are the mathematical conditions that specify whether or not a system is causal? To begin, the frequency response of the model is observed. If the Fourier transform of the †Don’t get the pulse response confused with the impulse response [h(t)]. Since pulse responses are more representative of a realistic digital driver, they are often used to analyze interconnects. REQUIREMENTS FOR A PHYSICAL CHANNEL 333 impulse response h(t) is equated, the frequency-domain equivalent H (ω) can be used to test for the conditions of causality: F{h(t)} = H (ω) The properties of the Fourier transform can be used to make some initial conclusions about the causality of a system. 1. If h(t) is real, H (−ω) = H (ω)∗, where ∗ indicates the complex conjugate [LePage, 1980]. Since time-domain waveforms are always real, this is a necessary requirement. 2. If h(t) is real and odd, H (ω) is imaginary and odd [O’Neil, 1991]. 3. If h(t) is real and even, H (ω) is real and even [O’Neil, 1991]. As a reminder, odd and even functions obey the following rules: Let f (t) be a real-valued function of a real variable. Then f is even if f (t) = f (−t) and odd if −f (t) = f (−t) This means that if h(t) is even or odd, there will exist a nonzero value for t < 0, which violates the deﬁnition of causality as expressed in equation (8-14a). Another useful property is that the vector space of all real-valued functions is the direct sum of the subspaces of even and odd functions. In other words, every function can be written uniquely as the sum of an even function and an odd function: f (x) = fe(x) + fo(x) = f (x) + f (−x) 2 + f (x) − f (−x) 2 Therefore, a causal function [where h(t) = 0 for t < 0] must be the sum of an even function he(t) and an odd function ho(t). h(t ) = he (t ) + ho (t ) = 1 2 h(t) + h(−t) + 1 2 h(t) − h(−t) (8-15) Consequently, for h(t) to be causal, h(t) must be composed of both odd and even functions. Therefore, based on Fourier transform properties 2 and 3 above, H (ω) must have both a real and an imaginary part. For an impulse response h(t) to be real and causal , H (ω) must be complex and satisfy the complex-conjugate rule shown in condition 1 above. 334 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS However, just because H (ω) is complex obviously does not guarantee causality. Equation (8-15) shows how the negative time components of the odd and even functions must cancel each other out to ensure a causal response. Since the odd and even values of h(t) are generated from imaginary and real parts of H (ω), respectively, causality requires a speciﬁc relationship between Re[H (ω)] and Im[H (ω)]. To demonstrate the relationship between Re[H (ω)] and Im[H (ω)], consider a simple causal system with the impulse response h(t) = us(t)e−pt , where us(t) is a unit step function with a value of 0 for t ≤ 0 and a value of 1 for t > 0 and p > 0. Following equation (8-15), the odd and even functions of h(t) can be written: he (t ) = 1 2 us (t )e−pt + 1 2 us (−t )ept ho (t ) = 1 2 us (t )e−pt − 1 2 us (−t )ept The functions he(t), ho(t), and h(t) and are plotted in Figure 8-15. Note that when ho(t) = he(t) for t > 0 and ho(t) = −he(t) for t < 0, the output h(t) is zero for t < 0 and therefore causal. This allows the odd function to be written in terms of the even component: ho(t) = sgn(t)he(t) (8-16) 1.0 0.8 0.6 he(t ) 0.4 0.2 −1.0 −0.5 0.0 0.5 −1.0 t 1.0 1.0 0.8 0.6 0.4 0.2 −0.5 −0.2 −0.4 ho(t ) t 0.5 1.0 1.0 0.8 0.6 h(t ) = he(t )+ho(t ) 0.4 0.2 t −1.0 −0.5 0.0 0.5 1.0 Figure 8-15 The odd and even functions are summed to produce the ﬁnal causal impulse response h(t). REQUIREMENTS FOR A PHYSICAL CHANNEL 335 where sgn(t) = 1 and sgn(−t) = −1. Now the causal impulse response can be written in terms of only the even component: h(t) = he(t) + sgn(t)he(t) (8-17) In the frequency domain, H (ω) can also be written in terms of the even function using the property that multiplication in the time domain is the same as convolution in the frequency domain and F{sgn(t)} = −j (1/πω) using the signal processing convention (a = 0, b = −2π) of equation (8-1a): H (ω) = F{he(t)} + (F{sgn(t)} ∗ F{he(t)}) = He(ω) − j 1 πω ∗ He(ω) (8-18) Equation (8-18) can be simpliﬁed by using the deﬁnition of a Hilbert transform, which is the convolution of a function g(ω) and 1/πω: gˆ(ω) = g(ω) ∗ 1 = 1 πω π ∞ g(ω ) −∞ ω − ω dω (8-19) Therefore, equation (8-18) can be written in terms of the Hilbert transform of the even function: Hˆe(ω) = 1 πω ∗ He(ω) H (ω) = He(ω) − j Hˆe(ω) (8-20) where Hˆe(ω) denotes the Hilbert transform of He(ω). Equation (8-20) demonstrates two very important properties of a system that will produce a real, linear, and causal response in the time domain. 1. The imaginary part of the frequency response is determined by the Hilbert transform of the real part. Knowledge of the real part is sufﬁcient to deﬁne the entire function. 2. Causality can be tested by performing the Hilbert transform of the real part and ensuring that it is identical to the imaginary part. Note that the real part of a causal signal can be derived from its imaginary part also. It is interesting to note that the Kramers–Kronig relations mentioned in Chapter 6 are another form of the Hilbert transforms that relate the real and imaginary parts of the complex dielectric permittivity to each other. ε (ω) = 1 + 2 π ∞ 0 ω (ω εr )2 (ω ) − ω2 dω ε (ω) = 2ω π ∞ 0 1 − εr (ω ) (ω )2 − ω2 dω (6-34a) (6-34b) 336 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS As described in Section 6.4.1, there is a speciﬁc relationship between ε (ω) and ε (ω) that must be enforced to ensure realistic behavior. If the dielectric models do not satisfy the Kramers–Kronig relations, the system will be noncausal. Example 8-3 Use the Hilbert transform to prove that the waveform shown in Figure 8-16a is noncausal and the waveform depicted in Figure 8-16b is casual. SOLUTION Step 1a: Calculate the Fourier transform of the noncausal waveform (Figure 8-16a) using the signal processing convention (a = 0, b = −2π) of equation (8-1a): f (t) = us(t + 1)us (−t + 2) F (ω) = cos πω sin 3πω − j sin πω sin 3πω πω πω Step 2a: Calculate the Hilbert transform of the real part of F (ω) and compare it to the imaginary part: Fˆ Re(ω) = Re[F (ω)] ∗ 1 πω = F−1 F cos πω sin 3πω πω F 1 πω = (3 + 2 cos 2π ω) sin(π ω)2 πω Since Im[F (ω)] = −FˆRe(ω), the waveform depicted in Figure 8-16a is noncausal. Of course, simple observation of the waveform for this example is proof of noncausality since it exhibits nonzero values for t < 0. f (t ) 1.0 0.8 0.6 0.4 0.2 f (t ) 1.0 0.8 0.6 0.4 0.2 −1 0 1 2 3 −1 0 1 2 3 Time (a) Time (b) Figure 8-16 (a) Noncausal and (b) causal waveforms for Example 8-3. REQUIREMENTS FOR A PHYSICAL CHANNEL 337 Step 1b: Calculate the Fourier transform of the causal waveform (Figure 8-16b): f (t) = us(t)us (−t + 3) F (ω) = sin 6π ω − j sin(3π ω)2 2π ω πω Step 2b: Calculate the Hilbert transform of the real part of F (ω) and compare it to the imaginary part: Fˆ Re (ω) = Re[F (ω)] ∗ 1 πω = F−1 F sin 6πω 2π ω F 1 πω = sin(3π ω)2 πω Since Im[F (ω)] = −FˆRe(ω), the waveform depicted in Figure 8-16b is causal. Of course, simple observation of the waveform for this example is proof of causality since it exhibits only zero values for t < 0. During bus design, commercial simulation tools are used to implement the models and generate a system response used to evaluate the signal integrity. Each simulator will have its own assumptions, approximations, and numerical methods that may affect causality. The methods detailed above should be used to verify that causal models are being generated. A practical method of judging causality is to observe the rising edge of a pulse response to see if a portion of the signal is arriving early, as demonstrated in Figure 8-14. If a portion of the pulse is arriving early, the system is noncausal. Although this method is less reliable than the rigorous methods deﬁned above, it will provide a general idea if the model is causal. Example 8-4 Use the Hilbert transform to determine the causality of a transmission line with a length of z = 2.0 in. (∼0.05 m) that is terminated perfectly with a relative dielectric permittivity of εr = 4.0, µr = 1, and a decay factor of α = 0.00000001|f |. SOLUTION Step 1: Calculate the transfer function of the transmission line. Since the transmission line is perfectly terminated, no reﬂections will be generated. Therefore, the loss-free voltage wave will behave as described by equation (3-29). However, since this is not a loss-free network, the voltage equation must be multiplied by the decay factor e−αz: vout = vine−αze−βz = vine−γ z Therefore, using equation (2-31) to simplify, the voltage is vout = vine−αz[cos(−βz) + j sin(−βz)] 338 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS From equations (2-45), (2-46), and (2-52), the propagation constant is calculated: f=c λ λ= c √ = 3.0 × 108 f εr 2f β = 2π λ = 4πf 3.0 × 108 and the transfer function is calculated: H (f ) = vout = e−0.00000001|f |z vin cos −4πf 3.0 × 108 z + j sin −4πf 3.0 × 108 z The real and imaginary parts of H (f ) are plotted in Figure 8-17a. H(f ) 1.0 0.5 Re{H(f )} Im{H(f )} −10 −5 f, GHz 5 10 −0.5 (a) 1.0 0.5 − HˆRe(f ) Im{H(f )} −10 −5 −0.5 f, GHz 5 10 (b) Figure 8-17 (a) Transfer function of the transmission line for Example 8-4; (b) Hilbert transform of the real part of H (f ) compared to the imaginary part of H (f ), showing that the model is noncausal. REQUIREMENTS FOR A PHYSICAL CHANNEL 339 Step 2: Calculate the Hilbert transform of the real part of H (f ) compare it to the imaginary part of H (f ). HˆRe(f ) = Re[H (f )] ∗ 1 πf = F−1 F e−0.00000001|f |z cos −4πf 3.0 × 108 F 1 πω The quantity −HˆRe (f ) and the imaginary part of H (f ) are plotted in Figure 8-17b. Since Im[H (f )] = −HˆRe(ω), the transmission-line model is noncausal. The noncausal nature of this model can also be observed by looking at the impulse response, which is calculated by taking the inverse Fourier transform of H (f ): F−1{H (f )} = h(t) and is plotted in Figure 8-18. Note how the impulse response rises prematurely. The theoretical delay of this transmission line can be calculated from the length, the speed of light, and the relative dielectric permittivity. √ εr = 6.66 × 10−9s/m c τd = 169.333 × 10−12s/in. Since the transmission line is 2 in. long, the pulse should arrive at τd = (169.333 × 10−12)(2.0) = 339 ps. It is obvious that the waveform has components arriving much earlier than 339 ps, indicating that the model is noncausal and is not obeying the limits placed on the speed of light. The causality problems associated with the transmission line in Example 8-4 are caused by the assumption of frequency-independent dielectric properties. In Chapter 6 we described numerous dielectric models that show how the dielectric h(t ) × 109 3.5 3.0 2.5 2.0 1.5 1.0 0.5 Theoretical arrival time for a causal model as dictated by the speed of light in the dielectric media −500 0 td 500 Time, ps 1000 Figure 8-18 Noncausal impulse response of the transmission line of Example 8-4. 340 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS permittivity is frequency dependent. Furthermore, the relationship between the dielectric permittivity and the dielectric losses must be maintained; otherwise, energy will propagate when it should be attenuated, inducting noncausal errors. The dielectric models described in Chapter 6 will produce causal responses when used to simulate transmission lines. Unfortunately, many commercially available simulators do not account properly for the frequency dependence of the dielectric. Consequently, the engineer should be very wary of prepackaged transmission-line models to ensure that physical behavior is being observed in the simulations. In reality, performing the Hilbert transform analytically is difﬁcult, and numerical methods have to be used. Also, with bandlimited frequency responses, the Hilbert transform may not be a good check for causality, due to aberrations in the time-domain waveform, which is discussed brieﬂy in Example 9-8. 8.2.2 Passivity A physical system is passive when it is unable to generate energy from within. For example, an n-port network is said to be passive if t vT(τ ) · i(τ ) dτ ≥ 0 −∞ (8-21) where vT(τ ) is the transpose of a matrix containing the port voltages and i(τ ) is a matrix containing the currents. The integral (8-21) represents the cumulative net power absorbed by the system up to time t. In a passive system, this quantity must be positive for all t. A more useful approach for digital designers would be to test the passivity in terms of the incident (ai) and exiting (bi) power waves at each port, as deﬁned in Section 9.3 and shown in Figure 8-19. |ai|2 = power incident to node i |bi|2 = power ﬂow out of node i This approach is particularly useful because in Section 9.3.1 it will be developed into a practical passivity test using S-parameters. Since power must be conserved, the power absorbed by the network (Pa) is equal to the power driven into the network minus the power ﬂowing out: |ai|2 − |bi |2 = Pa (8-22) where Pa ≥ 0 for a passive network. If Pa < 0, the network is generating power and the system would be considered nonpassive. When working in the frequency domain, the power waves ai and bi will be complex. Since power is real, equation (8-22) must be implemented using the REQUIREMENTS FOR A PHYSICAL CHANNEL 341 a1 b1 a3 b3 a2 b2 a4 b4 an−1 bn−1 an bn Figure 8-19 Power waves for an n-port system. Hermitian transpose, where the complex conjugate of each term is taken, and then the matrix is transposed. For example, consider matrix A: A= a − jb e + jf c + jd g − jh where the Hermitian transpose is AH: AH = a + jb c − jd e − jf g + jh This allows equation (8-22) to be written in terms of the power wave matrices that will produce a real value for the power absorbed by the network. A system is passive if aHa − bHb ≥ 0 (8-23) where a is a matrix that contains all the power waves incident to each port and b contains the power waves coming out of each port. The product of a matrix with complex values and its Hermitian transpose produces a real value. Therefore, equation (8-23) simply ensures that the total power absorbed in a network is greater than or equal to zero. In the frequency domain, (8-23) is evaluated at each frequency point. In the time domain, the passivity requirement is essentially the same, except that the function must be integrated: t a(τ )Ta(τ ) − b(τ )Tb(τ ) dτ ≥ 0 −∞ (8-24) where the transpose is used instead of the Hermitian transpose because timedomain signals are always real. Equations (8-21), (8-23), and (8-24) represent 342 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS the cumulative energy absorbed by the system. A system is passive only if the following requirements are met: 1. The system absorbs more energy than it generates. 2. Generation of energy happens after the absorption. A noncausal system that generates energy before it absorbs it would be considered to be nonpassive. Example 8-5 Determine if the transmission-line system depicted in Figure 8-20 is passive. SOLUTION Step 1: Calculate the power waves. The incident wave is a1(t). The reﬂected wave b1(t) is determined from the reﬂection coefﬁcient between the termination resistor Rt and the transmission-line impedance Z0. Using equation (3-102), the reﬂection coefﬁcient is calculated: = Rt − Z0 Rt + Z0 Therefore, the reﬂected wave is calculated as b1(t) = a1(t − 2τd ) which arrives at the input at t = 2τd , which is twice the electrical length of the transmission line as deﬁned in equation (3-107). Step 2: Test for passivity. Since this example is in the time domain, equation (8-24) is used: t [a1(τ )]2 − 2[a1(t − 2τd )]2 dτ ≥ 0 −∞ a1(t) Z0 b1(t) Rt ϳ Z0 Figure 8-20 Power waves for Example 8-5. REQUIREMENTS FOR A PHYSICAL CHANNEL 343 It is easy to see that the integral above will be positive as long as the square of the reﬂection coefﬁcient is less than or equal to 1 ( 2 ≤ 1). This puts a limitation on the termination resistance Rt . 2= Rt − Z0 Rt + Z0 2 ≤1 The only condition where 2 > 1 is when Rt is negative. Therefore, the system is passive as long as Rt ≥ 0. 8.2.3 Stability A model of a passive component such as a transmission line, via, connector, or package must remain stable in order to mimic real-world behavior successfully. For the purposes of this book, stability will be deﬁned so that the output of a system y(t) is stable for all bounded inputs x(t). Using this deﬁnition, the stability of a linear time-invariant system is guaranteed only if all the elements in the impulse response matrix [h(t)] satisfy [Triverio et al., 2007] ∞ |hij (t)| dt < ∞ −∞ (8-25) Example 8-6 Determine the conditions of stability of a mass–spring system similar to that used to derive the frequency dependence of the dielectric permittivity in Section 6.3.2. Assume that the system is subjected to a sharp impulse at time t = τ . The spring equation is given: d2x m dt2 + dx b dt + kx = δ(t − τ ) SOLUTION Step 1: Solve the differential equation and compute the impulse response. For simplicity sake, assume initially that m = 1, b = 3, and k = 2. Since the inverse Laplace transform of the transfer function is the impulse response, it is easiest to solve this problem by converting to the Laplace domain: s2Y (s) + 3sY (s) + 2Y (s) = e−τs where the Laplace transform of the input is L[δ(t − τ )] = e−τs. From equation (8-9b), Y (s) is solved, where X(s) = e−τs: Y (ω) = H (ω)X(ω) Y (s) = H (s)e−τs (8-9b) 344 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS and H (s) is the transfer function, which is determined from the roots of the characteristic equation: H (s) = (s + 1 1)(s + 2) = s 1 + 1 − s 1 + 2 The inverse Laplace transform of H (s) yields the impulse response in the time domain: h(t) = L−1[H (s)] = e−t − e−2t Step 2: Check the stability criteria. The system is stable of the integral of equation (8-25) converges: ∞ |e−t − e−2t | dt = −∞ 0 1 2 e−2(t −τ ) − e−(t−τ ) if − ∞ < t < τ if t ≥ τ At t = ∞, the integral is zero, so the integral converges and the system is stable. The impulse response to the system can be plotted by multiplying h(t) by the unit step function us(t − τ ), which occurs at time t = τ : y(t) = L−1[e−τsH (s)] = h(t − τ )us(t − τ ) Assuming that τ = 1, Figure 8-21a shows the response of this system. Now consider the case where the damping coefﬁcient b is negative, which means that the system must generate energy, which is clearly not possible for a damped mass–spring system. Assume that that m = 1, b = −3, and k = 2, which yields s2Y (s) − 3sY (s) + 2Y (s) = e−τs y (t ) y (t ) 0.25 ×106 3.0 0.20 2.5 0.15 2.0 0.10 1.5 1.0 0.05 0.5 246 Time, t (a) 8 10 246 Time, t (b) 8 10 Figure 8-21 Impulse response of a mass–spring system (Example 8-6) with (a) a positive damping coefﬁcient that is stable and (b) a negative damping coefﬁcient that is divergent and therefore unstable. PROBLEMS 345 The roots of the characteristic equation give the following transfer function: H (s) = (s − 1 1)(s − 2) = s 1 − 2 − s 1 − 1 which yields h(t) = L−1[H (s)] = e2t − et When plugged into equation (8-25) to test for stability, it clearly does not converge and therefore is unstable when the damping coefﬁcient b is negative. Assuming that τ = 1, Figure 8-21b shows the response of this system. During bus design, simulators such as HSPICE are generally used to implement the models and generate a system response used to evaluate the signal integrity. Each simulator will have its own assumptions, approximations, and numerical methods that will affect stability. Consequently, a practical method of testing stability is to generate a pulse response by driving the model with an ideal trapezoid with edge rates approximating those of realistic drivers. The transient simulation should be evaluated for a time period much longer than the time constant of the model. For example, if the propagation delay of a 10-in. bus is 1.5 ns, the reﬂections should have diminished to zero within two or three round trips (3 to 4.5 ns). However, the simulation should be evaluated for at least 10 round trips (30 to 45 ns) to ensure that the system remains stable. REFERENCES LePage, Wilbur P., 1980, Complex Variables and the Laplace Transform, Dover Publications, New York. O’Neil, Peter V., 1991, Advanced Engineering Mathematics, Wadsworth, Belmont, CA. Press, William H., Saul A. Teukolsy, William T. Vetterling, and Brian P. Flannery, 2005, Numerical Recipes in C++, 2nd ed., Cambridge University Press, New York. Triverio, Piero, Stefano Grivet-Talocia, Michel Nakhla, Flavio Canavero, and Ramachan- dra Achar, 2007, Stability, causality and passivity in electrical interconnect models, IEEE Transactions on Advanced Packaging, vol. 30, no. 4, Nov. Wolfram, Stephen, 2007, Mathematica 6.0 On-Line Manual , Mathematica, Champaign, IL. PROBLEMS 8-1 Derive an equation and draw a plot that approximates the spectral bandwidth for a digital waveform with 20 to 80% rise times. 8-2 If a perfect step was driven onto a 50- transmission line terminated in a 5-pF capacitor, what would the rise time be? 346 MATHEMATICAL REQUIREMENTS FOR PHYSICAL CHANNELS 8-3 How many harmonics are necessary to adequately represent the bandwidth of a 10-Gb/s pulse with 25-ps rise and fall times? 8-4 For a causal waveform, derive the real part of the spectrum from the imaginary part. 8-5 For an arbitrary time-domain waveform, devise a test to determine if the waveform is causal. Implement your causality tester in a program such as Mathematica or Matlab. Prove that your causality tester works using ideal square-wave inputs. 8-6 Create a causal and a noncausal microstrip model for a 50- transmission line built on FR4 (εr at 1 GHz = 3.9, tan δ = 0.019) with a 5-mil-wide smooth copper signal conductor (no surface roughness). Evaluate the causality of your models with the tester developed in Problem 8-5. 8-7 Would the noncausal model of Problem 8-6 cause signiﬁcant error for a 10-Gb/s bus design with transmission-line lengths of 15 in.? If so, how did you determine the impact of the causality error? 8-8 For the transmission line described in Problem 8-6, how does surface roughness inﬂuence the causality of the system? 8-9 For the transmission line described in Problem 8-6, what has more of an effect on the causality of the model, the conductor characteristics or the dielectric characteristics? Demonstrate how each affects the causality. 8-10 Is it possible for a real, causal time-domain waveform to have only real frequency components? Show proof of your answer. 8-11 For a loss free system, do the capacitance and inductance matrices need to be frequency dependent to guarantee causality? Show proof of your answer. 9 NETWORK ANALYSIS FOR DIGITAL ENGINEERS 9.1 High-frequency voltage and current waves 349 9.1.1 Input reﬂection into a terminated network 349 9.1.2 Input impedance 353 9.2 Network theory 354 9.2.1 Impedance matrix 355 9.2.2 Scattering matrix 358 9.2.3 ABCD parameters 382 9.2.4 Cascading S-parameters 390 9.2.5 Calibration and de-embedding 395 9.2.6 Changing the reference impedance 399 9.2.7 Multimode S-parameters 400 9.3 Properties of physical S-parameters 406 9.3.1 Passivity 406 9.3.2 Reality 408 9.3.3 Causality 408 9.3.4 Subjective examination of S-parameters 410 References 413 Problems 413 Historically, the techniques used to analyze signal integrity for digital designs required the use of equivalent circuits to describe components such as vias, connectors, sockets, and even transmission lines for low-data-rate applications. At low frequencies where the interconnects between the components of a digital system are small compared to the wavelength of the signal, the circuits can be described with lumped elements using resistors, capacitors, and inductors. In general, circuit theory works well for these types of problems because there is Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 347 348 NETWORK ANALYSIS FOR DIGITAL ENGINEERS negligible phase change in the voltage and current across the circuit. In other words, the signal frequency is low enough so that the electrical delay of the circuit is small compared to the switching rate of the digital waveforms. However, as system data rates increase, the delay of the interconnects becomes signiﬁcant. In fact, in many modern digital designs, such as in high-speed computers, the delay of the system interconnects is so long compared to the width of a single bit of digital information that many bits can be propagating on the bus simultaneously. In this case, the phase changes of the voltage and current across the interconnects are very signiﬁcant. As a result, digital engineers have turned to new techniques for describing and analyzing circuits at high frequencies, called network analysis . Network analysis is a method used traditionally by microwave and radiofrequency engineers to characterize devices such as waveguides, cables, couplers, and antennas. They are used to describe completely the behavior of linear time-invariant systems using only parameters evaluated at the input and output ports. Network analysis is a frequency domain methodology that allows discrete characterization of a linear network at each frequency. The question often arises: Why would a digital engineer use frequency-domain analysis when a digital system uses time-domain pulses? The answer is simple: It is often easier to analyze and characterize systems in the frequency domain. In Chapter 8 we discussed methods to describe a system in terms of its impulse response. Although this is a ﬁne theoretical concept, the problem remains that it is impossible to create or measure a true impulse physically. Furthermore, as described with equation (8-10), the impulse response can be calculated from the transfer function, which is a measurable frequency-domain parameter. Additionally, many of the concepts described in the book, such as skin effect resistance and loss tangents, are best analyzed in the frequency domain. In fact, the validity of a model in the time domain is sometimes judged using frequency-domain techniques, as described in Section 8.2. In short, network analysis is a useful tool for characterizing system interconnects, specifying component performance and creating portable, tool-independent models. Although general network theory is presented in this chapter, the main area of concentration will be on the derivation and use of the scattering matrix, more commonly known as S-parameters. S-parameters are quickly gaining acceptance in the electronics industry for specifying the performance of digital components such as transmission lines, CPU sockets, and connectors. Furthermore, methodologies have developed that allow the use of S-parameters as a portable “black box” model that can be included in the simulation environments of several commercial tools. The problem is that most digitally oriented engineers are not familiar with the concept of network theory or S-parameters because it is traditionally taught in microwave, electromagnetic interference, or radio engineering curricula. In this chapter we discuss the applications and usage models of network analysis that are most applicable to the design and validation of high-speed digital systems. HIGH-FREQUENCY VOLTAGE AND CURRENT WAVES 349 9.1 HIGH-FREQUENCY VOLTAGE AND CURRENT WAVES As a prerequisite to describing network theory, it is required to understand how voltage and current waves propagating on an interconnect will interact with different loads. Many of these concepts were covered partially in Chapter 3 when lattice diagrams and the reﬂection coefﬁcient at an impedance junction were discussed. In this chapter we build on those concepts to calculate the reﬂection coefﬁcient looking into a network, such as a transmission line terminated in a load that is not equal to the characteristic impedance. Similarly, the impedance looking into a terminated network is also calculated. These concepts are important for the development of network theory. 9.1.1 Input Reﬂection into a Terminated Network The reﬂection coefﬁcient looking into a network with a ﬁnite electrical length is different from the reﬂection coefﬁcient looking into an impedance junction because it has a phase component that will change with electrical length and frequency. Equation (3-102) from Section 3.5.1 deﬁnes the reﬂection coefﬁcient looking into an impedance junction: ≡ vr = Z02 − Z01 vi Z02 + Z01 (3-102) where vr and vi are the reﬂected and incident voltage values, respectively. In the case of equation (3-102) the reﬂection occurs immediately, so there is zero phase delay between the incident and reﬂected waves. However, consider the case shown in Figure 9-1, where there is a signiﬁcant distance between the point where the reﬂection is being evaluated and the impedance discontinuity. The reﬂection coefﬁcient at the load, (z = 0), can be calculated with equation (3-102): 0 = Rl Rl − + Z0 Z0 However, consider (z = −l), which is the reﬂection coefﬁcient looking into the input of the network. After a signal is driven onto the network, the reﬂection will not arrive back at the input until the signal propagates down the network, reﬂects off the impedance discontinuity at z = 0 (deﬁned by 0), and propagates back to the source. Depending on when the reﬂections arrive at the receiver, the incident and reﬂected waves will combine at speciﬁc frequencies and interact either constructively or destructively. If the incident and reﬂected waves interact destructively, the reﬂection coefﬁcient will be minimized (and vice versa). This means that the reﬂection coefﬁcient looking into the network will be inﬂuenced by propagation delay, characteristic impedance, termination impedance, length, and frequency . 350 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Incident wave V R = Z0 Reflected wave Z0 Rl z = −I z=0 Γ (z = −I) Γ (z = 0) Figure 9-1 The reﬂection looking into a network is dependent on the distance between the point where the reﬂection is being evaluated and the impedance discontinuity. The reﬂection coefﬁcient looking into a network can be derived from equations (6-49) and (3-102): v(z) = v(z)+e−γ z + v(z)−eγ z Let v(z)+ = vi and v(z)− = vr . Then v(z) = vi e−γ z + vr eγ z = vi(e−γ z + 0eγ z) = vie−γ z[1 + (z)] (9-1) where (z) ≡ vr eγ z vi e−γ z = 0e2γ z = Rl Rl − + Z0 Z0 e2γ z Equation (9-1) describes the reﬂection coefﬁcient looking into a transmission line with characteristic impedance Z0, length z, termination impedance Rl, and propagation constant γ . In Section 3.5 the concept of lattice diagrams was introduced to demonstrate how time-domain signals propagate on transmission lines. An important concept demonstrated was that the period of transmission-line “ringing” was dependent on the electrical length of the line. In frequency-domain analysis, the same principles apply; however, it is more useful to calculate the frequency when the reﬂection coefﬁcient is either maximum or minimum, which is dependent on both the electrical length of the structure and the frequency of the input stimulus. To demonstrate this concept, consider the case of a loss-free transmission line as HIGH-FREQUENCY VOLTAGE AND CURRENT WAVES 351 deﬁned by √ √ v(z) = v(z)+e−jzw LC + v(z)−ejzw LC √ √ γ = α + jβ = 0 + j ω LC → β = ω LC (3-29) (3-30) Beginning with equation (9-1), the reﬂection coefﬁcient looking into a terminated transmission line with a length of −l is calculated and expanded using equation (2-31): cos φ + j sin φ = ejφ 0e2γ (−l) = = √ 0ej 2β(−l) = 0e−j 2ωl LC √ √ 0(cos 4πf l LC − j sin 4πf l LC) (2-31) (9-2) where a negative length convention is chosen for convenience. Since the real and imaginary parts of (9-2) are periodic, the frequencies where the function is only real or imaginary can be calculated. The reﬂection looking in√to a nonperfectly terminated transmission line is purely imaginary when 4πf l LC = nπ/2 for odd n because it will force the cosine term to be zero. Solving for the frequency where the reﬂection is imaginary produces f (imaginary) = √n 8l LC n=1,3,5,. . . (9-3a) Similarly, the frequencies where equation (9-2) is pu√rely real can be calculated for the conditions where the sine term is zero: 4πf l LC = nπ. The frequency where the reﬂection is real is shown by f (real) = √n 4l LC n=1,2,3,. . . (9-3b) Equations (9-3) demonstrate that when the real part of the input reﬂection is zero, the imaginary portion is maximum, and vice versa, as plotted in Figure 9-2. This periodic behavior can be used to extract out useful information about the device under test. Example 9-1 Calculate the propagation delay, characteristic impedance, and the dielectric permittivity from the input reﬂection coefﬁcient plot in Figure 9-2, assuming a circuit similar to Figure 9-1 with a length of 2.28 in. (0.058 m) and termination impedance Rl of 50 . SOLUTION Step 1: To begin, consider the real portion of Figure 9-2. According to equation (9-3b), the frequencies will be purely real at multiples of n. Examination of 352 0.2 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Re{Γ(z = −l)} Im{Γ(z = −l)} 0.1 Γ(z = −l) Frequency, GHz 0.5 1.0 1.5 2.0 2.5 3.0 −0.1 0.775 GHz −0.2 2.29 GHz Figure 9-2 When the real part of the reﬂection coefﬁcient looking into a network is maximum, the imaginary part is zero. Figure 9-2 shows that the period of the real portion can be measured at 0.775 and 2.29 GHz. Since these peaks are the ﬁrst and the third frequencies, where the reﬂection coefﬁcient is purely real, they correspond to n = 1 and n = 3. Using equation (9-3b) and subtracting produces an equation in terms of the frequency difference between positive real peaks and the electrical delay: fn=3 − fn=1 = √3 − √1 = √1 = 2.29 GHz − 0.775 GHz 4l LC 4l LC 2l LC = 1.515 GHz √ where τd = l LC, from equation (3-107). Therefore, the propagation delta can be calculated: τd = √ l LC = 1 2(fn=3 − fn=1) = 1 2(1.515 GHz) = 330 ps Note that the propagation delay calculated using this technique is the average value between 0.775 and 2.29 GHz. Due to the frequency dependence of the dielectric permittivity described in Chapter 6, the actual value actually changes across the bandwidth. Step 2: To calculate the dielectric permittivity, the propagation delay must ﬁrst be translated into a velocity: νp = √1 l l LC = 0.058 330 ps = 1.75 × 108 m/s HIGH-FREQUENCY VOLTAGE AND CURRENT WAVES 353 Next, equation (2-52) with µr = 1 is used to calculate the relative dielectric permittivity. νp = 3.0 × 108 √ εr = 1.75 × 108 εr = 2.9 Step 3: Calculate the characteristic impedance using the peak values of the reﬂection coefﬁcient. When the imaginary term is zero, the real term will peak because the cosine term of equation (9-2) will equal 1 at frequencies predicted by (9-3b). Therefore, the easiest way to calculate the characteristic impedance is to use the value of the reﬂection coefﬁcient measured at a real peak. The ﬁrst real peak at 0.775 GHz shows a maximum reﬂection coefﬁcient of 0.2. The characteristic impedance can be calculated by setting the reﬂection coefﬁcient equal to equation (9-2) at 0.775 GHz and solving for Z0. cos 4πf √ l LC = cos[4π(0.775 × 109)(330 × 10−12)] ≈ −1 sin 4πf √ l LC = sin[4π(0.775 × 109)(330 × 10−12)] ≈ 0 0.2 = √ 0(cos 4πf l LC − j √ sin 4πf l LC) = Rl Rl − + Z0 [−1] Z0 = − 50 − Z0 50 + Z0 → Z0 = 75 Step 1 in Example 9-1 demonstrates a very useful relationship between the periodicity of the input reﬂection coefﬁcient looking into a network and the propagation del√ay. If the distance between peaks (fn=3 − fn=1) is represented as f and τd = l LC, the time delay can be calculated using τd = 2 1 f (9-4) The utility of equation (9-4) will become apparent when analyzing S-parameters in Section 9.2.2. In summary, the reﬂection coefﬁcient looking into a network is dependent on (1) the impedance discontinuities, (2) the frequency of the stimulus, and (3) the electrical length between discontinuities. 9.1.2 Input Impedance Not surprisingly, if the reﬂection coefﬁcient looking into a network is a function of length, impedance discontinuities, and frequency, the input impedance looking into the network must be a function of the same variables. Following a procedure similar to that used to derive equation (9-1), the impedance looking into a 354 NETWORK ANALYSIS FOR DIGITAL ENGINEERS 110 100 90 Zin 80 70 60 50 0.5 1.0 1.5 2.0 2.5 3.0 Frequency, GHz 0.775 GHz 2.29 GHz Figure 9-3 Impedance looking into a 75- 2.28-in. transmission line terminated in 50 . Z in varies signiﬁcantly with frequency due to the constructive and destructive combinations of the input stimulus and the reﬂected waves. transmission line of length z terminated with Rl, as depicted in Figure 9-1, is easily derived, where (z)is equated with equation (9-1): v(z) = vie−γ z + vr eγ z = vi(e−γ z + 0eγ z) = vie−γ z[1 + (z)] i(z) = 1 Z0 (vi e−γ z − vr eγ z) = 1 Z0 (vi e−γ z [1 − (z)]) Zin = Z(z) = v(z) i(z) = vi e−γ z[1 + 1/Z0(vi e−γ z[1 (z)] − (z)]) = 1 Z0 1 + − (z) (z) (9-5) Figure 9-3 shows the input impedance as a function of frequency for the terminated transmission line in Example 9-1. Note that although the characteristic impedance and the termination value are constant, the input impedance varies signiﬁcantly with frequency, due to the constructive and destructive combinations of the input stimulus and the reﬂected waves. At frequencies where the reﬂection is real and the imaginary term is zero, the reﬂected wave is aligned with the incident wave, causing the input impedance to peak. 9.2 NETWORK THEORY Network theory is based on the property that a linear time-invariant system can be characterized completely by parameters evaluated only at the input and output ports, without regard to the contents of the system. This allows the behavior of a system to be described fully in a frequency-dependent matrix that relates input stimuli to the output responses of the system. Networks can have any number of ports; however, consideration of a two-port network is sufﬁcient to explain NETWORK THEORY 355 the theory. The discussion begins with the most intuitive method of describing a network, which is the impedance matrix. 9.2.1 Impedance Matrix Consider the two-port network depicted in Figure 9-4. If the voltage and current are measured at the input and output ports, the system can be characterized in terms of its impedance matrix. The impedance from port 1 to port 2 is calculated by measuring the open-circuit voltage at port 2 when current is injected into port 1: Z21 = vopen,port2 iport1 iport2 =0 (9-6a) Similarly, the input impedance looking into port 1 is measured by injecting current into port 1 and measuring the voltage at port 1: Z11 = vopen,port1 iport1 iport2 =0 (9-6b) Using the deﬁnition shown in equations (9-6a) and (9-6b), a set of linear equations can be written to describe the network in terms of its port impedances: v1 = Z11i1 + Z12i2 v2 = Z21i1 + Z22i2 which is expressed more efﬁciently in matrix form: v1 v2 = Z11 Z21 Z12 Z22 · i1 i2 (9-7) More generally, the elements of an impedance matrix are described in equation (9-8) for an arbitrary number of ports, Zij = vi ij = open-circuit voltage measured at port current injected into port j i (9-8) i1 Port 1 v1+− Two Port Network i2 +−v2 Port 2 Figure 9-4 Two-port network used to generate the impedance matrix. 356 NETWORK ANALYSIS FOR DIGITAL ENGINEERS with ik = 0 for all k = j . If the impedance matrix of a system is known, the response of the system can be predicted for any input. Example 9-2 Calculate the impedance matrix at 1 GHz for the circuit shown in Figure 9-5a, where R1 = 50 , R2 = 50 , and C = 5 pF. SOLUTION Step 1: Calculate the input impedance (Z11) by injecting a current into port 1 and measuring the voltage at port 1, as shown in Figure 9-5b. The impedance of the capacitor at 1 GHz is Zc = 1/j 2πf C = 31.8 . v1 = i1(R1 + Zc) = i1(50 + 31.8) Z11 = v1 i1 = R1 + Zc = 81.8 Step 2: Calculate the through impedance (Z21) by injecting a current into port 1 and measuring the voltage at port 2, as shown in Figure 9-5c. v2 = v1 Zc R1 + Zc = i1(Zc + R1) Zc R1 + Zc = i1Zc Z21 = v2 i1 = Zc = 31.8 Step 3: Construct the impedance matrix at 1 GHz. Since the circuit is symmetrical, Z12 = Z21 and Z22 = Z11. 81.8 31.8 Z= 31.8 81.8 R1 R2 C (a) R1 R2 + i1 v1 C − (b) R1 R1 + i1 v1 − + C v2 − (c) Figure 9-5 (a) General circuit to be analyzed in Example 9-2; (b) calculating Z 11; (c) calculating Z 21. NETWORK THEORY 357 The admittance matrix is very similar to the impedance matrix except that it is characterized with short-circuit currents instead of open-circuit voltages. The admittance matrix is the inverse of the impedance matrix: Y = Z−1 (9-9) Although the impedance and admittance matrices are intuitive and relatively easy to understand, they have severe drawbacks for the high-frequency characterization of interconnects. The problem is that at high frequencies, open and short circuits are very difﬁcult to realize. Open circuits invariably have ﬁnite capacitances, and short circuits have inductance that can signiﬁcantly affect the accuracy of the measurements. Consequently, as a measurement technique, these methods are applicable only for low frequencies. Example 9-3 Calculate Z21 for the circuit analyzed in Example 9-2, assuming that the current was injected into port 1 and the voltage was measured at port 2 with a probe that has a capacitance of C = 0.3 pF, according to the instrument speciﬁcations. SOLUTION Step 1: Draw the equivalent circuit of the circuit and the probes, as depicted in Figure 9-6. From Example 9-2, R1 = 50 , R2 = 50 , C = 5 pF, and Zc = 1/j 2πf C = 31.8 . The impedance of the probe at 1 GHz is Zprobe = 2π(1 × 1 109)(0.3 × 10−12) = 530.8 Step 2: Solve the circuit for Z11 and Z21: Z11 = v1 i1 = (Zc||(R2 + Zprobe) + R1)||Zprobe = 69.6 Z21 = v2 i1 = v1 Zc Zc + R1 Zprobe = 24.7 Zprobe + R2 where v1 = Z11i1. Since the circuit is symmetrical, Z12 = Z21 and Z22 = Z11. + i1 v1 − R1 Cprobe R2 C + Cprobe v2 − Figure 9-6 Equivalent circuit used for Example 9-3 showing probe capacitance. 358 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Step 3: Compare the matrix measured with a realistic probe to the ideal case: 81.8 31.8 [Z]ideal = 31.8 81.8 69.6 24.7 [Z]probed = 24.7 69.6 Note that even for a relatively low capacitance value of the probe, it makes a signiﬁcant difference in the impedance matrix. This is because the voltage measured at port 2 is not across an open circuit; it is across a small capacitance. The fact is that at high frequencies, true short and open circuits do not really exist for small dimensions. There will always be a certain amount of parasitic capacitance or inductance with ﬁnite impedance. 9.2.2 Scattering Matrix In the previous section we discussed some of the problems associated with measuring high-frequency voltage and current waves, where short and open circuits do not practically exist due to parasitic inductance and capacitance values. The scattering matrix is the most common form of network parameters used in high-speed digital design. Instead of measuring voltages and currents at the ports, it relates the power waves incident on each port to those reﬂected from the ports. The scattering matrix, more commonly known as S-parameters, can be measured in the laboratory using a vector network analyzer (VNA). Once the S-parameters are known, conversion to other matrix parameters, such as the impedance or admittance matrices, can be done algebraically. Traditionally, S-parameters have been a tool used primarily by microwave and RF engineers to design antennas, waveguides, and other high-frequency narrowband applications. Higher speed data transmission on system buses is causing a convergence of two disciplines in industry today: microwave and digital engineering. Microwave engineers tend to concentrate mostly on high-frequency multi-GHz waveguides, resonators, and couplers, whereas digital engineers concentrate on binary signaling. Over the past decade, S-parameters have become much more common in the world of digital design and are often used for the dissemination of electrical models to design teams. In fact, most contemporary software suites used to design modern systems have built-in capabilities to handle S-parameter models. S-parameters can be very confusing, as unlike the impedance matrix, they are not intuitive. In this chapter we focus on the most important aspects of S-parameters that the digital engineer needs for signal integrity analysis of modern, high-speed buses. Emphasis is given to both the theoretical development of S-parameters and intuitive techniques that will allow the engineer to interpret data quickly, share models, and estimate channel performance. Deﬁnition Consider the two-port network in Figure 9-7. If a power wave is injected into port 1, the power must either be reﬂected back toward port 1, propagate through the network to port 2, or be dissipated as thermal or radiation NETWORK THEORY 359 a1 Port 1 b1 R Two Port Network a2 b1 R Port 2 Figure 9-7 Two-port model used to deﬁne S -parameters. losses. Similar to the derivation of the Poyting vector in Section 2.6, the power balance equation can be expressed as Pout = Pinput − Ploss − Pradiated (9-10) where Pinput is the total power injected into all ports, Pout is the total power ﬂowing out of all ports, Ploss is the power dissipated through ohmic losses (skin effect, loss tangent, etc.), and Pradiated is the power radiated into free space. The incident and reﬂected power waves are calculated from voltage and current waves. The voltage waves are obtained from equation (6-49), v(z) = v(z)+e−γ z + v(z)−eγ z and the current wave is derived by dividing the voltage wave by the characteristic impedance of the structure, i(z) = v(z)+e−γ z − v(z)−eγ z Z0 Z0 where v(z)+ is the voltage traveling in the +z-direction and v(z)− is the voltage traveling in the −z-direction. The power wave propagating on the network is calculated by multiplying the current and voltage waves: P (z) = [v(z)+e−γ z]2 + [v(z)−eγ z]2 Z0 Z0 (9-11) If it is deﬁned so that at port j , z = 0, the voltage at a port can be calculated where v(0) = vi and i(0) = ii, which are the incident voltage and current, respectively: v+ = 1 2 (vi + Rii ) v− = 1 2 (vi − Rii ) (9-12) where R is the termination values at the ports of the network. Since equation (9-10) says that the power must be balanced, the amount of power delivered to the network or radiated is deﬁned simply as the input power minus the output power: Pinput − Pout = Ploss + Pradiated 360 NETWORK ANALYSIS FOR DIGITAL ENGINEERS From equation (9-12), the power going into a node can be calculated using the power relation P = v2/R: Pinput = (v+)2 R (9-13a) and the power coming out can be calculated as (v−)2 Pout = R meaning that the power delivered to the network is calculated as (9-13b) Pinput − Pout = Pnetwork (9-13c) Equations (9-13) allow the deﬁnition of terms that describe the power wave propagating into port j and the power wave propagating out of port j : v(z)+ aj = √ = R v(z)− bj = √ = R Pinput Pout (9-14a) (9-14b) where aj is the square root of the power propagating into port j and bj is the square root of the power propagating out of port j , as shown in Figure 9-7 for a two-port network. Equations (9-14a) and (9-14b) are known as the scattering coefﬁcients. Since they are deﬁned in terms of the square root of power, ratios of the scattering coefﬁcients simplify into ratios of voltage as long as the termination impedance of each port (R) is the same. S-parameters are derived from the ratios of scattering coefﬁcients. For example, referring to Figure 9-7, the term S11 is calculated by the root of the reﬂected and incident power ratio at port 1, which is written in terms of the scattering coefﬁcients: S11 = b1 a1 (9-15a) Similarly, the term S21 is calculated by injecting power into port 1 and measuring at port 2. S21 = b2 a1 (9-15b) Using the deﬁnition shown in equations (9-15a) and (9-15b), a set of linear equations can be written to describe the network in terms of its scattering coefﬁcients: b1 = S11a1 + S12a2 b2 = S21a1 + S22a2 NETWORK THEORY 361 which is more efﬁciently expressed in matrix form as b1 b2 = S11 S21 S12 S22 · a1 a2 (9-16) More generally, the elements of a scattering matrix are described in equation (9-17) for an arbitrary number of ports, Sij = bi aj = power measured at port i power injected into port j (9-17) and an arbitrary-sized scattering matrix takes the form b = Sa (9-18) If the scattering matrix of a system is known, the response of the system can be predicted for any input. Return Loss Consider the circuit depicted in Figure 9-8. In this case there will be no reﬂections from the far end because the line is perfectly terminated with the characteristic impedance. However, the source impedance is not equal to the characteristic impedance, indicating that a portion of the power wave incident to port 1 will be reﬂected. This scenario allows the simplest deﬁnition of S11, which is simply the reﬂection coefﬁcient between the source resistor and the impedance of the transmission line. Note that a2 = 0 because there is no source at port 2. S11 = b1 a1 = a2=0 v1− v1+ √ /√R /R = v1− v1+ = vreﬂected vincident = 0 = Z0 Z0 − + R R (9-19) The term S11 is often referred to as the return loss, because it is a measure of power reﬂected, or returned to the source. The calculation of S11 becomes more complex when the far end of the network is not perfectly terminated because the reﬂection arriving at the source will have V R ∼ Z0 RI = Z0 S11 = Z0 − R Z0 + R Γ=0 Figure 9-8 Return loss for the special case when the network is perfectly terminated in its characteristic impedance. 362 NETWORK ANALYSIS FOR DIGITAL ENGINEERS V R ∼ Z0 R Zin = Z11 (a) V ∼R Zin S11 = Zin − R Zin + R (b) Figure 9-9 Return loss for the more general case where the network is not perfectly terminated in its characteristic impedance: (a) input impedance looking into the network; (b) equivalent circuit for the return loss. contributions from both the impedance discontinuity at the source and at the far-end termination. This means that the input impedance looking into the network from the source will be dependent on frequency. The return loss for a nonperfectly terminated structure such as the circuit shown in Figure 9-9a is calculated as S11(f ) = Zin(f ) Zin(f ) − + R R (9-20) where Zin(f ) is calculated for a transmission line with equation (9-5) for the general case. An intuitive understanding of the return loss can be achieved by constructing a simple equivalent circuit as shown in Figure 9-9b. Since Zin (or Z11) is dependent on both the propagating delay and the impedance of the structure, both can be calculated from S11, as demonstrated in Example 9-4. Example 9-4 Using the measured return loss of a transmission line shown in Figure 9-10, calculate the characteristic impedance and the propagation delay. Assume that the source and termination impedance values are 50 . SOLUTION Step 1: Determine the propagation delay. This is easily calculated from the periodic behavior of S11 using equation (9-4). The distance between peaks is NETWORK THEORY 363 Real Imaginary 0.3 0.2 0.1 S11 0.5 1.0 1.5 2.0 2.5 3.0 −0.1 −0.2 Frequency, GHz Figure 9-10 Return loss (S 11) for Example 9-4. used to calculate f . τd = 1 2f = 2(2.29 1 GHz − 0.775 GHz) = 330 ps Note that τd is the average propagation delay over the frequency range. Step 2: Calculate the input impedance using equation (9-20). When the return loss (S11) is maximum, the imaginary part is zero. Therefore, it is convenient to measure S11 at a peak to simplify the analysis: S11(0.775 GHz) = 0.38 = Zin Zin − + 50 50 Zin = 111.3 Step 3: Determine the polarity of the phase term. The polarity of the phase term e2γ z in equation (9-1) must be determined so that the input impedance can be properly related to the characteristic impedance. Since S11 is being evaluated at a peak, the imaginary term of e2γ z is zero, so the real part will either be 1 or −1. Since the propagation delay has been calculated, the polarity of the phase term can be evaluated using the real part of equation (9-2): Re(e−j √ 2ωl LC ) = cos 4πf √ l LC √ where τd = l LC = 330 ps and cos 4πf τd = cos[4π(0.775 × 109)(330 × 10−12)] ≈ −1 364 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Step 4: Calculate the characteristic impedance from equation (9-5): Zin = Z0 1 1 + − (z) (z) Since the phase term calculated in step 3 is −1, the input reﬂection coefﬁcient at 0.775 GHz is calculated from equation (9-1) with e2γ z = −1: (z) = Rl − Z0 (−1) Rl + Z0 Zin = Z0 [1 [1 + − (50 (50 − − Z0)/(50 Z0)/(50 + + Z0)](−1) Z0)](−1) = 111.3 = Z02 50 Z0 = 74.6 Insertion Loss When power is injected into port 1 and measured at port 2, the square root of the power ratio reduces to a voltage ratio. S21, the measure of the power transmitted from port 1 to port 2, is called the insertion loss: S21 = b2 a1 a2 =0 = vv21−+//√√RR = v2− v1+ = vtransmitted vincident (9-21) which is shown in Figure 9-11 for a transmission line. In digital system design, the insertion loss is the most commonly used parameter of the scattering matrix because it is a measure of both delay and amplitude as seen at the receiving agent. Vincident Vtransmitted V R Z0 R ∼ = S21 Vtransmitted Vincident Figure 9-11 Insertion loss (S 21) of a transmission line. NETWORK THEORY 365 The frequency-dependent delay can be calculated from the phase of the inser- tion loss, τp = θS21 360◦f (9-22) where θS21 = arctan Im(S21) Re(S21) (9-23) Note that equation (9-22), called the phase delay, has a slightly different meaning than that of the propagation delay of a digital pulse on a transmission line. Since a digital pulse is composed of numerous harmonics, and realistic dielectrics have values that vary with frequency, the propagation delay of a time-domain pulse will have numerous frequency components, each propagating with a unique phase velocity. The propagation delay of a digital pulse can be thought of as a group of harmonics propagating simultaneously and is sometimes called the group delay. The phase delay, as described with (9-22), is associated with only a single frequency. Figure 9-12a shows an example of the phase of S21 for a 1-in. transmission-line model constructed with a realistic frequency-dependent dielectric, as described in Chapter 6. To calculate the phase delay, the phase must be unwrapped, as shown in the ﬁgure. Figure 9-12b is the phase delay calculated from the unwrapped phase using equation (9-22). Note the frequency-dependent nature of the delay, which is required for a causal transmission-line model. For a loss-free network, the total power exiting the N ports must equal the total incident power. This means that for a two-port loss-free network, the power transmitted from port 1 to port 2 plus the power reﬂected from port 1 must be conserved: Preﬂected,port1 + Ptransmitted,port1−port2 = 1 Pincident Pincident (9-24) Essentially, equation (9-24) says that if the power is not transmitted from port 1 to port 2, it must be reﬂected. This allows us to write an equation that relates the Phase S21, degrees Phase Delay(tp), ps\inch 168 50 0 166 −50 −100 −150 −200 Unwrapped phase 164 162 −250 0.0 1.0 2.0 3.0 4.0 5.0 Frequency, GHz (a) 0.0 1.0 2.0 3.0 4.0 5.0 Frequency, GHz (b) Figure 9-12 (a) Phase of the insertion loss (S 21) for a 1-in. causal transmission line model; (b) phase delay. 366 1.0 S21 NETWORK ANALYSIS FOR DIGITAL ENGINEERS S11,S21 0.8 0.6 0.4 S11 0.2 td ≈ 1 2∆f 0.0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Frequency, GHz Figure 9-13 Insertion and return loss for a 2-in. loss-free transmission line. insertion loss and the return loss for a loss-free system†: S11S1∗1 + S21S2∗1 = 1 (9-25) where Si∗j represents the complex conjugate of the term. Since Sij is the square root of power measured at port i and injected at port j , the power ratio is Si2j . However, since power is a real quantity, the squaring function is replaced with the complex conjugate to ensure that the imaginary part is zero. Figure 9-13 shows the insertion and return loss for the loss-free transmission line used in Example 9-4 terminated with 50- reference impedances. Note that when the return loss (S11) peaks, the insertion loss (S21) dips, as described by (9-25). Since the imaginary part is zero when the terms peak and the real part is zero when the terms dip, it is easy to show that the power is conserved when the insertion loss is maximum and the return loss is minimum simply by squaring the terms, which is equivalent to the complex conjugate for these conditions. For example, at the ﬁrst S11 peak, S121 + S221 = (0.384)2 + (0.923)2 = 1 which proves that power is conserved. For the realistic case where the transmission line is lossy, an extra term, Ploss, is added to the power balance equation to account for conductor, dielectric, and †The terms insertion loss and return loss should not be confused with ohmic or radiation losses. When a system is said to be loss free, it means that the power lost through heat, such as conductor and dielectric losses, does not exist. It also means that no energy is being lost through radiation into free space. NETWORK THEORY 367 radiation losses: Preﬂected,port1 + Ptransmitted,port1−port2 + Ploss = 1 Pincident Pincident Pincident (9-26) Therefore, equation (9-25) can be rewritten to account for ﬁnite power losses: S11S1∗1 + S21S2∗1 = 1 − Ploss Pincident (9-27) Figure 9-14 shows an example of the insertion and return losses for a lossy transmission line. The power absorbed by the network can be calculated using (9-27). At the ﬁrst peak (770 MHz), S121 + S221 = (0.379)2 + (0.909)2 = 0.970 meaning that the power absorbed by the system or radiated into space is 1 − 0.970 = 0.03, or 3% of the total power. Of course, the absorbed power percentage will increase with frequency because both skin effect resistance and dielectric losses increase. For example, at 5.25 GHz, the total power absorbed by the network is 9.6%. S121 + S221 = (0.35)2 + (0.833)2 = 0.9035 1 − 0.9035 = 0.096 Since properly designed transmission lines are very inefﬁcient radiators, it is a valid assumption that very little or none of the power loss is due to radiation into free space. Of course, if there is signiﬁcant coupling (crosstalk) to adjacent structures, the power will be affected, which is covered in the next section. 1.0 S21 0.8 0.6 Pabsorbed = 3% 0.4 S11 Pabsorbed = 9.6% S11,S21 0.2 0.0 0 1 2 3 4 5 6 7 8 9 10 Frequency, GHz Figure 9-14 Insertion and return loss for a 2-in. lossy transmission line. 368 NETWORK ANALYSIS FOR DIGITAL ENGINEERS V R1 Z0 ∼ 2R R Z0 3 R 4 S31 = v3 v1 S41 = v4 v1 Figure 9-15 Two-transmission-line system for evaluating crosstalk using S -parameters. Forward (Far-End) Crosstalk When power is injected into port 1 and measured at port 4, as shown in Figure 9-15, it is called forward crosstalk , as described in Chapter 4 and deﬁned by S41 = b4 a1 a2=0 = vv41−+//√√RR = v4− v1+ (9-28) Note that forward crosstalk is often called far-end crosstalk . As detailed in Section 4.4, any bit pattern propagating on a bus with N signal conductors can be decomposed into N orthogonal modes. Section 4.4.2 describes how each mode will have a unique impedance and velocity. If a two-signal conductor system is considered, such as that shown in Figure 9-15, all digital bit patterns will be a linear superposition of the even and odd modes, which are described in Section 4.3. In Section 4.4.4, forward crosstalk for a two-signal conductor system was shown to be caused by the difference in propagation velocity between the even and odd modes. This knowledge can be used to predict the general behavior of the forward crosstalk in the frequency domain. To begin, consider the magnitude of S41 for a pair of coupled transmission lines, as shown in Figure 9-15. Using the concept of modal analysis, where the driving signal is decomposed into one-half even mode and one-half odd mode, the launch voltages at ports 1 and 3 can be constructed. For the case where port 1 is being driven with a signal and port 3 is quiet, the even- and odd-mode components are in phase at port 1 and 180◦ out of phase at port 3 and shown in Figure 9-16. Consequently, the sum of the modal voltages equals the driving voltage at port 1 and zero at port 3. If the transmission-line pair is constructed with a homogeneous dielectric, the even- and odd-mode propagation velocities are identical and will therefore arrive at the far end simultaneously. In this case, forward crosstalk will be zero because the odd- and even-mode components propagating on line 2 will still be 180◦ out of phase at node 4. However, if the transmission line is constructed with a nonhomogeneous dielectric such as a microstrip, the even and odd propagation NETWORK THEORY 369 Even Mode 1 2 veven Odd Mode 1 2 vodd Line 1 Line 2 1 3 1 3 1 2 veven 1 2 vodd Total (launch voltage) 1 2 (veven + vodd) = 1 1 2 (veven + vodd) = 0 Voltage launched onto port 1 Voltage launched onto port 3 Figure 9-16 Modal decomposition of launch voltages when port 1 is driving as shown in Figure 9-15. V R1 ∼ 2R 3 R 4R 1 2 (veven + vodd) ≠ 0 θ ≠ 180° tp,odd tp,even 1 2 veven 1 2 vodd Figure 9-17 Modal decomposition of voltages at node 4 when port 1 is driving, showing that the odd- and even-mode components of the signal arrive at different times when the dielectric media is nonhomogeneous, causing forward crosstalk to be ﬁnite. velocities will differ. Therefore, the odd and even components of the signal will arrive at node 4 at different times and will no longer be 180◦ out of phase. Consequently, for a nonhomogeneous dielectric, the forward crosstalk will be ﬁnite, as depicted in Figure 9-17. The propagation delay of the mode is calculated from the modal velocities using equation (4-81): νp,odd = √1 LoddCodd = c √εeff,odd νp,even = √1 LevenCeven = c √εeff,even 370 NETWORK ANALYSIS FOR DIGITAL ENGINEERS where the phase delay per unit length is simply τp = 1/νp. This means that the magnitude of the forward crosstalk is dependent on the difference in delay between the odd and even modes. Since equation (9-22) relates the delay to the phase, if the velocity of each mode is known, the frequency where the phase difference between the odd and even modes is 180◦ can be calculated: τp,even − τp,odd = 180◦ 360◦f180◦ f180◦ = 1 2(τp,even − τp,odd) (9-29) When f = f180◦, the odd- and even-mode voltage components will be perfectly out of phase on line 1 and in phase on line 2, which is the opposite of the launch conditions as depicted in Figure 9-18. This means that at f180◦, S21 = 0 and S41 = 1, and the coupling to the adjacent line is 100% . Example 9-5 Calculate the frequency where the insertion loss is minimum and the forward crosstalk is maximum for a 10-in. loss-free transmission line where εeff,even = 4.0 and εeff,odd = 3.5 for the circuit shown in Figure 9-15. SOLUTION Step 1: Calculate the propagation delay for the odd and even modes where 10 in. = 0.254 m and c = 3 × 108m/s: √ √ l τp,odd = εeff,odd c = 0.254 3.5 3 × 108 = 1.58 × 10−9 s τp,even = l√εeff,even c = √ 0.254 4.0 3 × 108 = 1.69 × 10−9 s Step 2: Calculate the frequency where the phase delay between odd and even modes is 180◦: f180◦ = 1 2(τp,even − τp,odd) = 2(1.69 × 1 10−9 − 1.58 × 10−9) = 4.41 × 109 Hz At 4.41 GHz, the insertion loss (S21) will be minimum and the forward crosstalk (S41) will be maximum. A simulation of this case is shown in Figure 9-19 for a 10-in. pair of coupled transmission lines with εeff,even = 4.0, εeff,odd = 3.5, Zodd = 25 , and Zeven = 100 . NETWORK THEORY ∼ f = f 180° 1 3 1 2 (veven + vodd) = 0 2 4 371 1 2 veven 1 2 vodd 1 2 (veven + vodd) = 1 1 2 veven 1 2 vodd Figure 9-18 Modal decomposition of voltages at nodes 2 and 4 when port 1 is driving, showing that at f = f180◦ the insertion loss (S 21) is zero and the forward crosstalk (S 41) is maximum. At this frequency, the coupling to the adjacent line is 100%. S21 – Insertion loss 1.0 S41 – Forward crosstalk 0.8 S21, S41 0.6 f180° 0.4 0.2 0.0 0 1 2 3 4 5 6 7 8 9 10 11 Frequency, GHz Figure 9-19 Insertion loss (S 21) and forward crosstalk (S 41) for the transmission-line pair in Figure 9-15, showing 100% coupling at f180◦ . 372 NETWORK ANALYSIS FOR DIGITAL ENGINEERS ∼ Odd mode 50 equivalent circuit Zodd 50 idrive 50 Zeven 50 Even mode equivalent circuit idrive Figure 9-20 Modal decomposition of launch voltages when port 1 is driving as shown in Figure 9-15. Reverse (Near-End) Crosstalk When the power is injected into port 1 and measured at port 3 for the circuit in Figure 9-15, it is called reverse crosstalk , as described in detail in Chapter 4. Reverse crosstalk is often referred to as near-end crosstalk . In terms of the scattering matrix, reverse crosstalk is deﬁned as S31 = b3 a1 a2=0 = vv31−+//√√RR = v3− v1+ (9-30) To explain how reverse crosstalk behaves in the frequency domain, consider a transmission-line pair built in a homogeneous dielectric that is perfectly terminated with its characteristic impedance, so the forward crosstalk and reﬂections can be neglected. At dc, the crosstalk is zero because the coupling mechanism is dependent on Lm(∂i/∂t) and Cm(∂v/∂t) as described in Section 4.1. However, as the frequency starts to increase, energy will be coupled onto a victim line. As described in Section 9.1, the peak will occur when the imaginary part is zero as described by f (real) = √n 4l LC n=1,2,3,. . . (9-3b) The peak value of the reverse crosstalk can be evaluated by decoupling the circuit into odd- and even-mode equivalents and driving the system with a current idrive as shown in Figure 9-20. The voltages propagating in the odd and even modes are calculated with the modal impedances: vodd = idriveZodd veven = idriveZeven For the case where port 1 is driven and both odd and even modes are perfectly terminated,† the line voltages propagating on each line when the imaginary part †This can be done with the appropriate T or pi termination network, as described by Hall [2000]. Another method is to choose the appropriate values of Zodd and Zeven, so the network is terminated. NETWORK THEORY 373 is zero are given by vline1 = 1 2 (veven + vodd) = 1 2 idrive(Zeven + Zodd) vline2 = 1 2 (veven − vodd) = 1 2 idrive(Zeven − Zodd) (9-31a) (9-31b) Since the voltage across the 50- resistor at port 2 is vport2 = 50idrive, the insertion loss can be calculated: S21 = vport2 vline1 = 100 Zeven + Zodd (9-32a) and the peak value of the reverse crosstalk is the ratio of the voltage coupled onto line 2 and the voltage propagating on line 1: S31 = vline2 vline1 = Zeven − Zodd Zeven + Zodd (9-32b) Example 9-6 Calculate the frequency where the insertion loss will be minimum and the reverse crosstalk will be maximum for a 10-in. loss-free homogeneous transmission line where εr = 4.0, with a circuit as shown in Figure 9-15. Assume that Zeven = 100 , Zodd = 25 , and all ports are terminated in 50 . SOLUTION Step 1: Calculate the propagation delay for the odd and even modes where 10 in. = 0.254 m and c = 3 × 108 m/s: √ √ τp = l εr c = 0.254 4.0 3 × 108 = 1.69 × 10−9 s Step 2: Use eq√uation (9-3a) to calculate the frequency of the ﬁrst hump in S31, where τp = l LC: f (real) = 1 4τp = 147 × 106 n=1 Hz Step 3: Calculate the maximum value of the crosstalk: S31 = Zeven Zeven − Zodd + Zodd = 100 − 25 100 + 25 = 0.6 Step 4: Calculate the minimum value of the insertion loss. Since this is a loss-free transmission line, the insertion loss (S21) will be minimum when the input reﬂections and crosstalk are maximum. Both the input reﬂections (S11) and the reverse crosstalk (S31) will peak when the frequency is equal to f (real), as 374 NETWORK ANALYSIS FOR DIGITAL ENGINEERS S21 – Insertion loss 1.0 S31 – Reverse crosstalk 0.8 S21,S31 0.6 0.4 0.2 0.0 0 0.2 0.4 0.6 0.8 1 Frequency, GHz 0.147 GHz Figure 9-21 Insertion loss and reverse crosstalk for Example 9-6. described by equation (9-3b). Consequently, when S11 and S31 are maximum, S21 must be minimum and is calculated with equation (9-32a): S21 = 100 Zeven + Zodd = 100 100 + 25 = 0.8 A simulation of this case is shown in Figure 9-21. Relationship Between S- and Z-Parameters Perhaps the most intuitive form of network analysis are the Z-parameters, which were explained in Section 9.2.1. The disadvantage of using an impedance (or admittance) matrix is simply that they are impossible to measure directly at high frequencies. Fortunately, it is relatively easy to convert Z-parameters into S-parameters and vice versa. The derivation is shown here. In matrix form, let’s begin with the voltage at port n of an N -port system, which is composed of an ingoing wave (v+) and an outgoing wave (v−): vn = vn+ + vn− where the current is calculated from the port voltage and the port (reference) impedance Zn: in = in+ − in− = (vn+ − vn−) 1 Zn NETWORK THEORY 375 In matrix form this becomes v = v+ + v− = Zi = Zv+ 1 − Zv− 1 Zn Zn (9-33) where Z is the impedance matrix. Assuming that the port impedance values are identical, equations (9-14) and (9-17) can be combined to show that the S-matrix can be calculated from v+ and v−: Sij = bi aj = vi− vj+ Therefore, the scattering matrix can be calculated by solving (9-33) for (v−)(v+)−1: v+ + v− = Zv+ 1 − Zv− 1 Zn Zn S = (v−)(v+)−1 = (Z + ZnU)−1(Z − ZnU) (9-34) where U is the identity or unit matrix and Zn is the termination impedance of each port. The derivation assumes that each port is terminated in the same value. Equation (9-34) converts Z-parameters to S-parameters. Solving (9-34) for Z allows the conversion from S-parameters to Z-parameters for an arbitrary-sized matrix: Z = Zn(U + S)(U − S)−1 (9-35) Equations (9-34) and (9-35) are solved for a two-port network and summarized in Table 9-1. Impulse Response Section 8.1 we introduced the concept of an impulse response matrix to completely describe the behavior of a system. If the systems impulse TABLE 9-1. S11 S12 S21 S22 Conversions Between S- and Z-Parameters for a Two-Port Network (Z11 − Zn)(Z22 + Zn) − Z12Z21 2Z12Zn (Z11 + Zn)(Z22 + Zn) 2Z21Zn − Z12Z21 (Z11 (Z11 + + Zn)(Z22 Zn)(Z22 + − Zn) Zn) − − Z12Z21 Z12Z21 (Z11 + Zn)(Z22 + Zn) − Z12Z21 (Z11 + Zn)(Z22 + Zn) − Z12Z21 Z11 Z12 Z21 Z22 Zn (1 (1 + − S11)(1 S11)(1 2Zn − S22 − S22 S21 ) ) + − S12S21 S12S21 (1 − S11)(1 − S22) − S12S21 2ZnS12 (1 Zn − S11)(1 − S22) − S12S21 (1 − S11)(1 + S22) + S12S21 (1 − S11)(1 − S22) − S12S21 376 NETWORK ANALYSIS FOR DIGITAL ENGINEERS matrix is known, it can be convolved with any arbitrary input (such as a pulse or a bit stream) and the signal integrity can be evaluated. It is impossible to measure the impulse response in the laboratory because it requires a driver capable of driving a Dirac delta function that has inﬁnitely fast rise and fall times. Furthermore, even when a fast pulse is generated in the laboratory, inductance and capacitive loading of the probes introduces into the measured response unwanted noise, ﬁltering, and resonances that are not associated with the electrical behavior of the device under test. Consequently, experimental evaluation of the impulse response in the laboratory using time-domain techniques is an impractical endeavor. For most practical purposes, the impulse response of the system interconnects can be measured indirectly using a vector network analyzer (VNA), which is a device used to evaluate the scattering matrix as a function of frequency in the laboratory. Standard techniques to remove the parasitic inductance and capacitance effects of the probes and test ﬁxtures from the measured scattering network are achieved through proper instrument calibration. Once the scattering matrix is measured, the impulse response can be calculated by taking the inverse Fourier transform, described in Section 8.1.4: h(t) = F−1{S(ω)} (9-36a) where S(ω) is the scattering matrix measured with a VNA. To calculate the impulse response, the scattering matrix must contain values for negative frequencies that obey the complex-conjugate rule to ensure a real-valued time-domain response, as described in Section 8.2.1. Since VNA measurements only provide values for the positive frequencies, the negative frequency values must be calculated from the positive values as S(−f ) = S(f )∗ (9-36b) When calculating the impulse response using the fast Fourier transform (FFT), the negative values are appended to the end of the positive values, as demonstrated in Example 9-7. Example 9-7 Calculate the impulse response from the measured values of S21 shown in Figure 9-22a. SOLUTION Step 1: Calculate the negative frequency values of S21 using (9-36b) as shown in Figure 9-22b. S(−f ) = S(f )∗ Step 2: Append the negative frequency values to the positive frequency values to create a continuous spectrum with both positive and negative frequencies. NETWORK THEORY 377 S21(f ) 1.0 0.5 Real S21(−f ) = S21(f )* Imaginary 1.0 0.5 S21 S21 0.0 0.0 −0.5 −0.5 −1.0 −1.0 1 2 3 4 5 −5 −4 −3 −2 −1 Frequency, GHz Frequency, GHz (a) (b) Figure 9-22 For Example 9-7: (a) measured values of S 21 for positive frequencies; (b) negative frequency values constructed from the complex conjugate of the positive frequencies. The magnitude of the continuous spectrum is shown in Figure 9-23a. Don’t get confused by the data format required by the FFT, which assumes that the input samples are periodic, resulting in a frequency response that looks more intuitive if proper windowing of the periodic response is chosen, as shown in Figure 9-23b. Step 3: Calculate the inverse FFT of the complex frequency spectrum to obtain the pulse response. In this case, the frequency data shown in Figure 9-22 were sampled at 100-MHz sample intervals for 1000 points for a bandwidth of 100 GHz. The ﬁnal spectrum with the negative frequency values appended to the positive values has a total of 2000 sample points. Mathematica was used to perform the inverse FFT on the frequency data to produce the impulse response shown in Figure 9-24. Although inﬁnite bandwidth is required to produce a true impulse response with inﬁnitely fast rise and fall times, the pulse response (i.e., single-bit response) as described in Section 8.1.5 can still be calculated with high accuracy if the harmonic bandwidth of the pulse is small compared to the bandwidth of the measurement. For example, if an 8-Gb/s single-bit response is calculated with rise and fall times of 35 ps, the bandwidth of the VNA used to evaluate the transfer function of the interconnects would need to be greater than 10 GHz, as calculated with Equation (8-8): fVNA,BW > 0.35 35 × 10−12 = 10 × 109 Hz Furthermore, properties of the FFT place speciﬁc requirements on the frequencydomain bandwidth to generate minimum granularity in the time domain. If a minimum granularity of t is required for the time-domain waveform, the maximum 378 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Magnitude of S21 1.0 0.8 0.6 Positive frequency 0.4 0.2 Negative frequency 0 500 1000 1500 2000 (Sample number) 0 50 +/−100 −50 0 (Frequency, GHz) (a) Magnitude of S21 1.0 0.8 0.6 Negative Positive 0.4 frequency frequency 0.2 0 500 1000 1500 2000 500 1000 1500 2000 (Sample number) (b) Figure 9-23 For Example 9-7: (a) magnitude of the frequency spectrum constructed by appending the measured positive frequency values of S 21 and the complex conjugate to create the negative frequency values; (b) FFT periodic treatment of the sampled data allows a more intuitive look at the magnitude of the frequency response when windowed properly. 2.0 1.5 Voltage 1.0 0.5 0.0 2.75 3.00 3.25 3.50 3.75 4.00 Time, ns Figure 9-24 Impulse response calculated from the inverse FFT of the S 21 data shown in Figure 9-23. NETWORK THEORY 379 positive frequency bandwidth of the measurement is deﬁned by fmax = 1 2t (9-37a) To support a time-domain step size of t, the frequency sample size must be f= 1 2n t and the maximum valid time-domain signal would be (9-37b) tmax = 2 1 f (9-37c) where n is the number of samples for the positive frequency values. For example, if a time-domain granularity of 5 ps is required, the maxi- mum bandwidth required using 1000 positive frequency sample points would be 100 GHz: fmax = 2(5 1 × 10−12) = 100 × 109 Hz Consequently, a time-domain waveform constructed from a frequency-domain measurement requires signiﬁcant bandwidth if ﬁne granularity is required. Unfortunately, it becomes very difﬁcult to perform VNA measurements above about 20 GHz. Both the calibration techniques and equipment costs become prohibitive. Fortunately, there are mathematic ways to sidestep the lack of high-frequency measured data without losing time-domain granularity. As long as the spectral bandwidth of the digital waveforms propagating on the system interconnects are signiﬁcantly lower than fmax, extrapolation or zero padding of the measured scattering matrix can be used to increase granularity with only a small degradation in accuracy. For example, consider a driving digital waveform with rise and fall times of 25 ps. The spectral bandwidth of this waveform is approximated by equation (8-8): f3dB ≈ 0.35 25 × 10−12 = 14 × 109 Hz Common bandwidths currently available on VNAs range from 20 to 110 GHz, which are equivalent to pulses with rise and fall times of about 3 to 18 ps. 0.35 20 × 109 = 17.5 × 10−12 s t10–90% = 0.35 110 × 109 = 3.18 × 10−12 s 380 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Therefore, measurements taken using a standard 20-GHz VNA would have adequate bandwidth to resolve rise and fall times as fast as 17.5 ps. Although a bandwidth of 20 GHz is sufﬁcient to resolve the edge rate, the granularity of the time-domain waveform calculated using the FFT is only 25 ps, as calculated by (9-37a): t = 1 2fmax = 1 2(20 × 109) = 25 × 10−12 s Consequently, extrapolation or zero padding is required to ensure reasonable granularity in the time-domain waveform, as demonstrated in Example 9-8. Example 9-8 Assume that the complex values of S21 used in Example 9-7 have been measured to 20 GHz. Calculate the impulse response with a resolution of 5 ps. SOLUTION Step 1: Calculate fmax using equation (9-37a): fmax = 1 2t = 1 2(5 × 10−12) = 100 × 109 Hz Step 2: Calculate the negative frequency values from −20 GHz to dc using equation (9-36b): S(−f ) = S(f )∗ Step 3 : Calculate the sample interval of the frequency-domain data assuming 1000 samples of the measurable positive frequency values: f = 1 2n t = 1 2(1000)(5 × 10−12) = 100 × 106 Hz Step 4: Calculate the number of zero points that need to be added to the spectrum. At a sample rate of 100 MHz, the number of samples up to 20 GHz is 100 20 × 109 Hz × 106 Hz/sample = 200 samples Therefore, 800 zero points need to be added to both the positive and negative spectra to achieve ±100-GHz bandwidth. Step 5: Append the positive and negative spectrums together as shown in Figure 9-25. Step 6: Perform an inverse FFT on the zero-padded spectrum to get the impulse response, as shown in Figure 9-26. NETWORK THEORY 381 200 Sample points Magnitude S21 1.0 S21, measured (S21, measured)* 0.8 0.6 800 Sample points 0.4 0.2 Zero padding 0 20 50 +/−100 −50 −20 0 Frequency, GHz Discontinuity caused by the zero padding Figure 9-25 Zero-padded spectrum. Bandwidh of measured response is 20 GHz and zero-padded to 100 GHz to ensure 5-ps resolution in the time domain. 2.0 1.5 Voltage 1.0 Ripple caused by zero padding 0.5 0.0 2.75 3.00 3.25 3.50 3.75 4.00 Time, ns Figure 9-26 Impulse response calculated from the inverse FFT of the 20-GHz padded to 100-GHz S 21 data shown in Figure 9-25. Note that the zero padding has introduced a small amount of ringing into the impulse response. The error is caused by the discontinuous spectrum where the measured data stops and the zero padding begins. This error can be minimized by increasing the bandwidth of the measured data, extrapolating the real and imaginary parts of the measured data instead of zero padding or smoothing the zero-padding discontinuity. 382 NETWORK ANALYSIS FOR DIGITAL ENGINEERS i1 + Port 1v1 − Two Port Network i2 + v2 Port 2 − Figure 9-27 Two-port network used to describe ABCD parameters. 9.2.3 ABCD Parameters Consider the two-port network depicted in Figure 9-27. If the voltage and current are measured at the input and output ports, the system can be characterized in terms of its ABCD matrix. The ABCD parameters have several advantages over other network parameters. They allow the full description of a network in terms of input and output voltage and current, which makes them convenient for cascading circuits; they are easily related to equivalent circuits; and they provide a convenient basis for writing specialized programs that allow voltage and current sources to drive channels constructed from cascaded ABCD elements. Two-port ABCD parameters are developed here. One signiﬁcant difference between the ABCD matrix and the impedance matrix is the direction of i2, which is pointing out of, not into, port 2. This allows easy cascading of networks (which is addressed in Section 9.2.4). The ABCD values are evaluated as A = v1 v2 i2=0 B = v1 i2 v2=0 C = i1 v2 i2=0 D = i1 i2 v2=0 (9-38) Using the deﬁnition shown in equations (9-38), a set of linear equations can be written to describe the network: v1 = Av2 + Bi2 i1 = Cv2 + Di2 which is more efﬁciently expressed in matrix form: v1 i1 = A C B D · v2 i2 (9-39) Consequently, if the ABCD matrix of a system is known, the response of the system can be predicted for any input. Since the ABCD parameters are evaluated with short and open circuits as shown in equation (9-38), they are not practical to measure directly. However, relationships exist that allow the ABCD matrix to be calculated directly from the S-parameters as will be shown later. NETWORK THEORY 383 i1 Z1 Port 1 v1 i2 Z2 Z3 v2 Port 2 Figure 9-28 Deriving the ABCD parameters for a T -topology equivalent circuit. Relationship to Common Circuit Parameters The ABCD parameters can be used to create equivalent-circuit models for common circuit topologies. For example, consider the T-circuit shown in Figure 9-28. To determine A, port 2 must be open (since i2 = 0) while port 1 is driven with v1. A = v1 v2 i2=0 v2 = v1 Z3 Z3 + Z1 A = Z3 + Z1 Z3 The term B is determined by shorting port 1 (since v2 = 0): (9-40a) B = v1 i2 v2=0 A current divider is used to calculate i2, which is substituted into the equation for B: i2 = i1 Z3 Z3 + Z2 B= v1 = v1 Z3 + Z2 i1[Z3/(Z3 + Z2)] i1 Z3 Since v1/ i1 is equal to the impedance looking into port 1, B can be simpliﬁed: B = (Z1 + Z2 ||Z3) Z3 + Z3 Z2 = Z1 + Z2 + Z1Z2 Z3 (9-40b) To determine C, port 2 must be open (since i2 = 0) while port 1 is driven with v1. C = i1 v2 i2=0 v2 = v1 Z3 Z3 + Z1 384 NETWORK ANALYSIS FOR DIGITAL ENGINEERS i1 = v1 Z1 + Z3 C = i1 v2 = v1/(Z1 + Z3) v1[Z3/(Z3 + Z1)] = 1 Z3 (9-40c) To determine D, port 2 must be short (since v2 = 0) while port 1 is driven: D = i1 i2 v2=0 i2 = i1 Z3 Z3 + Z2 D = i1 = Z3 + Z2 i2 Z3 (9-40d) Therefore, for a T-circuit like the one depicted in Figure 9-28, the ABCD matrix takes on the form Z3 + Z1 A C B D = T-circuit Z3 1 Z3 Z1 + Z2 + Z1Z2 Z3 Z3 + Z2 Z3 (9-41) The ABCD parameters of a lossy transmission line can be derived for the case where the line is terminated is its characteristic impedance. This conversion is particularly useful for extracting transmission-line parameters such as Z0 and γ from measurements. For the case where the termination impedance is not equal to the characteristic impedance, the matrix can be renormalized as described in Section 9.2.6. To begin the derivation, consider the transmission line in Figure 9-29: A = v1 v2 i2=0 Port 1 V R = Z0 Z = −I Z0 g Port 2 R = Z0 Z=0 Figure 9-29 Deriving the ABCD parameters for a lossy transmission line. NETWORK THEORY 385 Since i2 = 0, port 2 is open. Therefore, the voltage at port 2 will have an incident and a reﬂected component. For convenience, the total voltage at port 2 is set equal to 1: v2 = v+ + v− = 1 v+ = v− = 1 2 The voltage at port 1 is calculated using equation (6-49) with z = −l. v1 = v(z = −l) = v+eγ l + v−e−γ l = eγ l + e−γ l 2 The ratio of v1 and v2 is used to calculate A: A = eγ l + e−γ l = cosh γ l 2 The term C is also calculated with i2 = 0. (9-42a) C = i1 v2 i2=0 The current i1 is calculated at z = −l using the characteristics impedance. i1 = i(z = l) = 1 v+eγ l Z0 − 1 v−e−γ l Z0 = 1 Z0 eγ l − e−γ l 2 The ratio of i1 and v2 is used to calculate C: C= 1 eγ l − e−γ l = 1 sinh γ l Z0 2 Z0 Terms B and D are calculated with v2 = 0: (9-42b) B = v1 i2 v2=0 Since v2 = 0, port 2 is shorted: v2 = v+ + v− = 1 2 − 1 2 = 0 i2 = v+ Z0 − v− Z0 = 1/2 Z0 − −1/2 Z0 = 1 Z0 v1 = v(z = −l) = v+eγ l + v−e−γ l = eγ l − e−γ l 2 386 NETWORK ANALYSIS FOR DIGITAL ENGINEERS The ratio of v1 and i2 is used to calculate B: eγ l − e−γ l B = Z0 2 = Z0 sinh γ l D = i1 i2 v2=0 i2 = 1 Z0 i1 = i(z = l) = 1/2 eγ l Z0 − −1/2 e−γ l Z0 = 1 Z0 eγ l + e−γ l 2 (9-42c) The ratio of i1 and i2 is used to calculate D: D = eγ l + e−γ l = cosh γ l 2 (9-42d) Therefore, the ABCD matrix for a lossy transmission line. A C B D cosh γ l lossy transmssion line = 1 Z0 sinh γ l Z0 sinh γ l cosh γ l (9-43a) Using the same procedure, the ABCD parameters for a loss-free line can easily be derived where γ = jβ: A C B D cos βl loss-free transmssion line = j Z0 sin βl j Z0 sin βl cos βl (9-43b) Table 9-2 depicts the relationship between common circuits and the ABCD parameters. These common forms are useful for extracting equivalent circuits from S-parameter measurements. Of course, a methodology is needed to convert S-parameters into an ABCD matrix, which is covered in the next section. Relationship Between ABCD and S-Parameters To take advantage of the relationships between the ABCD matrix and common circuit forms, it is necessary to determine the relationship between the ABCD and S-parameters. The most straightforward derivation is ﬁrst to deﬁne the transformation of ABCD parameters into a two-port Z-matrix and then use equation (9-34) to get the S-parameters. Beginning with the deﬁnition of Z11 from equation (9-8), Z11 = v1 i1 i2=0 NETWORK THEORY 387 TABLE 9-2. Relationship Between Common Circuits and the ABCD Parameters Circuit i1 Z1 v1 Z3 V R=Z0 Z0 z=-l i1 Z v1 i2 Z2 v2 R=Z0 z=0 i2 v2 ABCD Matrix Z3 + Z1 Z3 1 Z1 + Z2 + Z1Z2 Z3 Z3 + Z2 Z3 cosh γ l 1 sinh γ l Z0 Z3 Z0 sinh γ l cosh γ l 1Z 01 i1 i2 v1 Y v2 i1 i2 Y3 v1 Y1 Y2 v2 10 Y1 1 + Y2 Y3 Y1 + Y2 + Y1Y2 Y3 1 Y3 1 + Y1 Y3 the ABCD equations reduce to v1 = Av2 i1 = Cv2 resulting in Z11 = v1 i1 = A C The deﬁnition of Z12 is Z12 = v1 i2 i1=0 so the ABCD equations reduce to v1 = Av2 + Bi2 0 = Cv2 + Di2 Solving the equations above for v1/i2 produces v1 = BC − AD i2 C (9-44a) 388 NETWORK ANALYSIS FOR DIGITAL ENGINEERS However, the convention for an ABCD matrix assumes that i2 is ﬂowing out of port 2, and a Z matrix assumes that it is ﬂowing into port 2. Consequently, the sign of i2 must be changed, which results in the deﬁnition of Z12 in terms of ABCD parameters: AD − BC Z12 = C (9-44b) The terms Z21 and Z22 are derived using a similar procedure. Z21 = v2 i1 i2=0 v1 = Av2 i1 = Cv2 Z21 = 1 C Z22 = v2 i2 i1=0 v1 = Av2 + Bi2 0 = Cv2 + Di2 D Z22 = C (9-44c) (9-44d) The ﬁnal relationship between a two-port Z-matrix and the ABCD matrix is shown as A AD − BC Z11 Z21 Z12 Z22 = C 1 C D (9-45) C C To derive the transformation of the ABCD to S-parameters, the results of (9-45) are substituted into equation (9-34). The ﬁnal solution is summarized in (9-46), where Zn is the termination impedance at the ports, which are all assumed to be equal. B − Zn(D − A + CZn) 2Zn(AD − BC) S11 S21 S12 S22 = B + Zn(D + A 2Zn + CZn) B B + − Zn(D Zn(A + − A D + + CZn) CZn) (9-46) B + Zn(D + A + CZn) B + Zn(D + A + CZn) Similar techniques are used to derive the transformation of S into ABCD parameters. The complete sets of transformations are summarized in Table 9-3. NETWORK THEORY 389 TABLE 9-3. Relationships Between Two-Port S and ABCD Parametersa S11 S12 S21 S22 AB CD B − Zn(D − A + CZn) B + Zn(D + A 2Zn + CZn) B + Zn(D + A + CZn) (1 + S11)(1 − S22) + S12S21 1 2S21 (1 − S11)(1 − S22) − S12S21 Zn 2S21 2Zn(AD − BC) B B + − Zn(D Zn(A +A −D + + CZn) CZn) B + Zn(D + A + CZn) Zn (1 + S11 )(1 + S22 2S21 ) − S12S21 (1 − S11)(1 + S22) + S12S21 2S21 aZn is the termination impedance at the ports. Port 1 Barrel Lbarrel Port 1 Cpad Cpad Port 2 Via pad (a) (b) Figure 9-30 Via and equivalent circuit for Example 9-9. Port 2 Example 9-9 Extract an equivalent circuit for the via shown in Figure 9-30a from the following S-parameter matrix measured at 5 GHz assuming port impedance values (Zn) of 50 : S11 S21 S12 S22 = −0.1235 − j 0.1516 0.7597 − j 0.6190 0.7597 − j 0.6190 −0.1235 − j 0.1516 SOLUTION Step 1: Transform the S-matrix into ABCD parameters using the relations in Table 9-3. A C B D = 0.790 j 0.01686 j 22.22 0.790 Step 2 : Choose the form of the equivalent circuit. A signal propagating through the via will experience the capacitance of the via pad, the inductance of the barrel, 390 NETWORK ANALYSIS FOR DIGITAL ENGINEERS and then the capacitance of the second pad. This conﬁguration ﬁts the pi model in Table 9-2 as shown in Figure 9-30b. Step 3: Use the relations in Table 9-2 to calculate the admittance of each segment in the pi model: B = 1 = j 22.22 Y3 Y3 = −j 0.045 A= 0.790 = 1+ Y2 Y3 = 1+ Y2 −j 0.045 Y2 = j 0.00945 Due to symmetry in the circuit, Y1 = Y2. Step 4: Calculate the circuit values: Y3 = −j 0.045 = 1 j ωL = 1 j 2π(5 × 109Lbarrel) → Lbarrel = 0.7 × 10−9 H Y1 = Y2 = j 0.00945 = j ωC = j 2π(5 × 109Cpad) → Cpad = 0.3 × 10−12 F 9.2.4 Cascading S-Parameters One of the most useful aspects of network analysis is the ability to cascade independently measured structures. For example, if the S-parameters are measured independently for a transmission line, via, package, and connector, the engineer has the ability to create the response of an entire channel from the individual measurements. This allows S-parameter ﬁles to be used as portable models, gives the designer the ability to evaluate different topologies, and provides the mechanism to deembed or calibrate out unwanted items in the measurements. The two most common methods of cascading S-parameters are with the ABCD parameters and the T-matrix. Cascading with the ABCD Matrix For two-port networks, the most common cascading methodology is to use the ABCD parameters. Cascading is achieved simply by multiplying the ABCD matrices. This is possible because of the current convention that ﬂows outward at port 2, as shown in Figure 9-27. For example, consider the ABDC parameters of the two cascaded circuits shown in Figure 9-31. The equations that describe the port voltage and currents of circuits 1 and 2 can NETWORK THEORY i1 i2 AB v1 v2 C D circuit1 391 i3 AB v3 C D circuit2 Figure 9-31 ABCD parameters are cascaded though multiplication. easily be written. v1 i1 = A C B v2 D circuit1 i2 v2 i2 = A C B v3 D circuit2 i3 Note that the output of circuit 1 is v2 and i2, which is the input of circuit 2. This allows the simple substitution of resulting in v2 i2 with AB v3 C D circuit2 i3 v1 i1 = A C B A D circuit1 C B v3 D circuit2 i3 Therefore, two-port S-parameters can be cascaded by converting to ABCD parameters and multiplying. The cascaded scattering matrix is then calculated by converting the cascaded ABCD matrix back to S-parameters using Table 9-3. The procedure is demonstrated in Example 9-10. Example 9-10 Using the two independently measured values of S-parameters for a via and a loss-free transmission line at 5 GHz, calculate the, equivalent S-parameters that would be measured if the two circuits were cascaded as shown Figure 9-32. Assume that the termination impedance is 50 . S11 S21 S12 S22 = via S11 S21 S12 S22 = t-line −0.1235 − j 0.1516 0.7597 − j 0.6190 0.00325 − j 0.00323 −1.00 − j 0.003 0.7597 − j 0.6190 −0.1235 − j 0.1516 −1.00 − j 0.003 0.00325 − j 0.00323 392 R NETWORK ANALYSIS FOR DIGITAL ENGINEERS R Z = −I Z0 b R Z=0 R (a) (b) R Z0 b R (c) Figure 9-32 For Example 9-10: (a) conﬁguration of the independently measured via; (b) transmission line; (c) via cascaded with the transmission line. SOLUTION Step 1: Convert to ABCD parameters using Table 9-3: A C B D = via A C B D = t-line 0.790 j 22.22 j 0.01686 0.790 −1 j 0.3228 j 0.000129 −1 Step 2: Multiply the ABCD matrices: A C B D = cascade A C BA D via C B D t-line = 0.790 j 0.01686 j 22.22 0.790 · −1 j 0.000129 j 0.3228 −1 = −0.790 −j 21.965 −j 0.01686 −0.795 Step 3: Convert A C B D back to S-parameters using Table 9-3 where cascade Zn = 50 (the termination impedance) S11 S21 S12 S22 = cascade −0.1259 − j 0.1553 −0.7645 + j 0.6182 −0.7635 + j 0.6186 −0.1200 − j 0.1565 The cascaded S-matrix is equivalent to what would be measured if the via was cascaded with the transmission line, as shown in Figure 9-32c. NETWORK THEORY 393 Cascading with the T -Matrix Another method commonly used to cascade S-parameters is the T-matrix , sometimes called transmission parameters. The T -parameters are derived simply by rearranging the equations for the S-parameters. Equation (9-18) describes the relationship between the incident power waves a and the exiting power waves b: b1 b2 = S11 S21 S12 S22 a1 a2 To facilitate the cascading of networks by simple matrix multiplication, the equation needs to be rearranged so that the ingoing and outgoing waves of port 1 can be described in terms of the waves at port 2. This is done with the T -matrix. A two-port T -matrix is a1 b1 = T11 T21 T12 T22 b2 a2 (9-47) If the output of circuit A is attached to the input of circuit B, the total response can be calculated simply by multiplying the T -matrices, because the power wave exiting circuit A is b2, which feeds into the input of circuit B, which is a3. Therefore, b2 = a3 and a2 = b3, as shown in Figure 9-33. a1 b1 = T11 T21 T12 b2 T22 A a2 a3 b3 = T11 T21 T12 b4 T22 B a4 a1 b1 = T11 T21 T12 T11 T22 A T21 T12 b4 T22 B a4 Therefore, S-parameters can be cascaded by converting to T -parameters and multiplying. The cascaded scattering matrix is then calculated by converting the product of the T -matrices back to S-parameters. Conversion between the T - and S-parameters (and vice-versa) requires simple algebraic manipulation of the equations, which can be done for any number of a1 b2 b1 [T ]circuit1 a2 a3 b4 b3 [T ]circuit2 a4 Figure 9-33 T -parameters are cascaded through multiplication. 394 NETWORK ANALYSIS FOR DIGITAL ENGINEERS ports. The conversion from S- to T -parameters is derived here for a two-port system to demonstrate the process. To begin, the equations that relate the incoming power waves to the outgoing are written b1 = S11a1 + S12a2 b2 = S21a1 + S22a2 and equation (9-47) is expressed in algebraic form: a1 = T11b2 + T12a2 b1 = T21b2 + T22a2 From the equations above, the individual T -parameters can be deﬁned in terms of the S-parameters: T11 = a1 b2 a2=0 b2 = S21a1 + S22(0) T11 = 1 S21 T22 = b1 a2 b2=0 0 = S21a1 + S22a2 → a1 = − S22a2 S21 b1 = S11a1 + S12a2 = S12 − S11S22 S21 a2 T22 = S12 − S11S22 S21 T12 = a1 a2 b2=0 0 = S21a1 + S22a2 a1 = − S22a2 S21 T12 = − S22 S21 (9-48a) (9-48b) (9-48c) NETWORK THEORY 395 TABLE 9-4. Relationships Between T - and S-Parameters for Two-Ports T11 T12 T21 T22 1 S21 S11 S21 − S22 S21 S12 − S11S22 S21 S11 S12 S21 S22 T21 T11 1 T11 T22 − T21T12 T11 − T12 T11 T21 = b1 b2 a2=0 b1 = S11a1 b2 = S21a1 T21 = S11 S21 (9-48d) A very similar analysis is used to transform the T -parameters back into S-parameters, which is left to an exercise for the reader. The transformations for two ports are summarized in Table 9-4. The four-port T -matrix takes the form a1 T11 T12 T13 T14 b2 ab31 = TT2311 T22 T32 T23 T33 T24 T34 ba42 b3 T41 T42 T43 T44 a4 (9-49) for the network shown in Figure 9-34. The transformations between T - and S-parameters for n-ports is developed in an identical manner, which is shown in Appendix B. 9.2.5 Calibration and De-embedding Very often in digital design it is desirable to measure a component such as a transmission line, a CPU socket, or a daughter-card connector using a vector network analyzer (VNA). The measurement can be used to create an equivalent-circuit model, validate the modeling methodologies used to simulate the channel, or estimate the performance of a component. Practicality dictates that the device under test (DUT) is usually mounted in a ﬁxture and connected to the VNA with a cable using probes or SMA connectors. Consequently, methodologies are 396 NETWORK ANALYSIS FOR DIGITAL ENGINEERS a1 Port 1 Port 3 b1 a3 b3 4 port network b2 Port 2 a2 b2 Port 4 a2 Figure 9-34 Four-port network. needed to separate the S-parameters of the DUT from the test ﬁxture, cables, or probes. In this section we introduce the basic calibration and deembedding methodologies so that the concept will be understood. Since each measurement setup requires a specialized calibration and deembedding procedure, the breadth of the topic is too wide to be covered here. The terms calibration and deembedding refer to the same concept: remstepoving an unwanted part of the measurement. More speciﬁcally, calibration refers to removing the effect of the VNA cables and probes, and deembedding refers to removing unwanted portions of the DUT, such as a via, a test ﬁxture, or a cable connector. A VNA measures S-parameters as ratios of complex voltage amplitudes. The reference for the measurement is some place within the VNA, not at the cable ends which are attached to the DUT, so the measurement will include the losses and phase delays of the cables, connectors, and probes used to connect the DUT to the analyzer. Calibration refers to the removal of these unwanted effects from the measured response so that only the measured response of the DUT remains. Figure 9-35 demonstrates this concept. For calibration purposes, it is convenient to work with ABCD parameters. For example, the measurement setup in Figure 9-35 can be represented by cascaded ABCD matrices: A C B D = measured A C B A D errorport1 C B A D DUT C B −1 D errorport2 (9-50) Note that the ABCD matrix for port 2 is inverted because the VNA drives into the port and measures the responses at all other ports. The inversion simply ensures that it represents current ﬂowing from port 2 into the DUT and not the other way around. Equation (9-50) suggests that if the ABCD matrices of the errors are known, the ABCD matrix (and thus the S-parameters) of the DUT can be calculated. In short, the calibrated measurement is simply the error-corrected S-parameters. The most straightforward method to remove the errors and calibrate a VNA is to use three or more carefully controlled loads, such as a short, open, and NETWORK THEORY 397 VNA DUT R ErrorPort1 DUT ErrorPort2 R Desired measurement reference planes Actual measurements are referenced internal to the VNA and include cable, probe and connector losses and phase delays Figure 9-35 Calibration removes the unwanted parts of the measurement, such as the cable and probe effects. a1 b2 b1 Error Port1 a2 ZL Γin L Figure 9-36 The simplest way to calibrate a VNA is to characterize the port error by driving different loads (Z L): an open, a short, and a resistive load. precision resistive load. For example, consider Figure 9-36, which represents one port of a VNA driving a load ZL. The equations for the S-parameters can be written in terms of the reﬂection coefﬁcient at the load where the incoming wave at port 2 (a2) is simply the reﬂected portion of the exiting wave (b2) a2 = b2 L. b1 = S11a1 + S12a2 = S11a1 + S12b2 L b2 = S21a1 + S22a2 = S21a1 + S22b2 L b2(1 − S22 L) = S21a1 → b2 = S21a1 1 − S22 L b1 = S11a1 + S12 1 S21a1 − S22 L L (9-51a) (9-51b) 398 NETWORK ANALYSIS FOR DIGITAL ENGINEERS The input reﬂection in is the ratio of the reﬂected wave b1 to the incident wave a1: in = b1 a1 = S11 + S12S21 1 − S22 L L If it is assumed that the error term is reciprocal, S21 = S12, the input reﬂection equation can be simpliﬁed: in = b1 a1 = S11 + S122 L 1 − S22 L (9-52) To calibrate the effect of the errors out of the measurement, the S-parameters must be found. This is done by considering three known values for the load, ZL: a short, an open, and a perfectly matched resistor. When the load is shorted, the reﬂection at the load is L = −1, which reduces (9-52) to in, short = S11 − 1 S122 + S22 (9-53a) When the load is open and perfectly impedance matched, the reﬂection at the load is L = 1 and L = 0 respectively, producing in, open = S11 + 1 S122 − S22 in, matched = S11 (9-53b) (9-53c) By measuring the open, short and matched loads, the three equations above can be solved simultaneously for S11, S12, and S22: S12 = S21 = ( in, matched − in,short)(1 + S22) 2 S22 = in, matched − in,short − in,short − in,open in, open S11 = in,matched (9-54a) (9-54b) (9-54c) The equations above are then used to create a scattering matrix for the errors at the ports: S11 S21 S12 S22 ⇒ short,open,load A C B D errorport (9-55) Finally, the measurements of the DUT are calculated by multiplying by the inverse of the error terms: A C B D = DUT A C B −1 A D errorport1 C B A D measured C B D errorport2 (9-56) NETWORK THEORY 399 The calibrated S-parameters are obtained by converting the ABCD matrix of the DUT calculated with (9-56) using Table 9-3. A C B D ⇒ DUT S11 S21 S12 S22 DUT (9-57) Basic deembedding principles utilize the same concepts to remove unwanted parts of the measurement. For example, Figure 9-32c shows a case where a via is cascaded with a transmission line. If it is desired to obtain only the measurements of the transmission line, the effects of the via must be deembedded. If the test board contains suitable structures to measure the S-parameters of the via in isolation, it can be effectively deembedded from the measurements using the same procedure shown in (9-56): A C B D = T-line A C B −1 A D via C B D measured (9-58) Using the concept of cascaded matrices, any number of structures can be deembedded from the measured data as long as the S-parameters for the structures are known. 9.2.6 Changing the Reference Impedance S-parameters are dependent on the reference impedance of the VNA. If the port impedance values change, the S-parameters change. It is generally standard to measure S-parameters assuming a reference impedance of 50 at each port. However, sometimes the port impedance values need to be adjusted after the measurements are performed. For example, perhaps the port impedance of the VNA was determined to be something other than 50 , or the engineer wished to examine the performance of the circuit referenced to an impedance consistent with the transmission lines used in a speciﬁc board design. When an S-parameter is measured with a reference impedance at the ports of Zn, it is said to be normalized to that impedance. To renormalize the S-matrix from Zn1 to Zn2, the deﬁnition of the Z-parameters from equation (9-35) is used: Z = Zn(U + S)(U − S)−1 (9-35) Since the impedance matrix is not dependent on the port impedance, it can be used to renormalize the S-matrix. Zn1(U + S1)(U − S1)−1 = Zn2(U + S2)(U − S2)−1 400 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Solving for S2 yields S2 = Zn1 Zn2 (U + S1)(U − S1)−1 + U −1 Zn1 Zn2 (U + S1)(U − S1)−1 − U (9-59) Example 9-11 Renormalize the following S-matrix measured with a 50- reference load to 75 : S50 = 0.385 j 0.923 j 0.923 0.385 SOLUTION Using equation (9-59), the S-matrix can be manipulated so that it looks like it was measured with a 75- reference impedance. S75 = 50 75 1 0 0 1 + 0.385 j 0.923 j 0.923 0.385 1 0 0 1 − 0.385 j 0.923 j 0.923 0.385 −1 + 1 0 0 1 −1 · 50 75 1 0 0 1 + 0.385 j 0.923 j 0.923 0.385 × 1 0 0 1 − 0.385 j 0.923 j 0.923 0.385 −1 − 1 0 0 1 = 0 −j −j 0 Therefore, the magnitude of the renormalized S-matrix is |S75 | = 0 1 1 0 meaning that the transmission line is loss free with a characteristic impedance of 75 since there is no reﬂection (S11 = 0) and the insertion loss is unity (S21 = 1). 9.2.7 Multimode S-Parameters In Chapter 4 differential signaling was explained. Since many of the high-speed buses being designed in modern computing systems consist of differential pairs, it is often convenient to describe the behavior of the interconnects in terms of multimode S-parameters. The multimode S-matrix breaks the signal on a differential pair into terms of differential (i.e., odd mode) and common (i.e., even mode) signaling states. A multimode matrix for two modes can be derived NETWORK THEORY ∼ R Port 1 401 Port 2 R R Port 3 R Port 4 Figure 9-37 A pair of coupled transmission lines is a common example of a four-port network that is conveniently described with a multimode S -matrix. directly from the four-port S-parameters. For example, the S-parameters of the four-port system pictured in Figure 9-34 are shown as follows: b1 S11 S12 S13 S14 a1 bb23 = SS2311 S22 S32 S23 S33 SS2344 aa23 b4 S41 S42 S43 S44 a4 (9-60) An example of a four-port system is two coupled transmission lines, as shown in Figure 9-37. To derive the multimode matrix, it is ﬁrst necessary to calculate the S-parameters for each mode. The differential S-parameters, where the energy is being transmitted in the odd mode, are calculated by driving the four-port network with +v on port 1 and −v on port 3 when ports 2 and 4 are not being driven. This allows (9-60) to be simpliﬁed. b1 S11 S12 S13 S14 v bb23 = SS2311 S22 S32 S23 S33 SS2344 0 −v b4 S41 S42 S43 S44 0 (9-61) The differential return loss at port 1 can then be calculated algebraically where bd1 = b1 − b3 and ad1 = a1 − a3 = v − (−v) = 2v. Sdd11 = bd1 ad1 = a2=a4=0 1 2 (S11 + S33 − S13 − S31) (9-62a) Sdd11 is a measure of the differential energy reﬂected, or returned to the source. Similarly, the differential insertion loss is calculated where bd2 = b2 − b4 and ad1 = a1 − a3 = v − (−v) = 2v. Sdd21 = bd2 ad1 = a2=a4=0 1 2 (S21 + S43 − S23 − S41) (9-62b) 402 NETWORK ANALYSIS FOR DIGITAL ENGINEERS Sdd21 is a measure of the differential energy transmitted across the network from port 1 to port 2. Identical analysis can be used to calculate the common-mode S-parameters, where the energy is being transmitted in the even mode, by driving the four-port network with +v on port 1 and +v on port 3 when ports 2 and 4 are not being driven. This allows (9-60) to be simpliﬁed where bc1 = b1 + b3, bc2 = b2 + b4, and ac1 = a1 + a3 = v + v = 2v. b1 S11 S12 S13 S14 v bb23 = SS2311 S22 S32 S23 S33 SS2344 v0 (9-63) b4 S41 S42 S43 S44 0 The common-mode S-parameters are easily obtained algebraically from (9-63): Scc11 = bc1 ac1 = a2=a4=0 1 2 (S11 + S33 + S13 + S31) Scc21 = bc2 ac1 = a2=a4=0 1 2 (S21 + S23 + S41 + S43) (9-64a) (9-64b) Equations (9-64a) and (9-64b) describe how much energy is being reﬂected and transmitted when the four-port system is being driven with a common-mode source (i.e., the system is being driven in the even mode). As described in Chapter 4, for a system with two signal conductors, the volt- ages at the ports are combinations of odd and even modes. Consequently, for a perfectly symmetric system (where each leg of the differential pair is electrically identical), if the system is driven differentially, all the energy will be contained within the odd mode. However, if the pair exhibits any asymmetry, a portion of the energy will be ﬂowing in the even mode. The multimode matrix also accounts for the differential-to-common mode conversion, which describes the amount of energy being transformed into the even mode when the system is being driven differentially and the common mode-to-differential conversion, which tells how much energy is being converted to the odd mode when driven commonly. Since most high-speed buses are driven differentially, the differential-to-common mode conversion is the most important parameter of the two. The differential-to-common mode S-parameters, where the energy is being transmitted in the odd mode and received in the even mode, are calculated by driving the four-port network differentially at the driver and sensing in common mode at the receiver. The differential-to-common mode coefﬁcients are bc1 = b1 + b3, bc2 = b2 + b4, and ad1 = a1 − a3 = v − (−v) = 2v. b1 S11 S12 S13 S14 v bb23 = SS2311 S22 S32 S23 S33 S24 S34 0 −v b4 S41 S42 S43 S44 0 (9-65) NETWORK THEORY 403 Scd11 = bc1 ad1 = a2=a4=0 1 2 (S11 − S13 + S31 − S33) Scd21 = bc2 ad1 = a2=a4=0 1 2 (S21 − S23 + S41 − S43) (9-66a) (9-66b) The common mode-to-differential S-parameters, where the energy is being trans- mitted in the even mode and received in the odd mode, are calculated by driving the four-port network in the common mode and sensing differentially at the receiver. The derivation is left to the reader. The multimode S-parameters for a four-port system can be combined to create the multimode matrix bd1 Sd d 11 bbdc12 = SSdcdd1211 bc2 Scd 21 Sd d 12 Sd d 22 Scd 12 Scd 22 Sd c11 Sd c21 Scc11 Scc21 Sdc12 ad1 SSdccc1222 aadc12 Scc22 ac2 (9-67a) Note that each quadrant of the multimode matrix represents unique parameters. For example, the upper left quadrant is the differential S-parameter matrix, bd1 bd2 = Sd d 11 Sd d 21 Sd d 12 Sd d 22 ad1 ad2 = [Sdd ] ad1 ad2 (9-67b) which assumes that all the energy is sourced and sensed in a differential manner. Consequently, the multimode matrix takes the form bd1 bbdc12 = bc2 [Sdd ] [Scd ] ad1 [Sdc] [Scc] aadc12 ac2 (9-67c) where each matrix describes a speciﬁc driving and receiving conﬁguration: Sdd : Driver and receiver are differential. Scc: Driver and receiver are common mode. Sdc: Driver is common mode and receiver is differential. Scd : Driver is differential mode and receiver is common mode. The format of (9-67c) allows the four-port S-parameters to be reduced to an equivalent simple two-port system for speciﬁc circumstances. For example, in high-speed differential buses, each pair is driven and received assuming that all 404 NETWORK ANALYSIS FOR DIGITAL ENGINEERS v ∼ 50 Zodd = 50 50 ∼ 50 −v Zodd = 50 Z0 = 50 50 ∆I Figure 9-38 Unbalanced differential pairs cause energy to be converted from the odd mode to the even mode, which looks like additional differential insertion loss. energy is in the odd mode, which means that for all practical purposes, equation (9-67b) is sufﬁcient for design purposes. Invariably, a question is asked: If there is signiﬁcant energy in the even mode, doesn’t the common mode matrix need to be accounted for? The answer is that it is already included in the differential matrix of (9-67b). If an asymmetrical four-port system is driven differentially, the energy that is converted to common mode will simply look like differential insertion loss. For example, consider the loss-free unbalanced differential transmission line pair being driven in the odd mode in Figure 9-38. Since the system is being driven differentially (in the odd mode), and the odd-mode impedance is equal to the termination, there will be no reﬂections. However, since it is unbalanced, it is expected that a portion of the energy will be transferred to the even mode. From a differential receiver’s point of view, this will look like extra channel loss. To demonstrate this concept, consider the case where the difference in length between the legs of the differential pair ( l) is 200 mils with a delay of 27.77 ps (refer to Figure 9-38). If the initial phase difference between v+ and v− at the driver is of 180◦, when the signals arrive at the receiver, the phase difference will no longer be equal to 180◦, due to the extra length on the second leg. The actual phase difference at the receiver can be calculated using equation (9-22) as a function of frequency. For example, at 1 GHz the phase difference at the receiver will be 180◦ + θ = 180◦ + (360◦)(27.77 × 10−12)(1 × 109) = 190◦ meaning that since the signals are no longer exactly 180◦ out of phase, and some of the energy is being dumped into the even mode. As the frequency increases, the phase difference due to the extra delay will approach 180◦. When θ = 180◦, the differential signal at the driver has been fully transformed into a common-mode signal at the receiver. When the delay difference between the legs of the differential pair is 27.77 ps, the frequency where the signal is 100% common mode at the receiver is 18 GHz: θ = (360◦)(27.777 × 10−12)(f180) = 180◦ ⇒ f180 = 18 × 109 NETWORK THEORY 405 1.0 Sdd21 Scd21 0.8 v v Sdd21, Scd21 0.6 −v −v q = 190° 0.4 q = 270° v q = 360° 0.2 −v 0.0 0 5 10 15 20 25 30 Frequency, GHz Figure 9-39 As the frequency increases, small asymmetries in a differential pair will cause larger phase differences between signals propagating on each leg, causing energy to be converted from odd to even mode. This shows up as differential insertion loss. In this case the differential signal is 100% converted to common mode at 18 GHz. So at 18 GHz, the phase difference at the receiver will be 180◦ + θ = 180◦ + (360◦)(27.77 × 10−12)(18 × 109) = 360◦ Therefore, at 18 GHz the differential insertion loss is zero because there is no longer any differential energy being transmitted to the receiver because all the energy has been converted to the common mode. Figure 9-39 shows the S-parameters for the loss-free perfectly terminated asymmetrical transmission line with a interleg delay difference of 27.77 ps. Note that the differential energy transferred from port 1 to port 2 (Sdd21) decreases until 18 GHz, after which it begins to increase again. At 1 GHz (where θ = 190◦), Scd21 = 0.1, which means that 10% of the differential energy is lost to the common mode. At 18 GHz, Scd21 = 1.0, so 100% of the energy is lost to the common mode. It is easy to see that the decrease in Sdd21 corresponds with the increase in Scd21. Do not falsely conclude that differential signals can be transmitted properly at frequencies above f180 (18 GHz in this example). Although Sdd21 increases, the phase difference between signals on each leg approaches 540◦ (3 · 180◦), not the ideal value of 180◦. For a digital signal, this means that the bit on line 1 is 180◦ out of phase with the next bit in the digital pulse train on line 2. Therefore, even though the common-mode conversion is small, the data are invalid. When loss is included in the transmission line, the Scd21 curve will peak at a lower frequency that does not correspond to the point where the differential-tocommon mode conversion is 100%, as shown in Figure 9-40. Be careful not to misinterpret the differential S-parameter data. The mode conversion is 100% when Sdd21 is zero, not necessarily when Scd21 is maximum. 406 NETWORK ANALYSIS FOR DIGITAL ENGINEERS 1.0 Loss curve (e −αz) 0.8 Scd21 (loss-free) Scd21, loss 0.6 0.4 0.2 Scd21 (lossy) 0.0 0 5 10 15 20 25 30 Frequency, GHz Figure 9-40 For a loss-free case, the differential-to-common mode conversion curve will be maximum when the phase difference between legs is 180◦. However, for a lossy case, the S cd21 curve will peak at a lower frequency, due to losses from the conductors and dielectrics. 9.3 PROPERTIES OF PHYSICAL -PARAMETERS S-parameters are a valuable tool in digital design. They can be used to analyze the performance of a component, extract equivalent-circuit models, or be employed as a tool-independent portable model. The problem arises that S-parameters are difﬁcult to generate accurately. For example, proper calibration of a vector network analyzer to high frequencies is both a science and an art. If probe effects are not removed or test ﬁxtures are deembedded improperly, the S-parameters could contain signiﬁcant errors that are difﬁcult to detect. Additionally, when component models (i.e., for a connector or CPU socket) are distributed to engineers in the form of a frequency-dependent S-parameter ﬁle, the assumptions of how the model was constructed are lost, so it is impossible to judge correctness or applicability. Fortunately, even though it is impossible to judge the accuracy of S-parameters without comparison to measurements or examination of the underlying model, the concepts outlined in Chapter 8 can be used to look for gross errors that blatantly violate the laws of physics. 9.3.1 Passivity As deﬁned in Section 8.2.2, a physical system is passive when it is unable to generate energy. For digital design, S-parameters are used to analyze the physical components of the bus, such as the transmission lines, sockets, and connectors, none of which generate energy. Therefore, in digital design, if an S-parameter measurement or model is shown to be nonpassive, either the VNA was not calibrated correctly or the underlying assumptions of the modeling methodology PROPERTIES OF PHYSICAL -PARAMETERS 407 are wrong. A practical test for the passivity of S-parameters can be developed directly from equations (8-22) and (8-23)[Ling, 2007]. Since power must be conserved, the power absorbed by the network (Pa) is equal to the power driven into the network less power ﬂowing out: (|ai |2 − |bi |2) = Pa (8-22) where Pa ≥ 0 for a passive network. If Pa < 0, the network is generating power and the system would be considered nonpassive. This allows equation (8-22) to be written in terms of the power wave matrices that will produce a real value for the power absorbed by the network. A system is passive if aHa − bHb ≥ 0 (8-23) where a is a matrix that contains all the power waves incident to each port, b contains the power waves coming out of each port, and xH indicates the conjugate transpose (sometimes called the Hermitian transpose), which is calculated by taking the transpose of x and then taking the complex conjugate of each entry. Using the deﬁnition of S-parameters from equation (9-18), the passivity requirement of (8-23) can be rewritten as b = Sa, bH = SHaH aHa − SHaHSa ≥ 0 aH(U − SHS)a ≥ 0 (9-68) where U is the unity (identity) matrix. Equation (9-68) leads to the general requirement for passivity: U − SHS ≥ 0 (9-69) If SHS is greater than 1, the requirement of (8-22) is violated and the system is not passive. A quick test to ensure passivity can be derived based on the eigenvalues of SHS. The eigenvectors ζ and eigenvalues λ are determined from the solution of the equation SHSξ = λξ Techniques for calculating the eigenvectors and eigenvalues are detailed by O’Neil [1991]; however, there are many software packages that can be used, such as Mathematica or Matlab. The eigenvectors can be formed into an N × N matrix, V = [ξ1 ξ2 ··· ξN ] = ζ11 ζ12 ... ζ21 ζ22 ··· ··· ··· ... ζN1 ζN2 ζ1N ζ2N · · · ζNN 408 NETWORK ANALYSIS FOR DIGITAL ENGINEERS with the property that VVT = U. The matrix SHS can be diagonalized into its eigenvalues with λ1 0 · · · SHS = VλVT = V 0 ... λ2 · · · ··· ... 0 0 ... VT 0 0 · · · λN (9-70) The deﬁnitions above allow (9-68) to be rewritten in terms of the eigenvalues: aH(VVT − VλVT)a ≥ 0 aHV(U − λ)aVT ≥ 0 (9-71) Equation (9-71) leads to a simple test for passivity [Ling, 2007]. Since U − λ must be greater or equal to zero and nonnegative, the system is passive if all the eigenvalues are between 0 and 1. The test for passivity is shown as 0 ≤ λi ≤ 1 (9-72) Some tools employ the strategy of applying correction algorithms for the nonpassive cases. Generally, the use of a passivity-corrected model or measurement is an unwise strategy because nonpassivity indicates an incorrect model or calibration. The best strategy is to reevaluate the method for equating the S-parameters and ﬁx the problem. 9.3.2 Reality Time-domain signals that are used to analyze digital interconnect are real. In other words, there is no imaginary part of a time-domain signal that can exist in nature. As mentioned in Section 8.2.1, the mathematical requirement for a frequency-domain response to ensure a real time-domain waveform is Sij (−ω) = Sij (ω)∗ (9-73) where∗ indicates the complex conjugate and j and i represent the driving and receiving ports of the S-parameters. Most S-parameter data are presented only with positive frequency-domain data, and the tools construct the negative frequency portion from (9-73). However, if the negative frequency values of the S-parameters are available, (9-73) is a useful check. 9.3.3 Causality As covered in Section 8.2.1, one seemingly obvious requirement of a model or measurement is that its output cannot precede its input. In other words, an effect cannot precede its cause. This fundamental principle is called causality. PROPERTIES OF PHYSICAL -PARAMETERS 409 Very often, VNA calibration problems or incorrect model assumptions generate S-parameters that are not causal. This error becomes especially critical when S-parameters are used to represent the behavior of physical channel components when simulating system performance. Causality errors will introduce delay and amplitude errors in the time-domain responses, as well as distort waveforms to such a degree that signal integrity may be falsely predicted, which can lead to false solution spaces, incorrect equalization settings, and incorrect understanding of bus performance. Many commercially available tools create models that are noncausal simply because of the assumptions of a frequency-invariant dielectric model where εr and tan δ are not properly related. Figure 8.14 is a good example of how a noncausal model can distort the waveform of a single bit. The errors depicted in Figure 8.14 will be ampliﬁed for very long channel lengths and high frequencies. Many of the digital designs created in the past were designed using noncausal models that assume frequency-invariant dielectric parameters. At low data rates (below about 1 to 2 Gbit/s) and short lengths (a few inches), causality errors do not appreciable affect signal integrity. However, at high frequencies and/or long lengths, they can completely destroy the model’s ability to predict realistic behavior. Causality requirements are often difﬁcult to judge by looking directly at S-parameters because the phase errors are small compared to the propagation delay of the structure. Furthermore, even if the causality errors of a single component such as a connector or via are small in isolation, when numerous models are cascaded in a complete channel simulation the errors become cumulative. Consequently, it is important to check the causality of each component as well as the system response for the entire channel. As described fully in Section 8.2.1 and demonstrated in Examples 8-3 and 8-4, causality can be tested by performing the Hilbert transform of the real part and ensuring that it is identical to the imaginary part. If it is not identical, the model is not causal. However, care should to be taken when performing this test because for bandlimited data, the Hilbert transform can exhibit truncation errors. Equation (8-19) is adapted to calculate the Hilbert transform of the real part of the S-parameter data, SˆRe,ij (f ) = Re[Sij (f )] ∗ 1 πf (9-74) where SˆRe,ij (f ) denotes the Hilbert transform of the real portion of the S-parameter, Sij (f ) is an S-parameter where port j is driving and port i is receiving, and ∗ denotes convolution. The S-parameter is causal when (9-75) is satisﬁed: SˆRe,ij (f ) = Im[Sij (f )] (9-75) To implement (9-74), it is often necessary to reconstruct the negative frequency components of the response using S(−f ) = S(f )∗ (9-36b) 410 NETWORK ANALYSIS FOR DIGITAL ENGINEERS 2.0 Noncausal Causal 1.5 h (t ) 1.0 0.5 Precursor error 2.5 5.0 7.5 10 Time, ns Figure 9-41 Classic impulse response behavior of a severely noncausal S -parameter. Note the precursor error, where part of the energy is arriving too early. The most straightforward way to calculate the Hilbert transform is to use the Fourier transform (usually in the form of the fast Fourier transform), as demonstrated in Examples 8-3 and 8-4. SˆRe,ij (f ) = F−1 F(Re[SRe,ij (f )])F 1 πf (9-76) If the fast Fourier transform is used, the negative frequency values must be appended to the end of the positive values, as demonstrated in Example 9-7. A less rigorous way to evaluate the causality is to transform the S-parameters into an impulse response, as described in 9.2.2. If there is no distinct point at which the pulse arrives, it is an indication of a noncausal response, as shown in Figure 9-41. 9.3.4 Subjective Examination of S-Parameters Often, it is desirable to recognize the trustworthiness of S-parameter data without the rigor of the mathematical analysis above. Understanding the basic properties of S-parameter data will allow an intuitive analysis. For example, a gross passivity violation can often be seen by observing the magnitude of the insertion loss. Figure 9-42 shows an example of measured insertion loss for a transmission line. Note that the scale often used with VNA measurements is, in decibels, dB = 20 log(mag) (9-77) where mag is the magnitude of the complex S-parameter value. Notice how S21 in Figure 9-42 rises above zero at the lower frequencies, indicating that the transmission line has gain and is therefore producing energy and nonpassive PROPERTIES OF PHYSICAL -PARAMETERS 411 Magnitude S21, dB 10 0 −10 −20 −30 −40 −50 −60 −70 −80 0 Passivity violation 5 10 15 20 25 30 35 40 45 50 Frequency, GHz Figure 9-42 Bad S -parameter measurement of a transmission line, showing passivity violations due to incorrect VNA calibration. Passivity violations Magnitude S21, dB 10 0 −10 −20 −30 −40 Calibration −50 looks marginal −60 −70 −80 0 5 10 15 20 25 30 35 40 45 50 Frequency, GHz Figure 9-43 Bad S -parameter measurement of a transmission line, showing passivity violations at high frequencies where the VNA calibration breaks down. (0 dB is equal to an insertion loss of 1 on a linear scale). In this case, the passivity violation was a result and an incorrect calibration of the vector network analyzer. Another example of a severe passivity violation is shown in Figure 9-43, where the calibration breaks down after about 17 GHz. Even below 17 GHz this measurement looks questionable, due to the signiﬁcant amount of nonperiodic ripples on the waveform. Periodic ripples would probably be a result of reﬂections or crosstalk, but nonperiodic noise on S21 indicates an inaccurate measurement. Figure 9-44 shows an example of a good measurement of a 5-in. microstrip transmission line on a FR4 substrate. Note that the insertion loss does not rise above 0 dB and that it does not contain any nonperiodic noise. It should be noted that the S-parameters could still be nonpassive even if S21 does not rise above 0 dB. Passivity can only be veriﬁed using the techniques outlined in Section 9.3.1; however, simple observation of S21 is often sufﬁcient to catch gross errors. The periodic behavior of S11 can also be used to get an intuitive feeling about the validity of the S-parameters by ensuring reasonable phase delays. If the structure under test is a transmission line, the return loss (S11) should be periodic and the distance between peaks (or dips) is related to the time delay with equation (9-4). The propagation delay can be double-checked at various 412 NETWORK ANALYSIS FOR DIGITAL ENGINEERS frequencies. From Figure 9-44 the propagation delays are calculated: τd1 = 2 1 f1 = 1 2(0.67 × 109) = 746 × 10−12 s ⇒ τd1 = 149 ps/in. 5 in. Magnitude S11, dB Magnitude S12, dB 0.0 0.0 S12 −2.0 −5.0 −10.0 −4.0 S11 −6.0 −15.0 −20.0 −25.0 −8.0 ∆f1 = 0.67 −10.0 ∆f2 = 0.7 −30.0 −35.0 −40.0 −12.0 −45.0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 Frequency, GHz Figure 9-44 Good S -parameter measurement with no obvious problems. Real 1.0 Imaginary When the real is at a max or min, the imaginary part is zero 0.8 0.6 0.4 0.2 S21, dB 0.0 −0.2 −0.4 −0.6 −0.8 −1.0 0.0 1.0 2.0 3.0 4.0 5.0 Figure 9-45 Good S -parameter measurement with no obvious problems, showing the proper relationship between the real and imaginary parts. PROBLEMS 413 Since the delay of a typical microstrip line on FR4 is about 140 to 160 ps/in., the result seems reasonable. At higher frequencies, the same test can be applied. From Figure 9-44, f2 = 0.7 GHz. τd2 = 2 1 f2 = 1 2(0.7 × 109) = 714 × 10−12 s ⇒ τd1 = 142 ps/in. 5 in. Note that the delay is less that it was at the lower-frequency test. This conﬁrms the validity of the measurement because the dielectric constant, and therefore the propagation delay, should decrease with frequency, as described in Chapter 6. Finally, the relationships between the real and imaginary parts of S21 can be examined. Figure 9-45 shows the real and imaginary parts of the insertion loss for the 5-in. transmission line measured in Figure 9-44. As discussed in Section 9.1.1, when the real part peaks, the imaginary part should be zero. REFERENCES Ling, Yun, 2007, Demystify S-parameters, I: The basics, class given at Intel Corporation. O’Neil, Peter V., 1991, Advanced Engineering Mathematics, Wadsworth, Belmont, CA. Hall, Stephen, Garrett Hall, and James McCall, 2000, High Speed Digital System Design, Wiley-Interscience, New York. PROBLEMS 9-1 Consider the measured S-parameters of a transmission line shown in Figure 9-46. What is the propagation delay of the transmission line? What is the characteristic impedance of the transmission line assuming that the port impedances are 50 ? Is the measurement valid? How can you tell? 9-2 Consider the S-parameters shown in Figure 9-47 of a transmission line. Are these S-parameters valid? How can you tell? 9-3 Consider the measured S-parameters of a transmission line shown in Figure 9-48. Are these S-parameters valid? How can you tell? 9-4 Derive the conversion from T -parameters back into S-parameters for two ports. 9-5 Derive the formulas to extract R, L, C, and G from a transmission line with arbitrary impedance. Prove that the method is accurate by extracting the frequency-dependent R, L, C, and G from a simulated transmission line. 9-6 Calculate the common-mode matrix and the common mode-to-differential conversion matrix of a four-port S-matrix. 414 NETWORK ANALYSIS FOR DIGITAL ENGINEERS S11S21 1.0 0.9 S21 0.8 0.7 S11 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 6 7 8 9 10 Frequency (GHz) Figure 9-46 Figure for Problem 9-1. Real S21(f ) 0.2 Imaginary 0.1 −3 −2 −1 0.1 Frequency, GHz 1 2 3 0.2 Figure 9-47 Figure for Problem 9-2. 9-7 If a differential pair has 10% of its energy transformed into the common mode when driven differentially, is 10% of its energy transformed into the differential mode when driven commonly? Show the math to support your answer. 9-8 Derive the ABCD matrix for the pi circuit shown in the bottom of Table 9-2. 9-9 For the differential pair in Figure 9-49, at what frequency will the ACCM be 100%? Assume that the effective permittivities are εr,odd = 3.5 and εr = 4.0. PROBLEMS 415 S11,S21 1.0 0.9 S21 0.8 0.7 0.6 0.5 S11 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 6 7 8 9 10 Frequency, GHz Figure 9-48 Figure for Problem 9-3. v ∼ 50 Zodd = 50 50 ∼ 50 −v Zodd = 50 7 inches Z0 = 50 50 I = 100 mils Figure 9-49 Figure for Problem 9-9. 9-10 Derive the equations to renormalize the S-matrix measured with 50loads at each port to arbitrary and different reference impedances at each port. 9-11 Derive the two-port S-parameters for a frequency-dependent, causal 10-in. transmission line from 10 Hz to 100 GHz. Using a tool such as Mathematica or Matlab, plot the S-parameters and prove that they are passive and causal. 9-12 From the S-parameters calculated from the transmission line in Problem 9-11, calculate the pulse (single-bit) response for a 15-Gbit/s data rate. 9-13 Create the S-matrices for 4- and 10-in. transmission lines using the method derived in Problem 9-11. Cascade the S-parameters to get the equivalent S-matrix for a 14-in. transmission line. Compare the cascaded S-matrix to that of a single 14-in. transmission line. 10 TOPICS IN HIGH-SPEED CHANNEL MODELING 10.1 Creating a physical transmission-line model 418 10.1.1 Tabular approach 418 10.1.2 Generating a tabular dielectric model 419 10.1.3 Generating a tabular conductor model 420 10.2 NonIdeal return paths 422 10.2.1 Path of least impedance 422 10.2.2 Transmission line routed over a gap in the reference plane 423 10.2.3 Summary 434 10.3 Vias 434 10.3.1 Via resonance 434 10.3.2 Plane radiation losses 437 10.3.3 Parallel-plate waveguide 439 References 441 Problems 442 So far, we have covered numerous topics that are required for the design of modern high-speed digital systems. There are a few important topics, however, that do not ﬁt nicely into the ﬂow of the other chapters. This section covers a few critical aspects of digital design that are important to understand. First, a methodology for creating frequency-dependent tabular transmission-line models is presented. The tabular format is a convenient way to model the frequency-dependent nature of the conductor and dielectrics described in Chapters 5 and 6. Next, the problem of nonideal current return paths is explored so that the engineer will understand how they affect signal integrity. Finally, the signal integrity impact of transitioning between layers using plated-through-hole vias is discussed. Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 417 418 TOPICS IN HIGH-SPEED CHANNEL MODELING 10.1 CREATING A PHYSICAL TRANSMISSION-LINE MODEL Digital design is typically performed using circuit simulators. Fortunately, most commercially available tools have transmission-line models that can be integrated with the digital circuitry to perform system-level bus simulations. Problems arise, however, when the assumptions built into the models break down or are simply not known by the engineer. For example, many simulators employ transmission-line models with frequency-invariant dielectric permittivity and loss tangents. Although these models are perfectly adequate for very short bus lengths or data rates below about 1 Gbit/s the incorrect relationship between εr and tan δ induces causality errors (such as the model in Example 8-4) that can render the simulations almost useless. Additionally, few commercial simulators account properly for such things as surface roughness or internal inductance. Consequently, it is usually desirable for the digital designer to create custom transmission-line models that use the methodologies presented in Chapters 3 through 6 to ensure accurate results. 10.1.1 Tabular Approach A convenient method of implementing user-deﬁned transmission-line models is to use a tabular approach, which is simply a lookup table that deﬁnes the transmission-line equivalent-circuit parameters at each frequency. Fortunately, user-deﬁnable tabular models are available in many commercial simulators. A tabular methodology allows calculation of the system transfer function in the frequency domain. An inverse Fourier transform can then be used to get a time-domain impulse response, which can be convolved with an arbitrary input waveform to evaluate signal integrity. A practical and efﬁcient approach to incorporate surface roughness, internal inductance, and wideband frequency-dependent dielectric properties into printed circuit board (PCB) transmission-line models is deﬁned here as a two-step procedure. In the ﬁrst step, quasistatic RLGC matrices in the transmission line model are generated at a single reference frequency (ωref), using a 2D transmission-line calculator or analytically with the methods outlined in Chapters 3 through 6. In this step it is important that the value of εr and tan δ be known at the reference frequency. This model will reﬂect the single-valued frequency response of the transmission-line geometry and conﬁguration but will not yet account for the frequency dependence of surface roughness losses, internal inductance, and dielectric properties. The matrices calculated during this step will be known as the reference values R(ωref), L(ωref), C(ωref), and G(ωref). In the second step, the reference values are modiﬁed to account for frequency-dependent material properties, internal inductance, and surface roughness [Liang et al., 2006]. CREATING A PHYSICAL TRANSMISSION-LINE MODEL 419 10.1.2 Generating a Tabular Dielectric Model The real portion of the dielectric permittivity, as calculated with ε (ω) ≈ ε∞ + m2 ε − m1 ln(ω2/ω) ln(10) (6-30a) is used to determine the frequency-dependent capacitance C(ω) by dividing C(ωref) by the value of the dielectric permittivity calculated at the reference frequency ε (ωref) and multiplying by the frequency-dependent value using equation (10-1), where boldfold type indicates a matrix for the multiconductor case: ε (ω) C(ω) = C(ωref) ε (ωref) → C(ω) = C(ωref)ε (ω)ε −1(ωref) (10-1) To implement equation (10-1), the effective dielectric permittivity must be calculated from the reference values. As detailed in equation (3-74), the effective permittivity can be calculated by dividing the capacitance of each value in C by the corresponding value in Cair, which is calculated by setting the dielectric permittivity to unity (εr = 1). Cεr,eff Cεr =1 = εr,eff (3-74) As described in Chapter 3, the values in Cair can be calculated directly from the inductance using equation (3-46), assuming that the magnetic permeability is unity (µr = 1), which is almost always the case because copper is usually the metal of choice. L = 1 c2Cεr =1 (3-46) Equations (3-74) and (3-46) lead to equation (10-2), which is the value of the dielectric permittivity at the reference frequency: ε (ωref) = C(ωref)C−air1 = c2[C(ωref)L(ωref)] (10-2) The frequency-dependent loss tangent is calculated as ε (ω) ≈ m2 ε − m1 −π/2 ln(10) (6-30b) 420 TOPICS IN HIGH-SPEED CHANNEL MODELING tan |δ| = ε ε The deﬁnition of G(ω) in shown in equation (6-45): (6-16b) G(ω) = tan δ ωC (6-45) The reference value of the loss tangent is calculated by rearranging equation (6-45): tan δ(ωref) = G(ωref) C(ωref)ωref → tan δ(ωref) = G(ωref)C−1(ωref) ωref (10-3) For a microstrip line, the values extracted would be the effective dielectric constant and effective loss tangent. This is an easy way to handle with nonho- mogeneous dielectrics, such as in a microstrip conﬁguration. For a stripline, the values extracted would be equal to the bulk properties of the material, including any nonhomogenous structures such as the ﬁber weave. Next, G(ωref) is scaled by the frequency-dependent loss tangent and permit- tivity, leading to (10-4), which is the frequency-dependent conductance of a transmission line. G(ω) = ε (ω) tan δ(ω) ω G(ωref) ε (ωref) tan δ(ωref) ωref → G(ω) = [tan δ−1(ωref) ε −1(ωref)]G(ωref)ε (ω)tan δ(ω) ω ωref (10-4) Using this procedure, equations (10-1) and (10-4) can be used to create the tabular transmission-line parameters that represent a frequency-dependent causal dielectric model that is valid for the conditions described during the derivation of the inﬁnite-pole model in Section 6.3.5. However, it should be noted that a transmission line using any dielectric model can be implemented using this technique, providing the frequency-dependent behavior of ε and ε are known. 10.1.3 Generating a Tabular Conductor Model When creating a frequency-dependent conductor model for a transmission line, Chapter 5 will remind the reader of the four effects that must be accounted for properly: 1. External inductance 2. Internal inductance 3. Dc resistance 4. Ac (skin effect) resistance CREATING A PHYSICAL TRANSMISSION-LINE MODEL 421 Quasistatic techniques, which are used by most commercial transmission-line calculators, provide reasonable values of the external inductance. Furthermore, many tools also provide realistic values of the skin effect and dc resistance as well as the internal inductance for ideal smooth conductors. Problems arise when creating models of realistic conductors, which are usually purposely roughened during the PCB manufacturing process to ensure adhesion of the metal traces to the dielectric substrate. Fortunately, the shortcomings of frequency-invariant quasistatic techniques can easily be overcome by using a tabular transmission-line model. Similar to the tabular model for the dielectric outlined in Section 10.1.2, the easiest methodology is to use a conventional 2D transmission-line calculator that assumes perfectly smooth conductors, and modify the output to establish the proper frequency-dependent relationships between values R(ω) and L(ω). The output of the transmission-line calculator will provide the values of the reference matrices R(ωref) and L(ωref). The reference values for each entry in the matrix are then scaled according to R(f ) = KSR(f )Rs f fref Rdc when δ < t when δ ≥ t (10-5) L(f ) = Lexternal + R(f ) 2πf Lexternal + R(fδ=t ) 2πfδ=t when δ < t when δ ≥ t (10-6) where the frequency-dependent surface roughness correction factor KSR(f ) is calculated with one of the modeling methods presented in Chapter 5 [i.e., Hammerstad equation (5-48), the he√mispherical model in equation (5-58), or the Huray model in equation (5-66)], Rs f is the skin effect resistance assuming perfectly smooth conductors, t is the conductor thickness, δ is the skin depth, fδ=t is the frequency where the skin depth equals the thickness of the conductor, and fref is the reference frequency where the seed values are calculated [Hall et al., 2007]. If the 2D transmission-line calculation does not include the internal inductance, the value of the reference frequency (fref) is immaterial because all the current is assumed to reside on the surface of the conductor and the internal inductance is zero. However, if the calculation includes the internal inductance, as described in Section 5.2.3, the reference frequency should be high enough that most of the current is conﬁned to a small layer near the surface, so that the internal inductance is minimized. This simpliﬁes the problem so that the reference inductance can be considered to be purely external and the internal portion can simply be added. Note that even at relatively low frequencies, the skin depth in copper is small compared to the thickness of the trace. For example, at 1 GHz, the skin depth is 2 µm, meaning that for a typical PCB trace thickness of about 1 to 2 mils 422 TOPICS IN HIGH-SPEED CHANNEL MODELING (25 to 50 µm), most of the current is conﬁned near the surface and the inductance is mostly external. 10.2 NONIDEAL RETURN PATHS So far, we have covered many of the advanced topics associated with high-speed digital design. Each topic has been rigorously covered analytically, beginning with fundamental theory and ending with a practical method that can be used for design. Now it is time to focus on what is often considered the most ambiguous concept of high-speed design: nonideal return paths. Many of the effects detailed in this section are very difﬁcult or impossible to model analytically, or even with circuit simulators. Although 3D electromagnetic simulators are capable of capturing full effects for small structures, they require too much computational power to simulate the effects reliably for an entire system. Subsequently, in this section we will focus less on speciﬁc modeling techniques and more on the general impact, physical mechanisms, and trends that nonideal return current paths have on signal integrity. As a general rule, great care should be taken to ensure that nonideal current return paths are minimized. 10.2.1 Path of Least Impedance As discussed in Chapter 3, a signal on a transmission line propagates between the signal trace and the reference plane in the form of an electromagnetic wave. Consequently, the physical characteristics of the reference plane are just as important as that of the signal trace. A very common mistake, even for experienced designers, is to focus on providing a very clean and controlled signal trace with no thought whatsoever of the reference plane. Remember that any current injected into a system must return to the source. It will do so through the path of least impedance. Figure 10-1a depicts a CMOS output buffer driving a microstrip line. The currents shown represent the instantaneous values that occur when the driver switches from a low state to a high state. Just prior to the transition (time t = 0−) the line is grounded through the NMOS. Immediately after the transaction (time t = 0+) the buffer switches to a high state and current ﬂows onto the line until it is charged up to VDD. As the current propagates down the line, a mirror current is induced on the reference plane, which ﬂows in the opposite direction as described in Section 3.2.3 and shown in Figure 3-8. To complete the loop, the current must ﬁnd the path of least impedance, which in the case of Figure 10-1a is the voltage supply VDD. A nonideal return path occurs when (1) a discontinuity in the reference plane causes return current to diverge from the ideal path, or (2) the return current must ﬂow through a region of increased impedance, such as bond wires, antipads in breakout areas, or socket pins. A few examples are shown in Figure 10-1b. When a physical discontinuity exists in the return path that forces the current to diverge from the ideal path, the total area of the current loop is increased. An NONIDEAL RETURN PATHS 423 VDD + − I Off at t = 0−, On at t = 0+ I On at t = 0−, Off at t = 0+ I (a) Bond wire & package pin VDD Signal line Ground plane Power island Driver Via field Return current Decoupling caps (b) Figure 10-1 (a) Example of an ideal return current path of a ground-referenced transmission line driven by a CMOS buffer; (b) examples of common nonideal return paths. increase in the loop area leads to an increase in inductance, which degrades the signal integrity. Subsequently, the most fundamental effect of a discontinuity in the return path is an effective increase in the series inductance with a magnitude that depends on the distance the current must diverge. When a high-impedance region such as a bond wire or inductive socket pin occurs in the return current path, it is similar to having the discontinuity in the signal path. 10.2.2 Transmission Line Routed over a Gap in the Reference Plane To demonstrate the effect of a nonideal return current path, refer to Figure 10-2, which depicts a microstrip line traversing a gap in the reference plane. This is a convenient return path discontinuity to analyze because it is a simple structure, the return current path is well understood, and it has similar trends in the more complicated structures. As the signal current travels down the transmission line, the return current is induced on the reference plane. When the signal reaches the gap, a small portion of the return current propagates across the gap through the gap capacitance, and the other portion is forced to travel around the gap. The extra path length of the return current causes the total current loop area to increase, which increases the inductance. This means that from the driver’s point of view, the gap looks inductive. To demonstrate this, consider Figure 10-3, which is a TDR measurement of a 65- microstrip transmission line traversing a 25-mil 424 TOPICS IN HIGH-SPEED CHANNEL MODELING D2 Driver Signal current Return current Receiver D1 w Figure 10-2 Driving and return currents when a signal passes over a gap in the ground plane. 50 Zo = 50 0.5 V TDR Cable Probe point Gap width = 25 mils 65Ω microstrip TDR 0.6 0.5 0.4 0.3 0.2 0.1 0.0 Volts 0 2 4 6 8 10 12 Time, ns Figure 10-3 Measured TDR proﬁle of a microstrip traversing a gap in the reference plane, showing the inductive nature of the nonideal return path. gap in the reference plane. Notice how the gap appears inductive in the TDR proﬁle. The receiver waveform will be ﬁltered by the low-pass nature of an inductor, and a portion of the signal will be delayed due to the ﬁnite time it takes the return current to propagate around the gap. A simulation of how a nonideal NONIDEAL RETURN PATHS 425 0.30 D = 37 mils 0.25 0.20 D = 93 mils 0.15 0.10 0.05 D = 185 mils Probe point D 50 Z0 = 50 0.5 V D 50 Volts 0.00 1.00 1.05 1.10 1.15 1.20 Time, ns Figure 10-4 Simulated step response of the waveform at the receiver of a microstrip traversing a gap in the reference plane, showing how different return path divergent lengths D affect the waveform received. return path can affect the receiving waveform is shown in Figure 10-4 using a loss-free transmission line. If the distance of the current return path (2D) is small compared to the edge rate, the gap will simply look like a series inductance in the middle of the transmission line. The extra inductance will ﬁlter out some of the high-frequency components of the signal, degrade the edge rate, and round the corners. When the electrical length of the return path, becomes longer than the rise or fall times, however, a ledge will appear in the waveform. The length of the ledge (in time) will be dependent on the distance the return current must travel around the discontinuity [Hall et al., 2000]. Since the width of the gap will govern the bridging capacitance and thus the portion of current shunted across the gap, the height of the ledge will be dependent on the gap width, as shown in Figure 10-5. The larger the gap width, the less the capacitive coupling and the lower the height of the ledge. Equivalent Circuit An equivalent circuit of a transmission line traversing a slot in the reference plane can be surmised by careful observation of the behavior described earlier. First, consider the measured TDR response shown in Figure 10-3. At the point where the transmission line crosses the slot, the TDR shows a voltage spike that represents the high-impedance, or inductive, section of the interconnect. As described earlier, the voltage spike is a function of the distance that the return current must diverge. Figure 10-6 shows how the return current ﬂows parallel to the slot until it reaches a low-impedance pathway, where it can complete the loop (such as a short or a bridging capacitor). When current ﬂows on one side of the slot, it induces a mirror current on the opposite side, an electric ﬁeld is established, and a transmission line is formed. This type of transmission line, called a slotline, is shown in Figure 3-2. Therefore, the equivalent circuit must contain a transmission line to represent the microstrip, a 426 TOPICS IN HIGH-SPEED CHANNEL MODELING 0.30 0.25 D = 278 mils w = 8 mils 0.20 w = 25 mils Probe point 0.15 Volts 0.10 0.05 0.00 1.00 D 50 Z0 = 50 0.5 V D w = gap width 1.05 1.10 1.15 Time, ns 50 1.20 Figure 10-5 Simulated step response, showing how the height of the ledge is dependent on the gap width. Signal current Return current Induced slotline transmission mode Driver Receiver Figure 10-6 The current that diverges around the slot causes energy to propagate, allowing the slot to be modeled as a transmission line. transmission line to represent the slot, and a coupling mechanism to model the energy transferred from the microstip to the slotline. Since a transmission-line analogy can be used to represent the slot instead of discrete L and C, the voltage spike seen in the TDR and ledge height at the receiver can be calculated by determining how much of the total current diverges around gap using a simple current divider. Consider the transmission line traversing a slot shown in Figure 10-6, where the driving transmission line is perfectly terminated in its characteristic impedance to simplify the analyses. The current ﬂowing around the gap is dependent on the slotline impedance and is calculated with a current divider as in igap = idrive 2Z0 2Z0 + 1 2 Zgap (10-7) NONIDEAL RETURN PATHS 427 where Z0 is the impedance of the transmission line, Zgap the impedance of the slotline, and idrive the current launched onto the transmission line: idrive = vi Z0 = v[Z0/(Z0 + Rs )] Z0 On both sides of the slot the desired current path exists that is perfectly terminated. The current that does not jump the slot will ﬂow around it, which represents the alternative return path. Since the transmission line is routed over the center section of the slot, current will ﬂow in both directions, and the effective gap impedance will be Zgap/2. The voltage spike seen in the TDR waveform is calculated using the portion of the current that jumped the gap: vspike = (idrive − igap)Z0 (10-8) The width of the voltage spike is equal to twice the delay of the distance the current must diverge, which can be calculated from the propagation delay of the slotline. Although some formulas exist to calculate the impedance and prop- agation delay of a simple slotline, the stackup geometry of real boards varies signiﬁcantly, so an electromagnetic simulator should be used. The ledge at the receiver is dependent on the current ﬂowing in the gap. The ledge will stay constant for a duration that equals twice the delay of the current return path divergence: vledge = igapZ0 (10-9) An equivalent circuit that mimics the behavior described above can be constructed as shown in Figure 10-7. The top circuit line represents the sourcing transmission line that crosses a gap. The voltage-controlled voltage source (VCVS) represents the physical location where the transmission line crosses the slot. The bottom circuit is a transmission line that represents the slot and the current-controlled current source (CCCS) is placed at the physical location on the slot where the transmission line crosses. The current in the VCVS is mirrored into the CCCS on the slotline, which represents the current diverging around the gap. In turn, the current sourced into the gap produces a voltage, which is mirrored back into the VCVS and accurately predicts both the inductive voltage spike seen at the driver and the ledge seen at the receiver. Although this simulation method is quite simple, it works very well and can be used to predict the signal integrity of one or more lines crossing a slot in a reference plane. An example of the model accuracy compared to a TDR measurement is shown in Figure 10-8. Example 10-1 Calculate the step response of a 50- microstrip transmission line traversing from a ground plane to a VDD island as shown in Figure 10-9. Assume that the gap acts like a slotline with an impedance of 90 . Since it is 428 TOPICS IN HIGH-SPEED CHANNEL MODELING Driving line TDR 50 Gain = 1 Vx [VCVS] ϩϪ Z0 lx V Slot location TDT Z0 50 D1 [CCCS] D2 Slot Zslot + Vx Zslot lx − G=1 Represents crossing of driver line Figure 10-7 Equivalent circuit of a transmission line routed over a slot in the reference plane. TDR 0.6 0.5 0.4 0.3 0.2 0.1 0.0 Model Measured Volts 0 2 4 6 8 10 12 Time, ns Figure 10-8 Comparison of a slot model compared to a measured response on a test board. generally not a good idea to short VDD to ground, 1-nF decoupling capacitors are used to create a high-frequency low-impedance path for the return current. SOLUTION Step 1: Estimate the electrical delays of the microstrip and the slotline assuming a board dielectric permittivity of 4. Since both the slotline and the microstrip will have part of the electric ﬁelds in the air and part in the dielectric, we estimate that the effective relative dielectric permittivity for both is εeff = 3.15. This value can be calculated rigorously using the techniques outlined in Chapter 3, NONIDEAL RETURN PATHS 429 TDR D = 1 inch 50 Z0 = 50 2 inch 0.5 V D = 1 inch Ground C VDD 2 inch C TDT 50 Figure 10-9 Transmission line routed onto a ﬂoating power island—a common occurrence. or an electromagnetic simulator can be used. However, for the purposes of this example, the estimation is adequate. The propagation delay is therefore calculated for both the microstrip and slotline: √ √ td = εeff c = 3.15 3 × 108 m/s ⇒ 150 × 10−12 s/in. Step 2: Create the equivalent circuit. Using the propagation delay, the electrical length of each segment is calculated. The total length of the slot is the segment between the decoupling capacitors. Since the capacitors are large, they will look like a short for high-frequency return current. In the equivalent circuit, they are placed at the end of the slot. The equivalent circuit is shown in Figure 10-10. Step 3: Simulate the circuit model. The step response is shown in Figure 10-11. The TDR is the waveform seen at the driver and the TDT is the waveform seen at the termination resistor at the far end of the line (as labeled in Figure 10-10). Driving line TDR 50 td = 300 ps Gain = 1 Vx +− 50 Ω Ix 0.5 V TDT [VCVS] td = 300 ps 50 Ω 50 1 nF td = 150 ps [CCCS] td = 150 ps 1 nF 90 Ω + 90 Ω 1 nF Slot Vx Ix − G=1 Figure 10-10 Equivalent circuit of the structure shown in Figure 10-9. 430 TOPICS IN HIGH-SPEED CHANNEL MODELING 0.35 0.30 0.25 TDR 0.20 0.15 0.10 0.05 300 ps 77.5 mV TDT 172.4 mV Volts 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 Time, ns Figure 10-11 Simulated waveform of a transmission line routed onto a ﬂoating power island. The width of the voltage spike and the ledge should be twice the return-path divergence length, which is 2 · 150 ps = 300 ps. The values of the initial voltage step in the TDR and the ledge in the TDT can be calculated to double-check the simulation using equations (10-7) through (10-9): idrive = vi Z0 = v[Z0/(Z0 + Rs )] Z0 = 0.5[50/(50 + 50)] 50 = 5×10−3 A igap = idrive 2Z0 2Z0 + 1 2 Zgap = 5 × 10−3 100 100 + 45 = 3.448 × 10−3 A The voltage of the spike in the TDR is vspike = (idrive − igap)Z0 = (5 × 10−3 − 3.448 × 10−3)50 = 77.5 × 10−3 V and the ledge in the receiver waveform is vledge = igapZ0 = 3.448 × 10−3 · 50 = 172.4 × 10−3 V which agrees with the simulated waveform in Figure 10-11. Nonideal Return Paths and Crosstalk Another consequence of many nonideal return paths is a dramatic increase in crosstalk. Conventional crosstalk is caused by mutual inductance and capacitance. For the case of a transmission line NONIDEAL RETURN PATHS 431 traversing a split in the reference plane, the coupling mechanism is the slot itself. A voltage is established between each side of the slot and propagates via the slotline transmission mechanism from the sourcing line to the victim, where it couples onto the line with much more efﬁciency than conventional crosstalk. Figure 10-12a shows an example of two transmission lines routed over a slot in the ground plane, and Figure 10-12b shows a measured response of a test board with a pair of 65- microstrip transmission lines 1.4 in. apart traversing a 25-mil gap. Note that the coupling coefﬁcient is approximately 20% (0.05 mV/0.25 mV), which is much higher than what would be expected between the two microstrips routed over a solid reference plane. The structure shown in Figure 10-12a can be modeled similar to the way it was done for the single line crossing a slot, as shown in Figure 10-13. The top circuit line represents the sourcing transmission line that crosses a gap. The voltage-controlled voltage source (VCVS) represents the physical location where the transmission line crosses the slot. The middle circuit is a transmission line that represents the slot and the current-controlled current sources (CCCSs) are placed at the physical location on the slot where the transmission lines cross. The current ﬂowing through the VCVS on the sourcing line is mirrored into the CCCS on the slotline, which represents the current diverging around the gap. The same circuit is duplicated for the victim transmission line, which effectively models the coupling by means of the slotline. Figure 10-14 shows a simulated example of loss-free slotline coupled crosstalk, where D = 0.462 in. and S = 1.0 in. with 50- microstrips. Note that the coupled voltage is about 22% of the voltage Slot Driving line D 50 Victim line S 500 mV 50 A 50 Near end D Far end 50 (a) 0.1 0.0 Volts at A −0.1 0 2 4 6 8 10 Time, ns (b) Figure 10-12 (a) Two transmission lines passing over a slot in the reference plane used to develop an equivalent circuit; (b) measurement of coupled voltage when S = 1.4 in., slot width = 25 mils, and 65 - microstrip transmission lines on FR4. 432 TOPICS IN HIGH-SPEED CHANNEL MODELING Driving line Slot TDR R td, 1 Z0 0.5 V Gain=1 Vx +− Ix [VCVS] TDT td, 2 Z0 R D td, gap1 Zgap [CCCS] S td, gap2 + Zgap Vx Ix − Gain = 1 [CCCS] D td, gap1 + Zgap Vy Iy − Gain = 1 Near- end Victim linecrosstalk R td, 3 Vy Gain = 1 [VCVS] td, 4 +− Z0 Iy Z0 Far-end crosstalk R Figure 10-13 Equivalent circuit of two transmission lines crossing over a slot in the reference plane. Volts 0.06 0.04 0.02 0.00 −0.02 −0.04 −0.06 0.7 0.8 0.9 0.1 1.1 1.2 1.3 1.4 1.5 Time, ns Figure 10-14 Crosstalk coupled via a slotline transmission line on a victim line 1.0 in. away. launched on the sourcing line (250 mV). If multiple lines are crossing a gap in the reference plane, a signiﬁcant amount of noise would be coupled onto the bus, which can destroy signal integrity. Differential Differential signaling is much more immune to the harmful effects of nonideal return paths because the adjacent transmission line provides a relatively low-impedance path for the current to return. From a mathematical point of NONIDEAL RETURN PATHS 433 Volts at driver 0.4 Single ended 0.3 0.2 Differential 0.1 0.0 −0.1 −0.2 −0.3 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Time, ns Figure 10-15 TDR of a single-ended and a differential transmission line traversing a gap in a reference plane, showing that differential routing signiﬁcantly reduces the negative impact of a nonideal return path. Note the decreased inductive spike seen in the differential response. view, the virtual reference plane that exists between the conductors in a differential pair can be thought of as the return path, as shown in Figure 7-7 and discussed in Section 7.3. The increased immunity of a differential signal to a nonideal return path is demonstrated in Figure 10-15, which shows both a single-ended and a differential TDR waveform for a signal traversing a 25-mil-wide gap in the reference plane. Note that the inductive spike is much smaller for the differential case. The step response seen at the receiver is shown in Figure 10-16, where the 0.4 Single ended 0.3 Volts at reciever 0.2 Differential 0.1 0.0 −0.1 −0.2 −0.3 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Time, ns Figure 10-16 Step response at the receiver of a single-ended and a differential transmission line traversing a gap in the reference plane, showing that differential routing signiﬁcantly reduces the negative impact of a nonideal return path. 434 TOPICS IN HIGH-SPEED CHANNEL MODELING differential response shows almost no ill effects. Differential routing will also reduce signiﬁcantly the crosstalk seen between pairs because much less current is forced to ﬂow around the gap. 10.2.3 Summary Section 10.2.2 we used the example of a transmission line traversing a gap in the reference plane to demonstrate the general behavior seen with any nonideal current return path. In summary, the signal integrity of a bus with nonideal return path(s) will exhibit the following attributes: 1. Appear as an inductive discontinuity in a TDR waveform 2. Slow the edge rate at the receiver 3. Severely distort the waveform at the receiver if the electrical delay of the current divergence path is longer than the rise and fall times 4. Create unwanted transmission paths that can increase coupling or loss sig- niﬁcantly Differential signaling is much more immune to the effects of a nonideal return path because the adjacent signal (or the virtual reference plane) provides a continuous reference for the ﬁelds. Hall et al. [2000] discuss numerous other nonideal current return paths. 10.3 VIAS For high data rates, vias can, if not designed, properly degrade the signal integrity of a binary bit stream signiﬁcantly. For single-ended signaling, if a suitable return path via is not placed in close proximity to every signal via, a nonideal return path is created. For differential signaling, if a ground via placement is not symmetrical with respect to each leg of the differential pair, asymmetry can be introduced, resulting in differential energy being converted to common mode (as described in Section 7.6.1). Finally, stubs associated with plated-through-hole vias can resonate, causing severe distortion of waveforms. In this section we cover the largest pitfalls of via design and discuss ways to mitigate the ill effects through proper design. 10.3.1 Via Resonance† To begin, consider the cross section of an eight-layer printed circuit board (PCB) shown in Figure 10-17. The ﬁgure shows a signal routed on signal layer 1 transitioning through a plated-through-hole (PTH) via to signal layer 4. The ground via †The authors would like to thank Guy Barnes of Ansoft Corporation for performing the 3D simulations used in this section. His technical knowledge, modeling expertise, and patience were invaluable. VIAS 435 Port 1 i −i Via pad Signal layer 1 Reference layer 2 Reference layer 3 Signal layer 4 Port 2 Signal via Stub length Ground via Signal layer 5 Reference layer 6 Reference layer 7 Signal layer 8 Back drilled portion Figure 10-17 Cross section a a backdrilled PTH via with a stub showing the nonideal current return path through the ground via. is included to provide a low-impedance path for the return current. Standard PTH vias are constructed by drilling a hole in the board and plating the walls with metal. A consequence of this manufacturing process is a “stub” that hangs off the via for all cases except when the signal transitions thorough the entire board (layers 1 to 8 in Figure 10-17). As will be shown later, the via stub has negative effects on the signal integrity. A common design practice is to shorten the length of the stub with precision depth backdrilling, as also depicted in Figure 10-17. If a current i is driven into port 1 in Figure 10-17, it will propagate down the transmission line on signal layer 1, pass through the via down to signal layer 4, and terminate at port 2. The return current −i will be mirrored on reference plane 2. Since the nearest return path from port 1 to port 2 is through the ground via, the return current must diverge away from the ideal path. The magnitude of the via inductance is therefore a function of the length of the via transition from signal layer 1 to signal layer 2 and the distance between the signal via and the ground via. The capacitance of the via will be dependent on the area of the via pads, the distance between the via pads and the closest reference layer, the capacitance of the via barrel to the adjacent planes it is passing through, and the capacitance of the stub. These observations make it possible to deduce two different forms of an equivalent circuit. Figure 10-18a shows the equivalent circuit of a via for the case where the stub length is zero and reference layers 2 and 3 are very far apart. In this scenario, which is almost always the case with standard four-layer PCBs, the via capacitance is dominated by the pads, so a very simple pi model can be used. Figure 10-18b shows an equivalent circuit where reference layers 2 and 3 are close in proximity and a stub may or may not exist. In this case, the barrel and stub capacitances are no longer negligible and are simply combined with the pad 436 TOPICS IN HIGH-SPEED CHANNEL MODELING Cpad Lbarrel Cpad Cvia Lbarrel (a) (b) Figure 10-18 Equivalent-circuit approximations of vias: (a) thick-core model, where the capacitance between reference layers is very low and can be ignored; (b) when the reference planes are close together, as reference layers 2 and 3 in Figure 10-17, the capacitance between layers can no longer be ignored and the equivalent circuit changes. capacitance. Regardless of the speciﬁc conﬁguration, the inductance and capac- itance of the via structures forms a tank circuit that will resonate at its natural frequency: f0 = √1 2π LbarrelCvia (10-10) For long via stubs, the resonant frequency can easily be low enough to inﬂuence signiﬁcantly the signal integrity of the digital waveforms. To demonstrate how vias behave at high frequencies, consider Figure 10-19, which shows the simulated insertion loss of a plated-through-hole via for a structure similar to Figure 10-17 with different stub lengths. The simulations were 0 No stub −1 −2 −3 18.9 mils −4 −5 37.8 mils S21, dB −6 44.1 mils −7 −8 56.6 mils −9 −10 0 5 10 15 20 25 30 Frequency, GHz Figure 10-19 Via resonance as a function of stub length for the cross section in Figure 10-17. VIAS 437 TABLE 10-1. HFSS-Predicted Resonance Compared to Tank Circuit Approximation Stub Length (mils) HFSS-Predicted Resonance (GHz) Q3DPredicted L (nH) Q3DPredicted C (pF) Tank Resonance (GHz) 56.6 12 0.139 0.636 16.9 44.1 15.5 0.141 0.586 17.5 37.8 22 0.141 0.419 20.7 performed using Ansoft’s HFSS 3D electromagnetic simulator with loss-free dielectrics and perfect conducting metals. The longer stub lengths exhibit more capacitance and therefore resonate at lower frequencies. The problem with simple circuit models is that it can be very difﬁcult to calculate the via capacitance and inductance for all but the simplest conﬁgurations. Consequently, it is necessary to extract the parasitic inductance and capacitance values from measurements or simulate them with a 3D electromagnetic simulator, such as Ansoft’s Q3D. Table 10-1 shows the extracted parasitic values using Q3D and the tank circuit resonance calculated with equation (10-10) for the structures simulated in Figure 10-19. Note that the tank circuit only approximates the actual resonance within a few gigahertz. The discrepancy between the HFSS and tank circuit results is due to the electrical delay of the physical geometry of the structure. Although the extracted inductance and capacitance are true values, a simple tank circuit approximation does not account for the physical delay of the return current diverging to the ground via. As the ground via is moved farther away, the inductance will increase. However, the peaks seen in the S-parameters created from the 3D simulation are due to both the resonance between the inductance and capacitance and the phase delay of the return current, so the resonant points will change accordingly. The approximation of the resonance using a tank circuit approach is extremely useful and accurate enough for most practical purposes. If a design has via stubs that are resonating anywhere near the frequency of operations, the primary concern is to ﬁx the design by decreasing the length of the stubs using layout rules, microvias, or backdrilling. 10.3.2 Plane Radiation Losses At the via stub resonance frequency, the stubs act like small antennas and “radiate” energy into the dielectric layers sandwiched between reference planes. For example, the cross section shown in Figure 10-17 will radiate a signiﬁcant amount of power into the dielectric layers conﬁned by the reference planes, where it will propagate in TEM parallel-plate waveguide mode. The energy transfer into the parallel-plate mode will be maximum at the resonant frequency. A simulation of the electric and magnetic ﬁelds between reference layers 2 and 3 in Figure 10-17 at the resonant frequency was performed in Ansoft’s HFSS. The 438 Z TOPICS IN HIGH-SPEED CHANNEL MODELING E H Figure 10-20 When a via does not have adequate current return paths, energy can radiate into the layers between reference planes in the TEM parallel-plate waveguide mode. Via stubs exacerbate this effect. ﬁelds are shown in Figure 10-20. Note how the electric and magnetic ﬁelds are orthogonal to each other (and therefore TEM) and are propagating outward in a circular pattern. From the signal’s point of view, the energy being sourced into the plane will look like loss. Additionally, as the energy propagates through the planes, it will be picked up by other vias similar to antennas, which can increase crosstalk dramatically. Note that this phenomenon is a direct result of a nonideal return path and is very similar to the slot example described in Section 10.2.2. The return current must diverge from the ideal path, so it can ﬂow through the ground via, increasing the loop area, leading to an increased inductance. In Section 10.2 it was shown that energy was transferred to the slotline, where it propagated and coupled to other transmission lines crossing the gap. Similarly, in this case, the energy is transferred to the parallel-plate mode, where it propagates and could be picked up by other vias, resulting in increased crosstalk. Figure 10-21 shows the simulated S-parameters of a signal propagating through a via with a 37.8-mil stub. Since both the dielectric and the conductors are loss-free, the percentage of the power radiated into the plane (Pplane) is calculated by rearranging equation (9-27) to yield VIAS 439 S11, S21 Pplane 0.1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 ~62% of the power is radiated into the plane 0.7 S21 0.6 Pplane 0.5 0.4 S11 0.3 0.2 0.1 10 20 30 40 Frequency, GHz 0.0 50 Figure 10-21 At resonance, a large percentage of the total power can be radiated into the layers between reference planes. This example shows the S -parameters and the power radiated into the plane for a 37.8-mil via stub, as shown in Figure 10-17. 1 − S11S1∗1 − S21S2∗1 = Ploss Pincident (10-11a) where Pplane = Ploss Pincident (10-11b) Note that as the frequency nears resonance, the energy transfer into the plane is most efﬁcient. Figure 10-22 shows an equivalent circuit and its response. The resistor represents the losses due to energy being sourced into the parallel-plate mode between reference layers 2 and 3. The L and C values were extracted from Ansoft Q3D, and the resistor value was varied until the value of S21 matched the HFSS result. Note that this circuit is only a crude approximation of the via behavior. The only methodology that can model this effect accurately is to use a 3D ﬁeld simulator such as Ansoft’s HFSS. 10.3.3 Parallel-Plate Waveguide To promote the understanding of how the energy propagates between reference layers after it has been sourced from a resonant via structure, it is useful to derive the equations that govern the parallel-plate waveguide. A parallel-plate waveguide is formed from two ﬂat plates, as shown in Figure 10-23. For this analysis it is assumed that the width w is much larger than the vertical separation between the plates h, so that any fringing ﬁelds at the edges can be ignored. Since 440 TOPICS IN HIGH-SPEED CHANNEL MODELING S11, S21 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 01 S21 S11 0.419 pF 2.5 Ω 0.141 nH 02 03 04 05 0 Frequency, GHz Figure 10-22 A simple equivalent circuit can crudely approximate the effect of the via resonance using the extracted L and C . More accurate models require a 3D electromagnetic simulator. x vs w y h z Figure 10-23 Dimension used to derive the electric and magnetic ﬁelds of a parallelplate waveguide. the propagation mode discussed in Section 10.3.2 is TEM, that mode is the focus of this derivation. As was done in Section 3.4.3 for the microstrip transmission line, we begin with Laplace’s equation: ∇2 = ∂2 ∂x2 + ∂2 ∂y2 =0 (3-48) The boundary conditions for this problem are given by (x = 0, y) = 0 (x = h, y) = vs (10-12a) (10-12b) REFERENCES 441 Since it is assumed that w h, the variation in the y-direction is assumed constant. Therefore, the variation with y drops out of (3-48), and the general solution is given by (x, y) = C1 + C2x (10-13) The electric ﬁeld is calculated using the electrostatic potential shown as discussed in Section 2.4.1. E = −∇ (x , y) = −ax vs h (10-14) Therefore, the electric ﬁeld propagating in the z-direction is E(x, y, z) = ax Ee−jβz = −ax vs h e−j β z (10-15) where the propagation constant is as deﬁned in Section 2.3.4: β = √ 2πf µε = √ ω µε rad/m (10-16) The magnetic ﬁeld is calculated by dividing (10-15) by the intrinsic impedance of the waveguide as described in Section 2.3.4. η≡ µ ε H (x, y, z) = ay 1 η vs h e−j β z (10-17) (10-18) The propagation velocity is calculated from νp = 1 √ µr µ0εr ε0 = c √ µr εr m/s (2-52) where µr is almost always unity for practical digital designs. REFERENCES Hall, Stephen, Garrett Hall, and James McCall, 2000, High-Speed Digital System Design, Wiley-Interscience, New York. Hall, Stephen, Steven G. Pytel, Paul G. Huray, Daniel Hua, Anusha Moonshiram, Gary A. Brist, and Edin Sijercic, 2007, Multi-GHz causal transmission line modeling using a 3-D hemispherical surface roughness approach, IEEE Transactions on Microwave Theory and Techniques, vol. 55, no. 12, Dec. Liang, Tao, Stephen Hall, Howard Heck, and Gary Brist, 2006, A practical method for modeling PCB transmission lines with conductor surface roughness and wideband dielectric properties, presented at IEEE MTT-S International, June, Microwave Symposium Digest, pp. 1780–1783. 442 TOPICS IN HIGH-SPEED CHANNEL MODELING PROBLEMS 10-1 For a transmission line in a bus speciﬁed to operate at 10 Gb/s with 50-ps rise and fall times, how high in frequency should the tabular model be constructed? 10-2 Derive an analytical formula to approximate the inductance and capacitance of a via. (Hint: Use the transmission-line techniques detained in Chapter 3.) 10-3 Derive the S-parameters directly from a tabular RLCG model. (Hint: Remember the telegrapher’s equations?) 10-4 Derive the characteristic impedance for a parallel-plate waveguide with a width of w. 10-5 Create a model to estimate the crosstalk between two differential pairs traversing a slot in the ground plane. 10-6 For an 8-bit-wide bus crossing a slot in the ground plane, what techniques can be used to minimize the nonideal return paths? Create a model for each option and simulate it in SPICE to demonstrate. 10-7 Can the slot be used as a bus? If so, what would the architecture and topology look like? What would the advantages of this type of bus be? Create a model of the bus and simulate the pulse response. 10-8 Approximate the voltage across a via 1 in. away if E and H are launched in the TEM mode. (Hint: E · dl.) 11 I/O CIRCUITS AND MODELS 11.1 I/O design considerations 444 11.2 Push–pull transmitters 446 11.2.1 Operation 446 11.2.2 Linear models 448 11.2.3 Nonlinear models 453 11.2.4 Advanced design considerations 455 11.3 CMOS receivers 459 11.3.1 Operation 459 11.3.2 Modeling 460 11.3.3 Advanced design considerations 460 11.4 ESD Protection circuits 460 11.4.1 Operation 461 11.4.2 Modeling 461 11.4.3 Advanced design considerations 463 11.5 On-chip termination 463 11.5.1 Operation 463 11.5.2 Modeling 463 11.5.3 Advanced design considerations 464 11.6 Bergeron diagrams 465 11.6.1 Theory and method 470 11.6.2 Limitations 474 11.7 Open-drain transmitters 474 11.7.1 Operation 474 11.7.2 Modeling 476 11.7.3 Advanced design considerations 476 11.8 Differential current-mode transmitters 479 11.8.1 Operation 479 11.8.2 Modeling 480 11.8.3 Advanced design considerations 480 11.9 Low-swing and differential receivers 481 11.9.1 Operation 481 11.9.2 Modeling 482 Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 443 444 I/O CIRCUITS AND MODELS 11.9.3 Advanced design considerations 483 11.10 IBIS models 483 11.10.1 Model structure and development process 483 11.10.2 Generating model data 485 11.10.3 Differential I/O models 488 11.10.4 Example of an IBIS file 490 11.11 Summary 492 References 492 Problems 494 So far we have discussed the behavior of high-speed interconnects and have provided techniques for analyzing and modeling the key physical phenomena that affect signal quality at multi-Gb/s data rates. To fully analyze and understand the behavior of high-speed signaling links, we must include the I/O circuits that transmit and receive the digital data. Design of high-performance links demands that the circuits and interconnects be jointly optimized as a uniﬁed system. To do so successfully, the signal integrity engineer must be able to communicate with the circuit designer. In this chapter we describe the operation and modeling of contemporary high-speed I/O circuits, including transmitters, receivers, and on-die terminations. We do not attempt to provide a complete treatment on how to design high-speed I/O; instead, we wish to give the SI engineer insight into the behavior and sensitivities of modern transceivers. This insight is fundamental to developing a sufﬁcient understanding of the interactions between I/O circuits and physical interconnects in order to optimize a signaling system. In this chapter we identify the design parameters of the circuits for use in analyzing and optimizing designs, and describe techniques for modeling I/O circuits. In addition, we introduce the Bergeron diagram, a useful tool analyzing the time-domain behavior of a complete signaling system. Finally, we acknowledge that transceivers can be designed using either bipolar or MOSFET devices, although throughout this chapter we focus on MOSFET-based circuits. 11.1 I/O DESIGN CONSIDERATIONS The function of a transmitter is to launch a signal representing digital data onto an interconnect for propagation to a receiver circuit. To maximize performance, engineers typically must use design techniques to provide controlled output impedance and rise and fall times. In addition, transmitters (often abbreviated as Tx) can be designed for either singled-ended or differential transmission, and to operate either as a voltage or a current source. In this section we use the term transmitter interchangeably with driver and output buffer, all of which are commonly used in the industry. I/O DESIGN CONSIDERATIONS 445 A range of complexity for modeling transmitters is possible. At its simplest, the model can be a simple transient voltage in series with an output resistance or a current source in parallel with an output resistance. Since they use simple circuit elements, variation of linear model parameters for the purposes of identifying a working solution and understanding sensitivity to design and process variation is extremely easy. In addition, they are easy to analyze and provide the fastest simulation times. As a result, linear models are very useful for the initial stage of the design process, when large numbers of simulations are performed in order to identify the potential solution space. More complex nonlinear behavioral transmitter models provide improved accuracy over linear models by comprehending the nonlinear relationship between the output voltage and output current, staged switching of the output devices to control rise and fall times, and parasitic capacitances. Parametric model variation is more difﬁcult than with linear models, but is still possible. Nonlinear behavioral models are widely used (via IBIS, the I/O Buffer Information Speciﬁcation) because they allow component suppliers to provide accurate models without divulging the speciﬁcs of the circuit design and manufacturing process. Finally, achieving maximum accuracy may require the use of full transistor models. These models are typically used only as a ﬁnal check of the design, since they are more complex to construct and they require signiﬁcantly more simulation time. In addition, suppliers are often reluctant to provide transistor models because they can divulge proprietary design and process information. As a result, in this chapter we focus on linear and nonlinear behavioral modeling rather than transistor-level modeling. Options for modeling receiver circuits follow the same progression as for transmitters. Simple receiver models include only the termination resistors and input capacitance. Nonlinear behavioral models include the voltage versus current characteristics of the ESD protection circuitry and of terminations that are implemented using transistor devices. Full transistor models incorporate all device effects. Requirements and trade-offs for the various model types are summarized in Table 11-1. In our modeling discussions in this chapter we focus heavily on linear models due to their extreme usefulness in the early design stages. We also discuss the TABLE 11-1. Summary of Modeling Approaches and Trade-offs Model Type Elements Intellectual Property Speed Sweepability Linear Voltage and/or current sources and resistors Nonlinear Current vs. voltage curves, voltage vs. time curves Full transistor All devices, including pre-driver, compensation circuits, etc. None Little Design and process Fastest Most Fast Some Slow Very limited 446 I/O CIRCUITS AND MODELS limitations of such models and give guidance for when to use nonlinear behavioral models. 11.2 PUSH–PULL TRANSMITTERS 11.2.1 Operation The simplest type of output circuit is a push–pull transmitter, which can be implemented using a simple CMOS inverter (paired with a receiver that is also an inverter in order to preserve the logic state), as shown in Figure 11-1. Push–pull transmitters were popular in the early days of CMOS ICs due to their ease of implementation and low power consumption. They can be used in interconnect systems without termination, with series termination, and/or with parallel termination. A comprehensive description of the operation of CMOS transistors and inverters is provided in a book by Rabaey et al. [2003]. We offer a brief overview here to make sure that the reader understands the fundamentals of push–pull transmitter operation. We start by providing the expression relating the current conducted by a MOS transistor, iD, as a function of the voltage potential applied across the terminal nodes: ID = 0 k k W L W 2L (vGS − vT )vDS − vD2 S 2 (1 + λvDS ) (vGS − vT )2(1 + λvDS ) vGS − vT < 0 0 ≤ vDS < vGS − vT 0 < vGS − vT ≤ vDS (subthreshold) (triode) (saturation) (11-1) VDD MP1 vin vout MN2 iout VSS Figure 11-1 CMOS inverter transmitter circuit. (From Dabral and Maloney [1998].) PUSH–PULL TRANSMITTERS 447 where k = process transconductance (A/V2) W = device width (µm) L = gate length (µm) vGS = potential difference between the transistor gate and source nodes (V) vT = transistor threshold voltage (V) vDS = potential difference applied across the source and drain (V) λ = channel length modulation parameter (V·m) The process transconductance is k = µεox tox (11-2) where µ is the device mobility (m2/V·s), εox the oxide permittivity (F/m), and tox the oxide thickness (m). The oxide permittivity εox is equal to 3.97ε0, and the oxide thickness tox is 5.7 nm for the MOSIS 0.25-µm process. Equation (11-1) describes three distinct regions of operation. In the subthreshold region, where vGS − vT ≤ 0, the device conducts only a very small amount of leakage current. In the triode region, the relationship between the device current and the drain–source potential is approximately linear. In the saturation region, the device enters a high-impedance state, acting like a (nearly) constant current source. Figure 11-2 shows iD versus vDS curves for various gate–source potentials (vGS) that were created from HSPICE simulation using the SPICE level 3 model for the MOSIS 0.25-µm process that is contained in Appendix F. The dimensions of the devices used to produce the ﬁgure are a width W of 222 µm and length L equal to 1 µm for the NMOS transistor and W/L = 845 µm/1 µm for the PMOS transistor. In the case of the PMOS device, vDS is less than zero since vout is less than VDD. As a result, the current ﬂows from the supply (source) to the drain, so that iD is also less than zero. For the NMOS transistor, the polarities for vDS and iD are both positive. Note from the ﬁgure that positive current is deﬁned as ﬂowing back into the transistor device. We follow this convention throughout this chapter, although it is equally valid to deﬁne positive current ﬂowing in the opposite direction, as long as consistency is maintained. The voltage transfer characteristic of the inverter shown in Figure 11-3a was created from HSPICE simulation using a SPICE level 3 model for the MOSIS 0.25-µm process with device dimensions identical to those in Figure 11-2. The ﬁgure shows how the output signal varies as a function of the input signal level. When the input signal vin is at ground (VSS), the NMOS transistor, MN2, does not conduct current, while the PMOS device, MP1, conducts current accord- ing to equation (11-1). As a result, the output signal vout is pulled up to VDD (2.5 V for the 0.25-µm process) through MP1. As the input is raised to VDD, the output drops to ground. At approximately one-half of VDD, the inverter enters a high-gain region in which vout changes rapidly as a function of vin. 448 iD (mA) iout 50 drain vout gate 40 vDS = vout iD = iout source VSS 30 Linear 20 I/O CIRCUITS AND MODELS vGS = 2.5 V Saturation vGS = 2.0 V vGS = 1.5 V 10 0 0.0 0.5 vGS = −0.5 V vGS = −1.0 V −5 1.0 1.5 vDS (V) (a) vGS = 1.0 V vGS = 0.5 V 2.0 2.5 vGS = −1.5 V −15 iD (mA) −25 vGS = −2.0 V Saturation Linear −35 source VDD −45 vGS = −2.5 V −55 −2.5 −2.0 −1.5 −1.0 vDS (V) (b) gate vDS = vout −VDD iD = iout drain vout iout −0.5 0.0 Figure 11-2 Example transmitter pull-up and pull-down i D versus vDS curves: (a) NMOS (W = 222 µm, L = 1.0 µm); (b) PMOS (W = 845 µm, L = 1.0 µm). Figure 11-3b shows the transient response of the inverter when driving a 1-pF load. 11.2.2 Linear Models The simplest push–pull transmitter model, shown in Figure 11-4, is the linear model. Here the transmitter behavior is modeled by a transient voltage source in series with a resistor (i.e., a The´venin equivalent circuit). The voltage source PUSH–PULL TRANSMITTERS 449 2.5 2.0 vout (V) 1.5 1.0 0.5 Voltage [V] 0.0 0.0 3 2.75 2.5 2.25 2 1.75 1.5 1.25 1 0.75 0.5 0.25 0 −0.25 −0.5 0.6 0.8 Tx Input Tx Output 0.5 1.0 1.5 vin (V) (a) 1 1.2 1.4 1.6 1.8 2 Time (ns) (b) 2.0 2.5 2.2 2.4 2.6 2.8 Figure 11-3 Example CMOS transmitter input–output characteristics: (a) voltage transfer characteristic; (b) transient response driving a 1-pF load. Rs vout Rs vout vs vs Cs vss (a) (b) Figure 11-4 Linear (The´venin) equivalent model for a CMOS transmitter (a) basic model; (b) with output capacitance, C S . 450 V Vs I/O CIRCUITS AND MODELS Vss t Figure 11-5 Example transient response for a linear transmitter model. is typically a pulse or a piecewise linear source, characterized by minimum and maximum voltage levels (VSS and vS in the ﬁgure) and rise and fall times, as illustrated by Figure 11-5. The resistor represents the effective output impedance of the transmitter. The ability to change the behavior of linear models by modifying the output resistance and/or rise and fall times makes them particularly useful for exploring the potential working solution space for a design. In addition, since they require only a couple of passive elements, simulations that make use of them execute with maximum speed. As a ﬁnal note, the model may include a capacitor at the output to account for the capacitance of the transmitter, as shown in Figure 11-4b. Limitations of the linear model The nonlinear relationship between output current and node voltages for real MOS transistors creates the potential for signiﬁcant behavioral differences between the linear model and a real transmitter, which we illustrate by example. Example 11-1 Load-Line Analysis Using a Linear Transmitter Model We compare the falling-edge behavior of interconnect circuit shown in Figure 11-6a with that of the linear model in Figure 11-6b. The current versus voltage characteristic for the pull-down transistor is described in Figure 11-2a for vGS = 2.5 V. To calculate the actual impedance of the transmitter, we use a technique known as a load-line plot [Rabaey et al., 2003] We start with a plot of output current versus output voltage for the NMOS transistor. Recognizing that the falling-edge transition for this circuit begins from a state in which no current ﬂows (i = 0) and the voltage is 2.5 V, we construct a load line for the transmission line that expresses the output current as a function of the output voltage using Ohm’s law: 2.5 V − vout = 50 · iout The current versus voltage plots for both the transistor and transmission line are shown in Figure 11-7. The intersection of the two lines gives the values for the PUSH–PULL TRANSMITTERS VDD = 2.5 V vGS = 2.5 V W/L = 845 vout 50 Ω W/L = 222 iout 0V 2.5 V 451 50 Ω vout iout 50 Ω (a) (b) Figure 11-6 Interconnect circuit for Example 11-1: (a) circuit with push–pull transmitter; (b) linear model. voltage and current at the transmitter output when driving the transmssion line for the falling edge (0.650 V, 37 mA). The NMOS transistor was sized to conduct 50 mA across a potential of 2.5 V, which would provide an output impedance RS of 50 . However, as the ﬁgure demonstrates, the transmitter actually conducts more than 95% of the maximum current by the time the output swing has reached one-half of the maximum value (1.25 V). As a result, the transmitter has an effective output impedance of approximately 0.65 V/37 mA = 18 as seen by the transmission line. For the linear model, we construct the voltage–current plot using Ohm’s law. vout = 50 · iout iout (mA) Transistor 50 (0.650 V, 37.0 mA) RS = 18 Ω 40 30 50 Ω transmission line 20 50 Ω linear model (1.250 V, 25.0 mA) 10 RS = 50 Ω 0 0.0 0.5 1.0 1.5 2.0 2.5 vout (V) Figure 11-7 Load-line plot for Example 11-1. 452 I/O CIRCUITS AND MODELS The linear model provides an initial output voltage and current of 1.250 V and 25 mA when driving the 50- transmission line. Thus, the linear model is a clear source of inaccuracy for modeling an interconnect system, although it may be good enough for low-speed designs or for use in the early design stages of a high-speed link. We have two options at our disposal if a better match between design and model is required. The ﬁrst is to modify the transmitter design to give it a more linear current–voltage relationship. This is most easily accomplished by placing a resistive element in series with the output transistor. From Example 11-1 we note that the transmitter actually conducts more than 95% of the maximum current by the time the output swing has reached one-half of the maximum value (1.25 V). As a result, the transmitter has an effective output impedance of approximately 25 across the output voltage range 0 to 1.25 V (the linear region of operation) and a very high impedance from 1.25 to 2.5 V (the saturation region). By adding a 25- resistor in series with the effective 25 of the transistor, we can better approximate the desired 50- impedance, as shown in Figure 11-8. Since it requires modifying the design, this approach is typically taken only when a constant impedance over the entire voltage swing is critical to the design [Esch and Chen, 2004]. An example is the AGP4X interface, which used a series-terminated interconnect to transmit graphics data at 266 Mb/s. The interface relied on the transmitter to provide the termination, which required linearization of the voltage–current relationship [Intel, 2002]. ID 50 VDS 40 W/L = 222 vGS = 2.5 V 25 W ID VDS W/L = 222 vGS = 2.5 V 30 20 50 W ID VDS ID (mA) 10 0 0 0.5 1.0 1.5 2.0 2.5 VDS (V) Figure 11-8 Comparison of push–pull transmitter current versus voltage characteristics to a linear model. PUSH–PULL TRANSMITTERS 453 11.2.3 Nonlinear models The other option for improving accuracy is to use a nonlinear model. In addition to the nonlinear current–voltage characteristic, this type of model can also comprehend the shape of the rising and falling edges of the output. The basic push–pull nonlinear model consists of an output current versus output voltage curve for both pull-up and pull-down devices and an output voltage versus time curve for both the rising and falling edges (Figure 11-9). In addition to the curves, iD (mA) 120 100 80 60 40 20 0 −20 −40 −60 −5 −4 −3 −2 −1 0 1 2 3 vDS (V ) (a) 50 0 −50 iD (mA) −100 −150 −200 −3 −2 −1 0 1 2 3 4 5 vDS (V ) (b) Figure 11-9 Nonlinear behavioral model components: (a) pull-up i out versus vout; (b) pull-down i out versus vout; (c) rising edge vout versus time; (d) falling edge vout versus time. (Continued) 454 I/O CIRCUITS AND MODELS Voltage (V) 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 −0.2 −1.0 −0.5 0.0 0.5 1.0 Time (ns) (c) 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 −0.2 −1.0 −0.5 0.0 0.5 1.0 Time (ns) (d) Figure 11-9 (Continued ) Voltage (V) a nonlinear behavioral model typically contains the load conditions under which the curves were constructed. Simulation tools use this information, typically in the form of an IBIS model (see Section 11.10), to adjust the model for the differing load conditions encountered when it is used in simulations of real interconnect systems. The device capacitance may also be included, if it is not already comprehended in the transient voltage versus time curves of the model. However, models created from either transistor-based simulations or from measurements typically include the effects of the capacitance, in which case it should not be explicitly called out in the model. Construction of the iout versus vout curves is discussed further in Section 11.10. PUSH–PULL TRANSMITTERS 455 In general, simulation tools create multiple i –v curves to account properly for the transient nature of the input signal to the model. Stated another way, a time-varying input signal, vin, to the circuit in Figure 11-1 causes the device gate-to-source potential, vgs, to also be transient, so that the output current–voltage relationship varies with time as well. This effect is shown in Figure 11-10 and Table 11-2 for an inverting transmitter. The rising-edge input signal starts at a value of 0.0 V, which corresponds to vgs = 0.0 V for the NMOS device and to vgs = −2.5 V for the PMOS device. As the input edge rises, the vgs values for each device change. For example, when vin is equal to 1.0 V, vgs is 1.0 V for the NMOS device and −1.5 V for the PMOS device. We see that for the case of a rising-edge input signal, as the PMOS device moves from the vgs = −2.5 V curve to the vgs = 0 V curve, the NMOS device is transitioning from the curve for vgs = 0 V to that for vgs = 2.5 V. As a ﬁnal note, we point out that the i –v data in the model should extend well beyond the expected operating range for the device to ensure proper operation in the event of signiﬁcant signal overshoot. For example, if the signal swing expected ranges from 0 to VDD, the IBIS speciﬁcation expects that the i –v data span a range from −VDD to 2VDD. 11.2.4 Advanced Design Considerations In this section and in the counterpart sections for other transceiver types, we touch brieﬂy on several of the issues that face circuit designers when developing transmitter circuits. For more extensive discussions of the issues and techniques, we refer the reader to books by Dabral and Maloney [1998] and Dally and Poulton [1998]. Overlap current control When the transmitter circuit of Figure 11-1 makes a rising or falling transition, the PMOS and NMOS devices will conduct current simultaneously for a brief period of time. The magnitude of this overlap current (also known as “crowbar” current) is sufﬁciently high that designers usually design the transmitter such that the initially conducting device turns off before the other device turns on. This is typically implemented with “pre-driver” control logic. Tristate function Systems in which multiple devices can drive a common signal, such as a multiprocessor bus, require that the transmitter be placed in a high-impedance state when it is not actively driving the system. This is also accomplished with pre-driver logic, as shown in Figure 11-11. The circuit uses the enable signals, en/en, to control whether the transmitter is connected and can actively drive the bus, or is disconnected and presents a high impedance, as described by Table 11-3. Process and environmental compensation When fabricating large numbers of components, the physical features (e.g., gate length, oxide thickness) and 456 I/O CIRCUITS AND MODELS vin (V) 3.0 t5 2.5 t4 2.0 t3 1.5 t2 1.0 t1 0.5 t0 0.0 −0.5 −0.05 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 Time (ns) (a) t5 −5 vGS = −0.5 V vGS = −1.0 V t4 t3 t0 −15 vGS = −1.5 V −25 t2 t1 −35 vGS = −2.0 V iD (mA) −45 vGS = −2.5 V −55 −2.5 −2.0 50 −1.5 −1.0 vDS (V) (b) −0.5 0.0 vGS = 2.5 V iD (mA) 40 t4 30 vGS = 2.0 V 20 t3 vGS = 1.5 V 10 vGS = 1.0 V t2 t5 vGS = 0.5 V t1 0 0.0 0.5 1.0 1.5 2.0 2.5 vDS (V) (c) Figure 11-10 Interaction between vin –t and i out –vout curves for an inverting push–pull driver: (a) transient input signal; (b) i –v curve progression for the PMOS pull-up; (c) i –v curve progression for the NMOS pull-down. PUSH–PULL TRANSMITTERS 457 TABLE 11-2. Relation Between vin and vGS for the Pull-up and Pull-down Devices of an Inverting Transmitter vGS (V) vin (V) NMOS PMOS 0.0 0.0 −2.5 0.5 0.5 −2.0 1.0 1.0 −1.5 1.5 1.5 −1.0 2.0 2.0 −0.5 2.5 2.5 0.0 TABLE 11-3. Logic Table for a Tristate Transmitter en/en vin vout 0/VDD VDD /0 VDD /0 X VDD 0 High Z 0 VDD en vin en MP1 vout MN2 Figure 11-11 Tristate push–pull driver. electrical characteristics will vary due to normal statistical variations in the manufacturing process. In addition, MOSFET currents are sensitive to environmental factors such as supply voltage (iD increases as the VDD increases) and device temperature (iD decreases with increasing T ). As a result, the current–voltage relationship for transistors can vary by a factor of 2 to 3 across the process and environmental extremes, leading to wide swings in the output impedance and rise–fall time. In the case of microprocessor-based systems, where yearly volumes reach into the hundreds of millions, excessive variation in the electrical characteristics can lead to failure of some of the systems to operate at full performance. Compensation refers to techniques that minimize the variation between different parts operating in different environments. Designers 458 I/O CIRCUITS AND MODELS can implement compensation using both digital and analog techniques, which are used most often to provide carefully controlled impedance and/or rise and fall times. Although a detailed discussion of those techniques is beyond our scope, we present an example of digital impedance compensation to illustrate application of the concept. Example 11-2 Digital Impedance Compensation The circuit shown in Figure 11-12 will provide controlled impedance for either the output of a transmitter or for on-die termination resistors (see Section 11.5). The circuit adjusts the impedance to closely match a desired value by comparing the strength of a binary weighted on-die replica circuit (PMOS transistors MP0 through MPN) with that of an external precision resistor (typically controlled to ±1%). Both are connected to identical reference current sources (iref), forming voltage dividers whose levels are fed into the clocked comparator at the bottom of the ﬁgure. The output from the comparator is used to increment/decrement an up/down counter. The output signals (S0 to S2) from the counter control the turn on–turn off of devices MP0 to MPN , which decreases or increases the impedance of the control network to achieve the desired impedance. Signals S0 to SN are also connected to the transmitter circuits, each of which contains a similar structure, thus providing the controlled impedance characteristic. Since it operates from a clock, the circuit can adjust the impedance dynamically to compensate for variation in the supply voltage and device temperature during operation. In this case the clock is a low-frequency clock that is updated about every millisecond. In addition, the design should also ensure that the control bits do not change while the circuit is actively driving in order to avoid noise on the signals due to changing impedance [Gabara and Nauer, 1992]. Achieving tighter impedance control is simply a matter of adding additional binary weighted devices. VDD Rref Off-chip MPN 2n−1Wref iref On-chip MP1 2Wref iref MP0 Wref + − Up/ Down Counter CLK S0S1 To Sn Transmitter Figure 11-12 Digital impedance compensation circuit. CMOS RECEIVERS 459 11.3 CMOS RECEIVERS The most basic receiver for interchip signaling circuits is the inverter. The simplicity, low power consumption, and ease of implementation of inverters made them the receiver of choice for full-swing CMOS-based interfaces through much of the 1990s. Receivers are characterized by timing parameters (setup and hold) and by logic thresholds, which inﬂuence system noise margin and noise immunity. Our discussion of the operation of merits of receivers for high-speed signal transmission will focus on these characteristics. 11.3.1 Operation The CMOS receiver is a low-gain inverting ampliﬁer that provides full rail-to-rail output swings, which allows for fairly large noise margins at speeds into the hundreds of Mb/s. Figure 11-13a illustrates the voltage transfer characteristic for an example inverting receiver, which speciﬁes the output signal as a function of the input. The input thresholds vil and vih are determined by the unity gain points (dvout/dvin = −1) of the transfer characteristic. The region in between vil and vih is a high-gain region in which the output signal level is extremely sensitive to variations in the input signal, making it a forbidden “keep out” area for steady-state signals. Large noise margins provide tolerance to noise sources, and so are desirable to guarantee robust operation. As Figure 11-13b demonstrates, the high-side noise margin, vNMh, is the difference between the minimum output signal when driving vout VSS Vil,min Vih,max VDD VDD Slope = −1 Weak PMOS VSS Strong NMOS Strong PMOS Weak PMOS Nominal PMOS Nominal NMOS voh,min vNMh Slope = −1 Keep Out vol,max vNMh vout VDD vih,min vil,min vSS vin Vin (a) (b) Figure 11-13 CMOS receiver response characteristics: (a) voltage transfer characteristics; (b) noise margin. 460 I/O CIRCUITS AND MODELS high and the minimum signal that the receiver recognizes as the high logic state. Conversely, the low-side noise margin, vNMl, is the difference between the maximum output signal when driving low and the maximum signal recognized by the receiver as the logic low state. Note from the ﬁgure that the receiver input specs must comprehend variations in the signal levels caused by variability in the fabrication process. Mathematically, the noise margins are expressed as vNMl = vil,min − vol,max vNMh = voh,min − vih,max (11-3a) (11-3b) 11.3.2 Modeling The CMOS inverter presents a high impedance to the input signal, limited only by the input capacitance of the gate. As such, we typically model a CMOS receiver as a simple capacitance to ground. 11.3.3 Advanced design considerations Though the relatively large swing results in high noise margins, voltage mode signaling systems possess multiple noise sources that degrade the noise immunity of the system [Dally and Poulton, 1998]. For example, process variations such as device thresholds and transconductance can cause inverter thresholds to vary by more than 10% of the signal swing (>20% if supply voltage variation is included), a phenomenon known as receiver offset. Other sources include power supply noise, crosstalk, reﬂections, and transmitter offset. These effects may be countered by designing additional noise tolerance into the receiver through hysteresis. An example is a Schmitt trigger, which we show in Figure 11-14a [Wang, 1989]. The MP1/MN2 and MP3/MN4 transistors form a sequential pair of inverters. The hysteresis is created by feeding the output vout back to the gates of MP5 and MN6, which shifts the voltage transfer characteristic, making it more difﬁcult for a noise pulse to put the circuit into the keep-out zone, as Figure 11-14b shows. 11.4 ESD PROTECTION CIRCUITS Transceiver designs include electrostatic discharge (ESD) protection circuits to prevent catastrophic failure due to breakdown of the MOSFET gates of the I/O and core circuits. ESD damage can occur at any point during the manufacture, assembly, test, and operation of a silicon chip, including handling and transport. An example would be a technician wearing rubber-soled shoes on a test ﬂoor. The insulating characteristics of the rubber can cause the technician to accumulate a substantial static charge, which may discharge into any component with which he or she may come into contact, damaging the component. An example ESD event ESD PROTECTION CIRCUITS 461 vout VDD low-to-high high-to-low MP1 MP5 vin vx MP3 vout ∆∆ MN2 MN6 MN4 VSS vin VSS VDD / 2 VDD (a) (b) Figure 11-14 CMOS receiver with hysteresis: (a) Schmitt trigger receiver circuit; (b) voltage transfer characteristic. based on a human body model can have a 3000 to 5000-V potential spike with a sub-10-ns rise time, 1 to 2-A peak current, and 100 to 200-ns duration [Dabral and Maloney, 1998]. The device oxide will break down at ﬁeld strengths in excess of about 7 × 108 V/m, which translates to voltages in excess of approximately 4.0 V for a 25-µm silicon process. So an ESD event can exceed the process limits by several orders of magnitude, necessitating the inclusion of ESD protection in the I/O circuitry. 11.4.1 Operation ESD protection circuits such as the one shown in Figure 11-15 protect active circuits by limiting the voltage excursions so that they do not exceed the breakdown voltage of the receiver MOSFET gates, and by limiting the amount of current that ﬂows into the transmitter device terminals. The function of the ESD diodes is to limit the voltage so that it does not exceed the gate oxide breakdown voltage and to steer the ESD current away from the internal circuits. If the voltage at the pad will be greater than a diode voltage drop (VD,on) beyond the VDD supply, the upper diode will turn on, shunting the current away from the I/O device and clamping it to VDD + VD,on. A negative excursion will be clamped to a value of VSS − VD,on. The series resistor limits the amount of current that ﬂows through the transceiver devices to prevent performance degradation due to device threshold shifts caused by “hot” electron tunneling [Dally and Poulton, 1998]. 11.4.2 Modeling The diodes add parasitic capacitance to the transceiver, which we include in our linear model. The series resistance is typically several hundred ohms, and 462 Primary ESD diodes Secondary ESD diodes I/O CIRCUITS AND MODELS VDD Input pad Current-limiting resistor Figure 11-15 CMOS receiver with ESD protection. does not require explicit modeling. Nonlinear models must take into account the current–voltage relationship of the diode, which is given by the ideal diode equation : idiode = iS (evdiode/φT − 1) (11-4) where idiode is the current through the diode; iS, the diode saturation current, is proportional to the diode area; vdiode is the bias voltage across the diode; and φT is the thermal voltage (26 mV at room temperature). Figure 11-16 presents an example current–voltage characteristic calculated using equation (11-4) with φT equal to 26 mV and a saturation current of 10 pA. We can further approximate the diode behavior for manual analysis by setting the diode current to zero below the “turn-on” voltage (approximately 0.6 V in the ﬁgure) and allowing it to approach inﬁnity at the turn-on voltage. This is shown iD (mA) 100 90 80 70 60 50 40 30 20 Ideal diode Quasilinear model 10 equation 0 −10 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 vD (V) Figure 11-16 Example diode current–voltage characteristic and quasilinear approximation. ON-CHIP TERMINATION 463 as the quasilinear model in the ﬁgure. We illustrate the use of the quasilinear model in Problem 11.6. 11.4.3 Advanced design considerations The circuit shown in Figure 11-15 is but a simple example of ESD protection. In reality, designers have many options for implementing ESD protection, and often go to great lengths to design structures that provide sufﬁcient protection while minimizing the parasitic capacitance for high-speed operation. We refer the reader to the books by Dally and Poulton [1998] and Dabral and Maloney [1998] for comprehensive treatment of the topic. 11.5 ON-CHIP TERMINATION On-chip (a.k.a. on-die) termination has become the method of choice as signaling speeds continue to increase, since it eliminates a source of reﬂections by removing the transmission-line stub that is required to connect off-die termination. 11.5.1 Operation Termination resistors are typically implemented using FETs. An example is shown in Figure 11-17, in which PMOS and NMOS devices are connected in parallel to create a 50- termination to VDD (2.5 V for the 0.25-µm process). The PMOS gate is connected to ground (vgs = −2.5 V) and NMOS to VDD (vgs = 2.5 V) to keep the transistors in the triode region for as long as possible in order to make the current–voltage relationship as linear as possible. As the ﬁgure shows, this conﬁguration provides a termination of approximately 46 to 58 over the entire range of operation for nominal device characteristics and operating conditions. Figure 11-17b shows the waveforms obtained when using the circuit to terminate a 50- transmission line when driven with a 2.5-V 12.5- linear transmitter. The FET termination provides nearly equivalent performance to a perfect 50- resistive termination. We also note that our choice to terminate to the positive supply rail rather than ground was also based on making the termination as linear as possible. FET termination is typically combined with the digital impedance compensation technique described in Section 11.2.4 to provide a controlled impedance termination across process, voltage, and operating temperature. 11.5.2 Modeling The basic model for on-chip termination is a simple resistor connected to the appropriate termination supply. However, as we showed earlier, on-chip termination can exhibit signiﬁcantly nonlinear response, in which case we need to model the termination as a table of current and voltage values using a nonlinear modeling format such as IBIS (refer to Section 11.10). 464 I/O CIRCUITS AND MODELS Reff (Ω) 60 58 56 W = 65 µm W = 80 µm 54 L = 0.25 µm L = 0.25 µm 52 vx 50 ix 48 46 44 42 40 0 0.5 1 1.5 2 2.5 vx (V) (a) 3.0 Transmitter 2.5 Receiver 2.0 Voltage (V) 1.5 1.0 12.5 Ω 0.5 0.0 V→2.5 V tr = 50 ps 50 Ω, 1.7 ns 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 Time (ns) (b) Figure 11-17 Example on-chip termination using parallel FETs: (a) FET termination circuit and resistance as a function of line voltage; (b) example waveform. 11.5.3 Advanced design considerations The primary motivation for implementing on-chip termination is to minimize reﬂections by eliminating the transmission line stub that is typically required when terminating on the printed circuit board. To fully realize the potential beneﬁt of on-chip termination, automatic impedance control similar to the impedance-matching technique described in Section 11.2.4 is commonly used. In addition, designers may implement techniques to improve the linearity of the on-chip termination [Dally and Poulton, 1997]. BERGERON DIAGRAMS 465 11.6 BERGERON DIAGRAMS At this point we take a slight detour from our survey of transceivers to introduce the Bergeron diagram, a technique that will allow us to analyze the behavior of interconnects with nonlinear transmitter and receiver characteristics by graphically solving the simultaneous current–voltage relationships of the components of a signaling system. We introduce this technique as it is useful for furthering our understanding of transmission-line basics. We demonstrate and develop the technique through a simple example. Example 11-3 Bergeron Diagram for a Linear Interconnect Circuit In this example we analyze the rising behavior for the circuit shown in Figure 11-18. The transmitter is a 2.5-V push–pull driver with symmetrical pull-up and pull-down impedances. The receiver has a resistive termination to a 2.5-V termination supply. We start the analysis by plotting the current versus voltage relationships for the transmitter and receiver. The easiest way to do so is to draw the equivalent circuits and write the Ohm’s law expressions. This is shown in Figure 11-19 along with the load-line plot. Note that we were careful to account correctly for the direction of the current ﬂow in all of the equivalent circuits and that we chose to use milliamperes as the units for the y-axis. Note from the plot that the intersection of the transmitter pull-down and the receiver load lines gives the voltage and current (0.357 V and 28.6 mA) for the circuit at the steady state when driven low. Since the circuit is at steady state, this is the potential and current ﬂow at all points on the transmission line. This gives us the starting point for our analysis, since we are studying a rising-edge transition. Our next step is to ﬁnd the initial voltage and current for the rising edge at the transmitter. We do this by drawing a load line representing the 50transmission line, starting at the intersection of the transmitter pull-down and receiver (steady-state low) and extending until it intersects with the load line for the transmitter pull-up. This load line has a slope equal to −1/Z0, since the transmission line also obeys Ohm’s law and the load line is a plot of current versus voltage. In effect, the load line for the transmission line is graphically depicting the Ohms’ law equation: v − v0 = Z0(i − i0) (11-5) 0 V 2.5 V tr = 100 ps 12.5 Ω i 50 Ω, 2 ns 2.5 V 75 Ω Figure 11-18 Interconnect circuit for Example 11-3. 466 I/O CIRCUITS AND MODELS 2.5 V 2.5 V v i 12.5 Ω i 12.5 Ω i 75 Ω v v − 2.5 V = (12.5 Ω)i (a) 0.0 V v = (12.5 Ω)i (b) v 2.5 V − v = (75 Ω)i (c) 200 Tx (t < 0) (0.357 V, 28.6 mA) 100 Rx 0 Tx Pull-Down Current [mA] −100 −200 Tx Pull-Up −300 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Voltage [V] (d) Figure 11-19 Equivalent circuits and voltage–current expressions for Example 11-3: (a) transmitter pull-up; (b) transmitter pull-down; (c) receiver; (d) load-line plot. where v0 and i0 are the steady-state voltage and current for the system when driven low. The slope of the line is negative because the rising-edge transition causes current to ﬂow from the transmitter into the transmission line, while we have deﬁned positive current as ﬂowing from the line back into the transmitter. The transmitter also follows Ohm’s law, so that the intersection between the transmission line and the transmitter pull-up lines gives the initial voltage and current at the transmitter for the rising edge. Figure 11-20a illustrates this, giving an initial voltage and current of 2.357 V and −11.4 mA. The next step is to draw another load line for the transmission line. The line starts at the previous point (2.357 V, −11.4 mA), has a slope equal to 1/Z0, and extends until it intersects with the load line for the receiver, as shown in Figure 11-20b. This gives us the voltage and current at the receiver after the ﬁrst propagation delay, and comprehends both the initial and reﬂected waves. It works because we are again satisfying simultaneous Ohm’s law expressions for the transmission line and the receiver, while taking into account the previous voltage and current levels. Ohm’s law gives us the relationships that allow us to solve for the signal values given the constraints imposed by Kirchhoff’s circuit laws. Since the transmission line is connected to the receiver, Kirchhoff’s circuit BERGERON DIAGRAMS 467 laws tell us that they will have equal voltages at the connection and equal currents ﬂowing through them. Thus, our Bergeron diagram gives us the solution to two simultaneous Ohm’s law equations in two unknowns (the voltage and current) via graphical means. By continuing to draw load lines for the transmission lines with slopes that alternate between ±1/Z0 we can ﬁnd the voltage and current levels at each end of the line as the waves propagate back and forth. For example, Figure 11-20c shows the extension of the next transmission load line back to the transmitter load line, which gives the signal levels at the transmitter after a round-trip propagation delay. The transient wave components are becoming small enough Current [mA] 200 Tx (t < 0) (0.357 V, 28.6 mA) 100 Rx 0 −100 −200 Tx Pull-Up Tx Pull-Down T-line Tx (t = 0) (2.357 V, −11.4 mA) −300 0.0 0.5 200 Tx (t < 0) (0.357 V, 28.6 mA) 100 Rx 0 −100 1.0 1.5 2.0 2.5 3.0 Voltage [V] (a) Tx Pull-Down T-line Tx (t = 0) Rx (t = td) (2.357 V,−11.4 mA) (2.757 V, −3.4 mA) −200 Tx Pull-Up Current [mA] −300 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Voltage [V] (b) Figure 11-20 Bergeron diagram construction sequence for Example 11-3: (a) initial wave at transmitter (t = 0); (b) receiver (t = t d ); (c) transmitter (t = 2t d ); (d) close-up: transmitter (t = 2t d ). (Continued) 468 I/O CIRCUITS AND MODELS Current [mA] 200 Tx (t < 0) (0.357 V, 28.6 mA) 100 Rx 0 −100 T-line −200 Tx Pull-Up Tx Pull-Down Tx (t = 2td) (1.4mA,2.517V) Tx (t = 0) Rx (t = td) (2.357 V, −11.4 mA) (2.757 V, −3.4 mA) −300 0.0 0.5 20 10 Rx 1.0 1.5 2.0 Voltage [V] (c) Tx (t = 2td) (2.517 V,1.4 mA) 0 −10 T-line −20 −30 −40 Tx Pull-Up Tx (t = 0) (2.357 V,−11.4 mA) −50 2.0 2.2 2.4 2.6 Voltage [V] (d) Figure 11-20 (Continued ) 2.5 3.0 Rx (t = td) (2.757 V, −3.4 mA) 2.8 3.0 Current [mA] at this point that they are difﬁcult to discern, so Figure 11-20d shows a close-up view of the diagram. The process of drawing load lines continues until the changes in voltage and current become small enough to suit our need. We can then construct voltage and current waveforms by reading the values from the Bergeron diagram. We must keep in mind that points that occur at intersections with the transmitter load line contribute to the transmitter waveform, and points that occur at intersections with receiver load line contribute to the waveform at the receiver. The full Bergeron diagram is shown in Figure 11-21 along with the voltage and current waveforms. At this point we note that we generated the voltage and current numbers shown in the waveforms by repetitively solving simultaneous Ohm’s law equations subject to initial voltages and currents—the analytical equivalent of the Bergeron BERGERON DIAGRAMS 469 Current [mA] 30 Tx (t < 0) Tx (t = 02td) 20 (0.357 V, 28.6 mA) (2.517 V, 1.4 mA) Rx (t = 3td) 10 T-line (2.469 V, 0.4 mA) 0 Tx Pull-Down −10 −20 Rx Tx (t = 0) (2.357 V,−11.4 mA) Rx (t = td) (2.757 V, −3.4 mA) −30 −40 Tx Pull-Up −50 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Voltage [V] (a) 3.0 Receiver 2.5 Transmitter 2.0 Voltage (V) 1.5 1.0 0.5 Current (mA) 0.0 −2 35 30 25 20 15 10 5 0 −5 −10 −15 −2 0 2 Receiver 4 6 Time (ns) (b) Transmitter 0 2 4 6 Time (ns) (c) 8 10 12 8 10 12 Figure 11-21 Complete Bergeron diagram and resulting waveforms for Example 11-3: (a) Bergeron diagram; (b) voltage waveform; (c) current waveform. 470 I/O CIRCUITS AND MODELS diagram. In practice, drawing a Bergeron diagram with better than 5 mV, 0.5 mA accuracy is difﬁcult, which limits its use primarily to initial, ﬁrst-order estimates. 11.6.1 Theory and method The Bergeron diagram works by using Kirchhoff’s circuit laws at the connections between the transmission line and the transmitter and receiver components, along with the current versus voltage relationships. For the transmission line, the current and voltage are related by Ohm’s law. The linear transmitter and receiver models that we used in our previous example also obey Ohm’s law, although nonlinear models will obey the more complex relationships described by equations (11-1) and (11-4). The Bergeron diagram also works with nonlinear transceivers, as we show in the next section. In describing the theory behind the Bergeron diagram, we use the generalized circuit shown in Figure 11-22 with a rising signal. We start at the steady state prior to the rising transition. Under steady-state conditions a lossless transmission line is a short circuit, so that the output of the transmitter is connected effectively directly to the receiver. In our discussion here, when we refer to the transmitter and receiver we mean the equivalent circuits for each, as shown in Figure 11-22. At steady state, we know that the current ﬂowing out of the equivalent circuit for the transmitter is equal to the current ﬂowing into the equivalent circuit for the receiver from Kirchhoff’s current law. From Kirchhoff’s voltage law we know that the voltage at the output of the transmitter equivalent circuit is equal to the voltage at the input of the receiver equivalent circuit. We also know that the relationship between the current and voltage at the output of the transmitter follows Ohm’s law, as does the current–voltage relationship at the input of the VDD VTT RTx,hi RTT i t=0 Z0, td RTX,lo VTT i RTT v i RTX,lo i v Transmitter pull-up Transmitter pull-down VDD RTx,hi v Receiver Figure 11-22 General linear circuit for Bergeron analysis. BERGERON DIAGRAMS 471 receiver. By equating the voltages and currents and applying Ohm’s law, we create a set of two simultaneous equations in two variables: v(t < 0) = i(< 0)RTX,lo VT T − v(t < 0) = i(t < 0)RT T (11-6) (11-7) The Bergeron diagram solves the equations graphically to give the steady-state current and voltage. To ﬁnd the voltage and current values for the initial transition at the transmitter end of the circuit, we recognize that the output of the transmitter is connected to one end of the transmission line. By applying the circuit laws in the same manner as above, we calculate the initial voltage and current wave magnitudes at the transmitter. However, the Ohm’s law expression for the transmission line must comprehend the initial current and voltage ﬂowing through the line: v(0) − VDD = i(0)RTX,hi v(t < 0) − v(0) = [i(0) − i(t < 0)]Z0 (11-8) (11-9) Note from equation (11-9) that the slope of the load line for the transmission line is equal to −1/Z0, just as we constructed it in our example. So the Bergeron diagram again solves the simultaneous equations while accounting for the steady-state potential and current ﬂow that existed on the line prior to the transition. When the initial wave reaches the connection between transmission line and receiver, we again apply the circuit laws and account for the current and voltage of the incident wave: VT T − v(t = td ) = i(t = td )RT T v(t = td ) − v(t = 0) = [i(t = td ) − i(t = 0)]Z0 (11-10) (11-11) The slope of this load line for the transmission line is 1/Z0 because it relates the current and voltage for the reﬂected wave, and since we must account for the current and voltage of the incident wave, it starts at the current and voltage, i(t = 0) and v(t = 0), that we calculated in the preceding step. The next step in the analysis returns back to the transmitter end of the system using a load line with slope −1/Z0 starting from the receiver voltage and current (i.e., the intersection between the previous transmission line load line and that of the receiver). In this case the load-line slope is negative because we are accounting for the reﬂected voltage and current waves at the transmitter. Since the waves ﬂow away from the transmitter and toward the receiver, the sign of the current wave is negative. The analysis continues in this fashion, alternating between transmitting and receiving load lines using transmission load lines with alternating slopes of 472 I/O CIRCUITS AND MODELS −1/Z0 and 1/Z0, until it approaches steady state. So we see that the Bergeron is simply a graphical technique for repeatedly solving the simultaneous equations arising from circuit laws that describe the transient voltage and current signals at both ends of a transmission line. We now illustrate the application of the technique to a system with nonlinear transceiver characteristics. Example 11-4 Bergeron Diagram with Nonlinear Transceivers In this example we analyze the falling edge for the circuit shown in Figure 11-23, which uses a CMOS push–pull transmitter with no termination at the receiver end. The system relies on the output impedance of the transmitter to provide source termination, and the transistors are sized to provide a 50- output impedance to match the target impedance of the transmission line. The load-line plot shows the nonlinear transistor behavior discussed in Section 11.2.1. The load line for the receiver is simply a zero-current (inﬁnite-impedance) line that represents the open circuit at that end. We know that the characteristic impedance of the transmission lines may vary by up to ±20% due to manufacturing tolerances. Thus, for our example we choose a 60- characteristic impedance for the transmission line. With this information, we step through the analysis as follows: 1. The Bergeron diagram begins at the load-line intersection of the transmitter pull-up and the receiver. The potential and current are 2.5 V and 0 mA, respectively. 2. From that point we draw the load line for a transmission line with slope equal to −0.0167 −1(−1/60 ), extending it until it intersects the nonlinear curve for the transmitter pull-down device at 0.80 V and 28.3 mA, which are the initial values at the transmitter after the falling-edge transition. 3. Draw a load line from the previous point, with slope equal to 0.0167 −1 until it intersects with the receiver load line at i = 0, which gives the voltage at the receiver (−0.90 V) after the ﬁrst incident wave reaches it. A simple means of checking the result at this point is to consider the magnitudes of the incident and reﬂected voltage waves. The incident-wave magnitude is equal to 0.80 V −2.50 V, or −1.70 V. The reﬂected wave is also equal to i 60 Ω, 1 ns iout (mA) 50 Tx pull-down 40 30 20 10 Rx 0 −10 −20 −30 −40 Tx pull-up −50−1.0 −0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 vout(V) Figure 11-23 Push–pull transmitter circuit and nonlinear i –v characteristics. BERGERON DIAGRAMS 473 −1.70 V (−0.90 V −0.80 V). Since the transmission line is open-circuited at the receiver end, this is exactly the result that we expect to ﬁnd. 4. Draw the next load line, starting at −0.900 V and 0 mA with slope equal to −0.0167 −1 until it intersects the transmitter pull-down curve, which yields the voltage (−0.25) and current (−10.8) at the transmitter after the ﬁrst reﬂected wave from the receiver reaches it. 5. Continue the analysis until reaching steady state (0.00 V, 0 mA). The resulting voltage waveforms at both ends of the line are shown in Figure 11-24b. 50 Tx pull-down 30 i (mA) 10 Rx −10 −30 Tx pull-up −50 −1.0 −0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 v (V) (a) 2.5 2 1.5 Voltage (V) 1 Tx 0.5 Rx 0 −0.5 −1 −1 0 1 2 3 4 5 6 7 8 9 10 Time (ns) (b) Figure 11-24 (a) Bergeron diagram and (b) transient waveforms for Example 11-4. 474 I/O CIRCUITS AND MODELS 11.6.2 Limitations As mentioned earlier, the accuracy of Bergeron diagrams is typically around 5 mV and 0.5 mA, limiting their use to the earliest stage of the design process. In addition, they can only handle impedance discontinuities at the ends of the lines, so they are not suitable for analysis of more realistic (and complex) topologies. Finally, they only work for lossless transmission lines, which limits their application to data rates in which losses may be ignored. Despite these limitations, Bergeron diagrams provide a means of evaluating the impact of I/O circuit nonlinearity, making them a helpful tool in the arsenal of the signal integrity engineer. 11.7 OPEN-DRAIN TRANSMITTERS A second type of transmitter circuit is the open-drain transmitter, which is shown in Figure 11-25. An example is Gunning transceiver logic (GTL) [Gunning et al., 1992] As the ﬁgure shows, open-drain systems are typically designed using only NMOS pull-down transistors, so that the PMOS pull-up device is eliminated. Ensuring proper function requires the addition of pull-up resistors. The resistors may be added as external components, but state-of-the-art designs typically include them on the silicon die. Note the presence of a termination supply, VT T , which is typically a low voltage in the range 1.2 to 1.5 V, as open-drain systems are usually implemented with signal swings in the range 800 to 1000 mV. 11.7.1 Operation When the open-drain transistor is turned off, the transmitter is effectively disconnected from the interconnect, which is pulled up to the termination supply through the termination resistor, RT T . No current ﬂows when the interconnect is in the high state, since it is open circuited. When the open-drain transistor is turned on, it effectively creates a voltage-divider circuit by creating a path for VTT vout vin iout RTT Figure 11-25 Open-drain signaling circuit. OPEN-DRAIN TRANSMITTERS 475 current ﬂow from the VT T supply through the termination resistor and the NMOS transistor. As a result, open-drain systems do not use “rail-to-rail” signal swings, since the signal level does not reach ground when driving low. Beneﬁts Open-drain designs offer multiple potential beneﬁts that have made them a popular design choice over the past two decades. The use of parallel termination reduces reﬂections for higher-speed applications, and elimination of the PMOS pull-up output transistors reduces the amount of die area consumed, which creates the potential to reduce the cost of the silicon. Since open-drain designs do not dissipate power in the high state, they are typically deﬁned as being active low , in order to achieve zero power consumption when idle. In addition, in contrast to push-pull circuits, the small signal swing reduces active switching power. These features combine to make open drain a popular choice for low-power designs. Another beneﬁt of open-drain systems is their ability to interface components that are manufactured on different processes that have differing maximum voltage supply limits. Use of the low-voltage external termination supply allows us to decouple the signaling values from the component supplies. For example, GTL uses a 1.2-V external supply, and signals swing from 0.4 V to 1.2 V. This would allow us to connect a component that uses a 3.3-V supply rail with one that uses a 2.5-V rail. Finally, open-drain systems provide a “wired-OR” function for multiprocessor systems, in which any one of multiple agents can assert an active signal on the bus interface (see Figure 11-26). An example is the MCERR# (machine check error) signal on the Intel Xeon processor system bus interface (a.k.a. front-side bus). MCERR# is asserted (driven low) to indicate an unrecoverable error. Since multiple agents may drive this signal at the same time, it is a wired-OR signal that must connect the corresponding pins of all processor front-side bus agents [Intel, 2005]. Wired-OR connections are susceptible to “glitches” that can occur when multiple agents drive the bus low simultaneously. In Problem 11-10 we explore this phenomenon in more detail. VTT VTT RTT RTT CPU CPU CPU CPU Memory Controller Figure 11-26 Multiprocessor system with open-drain wired-OR connection. 476 I/O CIRCUITS AND MODELS Limitations Despite the use of parallel termination, GTL is susceptible to ringing on rising-edge transitions. The reason for this is that when the device turns off, the transmission line is open circuited at the transmitting end. Any reﬂections on the interconnect that travel back to the transmitter will undergo a full reﬂection, as we show with an example. Example 11-5 Rising-Edge Reﬂections in an Open-Drain Interconnect The interconnect circuit in Figure 11-27 has an impedance discontinuity between boards 1 and 2. To analyze the behavior of the rising edge, we follow the current-mode analysis method [Hall et al., 2000]. We model the open-drain device as an effective resistance, RS, with a switch. The analysis starts with the circuit at steady state when driven low. Prior to t = 1 ns, the switch is closed, and a steady-state current ﬂows: iS,Slo = VT T RS + RT T = 12.5 1V + 50 = 16 mA The switch opens at t = 1 ns, creating an open circuit at z = 0. The open circuit requires that the net current ﬂow is zero at that point. The consequence of this is that a −16-mA current wave is launched onto the 75- line, creating a 1.2-V voltage wave. From that point we can carry out the analysis using a lattice diagram, which we shown in Figure 11-27b. Figure 11-27c shows the resulting waveform at the transmitter and at the receiver, which demonstrates the ringing at both ends of the system, despite the fact that the termination matches the impedance of the second line. 11.7.2 Modeling A linear model for an open-drain circuit requires a resistor and a switch, as shown in Figure 11-28. In addition, a termination supply and termination resistor are required to complete the open-drain system. The means for modeling the switch will vary between different simulation tools. In HSPICE, the easiest method is to use a voltage-controlled resistor [Synopsis, 2006]. Nonlinear models for open-drain transmitters look similar to those for push–pull transmitters, except that the current–voltage relationship for the pull-up is a straight line at i = 0. 11.7.3 Advanced Design Considerations As discussed above, the ﬁnite output impedance of an open-drain transmitter provides some damping of reﬂections when driving low. However, the transmitter will fully reﬂect any incoming signals on the rising edge as the open-drain device enters the high-impedance state. A straightforward way to address the ringing caused by the impedance mismatch is to add a termination resistor to provide impedance matching at the transmit end of the system, as shown in Figure 11-29. Although this technique will eliminate the excessive ringing that was caused by OPEN-DRAIN TRANSMITTERS t = 1 ns z=0 Rs = 12.5 Ω i 75 W, 0.5 ns Board 1 z = l1 477 VTT = 1.0 V RTT = 50 Ω 50 W, 1 ns Board 2 z = l2 (a) 1 ns 2 ns 3 ns 4 ns 5 ns 6 ns Γ1 T2 −0.2 0.2 0.8 1.2 v (z = 0) i (z = 0) 0.200 V 1.400 V 0.920 V 1.016 V 0.996 V 1.000 V 16.00 mA 0.00 mA 0.00 mA 0.00 mA 0.00 mA 0.00 mA 1.200V/16.0mA −0.240V/3.2mA −0.240V/3.2mA 0.048V/0.64mA 0.048V/0.64mA −0.010V/−0.13mA −0.010V/−0.13mA 0.002V/0.03mA 0.002V/0.03mA 0.000V/0.00mA 109..926m0 AV −−30.8.049m2 AV 00.7.073m8 AV −−00.1.060m8 AV 00.0.030m1 AV t 1.4 1.2 (b) Transmitter Receiver l2 z 0 1 v (z = l ) i (z = l ) 0.200 V 16.00 mA 1.160 V 3.20 mA 0.968 V −0.64 mA 1.006 V 0.13 mA 0.999 V −0.03 mA 1.000 V 0.00 mA 2.5 ns 3.5 ns 4.5 ns 5.5 ns 6.5 ns 1.0 Voltage (V) 0.8 0.6 0.4 0.2 0.0 0 1 2 3 4 5 6 7 Time (ns) (c) Figure 11-27 Open-drain interconnect and analysis for Example 11-5: (a) circuit; (b) lattice diagram; (c) waveform. 478 I/O CIRCUITS AND MODELS VTT RTT vout iout RS Figure 11-28 Open-drain transmitter linear model. VTT = 1.0 V RTT = 50 Ω t = 1 ns i Rs = 6.25 Ω z=0 75 Ω, 0.5 ns Board 1 z = l1 50 Ω, 1 ns Board 2 VTT = 1.0 V RTT = 50 Ω z = l2 Figure 11-29 Open-drain interconnect with termination at both ends. the open circuit, it does not come for free. To maintain the same signal swing we must reduce the output impedance of the transmitter, which increases the power dissipation and the die area consumption. Another source of ringing on the rising transition is transient current ﬂow through parasitic package inductance. Gunning et al. [1992] used a control circuit implementation to slow the rise time through the use of an analog feedback loop (Figure 11-30) in their open-drain interface, known as Gunning transceiver logic (GTL). VDD MP1 vin MN1 MN3 vout MN2 MN4 VTT RTT On-chip Off-chip Figure 11-30 GTL transmitter circuit with analog slew rate control. DIFFERENTIAL CURRENT-MODE TRANSMITTERS 479 11.8 DIFFERENTIAL CURRENT-MODE TRANSMITTERS 11.8.1 Operation Differential current-mode transmitters are often used for high-speed data transmission. In this case, the transmitter operates by injecting a current onto the transmission line. Figure 11-31a depicts a simple differential transmitter design. The transmitter uses complementary input signals, vin and vin, which ensure that only one side of the circuit is in the conducting state at any given time. Thus, the transmitter uses the differential input signals to steer the current from the constant current source, iS, to the desired side of the circuit. The ﬂow of current creates a voltage drop across the source termination resistor, RT T , on one side of the circuit, while the side that has no current ﬂow is pulled up to VDD, thereby creating the output signal levels vout and vout. The relationship between input and output signals is summarized in Table 11-4. The table demonstrates that the differential signal swing has a magnitude (2iTx RT T ) that is twice that of the singled-ended swing. VDD VDD RTT RTT RTT RTT vout vout vin vin is is vout vout is (a) (b) Figure 11-31 (a) Differential current mode simple transmitter design; (b) linear model. TABLE 11-4. Relationship Between Differential Transmitter Input and Output Signals vin/vin Low/High High/Low vout vout vdiff = vout − vout VDD VDD − iTx RT T iTx RT T VDD − iTx RT T VDD −iTx RT T 480 I/O CIRCUITS AND MODELS Beneﬁts As we demonstrated in Chapter 7, differential signal transmission improves signal-to-noise ratio (SNR), which offers a path to higher data rates. Part of the reason for the improved SNR is that the differential signal swing is twice that of a single-ended signal. In addition, the differential transmitter circuit draws a nearly constant current, which drastically reduces the simultaneous switching noise (SSN). In addition, differential receivers reject the vast majority of common-mode noise. All of these factors combine to provide substantial performance headroom over single-ended signaling. 11.8.2 Modeling As shown in Figure 11-31b, a linear model of a differential current source transmitter consists simply of a pair of complementary transient current sources connected to bias/termination transistors, along with any parasitic capacitance. Nonlinear models look similar to those for push–pull transmitters, except that the current–voltage relationships follow a saturated proﬁle rather than the more linear characteristic of the push–pull circuit. 11.8.3 Advanced Design Considerations In its simplest form, a current-mode transmitter is a MOSFET device operating in the saturation region (vDS ≥ vGS − vT ), as shown in Figure 11-31. However, to limit the variation in output current, current-mode transmitter designs typically include additional circuitry to compensate for process and environment effects. For an example, we refer to Figure 11-32, which shows a transmitter design for use in a low-voltage differential signaling (LVDS) interface [Granberg, 2004]. vpbias MP1ar MP1r Rlr C1 iref Rext = 100 Ω − 1.4 V + vin MP1a MP1 R1 vpbias MN4 MN5 vout vout MN5r MN1ar MN1r C2 + − Reference Circuit vin 1.0 V C3 C4 vnbias MN3 MN2 MN1a MN1 Output Buffer Figure 11-32 LVDS current mode transmitter circuit with controlled current reference. (From Gabara [1997].) LOW-SWING AND DIFFERENTIAL RECEIVERS 481 The circuit contains a reference current generator on the left-hand side of the ﬁgure that is used to bias the output buffer on the right-hand side of the circuit in order to produce a tightly controlled output current. The external resistor connects to the on-die reference circuit to generate a 4-mA reference current which establishes the bias levels, vp,bias and vn,bias. The bias voltages feed the output buffer in order to produce the LVDS output swing (1.0 to 1.4 V) when connected by a 100- load resistor, when the differential input signals are applied. It is worth pointing out that current-mode transmission does not require differential signaling but can be applied to high-speed single-ended signals as well. Low-swing current-mode transmission systems in general provide better noise immunity and power dissipation than does voltage mode rail-to-rail signaling [Dally and Poulton, 1998]. An example of a single-ended current-mode interface is the Direct Rambus DRAM technology [Lau et al., 1998; Granberg, 2004]. 11.9 LOW-SWING AND DIFFERENTIAL RECEIVERS We treat the input receivers for low-swing and differential signaling techniques in the same section, as they both typically employ differential ampliﬁers, which provide fast transient response to small signal swings. 11.9.1 Operation In principle, the differential ampliﬁer design is the same for single-ended low-swing and differential signaling applications. In the single-ended application, one input is connected to a reference signal, vref, while the other is connected to the data signal. For differential signals, the ampliﬁer inputs are connected to the complementary data signals. Examples of each are shown in Figure 11-33. VDD VDD vout vout vin vref vin vin (a) (b) Figure 11-33 Example receiver circuits for low-swing and differential signaling: (a) single-ended; (b) differential. 482 I/O CIRCUITS AND MODELS The single-ended receiver is the original design for use with GTL signaling and was designed to switch within vref ± 50 mV across process, voltage, and operating temperature in order to provide a high noise margin with an 800-mV swing [Gunning et al., 1992]. The differential receiver is a self-biasing Chappell ampliﬁer that uses an internal feedback signal to adjust the bias voltage for proper operation [Chappell et al., 1998]. As mentioned above, differential ampliﬁers respond to small changes in input signals, as their symmetry gives low-input offset voltages and makes them relatively insensitive to power supply ﬂuctuations. Differential receivers also reject common-mode noise, with typical common-mode rejection ratios (CMRRs) of −20 dB or more. 11.9.2 Modeling We model differential ampliﬁer–based receivers in the same fashion as we model single-ended receivers. The simplest model is simply a capacitance to ground in conjunction with the termination. With differential signals, we have multiple options for termination, as Figure 11-34 shows. Note that the termination for differential signals is typically implemented on the silicon die, resulting in nonlinear current versus voltage characteristics that we may need to include in a behavioral model. In addition, we need to include the parasite effects and nonlinear characteristics of the ESD protection devices. + Zodd Zodd − (a) + 2Zodd − (b) Figure 11-34 Termination options for differential signals: (a) single-ended termination; (b) differential termination. IBIS MODELS 483 11.9.3 Advanced Design Considerations The examples in Section 11.9.1 are intended to illustrate the concepts, and represent only two of a wide range of choices for high-performance receiver designs. In practice, designers have multiple options for improving the various performance aspects of the receiver, such as common-mode range and input offset voltage. We refer interested readers to Dally and Poulton [1997] for more information on receiver design options and techniques. 11.10 IBIS MODELS As we mentioned in Section 11.1, silicon suppliers do not like to provide transistor models for their I/O circuits. However, simple linear models often do not satisfy the accuracy requirements of high-speed signaling links. To meet this need, the industry has developed the I/O Buffer Information Speciﬁcation (IBIS). As the name suggests, IBIS is a format for specifying I/O circuit information. Created in the early 1990s, it is now an industry standard owned by the IBIS Forum and is supported by approximately 60 companies. The diverse membership has allowed IBIS to evolve to meet the changing needs of signal integrity and I/O design engineers. In this section we give a brief overview of the IBIS standard, highlighting the major components and providing a high-level description of the model development process. The major features of IBIS include: • Nonlinear current versus voltage curves for transmitters, ESD devices, and on-chip termination • Separate nonlinear voltage versus time curves for transmitter pull-up and pull-down devices • Pad capacitance for I/O circuits • Models for minimum, typical, and maximum cases within a single model • Description for multiple types of I/O, including differential pins, open-drain output, tristate outputs, and receivers with hysteresis • Inclusion of signal quality speciﬁcations, including input logic thresholds, overshoots, and so on • The Golden Parser, a tool that checks model syntax for conformance to the standard • Backward compatibility with models created under previous revision of the standard For further details we refer readers to the IBIS speciﬁcation [IBIS, 2006] and model development “cookbook” [IBIS, 2005]. 11.10.1 Model Structure and Development Process The basic structure of an IBIS-compliant model is shown in Figure 11-35. The i –v and v –t curves are speciﬁed in table format, with columns for the minimum, 484 Pull-up i vs. v Pull-up v vs. t Power clamp i vs. v I/O CIRCUITS AND MODELS Power clamp i vs. v Pull-down Pull-down i vs. v v vs. t Transmitter Grount clamp i vs. v Cpad Cpad Figure 11-35 Basic structure of an IBIS model. Grount clamp i vs. v Receiver typical, and maximum case. The power clamp and ground clamp tables contain both ESD clamp information and the behavior of any parasitic diodes which are part of the pull-up and pull-down transistor structures. Since the clamp circuitry is always active, the i –v data for the power and ground clamp models can be combined with the i –v data for the transmitter or receiver circuit to provide a single data table. This same property allows on-die termination resistance to be incorporated into the clamp models for transmitters and receivers, as well. The development process includes four major steps [IBIS, 2005]: 1. Determine required model features, complexity, and operating range. 2. Obtain the simulated or measured model data (i –v and v –t curves, parasitic capacitances). We treat these in more detail in subsequent sections. 3. Put it into IBIS format and check the ﬁle using the Golden Parser. This tool checks the model for syntax compliance with the IBIS standard and is available from the IBIS Forum website. 4. Validate the model. Initially, this is typically done by comparing the transient response of the IBIS model to the original transistor model when driving a reference load. Eventually, the model should be correlated against real silicon. The data extraction and IBIS formatting can be done manually, though automated tools are available [Varma et al., 2003]. The model process works for both single ended and differential I/O, though it must be modiﬁed somewhat to handle the complementary outputs, and to extract both common mode and differential mode data, including capacitances. IBIS MODELS 485 11.10.2 Generating Model Data Current Versus Voltage (i –v) Curves In constructing the i –v curves for a transmitter circuit, the output is connected to an independent voltage source, as shown in Figure 11-36. The current ﬂow into the pad is then measured as the source is swept across a range from −VDD to 2VDD. IBIS uses the convention that current ﬂow into the transmitter is positive. Separate curves are required for pull-up and pull-down devices. For example, the input to an inverter-based push–pull transmitter would be set to ground to obtain the curve for the pull-up device. The circuit setup for characterizing a receiver or ESD clamp is identical except that the independent voltage source is connected to the input node. As mentioned earlier, incorporating on-chip termination in an IBIS model is accomplished by including it in the i –v curves for the clamp circuits. This can be accomplished by including it with the clamp circuit during the data extraction process or by extracting the curves separately and adding them together. The data tables should use enough data points around sharply curved areas of the i –v characteristics to describe the curvature accurately. IBIS does not require equally spaced points in the models, so in linear regions there is no need to include unnecessary data points. Also, as Figure 11-37 demonstrates, the i –v curves for pull-ups and power clamps are VDD referenced, while pull-downs and ground clamps are referenced to ground. Voltage Versus Time Curves Transmitter output waveforms are described with tables that specify the output voltage versus time (v –t). These v –t tables are created when the transmitter is driving into a test load, which is usually a 50resistor that is connected to the appropriate supply, as described below. The actual v –t data are generated by simulating rising and falling edges, with the output of the transmitter to each rail through the test load, as shown in Figure 11-38. For a i Pull-up Sweep from −VDD to 2VDD i Pull-down Sweep from −VDD to 2VDD i Clamp Sweep from −VDD to 2VDD VDD i v v i i v Figure 11-36 i –v curve extraction process. 486 Pull-up I/O CIRCUITS AND MODELS VDD - v(extracted) = v(relative) VDD v VDD v i Extracted i-v curve i VDD relative i-v curve Figure 11-37 Construction of V DD relative i –v curves. 50 Ω VDD 50 Ω Voltage (V) Voltage (V) 2.5 2.0 1.5 1.0 0.5 0.0 −0.5 0.0 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 Pull-up turns off Pull-up turns on 2.0 4.0 6.0 8.0 10.0 Time (ns) Pull-down turns on Pull-down turns off 2.0 4.0 6.0 8.0 10.0 Time (ns) Figure 11-38 Voltage versus time curves for a push–pull IBIS model. push–pull transmitter, a minimum of four v –t curves is desired, as shown in the ﬁgure. All of the curves must be time correlated, which means that they start from the same reference point in time. By specifying the curves in this manner, IBIS models are able to capture the effects of separate switching for the pull-up and pull-down devices. This is particularly useful for modeling “break-before-make” circuits, such as the one studied in Problem 11.1, and also allows the support of “multistaged” output drivers, which are described below. Multistage drivers control the rise and fall times of the output by spreading out switching of the pull-up and pull-down devices over time (Figure 11-39). These models break the circuit into multiple pull-up and pull-down devices, which IBIS MODELS 487 P1 vin N1 P2 P3 vout N2 N3 Figure 11-39 Multistage controlled-rise-time transmitter. requires multiple sets of v –t and i –v curves to properly model the behavior of the circuit. IBIS provides a keyword to allow scheduling of the application of the multiple curves to reﬂect the staged switching behavior. IBIS supports up to 1000 v –t points per table, and the time step used in the table should be minimized to provide maximum resolution. The test load that is used in generating the v –t tables must be described in the model. The signal integrity simulation tool will use this information from the model to reconcile the output waveform to reﬂect the behavior when driving the actual system load. As with the i –v curves, IBIS accommodates separate v –t curves for minimum, typical, and maximum cases. I/O Capacitance The capacitance of the I/O circuit includes the transistors, on-die interconnect, and die pad, and can have separate values for minimum, typical, and maximum cases. Designers have multiple options at their disposal for extracting the I/O capacitance values. We describe one of them here. We connect the circuit to an ac voltage source at the I/O pad so that we capture all the contributors to the capacitance, as shown in Figure 11-40. By measuring the current ﬂow into the circuit, we can calculate the capacitance using Ccomp = −Im(iac) 2πf vac (11-12) where f is the frequency of the ac source, vac is the amplitude of the source, and Im(iac) is the imaginary portion of the current ﬂowing into the ac source. If the I/O capacitance may be affected by the bias (dc) voltage of the circuit, we must include a dc offset in the ac source. In addition, I/O capacitance typically varies with frequency and applied voltage, so performing multiple extractions while sweeping both of those parameters is recommended. For a given model, simply choose the capacitance value that corresponds to expected switching frequency and bias voltage of the application [IBIS, 2005]. 488 I/O CIRCUITS AND MODELS iac vac Figure 11-40 Circuit for extracting I/O capacitance. 11.10.3 Differential I/O Models The IBIS standard includes the ability to specify models for differential transceivers, although the process is more complex than for singled-ended circuits. A differential model is speciﬁed as using a driver pair along with a series model, in the manner shown in Figure 11-41a. This structure allows the models to comprehend the differential and common-mode characteristics of a differential I/O circuit. Proper modeling requires speciﬁcation of the common-mode and differential-mode i –v tables. As a result, the common-mode and differential-mode currents must be must be extracted and separated. Extraction of the v –t tables uses the same techniques as for single-ended transmitters, with the addition of a current source in between the output pins to cancel differential currents inside the transistor model, as shown in Figure 11-41b. The I/O capacitance must include the differential capacitance between the two signals, as Figure 11-41c shows, with the differential and total capacitance expressed as Cdiff = −Im(id c ) 2πf vac Ccomp = −Im(idc) − Im(iac) 2πf vac (11-13) (11-14) where idc is the current measured through the dc voltage source, iac is the current measured through the ac voltage source, and vac is the amplitude of the ac source. The current through the dc source will have an imaginary portion only if there is a reactive path between the two pads of the differential signal. As a ﬁnal thought, we note that multi-Gb/s differential signaling is still in the early stages of deployment and that the IBIS speciﬁcation has not yet fully comprehended the needs of those high-performance links, although the standard will continue to evolve to do so in the near future. IBIS MODELS 489 Driver [Model] out Series [Model] Driver [Model] out (a) idiff Rfixture Rfixture vfixture (b) iout iout vdc vac + vdc (c) Figure 11-41 (a) IBIS differential model structure; (b) v –t extraction ﬁxture; (c) capacitance extraction ﬁxture. 490 I/O CIRCUITS AND MODELS 11.10.4 Example of an IBIS File In this section we provide an example of a very basic IBIS model ﬁle, which contains a model for a push–pull transmitter. The IBIS keywords are enclosed in brackets. Note that IBIS also includes provision for specifying package information and to enable post-layout analysis by including both package and pinlist associations with buffer data, although we have elected not to specify it in this example. |****************************************************** [IBIS Ver] 2.1 [File name] sample.ibs [File Rev] 0.0 [Date] August 31, 1999 [Source] Data Book [Notes] Default model for source. [Disclaimer] This information is modeling only. |****************************************************** [Component] Driver [Manufacturer] Generic [Package] | typ min max R pkg 0 NA NA L pkg 5.0nH NA NA C pkg 2.0pF NA NA [Pin] signal name model name R pin L pin C pin 1 UNKNOWN out NA NA NA |****************************************************** [Model] out Model type Output Polarity Non-Inverting Vmeas = 1.5V Cref = 15pF Rref = 500 Vref = 0.0 | typ min max C comp 5.5pF 2.0pF 8.0pF |****************************************************** | typ min max [Voltage range] 3.3V 3.0V 3.6V |****************************************************** [Pulldown] IBIS MODELS 491 | Voltage I(typ) I(min) I(max) 00 0 0 1 0.150 0.075 0.225 2 0.230 0.115 0.345 ... ... ... ... 3 0.270 0.135 0.405 [Pullup] | Voltage I(typ) I(min) I(max) 00 0 0 1 −0.240 −0.120 −0.360 2 −0.320 −0.160 −0.480 ... ... ... ... 3 −0.340 −0.170 −0.510 [GND clamp] | Voltage I(typ) I(min) I(max) −1.0 −0.100 −0.050 −0.150 −0.5 −0.020 −0.010 −0.030 −0.4 0 0 0 ... ... ... ... 0 0 0 0 [POWER clamp] | Voltage I(typ) I(min) I(max) −1.0 0.100 0.050 0.150 −0.5 0.020 0.010 0.030 −0.4 0 0 0 ... ... ... ... 0 0 0 0 |****************************************************** [Ramp] | variable typ min max dV/dt r 1.980/3.300n 1.800/3.750n 2.16/2.900n dt/dt f 1.980/3.250n 1.800/3.550n 2.16/2.860n R load = 50 | [Falling Waveform] | typ min max R fixture = 50 V fixture = 3.3 V fixture min = 3.15 V fixture max = 3.45 0.000ns 3.300 3.150 3.450 0.300ns 3.250 3.110 3.390 ... ... ... ... 492 I/O CIRCUITS AND MODELS 5.000ns 0.000 0.000 0.000 | [Rising Waveform] typ min max R fixture = 50 V fixture = 3.3 V fixture min = 3.15 V fixture max = 3.45 0.000ns 0.000 0.000 0.000 0.300ns 0.050 0.040 0.060 ... ... ... ... 5.000ns 3.300 3.150 3.450 | |****************************************************** [End] 11.11 SUMMARY In this chapter we described the operation and modeling of contemporary high-speed I/O circuits, including transmitters, receivers, and on-die terminations. Insight into the behavior of these circuits is critical to designing successful high-speed signaling solutions. The signal integrity engineer who gains sufﬁcient understanding to interact successfully with his or her I/O circuit counterpart will have a key tool for optimizing a signaling system design for high-speed operation. REFERENCES I/O circuit design remains an area of active research, with dozens of papers published in conference proceedings and technical journals each year. We do not attempt here to provide an exhaustive survey of the published literature. For more complete treatments of I/O and ESD circuits, we refer the reader to Dabral and Maloney [1997] and Dally and Poulton [1997]. The work by Dabral and Maloney focuses more on basic techniques, whereas Dally and Poulton offer a more comprehensive approach. Granberg [2004] compiled a comprehensive reference that includes technical data on a wide variety of I/O techniques and standards, including memory and multi-Gb/s serial links. Boni, Andrea, Andrea Pierazzi, and Davide Vecchi, 2001, LVDS I/O interface for Gb/s-per-pin operation in 0.35- µm CMOS, IEEE Journal of Solid-State Circuits, vol. 36, no. 4, Apr., pp. 706–711. Chappell, Barbara, et al., 1998, Fast CMOS ECL receivers with 100-mV worst-case Sensitivity, IEEE Journal of Solid-State Circuits, vol. 23, No. 1, Feb., pp. 59–67. Dabral, Sanjay, and Timothy Maloney, 1998, Basic ESD and I/O Design, Wiley-Interscience, New York. REFERENCES 493 Dally, William, and John Poulton, 1998, Digital Systems Engineering, Cambridge University Press, Cambridge, UK. Esch, Gerald, Jr., and Tom Chen, 2004, Near-linear CMOS I/O driver with less sensitivity to process, voltage and temperature variations, IEEE Transactions on Very Large Scale Integration Systems, vol. 12, no. 11, Nov. Gabara, Thaddeus, and Scott Nauer, 1992, Digitally adjustable resistors in CMOS for high performance applications, IEEE Journal of Solid-State Circuits, vol. 27, no. 8, Aug., pp. 1176–1185. Gabara, Thaddeus, and David Thompson, 1998, Ground bounce control in CMOS integrated circuits, Proceedings of the 1988 IEEE International Solid-State Circuits Conference, pp. 88–90. Gabara, Thaddeus, et al., 1997, LVDS I/O Buffers with a Controlled Reference Circuit, IEEE Publ. 1063-0988/97, IEEE Press, Piscataway, NJ, pp. 311–315. Granberg, Tom, 2004, Digital Techniques for High-Speed Design, Prentice Hall, Upper Saddle River, NJ. Gunning, Bill, et al., 1992, A CMOS low-voltage-swing transmission-line transceiver, Proceedings of the 1992 IEEE International Solid-State Circuits Conference, pp. 58 – 59. Hall, Stephen, Garrett Hall, and James McCall, 2000, High-Speed Digital System Design, Wiley-Interscience, New York. IBIS Open Forum, 2005, IBIS Modeling Cookbook for IBIS Version 4.0 , http://www.eigroup.org/ibis/, Sept. 15. IBIS Open Forum, 2006, IBIS (I/O Buffer Information Speciﬁcation), Version 4.2 , http://www.eigroup.org/ibis/, June. Intel Corporation, 2002, AGP V3.0 Speciﬁcation, revision 1.0, Intel Press, Hellsboro, OR, Sept. Intel Corporation, 2005, 64-Bit Intel Xeon Processor with 2MB L2 Cache, Document 306249-002, Intel Press, Hillsboro, OR, Sept. Lau, Benedict, et al., 1998, A 2.6-Gbyte/s multipurpose chip-to-chip interface, IEEE Journal of Solid-State Circuits, vol. 33, no. 11, Nov., pp. 1617–1626. Rabaey, Jan, Anantha Chandrakasan, and Borivoje Nikolic´, 2003, Digital Integrated Circuits: A Design Perspective, 2nd ed., Prentice Hall, Upper Saddle River, NJ. Synopsis, Inc., 2006, HSPICE Simulation and Analysis User Guide, Version Y-2006.3 , Synopsis, Mountain View, CA, Mar. Texas Instruments, 1996, The Bergeron Method: A Graphic Method for Determining Line Reﬂections in Transient Phenomena, Document SDYA014, Texas Instruments, Dallars, TX, Oct. Varma, Ambrish, et al., 2003, The development of a macro-modeling tool to develop IBIS models, Proceedings of the IEEE 12th Topical Meeting on Electrical Performance of Electronic Packaging, Oct. 27–29, pp. 277–280. Wang, Niantsu, 1989, Digital MOS Integrated Circuits, Prentice Hall, Upper Saddle River, NJ. 494 I/O CIRCUITS AND MODELS PROBLEMS Use the device parameters in Table 11-5 when solving these problems. TABLE 11-5. Device Parameters for a 0.25 µm Process NMOS PMOS VT 0 (V) γ (V1/2) VD,SAT (V) k (A/V2) λ (V−1) 0.43 0.4 0.63 115 × 10−6 0.06 −0.4 −0.4 −1 −30 × 10−6 −0.1 Source: Rabaey et al. [2003]. 11-1 Describe operation of the break-before-make circuit shown in Figure 11-42. vin vout Figure 11-42 Break-before-make push–pull transmitter circuit. 11-2 Construct the i –v curve(s) and use them to calculate output impedance for the push–pull transmitter shown in Figure 11-43. 11-3 Explain how the staged circuit in Figure 11-39 provides a controlled rise time at the output of the circuit. Sketch the output waveform expected. 11-4 Describe the operation of the Schmitt trigger receiver shown in Figure 11-14. 11-5 Use a Bergeron diagram to analyze the falling-edge response of the circuit in Figure 11-18. 11-6 Use a Bergeron diagram to analyze the rising edge for the circuit shown in Figure 11-44. PROBLEMS 495 VDD W = 440 mm L = 1 mm iout vin vout W = 100 mm L = 1 mm Vss Figure 11-43 Push-pull transmitter for Problem 11-2. 2.5 V 2.5 V 50 W 100 W, 1 ns vd = 0.5 V vd = 0.5 V 0V Figure 11-44 Diode-terminated circuit for Problem 11-6. 11-7 11-8 11-9 11-10 11-11 Describe how the GTL transmitter circuit shown in Figure 11-30 slows the rising-edge transition to reduce ringing due to parasitic package inductance. Use a Bergeron diagram to analyze the far-end-terminated open-drain circuit in Figure 11-45. Analyze the two open-drain circuits in Figure 11-45. How do the resulting waveforms differ? Which is likely to be capable of supporting a higher data transfer rate? Wired-OR glitch: The circuit in Figure 11-46 contains transmitters at each end, with a receiver circuit near the middle. Initially, the right-hand transmitter pulls the circuit low while the transmitter on the left is turned off. At time t = 0 the left-hand transmitter turns on and the right-hand transmitter turns off. Sketch the waveform at the receiver. Use the device parameters in Table 11.5 to calculate the current versus voltage curve for the on-die termination circuit shown in Figure 11-47 and estimate the effective termination resistance. 496 1.5 V I/O CIRCUITS AND MODELS 1.5 V 100 W 50 W, 1 ns 100 W 10 W Termination at both ends 1.5 V 100 W 50 W, 1 ns 10 W Far end termination Figure 11-45 Open-drain circuits for Problem 11-8. 1.5 V 50 W t=0 10 W 50 W, 1.0 ns 50 W, 1 ns 1.5 V 50 W t=0 10 W Figure 11-46 Wired-OR circuit for Problem 11-10. 11-12 Discuss the potential advantages and disadvantages of the single-ended and differential termination schemes shown in Figure 11-34. PROBLEMS 497 W = 300 µm L = 1.0 µm W = 70 µm L = 1.0 µm vx ix Figure 11-47 FET termination circuit for Problem 11-11. 11-13 Complete a Bergeron diagram using the transmitter and receiver load lines shown in Figure 11-48, which are used in conjunction with a 70transmission line. 0.30 0.20 Receiver 0.10 Transmitter (high) I [A] 0.00 −0.10 Transmitter (low) −0.20 −0.30 −3.0 −2.0 −1.0 0.0 1.0 2.0 3.0 4.0 5.0 V [V] Figure 11-48 Load lines for Problem 11-13. 12 EQUALIZATION 12.1 Analysis and design background 500 12.1.1 Maximum data transfer capacity 500 12.1.2 Linear time-invariant systems 502 12.1.3 Ideal versus practical interconnects 506 12.1.4 Equalization overview 511 12.2 Continuous-time linear equalizers 513 12.2.1 Passive CTLEs 514 12.2.2 Active CTLEs 521 12.3 Discrete linear equalizers 522 12.3.1 Transmitter equalization 525 12.3.2 Coefﬁcient selection 530 12.3.3 Receiver equalization 535 12.3.4 Nonidealities in DLEs 536 12.3.5 Adaptive equalization 536 12.4 Decision feedback equalization 540 12.5 Summary 542 References 545 Problems 546 We have already discussed the impact of Moore’s law, which drives the interchip data bandwidth to continually increasing performance levels. We have also shown that nonideal aspects of transmission lines, such as crosstalk and losses, can have a signiﬁcant impact on signal integrity and timing. These impacts dominate at multi-Gb/s speeds, causing “smearing” of signals so that their energy is spread over multiple bit positions, a phenomenon known as intersymbol interference (ISI). The impact of ISI is an increase in the jitter that degrades the timing margin and a distortion in the signal levels that degrades the voltage margin of the interchip signaling link. Equalization is a circuit technique that reduces the Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 499 500 EQUALIZATION ISI-induced timing jitter and voltage margin loss by compensating for nonideal aspects, in particular the loss of interconnects at high speed. In this chapter we adopt a communications channel–based approach to analyzing our signaling interfaces. Communications engineers view the I/O circuits and the interconnect (also known as the channel ) as ﬁlters, as shown in Figure 12-1. In previous chapters we noted the low-pass ﬁltering effect of the I/O and interconnect. By viewing the system as a series of ﬁlters, including the equalizer, we can determine the desired characteristics for a given equalizer. 12.1 ANALYSIS AND DESIGN BACKGROUND Before considering equalization analysis and design, we must cover some necessary background material. We ﬁrst offer some motivation for employing equalizers in high-speed signaling systems by examining the requirements for maximizing data transfer rates. We then present the notion of linear time-invariant (LTI) systems and show how to use their characteristics to analyze the behavior of high-speed signals. Finally, we contrast the characteristics of an ideal interconnect with that of a physically realizable interconnect. 12.1.1 Maximum Data Transfer Capacity Shannon’s capacity theorem [Shannon, 1949] describes the upper limit on the information rate that can be transmitted over a communications channel. The theorem is widely accepted in the scientiﬁc community and has never been exceeded in practice. We provide a heuristic derivation of the capacity equation with the motivation of developing an understanding of how closely we can approach the theoretical maximum with conventional interconnects, and we then demonstrate how equalization techniques allow us to come closer to achieving the maximum rates. We start by deﬁning the data transfer rate in bits per second as the product of the number of symbols transmitted per second (S) and the number of bits per symbol (B): D = SB (12-1) The symbol transfer rate is related directly to the channel bandwidth by the Nyquist rate [Nyquist, 1928]: S = 2BW (12-2) where BW is the bandwidth in hertz. To develop an intuitive feeling for (12-2), consider a simple binary nonreturn-to-zero (NRZ) signaling scheme. In Figure 12-2 we show a periodic pulse train with a repetition frequency f and a random data sequence with the same fundamental frequency. A single symbol is contained within successive edge positions, each positioned a width of Tsymbol apart, and the full cycle is twice the symbol width. For the periodic signal, the ANALYSIS AND DESIGN BACKGROUND 501 Digital data T Transmitter Channel {xk} ht (t) x (t ) hc (t) Analog signal Noise n(t) Σ y(t) Digital data Receiver Detector hr (t) {x’k} Figure 12-1 Communications channel view of a high-speed signaling interface. 1/f Periodic Signal Tsymbol Tsymbol Random Data 1 0 0 0 1 1 0 1 0 1 1 Signal Figure 12-2 Symbol rate illustration for a binary NRZ signal. bandwidth of the signal is equal to the repetition frequency.† Thus, we have two symbols per cycle. One of the outcomes of Shannon’s work states that the maximum number of bits per symbol, B, that can be transmitted without error is given by B = 1 2 log2 1 + Ps Pn (12-3) where Ps is the average signal power and Pn is the noise power. The quantity Ps/Pn is also known as the signal-to-noise ratio (SNR). Equation (12-3) assumes that the noise is Gaussian, meaning that it is constant at all frequencies within the channel bandwidth, which is a reasonable approximation for digital systems [Sklar, 2001]. Combining the preceding equations gives the Shannon–Hartley theorem, which expresses the maximum data transfer rate in bits per second (b/s) as a function of the interconnect channel bandwidth and the SNR: D = BW log2(1 + SNR) (12-4) Equation (12-4) shows that we can increase throughput across an interchip interconnect either by increasing the signal-to-noise ratio or by increasing the †A real digital signal contains energy at harmonic frequencies above the fundamental, which we can estimate from the rise time (BW ∼= 0.35/tr ) as derived in Section 8.1.3. In this analysis, however, we are considering only the fundamental frequency. 502 EQUALIZATION bandwidth of the signal. However, the lossy nature of PCB transmission lines will tend to nullify the beneﬁt of increasing the signal bandwidth since the content at high frequencies will be attenuated by the interconnect, thereby limiting the usable bandwidth. We look to equalization as a technique to counter the low-pass effect of the transmission-line system so that we can realize the performance beneﬁt of increasing signal bandwidth. 12.1.2 Linear Time-Invariant Systems Before discussing the design and analysis of equalizers we need to introduce the notion of linearity and time invariance. Since our stated intent is to use equalization to counter the low-pass effect of the transmission lines, we examine the behavior of the equalizer and the interconnect in the frequency domain. The assumptions of linearity and time invariance give us the ﬂexibility to analyze the system in either the time or frequency domain to suit our needs. In a linear system, the output of the system, y(t), depends linearly on the input, x(t). Mathematically, this is expressed as y(t) = f [cx(t)] = cf [x(t)] (12-5) In addition, linearity also means that the relationship between the input and the output of the system satisﬁes the superposition property. That is, if the system input can be expressed as the sum of multiple input components, x(t) = i xi(t), the output is equal to the sum of the output values that are obtained for each of the input components: y(t) = f [xi(t)] = yi(t) i i (12-6) These are interesting results that we will use many times in analyzing multi-Gb/s systems, but we have only begun to scratch the surface. Additional aspects of LTI systems that we will exploit are the notions of the impulse response, transfer function, and equivalence between the time and frequency domains. In the time domain, the impulse response of the system, h(t), relates the input to the output via convolution: ∞ y(t) = h(t − τ )x(τ ) dτ = h(t) ∗ x(t) t =−∞ (12-7) where ∗ signiﬁes the convolution operation. The impulse response is the response of the system to an impulse function. The impulse function δ(t) has the properties that it is zero everywhere except at t = 0 and that the area under it is equal to 1 (see Figure 12-3): δ(t) = 0, t = 0 ∞, t = 0 and ∞ δ(t) dt = 1 t =−∞ (12-8) ANALYSIS AND DESIGN BACKGROUND 503 h(t ) t 0 Figure 12-3 Impulse function. Examination of equation (12-7) reveals the convolution operation to be an integral equation that may be difﬁcult to solve. A simpler alternative is to perform the operation in the frequency domain, where it becomes a simple multiplication: Y (f ) = H (f )X(f ) (12-9) where Y (f ), H (f ), and X(f ) are the frequency-domain representations of y(t), h(t), and x(t). We obtain the frequency-domain forms by applying the Fourier transform: ∞ H (f ) = h(t)e2πjf t dt t =−∞ (12-10) where f is the frequency in hertz. Notice that the units of H (f ) will be the units of h(t) multiplied by seconds. For example, if h(t) is dimensionless, H (f ) has units of seconds. Restoring results back to the time domain requires application of the inverse Fourier transform: ∞ h(t) = H (f )e−2πjf t df f =−∞ (12-11) The time–frequency equivalence and the relationships between input and output are summarized in Figure 12-4. Note that the frequency-domain representation of the impulse response H (f ) is known as the transfer function. In general, the frequency-domain representations will be complex quantities, whereas time-domain quantities are real. Example 12-1 Transfer Function for a PCB Differential Transmission-Line Pair Figure 12-5 shows the transfer function for a 0.381-m-long differential pair on a printed circuit that has the following characteristics at a reference frequency, f0, of 1 GHz: L= 3.299 0.407 0.407 3.299 nH/cm C= 1.098 0.085 0.085 1.098 pF/cm R= 509.1 60.64 60.63 509.1 m /cm G= 0.131 0.012 0.012 0.131 mS/cm 504 x (t ) * Impulse Response h (t ) EQUALIZATION ∞ y(t ) = h(t − t)x (t)d t = h (t )*x (t ) t = −∞ Fourier Transform Fourier Transform Fourier Transform Transfer X (f ) Function H (f ) Y (f ) = X (f ) · H (f ) Figure 12-4 Time/frequency-domain relationship for LTI systems. Magnitude of the PCB Transfer Function 0 HPCB (dB) −10 −20 0 5 2 4 6 8 10 Frequency (GHz) (a) HPCB phase angle (rad) 0 −5 0 2 4 6 8 10 Frequency (GHz) (b) Figure 12-5 Frequency response for a PCB-based lossy differential transmission-line pair: (a) magnitude; (b) phase. The differential transfer function is calculated from equation (6-49) assuming no reﬂections and the odd-mode values for the propagation constant using the method described by Johnson and Graham [2003], which gives HPCB(f ) = vout vin = e−γ (f )l (12-12) ANALYSIS AND DESIGN BACKGROUND 505 0.01 PCB impulse response 0.005 0 27 27.5 28 28.5 29 Time (ns) Figure 12-6 Impulse response for a PCB-based lossy differential transmission-line pair. where γ (f ) = [Rodd(f ) + j 2πf Lodd(f )][Godd(f ) + j 2πf Codd(f )] (12-13) The frequency-dependent transmission-line parameters in equation (12-13) are calculated using the causal modeling methodology presented in Chapter 10. Equation (12-12) assumes that the line is perfectly terminated so that there is no reﬂected wave.† The plot in Figure 12-5a shows the magnitude of the transfer function plotted in decibels: |HPCB| = 20 log[ Re(HPCB)2 + Im(HPCB)2] dB Figure 12-5b shows the phase angle formed by the imaginary and real compo- nents: ∠HPCB = tan−1 Im(HPCB) Re(HPCB) rad Figure 12-6 shows the impulse response for the differential pair that was calculated from equation (12-11). †The transfer function for a terminated transmission line is equal to 1 + RTx RRx 1 eγ l +e−γ l 2 + Z0 RRx + RTx Z0 eγ l −e−γ l 2 where RTx and RRx are the termination values at the transmitter and receiver, respectively, and Z0 is the frequency-dependent characteristic impedance. If we assume that RTx = RRx = Z0, the transfer function becomes 1 2 e−γ l . The differential transfer function will have twice the swing, so that H (f ) = e−γ l . 506 EQUALIZATION 12.1.3 Ideal Versus Practical Interconnects Now that we can analyze the behavior of a signaling system in either the time or frequency domain, we are ready to consider the impacts of transmission-line losses on their performance. Once we have done so, we will be ready to begin our study of equalization concepts and designs. We begin by examining the characteristics of an ideal interconnect. An ideal interconnect channel passes signals from source (transmitter circuit) to destination (receiver circuit) without distortion. Signal distortion takes two forms, amplitude distortion and phase distortion. Amplitude distortion results from the unequal attenuation of the different frequency components of the signal. Phase distortion is induced by frequency dependence of the propagation velocity, which causes the phase relationship of the frequency components of the signal to change as it propagates on the transmission line. Amplitude distortion shows up clearly in a plot of the transfer function magnitude, while phase distortion is evident from a plot of the phase of the transfer function versus frequency. Distortion-free transmission is demonstrated in Figure 12-7, which shows the transfer function and impulse response for the differential transmission lines of the example from the preceding section, except that conductor and dielectric losses have been eliminated and the dielectric permittivity is constant across all frequencies. Since losses have been eliminated, the magnitude of the transfer function is equal to 1 (or to 0 dB) at all frequencies. In addition, the impulse response shows a very sharp peak with almost no tail. Example 12-2 Maximum Data Rate Capacity of an Ideal Interconnect In this example we use the Shannon–Hartley theorem to calculate the theoretical maximum transfer capacity of a channel possessing ideal transmission characteristics up to a frequency of 10 GHz, with the assumption that the signal transmitted contains no energy above 10 GHz. Recalling the Shannon–Hartley theorem, D = BW log2 1 + Ps Pn we see that we have the bandwidth information, but we also need the power spectral densities for the signal and noise. We assume that the noise on the channel is white noise, so that it is spread uniformly across the entire frequency band [Sklar, 2001]. A reasonable ﬁgure for the noise spectrum in terms of the voltage is Vnoise = 10−7 V · s0.5. The power spectrum can then be approximated as Pnoise = Vn2oise = 10−14 V2 · s (since power is proportional to V2, P = V 2/R). Since our goal is to estimate the theoretical maximum data rate, we assume that the signal is spread uniformly across the entire band with a 1-V amplitude. The signal spectrum is equal to Vs = √ 1 V = 10−5V · s0.5 10 GHz ANALYSIS AND DESIGN BACKGROUND 507 5 HPCB (dB) 0 −5 0 2 4 6 8 10 Frequency (GHz) (a) 5 HPCB phase angle (rad) 0 −5 0 2 4 6 8 10 Frequency (GHz) (b) 1 PCB impulse response 0.5 0 −0.5 27 27.5 28 28.5 29 Time (ns) (c) 1 PCB impulse response 0.5 0 −0.5 27.55 27.56 27.57 27.58 Time (ns) (d) 27.59 Figure 12-7 Transfer function and impulse response for a lossless, dispersionless PCB-based differential transmission line pair: (a) transfer function magnitude in decibels; (b) transfer function phase in radians; (c) impulse response; (d) impulse response (close-up). 508 EQUALIZATION The power spectrum of the signal is proportional to Ps = Vs2 = 10−10 V2 · s. The maximum data transfer capacity is then D = (10 GHz) log2 1 + 10−10V2 10−14V2 · · s s ∼= 133 Gb/s The assumption of ideal interconnect in this example means that the maximum data rate is independent of physical length. Since a real system will experience per unit length signal attenuation that degrades the signal-to-noise ratio, the theoretical limit will depend on the length of the transmission line. In addition, since the attenuation increases with frequency, the calculated capacity can also be increased by dividing the signal band into subbands, each of which has a unique SNR. Doing so allows realization of the full beneﬁt of the higher SNR at low frequencies while using reduced-SNR, high-frequency subbands to carry some of the data. The frequency- and time-domain responses shown in Figure 12-7 represent the desired (ideal) behavior of a high-speed interconnect. Real interconnects, such as FR4 printed circuit boards, do not resemble the ideal behavior, as Figure 12-8 illustrates. The magnitude plot in Figure 12-8a† indicates that the lossy line from our previous example attenuates the signal by 50% (−6 dB) at approximately 2.1 GHz, which is the fundamental frequency for a 4.2-Gb/s signal. In the next section we demonstrate that the loss on this interconnect channel will completely close the eye of a 4.2-Gb/s data signal. The phase response in Figure 12-8b also indicates that phase distortion is signiﬁcant, and a plot of the propagation velocity in Figure 12-8c reveals an approximate 9% variation from low frequency up to 4 GHz. Both of these effects are apparent in a comparison of the impulse response for the ideal and real cases. Notice that the peak of the ideal impulse response from Figure 12-8d is many times greater than that of the lossy response shown in Figure 12-8e and the tail is much shorter in duration. In essence, the ideal interconnect gives a result that much more closely resembles an impulse function than does the nonideal interconnect, which attenuates and spreads the signal energy. Two other aspects of the impulse response plots are worth noting. First, the area under the ideal impulse response is greater than that of the nonideal response. Numerical integration of the lossless impulse response gives a value of 1 for the area. This is the result expected, since the deﬁnition of the impulse function speciﬁes that it has an area equal to 1. On the other hand, numerical integration of the nonideal interconnect gives a value that is less than 1, which we expect since the lossy transmission line for the nonideal interconnect attenuates the impulsive input signal. †The ﬁgure plots the magnitude of the transfer function as a power ratio in decibels as 20 log[H (f )/H0], where H (f ) is the voltage transfer function at frequency f and H0 is equal to 1. ANALYSIS AND DESIGN BACKGROUND 509 Magnitude (dB) 1 Ideal −1 −3 Nonideal −5 −7 −9 50% amplitude attenuation −11 −13 −15 0 0.5 1 1.5 2 2.5 3 3.5 4 Frequency (GHz) (a) 4 Ideal 3 2 1 0 −1 −2 −3 Nonideal −4 0 0.5 1 1.5 2 2.5 3 3.5 4 Frequency (GHz) (b) 2 · 108 Phase angle (rad) Propagation velocity (m/s) 1.9 · 108 1.8 · 108 1.7 · 108 Ideal Nonideal 1.6 · 1080 1 2 3 4 Frequency (GHz) (c) Figure 12-8 Comparison of ideal versus practical interconnect response in both the time and frequency domains: (a) transfer function magnitude; (b) transfer function phase; (c) propagation velocity; (d) ideal interconnect impulse response; (e) nonideal interconnect impulse response. (Continued) 510 EQUALIZATION Impulse response 1.000 0.800 0.600 0.400 0.200 0.000 −0.200 −0.400 27.4 27.45 27.5 27.55 27.6 27.65 27.7 27.75 27.8 Time (ns) (d) 0.006 0.005 Impulse response 0.004 0.003 0.002 0.001 0.000 27.4 27.45 27.5 27.55 27.6 27.65 27.7 27.75 27.8 Time (ns) (e) Figure 12-8 (Continued ) The second feature that we note is the oscillatory nature of the impulse response for the ideal interconnect. This is due to the fact that the ideal interconnect essentially behaves like an ideal rectangular ﬁlter with a minimum frequency at zero and a maximum frequency fmax. The impulse response of an ideal rectangular ﬁlter is given by the sinc function [Sklar, 2001], which is shown below and in Figure 12-9. h(t ) = 2fmax sin[2πf (t − t0)] 2πf (t − t0) = 2fmax sinc 2πf (t − t0) ANALYSIS AND DESIGN BACKGROUND 511 1 0.5 sinc(x-x0) 0 −0.5 x Figure 12-9 Sinc function. 12.1.4 Equalization Overview Our study of high-frequency interconnects has established that they typically impart a low-pass effect on high-speed signals, which causes amplitude and phase distortion. The typical result is that the interconnect channel closes the data “eye.” An eye diagram is constructed by superimposing numerous consecutive bits in a data stream. The opening of the eye is a metric often used to judge the quality of the signal integrity; an open eye generally indicates a recoverable bit stream, and a closed eye indicates a distorted bit stream that is not recoverable. The eye diagram is discussed fully in Chapter 13. Figure 12-10a shows an open eye at the output of a transmitter. Figure 12-10b shows the eye at the receiver after the data stream has propagated across a lossy interconnect. Note that the interconnect has reduced the area of eye opening dramatically. To develop our conceptual understanding further, we consider the waveforms in Figure 12-11. Figure 12-11a shows the transmitter output for a bit pattern that contains sections with alternating logic levels for each consecutive bit, along with sections that maintain a given logic level for multiple bit positions. As a result, the signal power spectrum spans a wide range of frequencies, with signiﬁcant content from dc to beyond the 5-GHz fundamental frequency. The high-frequency components get attenuated much more severely than do the low-frequency and dc components, as Figure 12-11b demonstrates. A bit pattern of alternating logic states (101010) will have a fundamental frequency that is higher than that of bit patterns that have fewer transitions (such as 110011001100). Since losses increase with frequency, bit patterns that have a higher frequency content will be attenuated more. In other words, for fast bit patterns, the time it takes for the signal to “charge up the interconnect” and transition to its maximum value is greater than the switching rate of the transmitter. This is the intersymbol 512 EQUALIZATION Voltage (mV) 300 250 200 150 100 50 0 −50 −100 −150 −200 −250 −300 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) (a) Voltage (mV) 250 200 150 100 50 0 −50 −100 −150 −200 −250 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) (b) Figure 12-10 Example interconnect channel impact on a 10-Gb/s data eye: (a) transmitter output; (b) receiver input. interference (ISI). Since the disparity of losses between low and high frequencies is what causes ISI, the slope of the loss curve tends to outweigh its magnitude. We want the equalizer to counter the low-pass effect of the interconnect, which causes the high-frequency signal content to be attenuated much more severely than the low-frequency content. Thus, the desired operation of the equalizer is to amplify the signal in such a way as to perfectly counter the attenuation at each frequency. This is expressed mathematically as Heq(f ) = Hc−ha1nnel(f ) (12-14) Simply stated, the ideal equalizer has a transfer function that is the inverse of the channel transfer function, making it a high-pass ﬁlter. Figure 12-12 illustrates this concept, showing that the ideal equalizer counters not only the amplitude distortion caused by the interconnect, but the phase distortion as well. As CONTINUOUS-TIME LINEAR EQUALIZERS 513 Voltage (mV) 300 250 200 150 100 50 0 −50 −100 −150 −200 −250 −300 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Time (ns) (a) 300 250 200 150 100 50 0 −50 −100 −150 −200 −250 −300 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Time (ns) (b) Figure 12-11 Example interconnect channel impact on a 10-Gb/s signal waveform: (a) transmitter output; (b) receiver input. Voltage (mV) Figures 12-13 and 12-14 demonstrate, an idealized equalizer would completely reverse the effects of the interconnect, restoring the signal to its original form, exactly matching the waveform and eye at the transmitter output. In practice, power and device count limitations make the design of a perfect equalizer impractical. However, we do not require ideal implementation in order to realize the beneﬁts of equalization, as we shall see in the remaining sections of this chapter. In subsequent sections we explore the design, operation, and limitations of the different types of equalizers that ﬁnd use in multi-Gb/s signaling systems now and in the future. Equalization is an area in which there is considerable ongoing research and one that has many possible implementations. Although we present some representative examples, the nuances of equalizer design implementations are beyond our scope. Instead, we focus on developing a solid understanding of their behavior in terms of fundamental building blocks so that we may use them effectively to maximize the performance of our signaling systems. 12.2 CONTINUOUS-TIME LINEAR EQUALIZERS Continuous-time linear equalizers (CTLEs) are analog in nature, operating continuously as the name implies. This is in contrast to discrete-time equalizers, 514 EQUALIZATION 15 10 Ideal equalizer 5 Magnitude (dB) 0 −5 Phase angle (rad) −10 −15 0 4 3 2 1 0 −1 −2 −3 −4 0 Nonideal channel 0.5 1 1.5 2 2.5 3 3.5 4 Frequency (GHz) (a) Ideal equalizer Nonideal channel 1 2 3 4 Frequency (GHz) (b) Figure 12-12 Transfer function of an ideal equalizer: (a) magnitude; (b) phase. which we discuss in subsequent sections. CTLEs do not require digital devices for implementation but are constructed from simple analog components, both passive (resistors, capacitors, inductors) and active (ampliﬁers). In this section we use the CTLE to expand on our previous discussion of equalizer behaviors, and we develop an understanding of their operation through the presentation of some representative designs. 12.2.1 Passive CTLEs An example of a passive CTLE is shown in Figure 12-15. As the name suggests, this type of equalizer does not amplify any components of the signal that pass through it. Instead, to achieve the desired high-pass effect the equalizer attenuates the low-frequency signal components. The main sections of the equalizer are the CONTINUOUS-TIME LINEAR EQUALIZERS 515 Voltage (mV) 300 250 200 150 100 50 0 −50 −100 −150 −200 −250 −300 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 Time (ns) Figure 12-13 Waveform showing the restoration of the lossy signal from Figure 12-11 after processing with an ideal equalizer. Voltage (mV) 300 240 180 120 60 0 −60 −120 −180 −240 −300 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) Figure 12-14 Impact of an ideal equalizer on the data eye from Figure 12-10. CHP RTT Z0 RHP RL CL Figure 12-15 Passive CTLE. 516 EQUALIZATION termination, high-pass ﬁlter, and dc power-limiting ﬁlter. Termination is provided by the single resistor RT T , which is matched to the transmission-line impedance. The high-pass ﬁlter is composed of the parallel capacitance and resistance, CHP and RHP, whose values are chosen to give the desired frequency response. Finally, CL and RL prevent the system from dissipating excess power at dc while providing a high-frequency path for the termination. The transfer function for this circuit is Heq(f ) = RT T ZL + ZHP + ZL (12-15) where ZL(f ) = 1 + RL j 2πf RLCL ZHP(f ) = 1 + RHP j 2πf RHPCHP (12-16) (12-17) From (12-15) through (12-17) we can plot the frequency response of the equalizer, which is shown in Figure 12-16 for RT T = 100 , RHP = 5 k , CHP = 100 fF, RL = 2.5 k , and CL = 20 fF. In addition to the equalizer, the ﬁgure shows the transfer function for the printed circuit board interconnect from Example 12-1 and the combined response of the PCB–equalizer system. From the transfer function we see that compared to the PCB alone, the passive equalizer reduces the overall loss variation between dc and 10 GHz by 5.2 dB. The signiﬁcance of this reduced variation between minimum and maximum signals becomes apparent when we consider that 6 dB of loss at the fundamental frequency is sufﬁcient to close the eye completely (see Figure 12-17). This means that if the slope of the loss curve is such that high-frequency bit patterns (101010) are attenuated 6 dB or more relative to low-frequency bit patterns (111000111000), the eye will close. To comprehend the beneﬁt of equalization, consider Figure 12-16a, which shows that the magnitude of the equalized channel is more lossy, but the total variation in the loss decreased from −18.3 dB to −23.5 dB. The reduction in the loss variation in our example doubles the frequency at which the eye closes completely from approximately 2 to 4 GHz. The nonequalized eye will close at −6 dB, and the equalized eye will close when there is −6 dB of variation from the −7-dB peak. Thus, the equalizer provides an increase in usable bandwidth simply by ﬂattening the slope of the loss curve that will allow us to extract more performance from the system by running at a substantially higher data rate. Assuming that the noise power spectrum is unaffected by the equalizer, the Shannon–Hartley theorem predicts that the equalizer would allow a doubling of the maximum data rate. In terms of the expected performance demand trend, an equalizer can extend the useful life of a given interconnect by approximately two years. (Of course, the exact beneﬁt for a given system will depend on the speciﬁcs of the channel loss characteristics, the equalizer design, and actual performance demand.) CONTINUOUS-TIME LINEAR EQUALIZERS 517 Magnitude (dB) 0 −3 CTLE −6 100% eye closure −9 −12 −23.5 dB −15 −18.3 dB −18 −21 −24 100% eye closure PCB PCB + CTLE −27 −30 0 1 2 3 4 5 6 7 8 9 10 Frequency (GHz) (a) 2 Phase (rad) 0 −2 0 0.5 1 1.5 2 Frequency (GHz) PCB (b) CTLE PCB + CTLE Figure 12-16 Passive CTLE transfer function: (a) magnitude; (b) phase. The other impact of the equalizer is that the overall swing is reduced in order to achieve the reduction in minimum-to-maximum loss for the system. This “ﬂattening” of the magnitude of the frequency response is a good qualitative way to view the beneﬁt of equalization. Even though we reduce the maximum signal swing of the system, we realize a net gain because we have reduced the variation between the maximum signal swing (low frequency) and the minimum signal swing (high frequency). Figures 12-17 through 12-19 each illustrate this concept in slightly different ways. In Figure 12-17 we see that a 6-dB loss is sufﬁcient to completely 518 EQUALIZATION Voltage (V) 0.8 0.6 1.0 V 0.4 1.0 V 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Time (ns) (a) Voltage (V) 0.8 0.6 1.0 V 0.4 0.0 V 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Time (ns) (b) Figure 12-17 Impact of 6-dB loss on the eye for a sinusoidal signal: (a) lossless; (b) 6-dB loss. close the eye of a sinusoid. Note that we chose a 1-V swing, 1-GHz signal for illustration, but that the 6-dB rule is independent of swing and frequency. What does depend on frequency is where the 6-dB loss point actually occurs. Figure 12-18 portrays the eye diagram for a 10-Gb/s signal traveling across the 0.381-m-long differential PCB transmission-line pair from Example 12-1. Given a 500-mV signal swing and a 300-bit random sequence at the transmitter output, we ﬁnd that the eye at the receiver in Figure 12-18a is completely closed. Figure 12-18b shows that the equalizer improves the situation considerably, providing an eye opening of approximately 55 ps and 40 mV. Comparison of the eyes shows that the equalizer reduces the maximum signal swing from 430 mV at the receiver (Figure 12-18a) input to 250 mV at the output of the equalizer (Figure 12-18b). CONTINUOUS-TIME LINEAR EQUALIZERS 519 Voltage (mV) 250 200 150 100 50 0 −50 −100 −150 −200 −250 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) (a) 150 120 90 60 30 0 −30 −60 −90 −120 −150 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) (b) Voltage (mV) Figure 12-18 Received eye diagram with passive CTLE (a) before and (b) after equalization. Finally, Figure 12-19a shows the signal waveform for the same bit pattern as in Figure 12-11a. The waveform at the receiver in Figure 12-19b shows the nature of the behavior that leads to the closed data eye. The signal shows a maximum range of approximately −210 mV to approximately 200 mV. However, the minimum swings for the low-to-high and high-to-low transitions reach only 5 mV and −5 mV, respectively. So for this abbreviated bit pattern, there is only a 10-mV eye opening. A longer bit sequence will further degrade the eye to the point where it is completely closed. Moving to the passively equalized signal in Figure 12-19c reveals a maximum range of approximately −105 mV to approximately 125 mV. The minimum swings for the low-to-high and high-to-low transitions reach 30 mV and −20 mV, respectively. So while the equalizer reduces the maximum signal swing from 410 mV to 230 mV, it increases the minimum eye from 10 mV to 50 mV. Clearly this is a favorable trade-off. Before moving to discussion of additional CTLE design options, we pause to note that the transfer function for our passive equalizer does not come close 520 EQUALIZATION Voltage (mV) 300 250 200 150 100 50 0 −50 −100 −150 −200 −250 −300 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 Time (ns) (a) Voltage (mV) 300 250 Max low-high swing 200 150 100 Min low-high swing 50 0 −50 −100 Min high-low swing −150 −200 −250 Max high-low swing −300 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 Time (ns) (b) 300 250 Max low-high swing 200 150 Min low-high swing 100 50 0 −50 −100 −150 −200 −250 Max high-low swing Min high-low swing −300 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 Time (ns) (c) Voltage (mV) Figure 12-19 Example signal waveforms with passive CTLE: (a) transmitter output; (b) received input; (c) equalized output. to providing perfect equalization (compare Figures 12-12 and 12-16). However, the equalizer compensates sufﬁciently for the loss of the 0.381-m PCB trace to increase the eye height by more than 400% for a 10-Gb/s data signal (from 10 to 50 mV). Another example of a passive equalizer from Sun et al. [2005] based on a high-pass RLC ﬁlter is shown in Figure 12-20. In this case we show versions CONTINUOUS-TIME LINEAR EQUALIZERS 521 C R1 In R1 Out C L R1 R1 In Out R2 R2 L L In Out R1 R1 C (a) (b) Figure 12-20 Alternate passive CTLE implementation: (a) single-ended; (b) differential. for both singled-ended and differential signals. Analysis of this equalizer is left as a problem at the end of the chapter. Passive equalizers offer the advantage of improved performance with no additional power consumption. In addition, multi-Gb/s data rates drive smaller, higher-frequency passive devices, making integration into the silicon an attractive design option. However, they demand tighter control of the components values than for typical digital applications. In addition, the frequency response of passive equalizer circuits is not directly tunable without additional active control circuitry, which will tend to degrade the power beneﬁt. 12.2.2 Active CTLEs Equalizers can also be constructed with active components (ampliﬁers) to provide some signal gain. This type of equalizer is often done using a split-path approach, as shown in Figure 12-21 [Liu and Ling, 2004]. The incoming signal is fed into a unity-gain path and a high-frequency boost path, which are then summed to create the output. The transfer function for this equalizer is H (f ) = 1 + R2/R3 + 1 1 + 1/2πf R1C1 (12-18) The high-pass ﬁlter has a voltage gain of 1 + R2/R3, with a corner frequency equal to 1/R1C1. The equalizer transfer function for a high-pass ﬁlter voltage gain of −3.5 dB and a 5-GHz corner frequency is plotted along with that of the passive equalizer in Figure 12-22. The magnitude plots for the two equalizers show a very similar shape, with an offset of approximately 5 dB. The resulting 522 EQUALIZATION data eye for the equalized 200-bit random pattern is shown in Figure 12-23. As expected, the equalizer gives an open data eye with approximately twice the signal amplitude and eye height than those of the data eye from the passive equalizer in Figure 12-18. A potential source of confusion that merits discussion is the use of power gain versus that of voltage gain. In our transfer function plots, we use power gain. Since power is a function of the square of the voltage (or current) signal, gainpower(dB) = 10 log Pout Pin (12-19) Conversely, ampliﬁer manufacturer datasheets often specify gain in terms of the voltage: gainvoltage(dB) = 20 log Vout Vin (12-20) The moral here is that when working with gain ﬁgures, be sure to know whether you are dealing with voltage or power gains, as mixing the two can lead to erroneous results. Active CTLEs provide gain, so they must dissipate some power, although careful design can keep the power to less than 10 mW. In addition, active CTLEs can be designed using higher-order high-pass ﬁlters, but the performance gain typically may not justify the additional power consumed depending on the channel response of the speciﬁc application. The active CTLE has some fundamental limitations, notably the limited bandwidth of the ampliﬁer and phase mismatch between the two ampliﬁer paths. For example, Kudoh et al., [2003] show −3-dB bandwidths of 3 GHz for a conventional feedback ampliﬁer and 10 GHz for an improved design. Finally, tailoring the frequency response and the gain of the ﬁlter requires the inclusion of additional control circuitry, just as it did with the passive equalizer. 12.3 DISCRETE LINEAR EQUALIZERS In Section 12.2 we dealt with equalizers that were completely analog in nature. Contemporary high-speed components such as microprocessors, graphics processors, or memory controllers are manufactured using processes that are highly optimized for digital circuit applications. Whereas the equalization problem is very analog in nature, the discrete linear equalizer makes use of a combination of digital and analog techniques. This allows us to better utilize the economies of scale provided by Moore’s law, thus satisfying the performance demand at minimum cost. As such, they ﬁnd wider use in computing equipment than do their analog counterparts. The linear time-invariant nature of our signaling system has another aspect that we have not yet covered that will give us ﬂexibility in implementing equalizer solutions. In particular, the response of an LTI system does not depend on the DISCRETE LINEAR EQUALIZERS 523 C1 Vin R1 + − R2 R3 + Vout Figure 12-21 Example active equalizer using a ﬁrst-order high-pass ﬁlter. 5 4 Active CTLE 3 2 Magnitude (dB) 1 0 −1 Passive CTLE −2 −3 −4 −5 0 2 4 6 8 10 Frequency (GHz) (a) 0.5 0.44 0.38 Passive CTLE 0.32 Phase (rad) 0.26 0.2 0.14 0.08 Active CTLE 0.02 −0.04 −0.1 0 2 4 6 8 10 Frequency (GHz) (b) Figure 12-22 Transfer function for the active equalizer: (a) magnitude; (b) phase. 524 EQUALIZATION Voltage (mV) 300 240 180 120 60 0 −60 −120 −180 −240 −300 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) Figure 12-23 Eye diagram results for the active equalizer. H Tx(f ) H(f ) = S21 H eq(f ) (a) H Rx(f ) H Tx(f ) H (f ) = S 21 (b) HRx(f ) × Heq(f ) H Tx(f ) × H eq(f ) H (f ) = S 21 H Rx(f ) (c) Figure 12-24 Equalizer location options: (a) in the interconnect channel; (b) integrated into the receiver; (c) integrated into the transmitter. order in which the ﬁltering operations are carried out. This allows us to locate the equalizer within the system to meet our needs, as Figure 12-24 shows. To minimize the cost of the system, the equalizers are typically integrated with the transmitter circuits or receiver circuits, as demonstrated by Figure 12-24b and c. Although the location of the equalizer does not matter in theory, we will see that there are trade-offs associated with placement that affect design decisions. DISCRETE LINEAR EQUALIZERS 525 xk T T T T C−N C−N + 1 CN − 1 CN S yk Figure 12-25 Finite impulse response ﬁlter. 12.3.1 Transmitter Equalization The basic architecture of a discrete linear equalizer is the transversal ﬁlter, also known as a ﬁnite impulse response (FIR) ﬁlter, which is shown in Figure 12-25. In the ﬁgure the rectangles represent delay elements, such as the stages of a shift register. The circles represent the ﬁlter taps. In this ﬁlter, input samples (typically, voltage samples), xk, propagate through the delay elements, each of which has a delay value of T , which is also known as the tap spacing. At each stage, the input samples are multiplied by the ﬁlter tap coefﬁcient, Ci , where i is simply the index into the tap subscripts. With each cycle the outputs from the taps are then summed to provide the ﬁlter output, yk. In effect, the current and past values of the signal are linearly weighted with the equalizer coefﬁcients (also known as tap weights) and then summed to produce the output. The ﬁgure shows a total of 2N taps in the ﬁlter, numbered from −N to N . The main contribution comes from the cursor tap, C0. This tap is intended to amplify the main portion of the signal. Filter taps with negative coefﬁcients are known as precursor taps; those with positive coefﬁcients are called postcursor taps. Figure 12-25 shows symmetry in the number of precursor and postcursor taps, but equalizers are typically designed with unequal numbers of precursor and postcursor taps. Precursor taps compensate for dispersion-induced phase distortion, which typically requires only a single tap. Postcursor taps compensate for the ISI caused by amplitude distortion and may require multiple taps, depending on the length of the channel with respect to the width of a data bit. The output of the equalizer, y(k), is expressed as the discrete convolution of the input signal, x(k), with the equalizer ﬁlter coefﬁcients: N y(k) = x(k − n)cn k=−N (12-21) where k is the sample number of the discretely sampled signal (i.e., the time position of a given sample is tk = kT , where T is the tap spacing of the equalizer). 526 EQUALIZATION Example 12-3 Transmit DLE Operation To further develop our understanding, we step through operation of the discrete linear equalizer shown in Figure 12-26 for the case in which it is presented with a lone pulse input of 600 mV amplitude and 1 ns width. For this example we use a tap delay T of 1 ns and the following values for the tap weights: C−1 = −1/12, C0 = 2/3, C1 = −1/6, and C2 = 1/12. 1. The pulse arrives at the input to the ﬁlter, where it is multiplied by C−1, generating a precursor pulse of −50 mV amplitude and 1 ns duration. 2. After a delay of 1 ns, the pulse appears at the second tap and is multiplied by C0, generating a 400-mV 1-ns cursor pulse. 3. One nanosecond later, the pulse is at the third tap, where it is weighted with C1, generating the ﬁrst postcursor pulse of −100 mV amplitude and 1 ns duration. 4. After an additional 1-ns delay, the input comes to the ﬁnal tap, where it undergoes multiplication with C2, creating a +50-mV, 1-ns postcursor pulse. The summing element operates during the entire sequence. Since our input is a lone pulse, the signal inputs to the ﬁlter taps on either side of our pulse do not generate any “echoes,” so that the output from the equalizer is a linearly weighted time-delayed version of the original input, as Figure 12-26b shows. Equalization at the transmitter is often called transmitter preemphasis to reﬂect the effect of the ﬁlter operation. As we have discussed, the function of the equalizer is to provide a high-pass ﬁltering effect to the signal. In a digital signal, the highest-frequency content is contained in rapid transitions between logic states, while low-frequency content is contained in portions of the signal that do not make transitions. The manifestation in a discrete linear equalizer is increased amplitude for the ﬁrst bit after a logic transition relative to successive bits. This effect is shown in Figure 12-27 for a 500-mV transmitter with 20% equalization: vmax.swing − vmin.swing × 100 = 500 mV − 300 mV × 100 vtotal swing 500 mV = 20% = −13.97 dB The full signal swing occurs for the case when the signal transitions between logic states for at least two successive bit positions. It ranges from −100 mV on the low side to 400 mV on the high side, for a total of 500 mV. To ﬁnd the minimum swing we look at the waveform regions that contain multiple consecutive zeros (0 mV) and 1’s (300 mV). The variation in swing from 500 mV maximum to 300 mV minimum corresponds to 20% equalization. High-speed signaling systems typically use as much drive current as possible in order to maximize speed, the limitation being the maximum voltage swing that DISCRETE LINEAR EQUALIZERS 527 1 ns 1 2 3 4 600 mV xk T T T C−1 C0 C1 C2 50 mV 4 400 mV 3 2 −100 mV 1 S yk −50 mV (a) 1 2 + 400 mV −100 mV 3 4 ns 4 T (b) Figure 12-26 Transmitter equalization ﬁlter operation: (a) propagation through the ﬁlter; (b) output signal creation. the process will support. Leading-edge digital silicon manufacturing processes are pushing the maximum swing below 1.0 V. As a result, designers most often implement transmit equalization by attenuating the low-frequency bits rather than by amplifying the high-frequency bits. This is often called deemphasis rather than preemphasis, to reﬂect the fact that the equalization is done by reducing the amplitude of the repetitive bits. The maximum signal swing limit (for both pre- and deemphasis) places a constraint on the coefﬁcient settings for an equalizer according to |ci| = 1 i (12-22) 528 EQUALIZATION Voltage (mV) 450 400 350 300 500 mV 250 200 150 00001010101000001000111101110 1 0 1 0 1 0 1 0000 100 50 0 300 mV −50 −100 −150 0 0.5 1 1.5 2 2.5 3 3.5 Time (ns) Figure 12-27 Example preemphasized transmitter output. Satisfying the equation guarantees that the output signal swing does not exceed the maximum swing that the manufacturing process can accommodate. Having developed an understanding of transmitter preemphasis operation, we can turn to the questions of how many taps are needed to equalize a given system and how to determine the coefﬁcient settings. We begin the discussion with an example. Example 12-4 Effect of the Number of Taps for a 10-Gb/s Interface We compare the performance of the 0.381-m-long differential pair operating at 10 Gb/s for different numbers of equalizer taps. Once again we use the PCB interconnect from Example 12-1, which has the following odd-mode quantities at a reference frequency, f0, of 1 GHz: C = 1.184 pF/cm, L = 2.892 nH/cm, R = 448.2 m /cm, and G = 0.144 mS/cm. We drive the interconnect using a ±2.5-mA 0.5-pF differential transmitter with perfect termination. Without equalization the worst-case eye at the receiver for the system shown in Figure 12-18b is closed completely (eye height = −34 mV), as calculated using the peak distortion analysis method (see Chapter 13). Simulated results for a number of equalizer conﬁgurations are shown in Figure 12-28. Conservative timing and voltage specs for a 10-Gb/s differential system would require approximately 65 ps and 80 mV minimum at the receiver, and we use them to assess the adequacy of the equalizer designs. In calculating the response of the equalizer, we use the expression for the transfer function of the equalizer (the derivation of which is left as a problem at the end of the chapter): Npost H (f ) = ck e−j 2πf (k−Npre)T k=−Npre (12-23) DISCRETE LINEAR EQUALIZERS 529 Minimum spec Eye height (mV) 80 85 90 85 80 C1 = −0.25 C2 = −0.03 Small benefit from 2nd postcursor tap C1 = −0.27 C1 = −0.26 75 95 100 105 110 Precursor tap improves eye C−1 = −0.06 C1 = −0.25 C−1 = −0.08 C1 = −0.25 C−1 = −0.08 C3 = 0.02 C1 = −0.25 C1 = −0.28 C1 = −0.30 C−1 = −0.02 C1 = −0.30 C−1 = -0.04 C1 = -0.30 Benefit from 3rd postcursor C1 = −0.29 C1 = −0.31 C−1 = −0.06 tap C1 = −0.30 C1 = −0.32 C1 = −0.25 Best 2-tap design C1 = −0.33 Eye width (ps) 70 ZFS C−1 = −0.09 C1 = −0.25 65 C3 = 0.04 Minimum spec Figure 12-28 Worst-case received eyes as a function of equalizer design for Example 12-4. The results for two tap designs (cursor tap, co, plus one postcursor tap, c1) are connected via lines on the ﬁgure and show that a two-tap design will meet the specs proposed when the postcursor tap coefﬁcient (c1) is −0.25. In fact, the effectiveness of the equalizer increases up to a value of approximately −0.30 to −0.32. Beyond that, the eye margin begins to degrade. The question of which coefﬁcient value works best will depend on whether voltage margin or timing margin is more important to the design. If we were to assume that they are equally important, we could evaluate the results by calculating the product of the eye height and width. This trend is plotted in Figure 12-29 for the two-tap equalizer, which shows that −0.31 gives the best results for this criterion, providing a minimum eye height of 101 mV and width of 76 ps. Figure 12-28 also shows that adding taps can improve the voltage and timing margin, although it will not necessarily do so. In particular, the addition of a second postcursor (c2) tap offers very little beneﬁt, as shown in the ﬁgure. Adding a precursor tap (c−1) provides incremental improvement in the eye size (101 mV, 83 ps). With a third postcursor tap the eye increases to 109 mV and 81 ps. All of these results assume that we can optimize the equalizer coefﬁcients with a step size (granularity) of 0.01. Note that all equalizer conﬁgurations meet the criteria speciﬁed by equation (12-22). 530 EQUALIZATION Height·width (mV·ps) 8000 7800 7600 7400 7200 7000 6800 6600 6400 6200 6000 −0.34 −0.32 −0.3 −0.28 C1 −0.26 −0.24 Figure 12-29 Worst-case eye trend for the two-tap equalizer for Example 12-4. In summary, the two-tap equalizer is capable of meeting the voltage and timing specs proposed, with some room for adjustment. Addition of a precursor improves the margins by 7.5% for eye height and 3.8% for eye width. Addition of the additional postcursor taps provides an additional 2.4% increase in eye width but no improvement in eye height. The ﬁnal step in this example is to use the results to decide how many taps to include in the equalizer design. The two-tap design provides a sufﬁcient eye to meet the design specs, and it can be implemented with a very simple design. In general, designers will choose the simplest equalizer design that meets the requirements for a given application. In our case, the two-tap equalizer will work just ﬁne. Longer channels will typically have more attenuation, causing the ISI to intrude on additional postcursor bit positions and may therefore require additional equalizer taps. Higher data rates may also require more taps. 12.3.2 Coefﬁcient Selection At this point the question of how to determine the tap coefﬁcients seems appropriate to consider. In this section we present a method that can be used to establish coefﬁcient values based on speciﬁc performance criteria. The method, called a zero forcing solution (ZFS), sets the tap coefﬁcients to force the equalizer output to match the values desired at all sample points. [Qureshi, 1985; Sklar, 2001]. Development of the algorithm follows. Given a pulse response input to the equalizer, we start by extracting samples from the input stream xi, where i runs from −npre to npost, with npre being the number of precursor taps and npost the number of postcursor taps in the equalizer. This gives a total of npre + npost + 1 samples. We can express the DISCRETE LINEAR EQUALIZERS 531 input samples in matrix form: x(0) x = x x(1) ... (npost 0 ... 0 ) 0 x(−1) x(0) ... x(npost − 1) x (npost ) ... ··· ··· ··· ··· x (npost ) 0 x(−npre) x(−npre + 1) ... ... ... ... x(npost − 1) x (npost ) 0 x(−npre) ... ··· ··· ··· 0 ... x(−npre + 1) x(−npre + 2) ... x(0) x(1) 0 x(x−x(n−(p−0......nre1p+r)e)1) x(0) (12-24) The columns of x represent the taps of the equalizer and the rows represent consecutive time steps with an interval between steps that is equal to the tap spacing of the equalizer. The matrix is square with npre + npost + 1 rows and columns. In this form, x shows the propagation of the input samples through the equalizing ﬁlter. For example, x(0) appears in the ﬁrst row and the ﬁrst column, which indicates that it is at the ﬁrst tap at the ﬁrst time sample. It also appears in the second row and second column, which corresponds to the second tap and second time sample. This is exactly what we expect for a discrete linear equalizer. In matrix form, equation (12-21) is written y = xc (12-25) where y is the vector containing the output from the equalizer and c is the vector of equalizer tap coefﬁcients. The number of elements in both y and c is npre + npost + 1. We have essentially expressed the discrete convolution of the data stream and equalizer as a matrix multiplication, which we now apply to our problem. By transmitting a lone pulse, we deﬁne the expected output results from the equalizer according to Nyquist’s ﬁrst method for the elimination of ISI [Nyquist, 1928; Couch, 1987]: ytarget = 0 1 for k = 0 for k = 0 (12-26) In (12-26) the expected value is 1 for the cursor sample and zero for all others. The equalizer coefﬁcients that give the zero forcing solution are then cZFS = x−1ytarget (12-27) The coefﬁcients calculated are not constrained by power or by the maximum signal swing that can be achieved. Application of such constraints requires that the coefﬁcients be adjusted using equation (12-22), as we illustrate next. 532 EQUALIZATION Example 12-5 Zero Forcing Solution We wish to ﬁnd the coefﬁcients for an equalizer with a single precursor tap and three postcursor taps that give the zero forcing solution for the PCB differential pair from Example 12-1 when operating at 10 Gb/s. In Example 12-3 we saw that the worst-case eye height without equalization is −34 mV for this interconnect. Using the 5-mA transmitter from our earlier example gives a differential swing of ±250 mV. The driving pulse is 100 ps wide, corresponding to a single bit of data in a 10-Gb/s data stream. From the simulated differential pulse response in Figure 12-30, we extract the input sample vector for a tap spacing of 100 ps: −214 xin = −−−12251023 mV −232 The expected output from the equalizer with npre = 1 and npost = 3 is 0 ytarget = 100 0 In order to apply equations (12-26) and (12-27), we must apply a level shift to the differential signal so that the minimum output voltage at the transmitter is zero. We do this by adding one-half of the differential swing (250 mV) to the input vector to produce an adjusted input sample vector: 0 −50 Voltage (mV) −100 −150 −200 −250 −300 6800 7300 7800 Time (ps) 8300 Figure 12-30 Differential pulse response before equalization for Example 12-5. DISCRETE LINEAR EQUALIZERS 533 36 xin,adj = 230 97 37 mV 18 We then construct the input sample matrix and calculate the equalizer coefﬁcients: 230 36 0 0 0 x = 97 37 18 230 97 37 36 230 97 0 36 230 0 0 36 mV 0 18 37 97 230 −0.77 c = x−1ytarget = −410...991384 −0.13 We apply the maximum signal swing constraint by using equation (12-22) to adjust the equalizer coefﬁcients: cZFSi = −0.097 ci |ci | = −000...620251407 −0.016 Figure 12-31 shows the resulting equalized pulse response, which demonstrates the effect of the zero forcing equalizer. Since the width of the nonequalized pulse in Figure 12-30 is much wider than a single bit (which is 100 ps wide), signiﬁcant ISI is expected. The equalized pulse response is much more narrow and the ISI has been removed (i.e., forced to be zero) at the equalizer sample points. However, the ﬁgure also reveals the fact that the ZFS approach eliminates the ISI only at the sampling points that correspond to the equalizer taps. The equalized pulse shows ISI in the intervals between the sample points and at sample points outside the equalizer. The simulated output from the ZFS equalizer in Figure 12-32 shows an eye opening of approximately 110 mV and 85 ps for a 300-bit pseudorandom bit sequence. In addition, the worst-case data eye calculated from peak distortion analysis on the equalized pulse response is 107 mV high and 80 ps wide. As we can see, the zero forcing equalizer provides signiﬁcant improvement in this case over that of the nonequalized channel. 534 EQUALIZATION Voltage(mV) 50 30 10 −10 −30 −50 −70 −90 6800 7300 7800 Time (ps) 8300 Figure 12-31 Differential pulse response with equalization for Example 12-5. Voltage (mV) 150 120 90 60 30 0 −30 −60 −90 −120 −150 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Time (ps) Figure 12-32 Equalized differential data eye for Example 12-5. The chief limitation of the zero forcing approach is that for a ﬁnite-length equalizer, minimization of the ISI is guaranteed only if the eye is initially open, a condition that is not necessarily true for long, lossy channels such as backplanes. An approach that avoids the limitation of the ZFS algorithm is the minimum mean-square error (MMSE) algorithm. This approach sets the equalizer coefﬁcients such that the mean-square error of the ISI is minimized at the output of the equalizer. Examination of the MMSE equalizer is left for Problem 12-7, and a thorough treatment of both the ZFS and MMSE methods is given by Proakis [2001]. DISCRETE LINEAR EQUALIZERS 535 12.3.3 Receiver Equalization As we showed in Section 12.1.2, the equalizer may be located anywhere in the channel under the LTI assumption. Although equalization is often implemented at the transmitter, it may also be done at the receiver. The main reason that discrete linear equalizers are most often located at the transmitter is ease of implementation. Figure 12-33 provides a block-level diagram of a discrete linear equalizer implementation at the receiver which shows that the equalizer looks like the transmit equalizer with the addition of a sample-and-hold circuit on the front of the equalizer (represented by the switch in the ﬁgure). The input to the receiver equalizer is an analog voltage waveform rather than a binary data pattern. In addition, transmitter equalization can be accomplished in a straightforward manner by using a multiplexer to multiply the tap coefﬁcient by the bit value. 600 mV 1 2 3 4 xk T T T C−1 C0 C1 C2 4 3 2 1 S yk (a) 1 −50 mV 400 mV 2 + 3 −100 mV 4 −50 mV t T (b) Figure 12-33 Receiver equalization ﬁlter operation: (a) propagation through the equalizer; (b) output signal creation. 536 EQUALIZATION Insertion loss [dB] Linear Equalizer Equalized Channel Enhanced Noise Noise Channel f [GHz] Figure 12-34 Noise enhancement in a linear equalizer. 12.3.4 Nonidealities in DLEs Thus far we have assumed that the equalizers operate with ideal characteristics in our analysis. Of course, since real implementations will not behave in an ideal fashion, we offer a brief discussion of the limitations of discrete linear equalizers. We begin by pointing out that a practical equalizer implementation has limited resolution on the tap coefﬁcients. Our previous analysis has assumed that we can set the tap coefﬁcients with resolution down 0.001. Tap coefﬁcient values are often set using current digital-to-analog converters (DACs) that bias the transmitter tail currents [Dally and Poulton, 1997]. Achieving a tap resolution of 0.001 would entail using a 10-bit binary weighted DAC, which is likely to consume excessive silicon area and power. As a point of comparison, Jaussi et al. [2005] used a 6-bit DAC for a four-tap equalizer that achieved 8 Gb/s over a 102-cm PCB-based channel. Other nonidealities include errors in the sampled voltage due to sampling jitter and charge leakage, quantization noise of the analog-to-digital conversion, nonlinearity of the equalizer taps and summing circuits, and offset currents due by device mismatch. [Jaussi et al., 2005]. Finally, discrete linear equalizers do not distinguish between signal and noise, so they ﬁlter both the signal and noise, as shown in Figure 12-34. As a result, DLEs do not improve signal/noise ratio. Instead, the performance gain is due to the increase in usable bandwidth provided by the ﬂattening of the frequency response, as discussed earlier. 12.3.5 Adaptive Equalization At intermediate data rates, the equalizer coefﬁcients are often set based on the average characteristics of the interconnect channel. For example, the PCI Express interface calls for −3.5 dB of equalization for PCB-based interconnects of up DISCRETE LINEAR EQUALIZERS 537 to 15 in. in length [Coleman et al., 2004; PCI-Sig 2005]. However, effectively equalizing channels that may exhibit wide variation in frequency-dependent loss (e.g., a range of PCB lengths) requires that the equalizer design contain the ﬂexibility to set the equalizer coefﬁcients adaptively to minimize the ISI. Such an equalizer, called an adaptive equalizer, was invented by Lucky in 1964 to improve data transmission rates over telephone lines from 1200 b/s (using nonadaptive equalizers based on average channel characteristics) to 9600 b/s [Lucky, 2006]. The beneﬁt of adaptive equalization is the ﬂexibility to accommodate a range of interconnect lengths and/or data rates. It does, however, add signiﬁcant complexity to the design and consumes more power and chip real estate. A high-level schematic of the adaptive equalizer structure is shown in Figure 12-35. The intent of the adaptation is illustrated in Figure 12-36. The ﬁgure depicts the deviation of the equalized signal from desired value as a function of equalizer coefﬁcients. The deviation is plotted as a set of error contours in which the error is the difference in the equalized output, y(t), from the training data, yˆ(t). The error is a convex function of the equalizer coefﬁcients, so that it has a global minimum. The goal of the adaptive algorithm is to converge on a set of coefﬁcient values that minimize the error in a small number of iterations. The general approach of an adaptive equalizer to updating the equalizer tap coefﬁcient is cnew = cold + (step size)(error function)(input function) (12-28) The error function is typically based on the difference between the actual equalized signal y and the desired equalizer output y. The input function is based on the signal at the input to the equalizer, and step size is a design parameter. Designers have many options for implementing adaptive equalizers, the range of which extends beyond our scope. However, we present a pair of examples to provide some insight into the operation of adaptive equalizers. The ﬁrst example is the adaptive implementation of the zero forcing equalizer. In this approach a known data pattern (a.k.a. training sequence) of equal or larger length than that of the equalizer is transmitted and equalized. The coefﬁcients are updated from the equalized results using ck(n + 1) = ck(n) + k[yˆ(n) − y(n)]x(n − k) (12-29) x (t ) Linear Equalizer Coefficient Update y (t ) in y∧(t ) + Training Data Figure 12-35 Adaptive linear equalizer. 538 EQUALIZATION error n=1 n=2 n=3 n=0 0.2 0.16 0.12 0.08 c2 0.04 c1 0 0.0 0.1 0.2 0.3 0.4 0.5 Figure 12-36 Adaptive equalization error “contour” and coefﬁcient convergence. where ck(n) is the value of the kth coefﬁcient at time t = nT , y(n) the equalized signal, yˆ(n) the training signal, and k a scaling factor that controls the rate of coefﬁcient adjustment. The adaptive ZFS approach has the advantage that it is very easy to implement but has the drawback that it does not comprehend ISI that occurs outside the length of the equalizer. Another class of adaptive algorithms is the least mean square (LMS) set of algorithms. They attempt to minimize the mean square error of the equalizer output at all times and are commonly used in adaptive equalizers because they typically achieve better performance than ZFS algorithms and are relatively easy to implement. An example is the sign–sign least mean square approach [Kim et al., 2005]. Sign–sign LMS updates the equalizer coefﬁcients based on the sign of the error of the equalized signal and the sign of the input signal, as described by ck(n + 1) = ck(n) + µ sign[y(n) − yˆ(n)] sign[x(n − kT )] (12-30) where y(n) is the estimated signal (the output from the equalizer), yˆ(n) the reference signal, x(n − kT ) the input to the equalizer, and µ the scaling factor. Example 12-6 Adaptive Equalizer Operation We now compare the behavior of the adaptive ZFS and sign–sign LMS algorithms for a two-tap equalizer (cursor plus postcursor) as they apply to the same differential PCB channel that we DISCRETE LINEAR EQUALIZERS 539 have studied in previous examples in this chapter. The input samples and desired output values are x = [6, 36, 233, 99, 37, 18, 11, 7, 5, 4, 3, 2, 2, 2, 1] mV ytarget = [0, 0, 150, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] mV For the ZFS equalizer we choose = 12, and for the sign–sign LMS equalizer we set µ equal to −0.025. Applying equations (12-29) and (12-30) iteratively gives the results shown in Figure 12-37, which shows the trajectories of each algorithm in terms of the mean square error of the equalizer output. The mean MSE (V2) 0.020 0.018 Start 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 End 0.000 0.50 0.60 0.70 0.80 0.90 1.00 C0 (a) MSE (V2) 0.020 0.018 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.000 0.5 End 0.6 0.7 0.8 C0 (b) Start 0.9 1 Figure 12-37 Adaptive algorithm convergence for Example 12-6: (a) adaptive ZFS; (b) adaptive sign–sign LMS. 540 EQUALIZATION square error is calculated as n MSE = [y(i) − yˆ(i)]2 i=0 (12-31) Note that by applying the maximum voltage swing constraint of equation (12-22), which makes C0 and C1 interdependent for the two-tap equalizer, we make the optimization problem a function of a single variable. 12.4 DECISION FEEDBACK EQUALIZATION As mentioned in Section 12.3, the chief drawback of linear equalizers is that they are not effective at dealing with most types of noise. In this section we describe brieﬂy a technique, decision feedback equalization, which is effective at minimizing sources of noise. The decision feedback equalizer (DFE), as shown in Figure 12-38, is a nonlinear ﬁlter that uses previously detected symbols to subtract ISI from the input stream. As the ﬁgure shows, the DFE uses a linear feedforward ﬁlter in conjunction with a feedback ﬁlter. The input to the feedback ﬁlter consists of the sequence of decisions from previously detected symbols, which it uses to remove the portion of the intersymbol interference that was caused by those symbols. This implies that the DFE can only remove postcursor ISI. In addition, at high data rates the feedback loop may not be able to respond quickly enough to cancel the ISI of the ﬁrst postcursor sample, although the “loop unrolling” technique can alleviate the problem [Kasturia and Winters, 1991]. Another limitation of the DFE is that it assumes that past symbol decisions are correct. Incorrect decisions from the symbol detector corrupt the ﬁltering of the feedback loop. This is the reason for inclusion of the DLE on the front end, which helps to minimize the probability of error. Using a DFE, the equalized output is expressed as 0 nFBF yk = cDLEi xk−i + cFBFi bk−i i=−nDLE i=1 (12-32) xk DLE + yk Symbol Detector bk FBF Figure 12-38 Decision feedback equalization. DECISION FEEDBACK EQUALIZATION 541 where yk is the input to the symbol detector, cDLEi the coefﬁcients of the feedforward ﬁlter, cFBFi the coefﬁcients of the feedback ﬁlter, and xk the input to the DFE. The previously detected symbols, bk, are expressed as bk = 0 1 if yk < ythreshold if yk > ythreshold (12-33) where ythreshold is the decision threshold. Decision feedback equalization is often used to remove ISI that is caused by reﬂections due to impedance mismatch and may also use zero forcing and LMS adaptive algorithms [Proakis, 2001]. Note that because of the nonlinearity of the DFE response, it must be modeled in the time domain. Example 12-7 DFE Operation for a 10-Gb/s Signaling Link We now apply a DFE to the differential transmission lines that we have used throughout this chapter. To guarantee successful operation at 10 Gb/s, the system must meet minimum differential received specs of 80 mV and 80 ps for the height and width, respectively. Our system, shown in Figure 12-39, uses 5.057-mA current-mode transmitters with 5 M output resistance and 0.5 pF output capacitance. The lines are terminated to ground in 50 , and the receiver has an input capacitance of 0.5 pF. The model for the differential pair is generated using the methodology presented in Chapter 10 from the distributed transmission-line parameters listed below that were calculated at a reference frequency of 1 GHz: L= 3.299 0.407 0.407 3.299 nH/cm C= 1.098 0.085 0.085 1.098 pF/cm R= 509.1 60.64 60.63 509.1 m /cm G= 0.131 0.012 0.012 0.131 mS/cm As the ﬁgure shows, the system employs the linear equalizer at the transmitter. The transmit equalization does not include adaptive capability, and is set based on an expected channel loss characteristic. However, while the transmit equalizer coefﬁcients open the data eye, they are too strong for this particular system, iTx CDLE = 0.65 −0.35 iTx 5M Ω 0.5 pF 5M Ω 0.5 pF 50 Ω 0.5 pF 0.381 m 50 Ω 0.5 pF ++ +− FBF CFBF = 0.020 0.008 Figure 12-39 Differential signaling system for Example 12-7. 542 EQUALIZATION leading to overequalization. The DFE taps, on the other hand, are adaptive, based on the zero forcing criteria for the ﬁrst two postcursor samples. Figure 12-40 compares the pulse responses and the worst-case data eyes for the system without equalization, with the transmitter preemphasis only, and with DFE. As before, the worst-case data eyes are calculated using the peak distortion analysis method. The ﬁgure shows that without equalization, the data eye is closed completely. Transmit equalization opens the eye, providing a 96-mV minimum height and 76-ps minimum width, which meets the specs for eye height but not for width. The DFE allows the system to meet the speciﬁcation requirements by trading off eye height margin by approximately 6% in order to obtain an 11.5% improvement in eye width. This results in minimum eye height and width of 90 mV and 88 ps. As in the case of the linear equalizer, further improvements in both eye height and width are possible with the addition of taps to the DFE. For example, a 10-tap zero forcing DFE provides a minimum eye of 95 mV and 91 ps. However, since the system meets the speciﬁcations, the additional taps are not likely to justify the additional design complexity and power that they would add to the design. 12.5 SUMMARY In this chapter we described the application of equalization to signaling systems whose performance is limited by intersymbol interference caused by frequency-dependent interconnect losses. Passive and active continuous linear equalizers and discrete linear equalizers all reduce ISI by ﬂattening the frequency response of the interconnect system. By ﬂattening the frequency response, equalization increases the usable bandwidth of the interconnect, making it possible to increase data transfer rates. Decision feedback equalizers can further cancel ISI by using information from symbol decisions as input to a feedback ﬁlter, and adaptive equalization offers a way to improve performance further by optimizing ﬁlter performance based on the data rate and the characteristics of the interconnect system. Equalization may be incorporated into the transmitter or the receiver, or is split between both. SUMMARY 543 100 50 Rx with no equalization 0 Differential voltage [mV] −50 −100 −150 −200 −250 −300 2 3 4 5 6 7 8 9 10 Time (ns) 100 50 Rx with DLE 0 Differential voltage [mV] −50 −100 −150 −200 −250 −300 2 3 4 5 6 7 8 9 10 Time (ns) 100 50 Rx with DLE and DFE 0 Differential voltage [mV] −50 −100 −150 −200 −250 −300 2 3 4 5 6 7 8 9 10 Time (ns) (a) Figure 12-40 (a) Pulse responses and (b) worst-case data eyes for Example 12-7. (Continued) 544 EQUALIZATION Differential voltage (mV) Differential voltage (mV) Differential voltage (mV) 100 80 60 40 20 0 −20 −40 −60 −80 −100 −20 −10 0 100 80 60 40 20 0 −20 −40 −60 −80 −100 −20 −10 0 Nonlinearity caused by DFE 100 80 60 40 20 0 −20 −40 −60 −80 −100 −20 −10 0 Rx with no equalization Eye Height = -22.1 mV Eye Width = 0 ps 10 20 30 40 50 60 70 80 90 100 110 120 Time [ps] Rx with DLE Eye Height = 96.3 mV Eye Width = 76.5 ps 10 20 30 40 50 60 70 80 90 100 110 120 Time [ps] Rx with DLE and DFE Eye Height = 89.9 mV Eye Width = 88.2 ps 10 20 30 40 50 60 70 80 90 100 110 120 Time [ps] (b) Figure 12-40 (Continued ) REFERENCES 545 REFERENCES With data rates moving into the multi-Gb/s range, dozens of research papers are being published in conference proceedings and technical journals each year. We make no attempt to provide an exhaustive survey of the publish literature here. For tutorial treatments of equalization, we refer the reader to books by Couch [1987], Dally and Poulton [1997], and Liu and Ling [2004]. Books by Qureshi [1985] and Proakis [2001] give adaptive equalization a thorough and rigorous treatment. Finally, Lucky [2006] gives a historical account of the invention of the adaptive equalizer that provides insight into the nature of the discovery and development of a groundbreaking advance in communications technology. Coleman, Dave, Scott Gardiner, Mohammad Kolbehdari, and Stephen Peters, 2004, PCI Express Electrical Interconnect Design, Intel Press, Hillsboro, OR. Couch, Leon, 1987, Digital and Analog Communication Systems, 2nd ed., Macmillan, New York. Dally, William, and John Poulton, 1997, Transmitter equalization for 4-Gbps signaling, IEEE Micro, Jan.–Feb., pp. 48–56. Jaussi, James, et al., 2005, 8-Gb/s source-synchronous I/O links with adaptive receiver equalization, offset cancellation, and clock de-skew, IEEE Journal of Solid-State Circuits, vol. 40, no. 1, Jan., pp. 80–88. Johnson, Howard, and Martin Graham, 2003, High-Speed Signal Propagation: Advanced Black Magic, Prentice Hall, Upper Saddle River, NJ. Kasturia, Sanjay, and Jack Winters, 1991, Techniques for high-speed implementation of nonlinear cancellation, IEEE Journal on Selected Areas in Communications, vol. 9, no. 5, June, pp. 711–717. Kim, Jinwook, et al., 2005, A four-channel 3.125-Gb/s/ch CMOS serial-line transceiver with a mixed-mode adaptive equalizer, IEEE Journal of Solid-State Circuits, vol. 40, no. 2, Feb., pp. 462–471. Kudoh, Yoshiharu, Muneo Fukaishi, and Masayuki Mizuno, 2003, A 0.13 µm CMOS 5-Gb/s 10-m 28AWG cable transceiver with no-feedback-loop continuous-time post-equalizer, IEEE Journal of Solid-State Circuits, vol. 38, no. 5, May, pp. 741–746. Liu, Jin, and Xiaofen Ling, 2004, Equalization in high-speed communication systems, IEEE Circuits and Systems Magazine, vol. 4, no. 2, pp. 4–17. Lucky, Robert, 2006, The adaptive equalizer, IEEE Signal Processing Magazine, May, pp. 104–107. Nyquist, H., 1928, Certain topics in telegraph transmission theory, Proceedings of the AIEE , vol. 47, Apr., pp. 617–644. Reprinted as a classic paper in Proceedings of the IEEE , vol. 90, no. 2, Feb. 2002. PCI-SIG, 2005, PCI Express Base Speciﬁcation, revision 1.1, PCI, Wakeﬁeld, MA, Mar. 28. Proakis, John, 2001, Digital Communications, 4th ed., McGraw-Hill, New York. Qureshi, Shahid, 1985, Adaptive equalization, Proceedings of the IEEE , vol. 73, no. 9, Sept., pp. 1349–1387. Shannon, Claude, 1949, Communication in the presence of noise, Proceedings of the Institute of Radio Engineers, vol. 37, Jan., pp. 10–21. Sklar, Bernard, 2001, Digital Communications: Fundamentals and Applications, Prentice Hall, Upper Saddle River, NJ. 546 EQUALIZATION Sun, Ruifeng, Jaejin Park, Frank O’Mahony, and C. Patrick Yue, 2005, A low-power, 20-Gb/s continuous-time adaptive passive equalizer, IEEE Symposium on Circuits and Systems, May 23–26, pp. 920–923. PROBLEMS 12-1 Given the channel transfer function shown in Figure 12-41, calculate the maximum achievable data rate, assuming a signaling scheme that uses a 10-GHz bandwidth. Use the same signal and noise power spectra ﬁgures as in Example 12-2. Magnitude (dB) 0 −5 −10 −15 −20 −25 0 2 4 6 8 10 12 14 Frequency (GHz) Figure 12-41 Channel transfer function for Problem 12-1. 12-2 Given the sampled differential voltages at the receiver input and the coefﬁcients for the discrete linear equalizer shown below, calculate the values of the output samples: −246 x = −190 −31 −165 −218 −235 −242 −245 −247 mV and −0.05 c = −00..2700 −0.05 −248 12-3 Given the received pulse response below, use the ZFS algorithm to calculate the coefﬁcients for a ﬁve-tap discrete linear equalizer that has one PROBLEMS 547 precursor and three postcursor taps. Calculate the sample values of the resulting equalized response. −75 y = −74 −74 −68 18 −41 −62 −68 −71 −72 −72 −73 −73 −74 mV −74 12-4 Using the received pulse response from Problem 12-3 and the coefﬁcients for the DLE and DFE below, calculate the sample values of the equalized output response. CDLE = 0.65 −0.35 CDFE = 0.010 0.005 12-5 Derive the transfer function for a discrete linear equalizer, starting from equation (12-21). The time-shift property of the Fourier transform will prove useful: If y(t) has transform F [y(t)](f ) = Y (f ), then ∞ y(t − t0)e−2πjf t dt = e−2πjf t0 Y (f ) −∞ 12-6 Derive the transfer function of the alternate passive equalizer shown in Figure 12-20. 12-7 An alternative to the ZFS approach is the minimum mean square error (MMSE) equalizer. Finding the coefﬁcients using the MMSE requires ﬁrst calculating the autocorrelation matrix, Rxx = xTx, and the cross-correlation matrix, Rxz = xTy. The equalizer coefﬁcients are then calculated as c = R−xx1Rxz. Use the pulse response input samples from the Problem 12-3 to calculate the coefﬁcients for a linear equalizer with one precursor and three postcursor taps using the MMSE algorithm. Calculate the sample values of the resulting equalized response. 12-8 Given the pulse response in Problem 12-3, calculate the equalizer coefﬁcient progression obtained using the adaptive ZFS algorithm. 548 EQUALIZATION 12-9 12-10 Given the pulse response in Problem 12-3, calculate the equalizer coefﬁcient progression obtained using the adaptive sign–sign LMS algorithm. Determine the output from a DFE applied to a 1-V signaling system whose sampled voltages at the receiver are −0.12 x = −000...9105 V and −0.090 cDFE = 0.030 −0.001 0.02 13 MODELING AND BUDGETING OF TIMING JITTER AND NOISE 13.1 Eye diagram 550 13.2 Bit error rate 552 13.2.1 Worst-case analysis 552 13.2.2 Bit error rate analysis 555 13.3 Jitter sources and budgets 560 13.3.1 Jitter types and sources 561 13.3.2 System jitter budgets 568 13.4 Noise sources and budgets 572 13.4.1 Noise sources 572 13.4.2 Noise budgets 579 13.5 Peak distortion analysis methods 583 13.5.1 Superposition and the pulse response 583 13.5.2 Worst-case bit patterns and data eyes 585 13.5.3 Peak distortion analysis including crosstalk 594 13.5.4 Limitations 599 13.6 Summary 600 References 600 Problems 601 This chapter ties together the concepts presented in previous chapters with the intent of providing the reader with a method for managing timing noise and voltage noise successfully in order to create a successful multi-Gb/s design. We start by introducing the eye diagram as a tool for evaluating the performance of a signaling interface, introducing eye width and eye height as key metrics. The eye diagram is also a fundamental prerequisite to understanding the bit error rate (BER) of a link. Modern high-speed interfaces are designed to a speciﬁed BER Advanced Signal Integrity for High-Speed Digital Design, By Stephen H. Hall and Howard L. Heck Copyright 2009 John Wiley & Sons, Inc. 549 550 MODELING AND BUDGETING OF TIMING JITTER AND NOISE and data rate, which often determines the amount of voltage and timing noise allowed on a link. Accordingly, the chapter is centered on the concept of bit error rate, and we relate each of the analysis techniques described in the chapter to the BER to provide a useful design methodology. Following the introduction to bit error rate, we discuss sources of timing variation, known as jitter, that degrade the eye width, and describe the method creating a jitter budget. We follow that with a description of the sources of voltage noise that degrade the eye height, and describe a method for analyzing and creating a system noise budget. We then present the peak distortion analysis (PDA) technique for comprehending voltage and timing noise in determining the worst-case received eye from the pulse response of a high-speed interconnect channel. In providing a method for determining the maximum jitter and voltage noise, PDA accounts for intersymbol interference (ISI) caused by losses and reﬂections, and for degradation of the data eye due to crosstalk. 13.1 EYE DIAGRAM Most high-speed designs use the eye diagram to evaluate system performance. We show an example eye diagram for a 10-Gb/s 100-bit data sequence in Figure 13-1. An eye diagram is constructed by slicing the time-domain signal waveform into sections that are a small number of symbols in length, and overlaying them. The horizontal axis of the eye diagram represents time and is typically one or two symbols wide, while the vertical axis represents the amplitude of the signal. Figure 13-2 illustrates the eye diagram construction process for both a “perfect” eye and one that is distorted by losses and/or reﬂections. As the ﬁgures show, distortion of the signal causes the data eye to close. Conceptually, we want the eye to be as “open” as possible, as a larger eye opening implies that we have more margin to the voltage and timing requirements. From a Voltage noise Voltage (mV) Maximum signal swing 250 200 150 100 50 0 −50 −100 −150 −200 −250 70 Unit Interval Eye width Eye height 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 Timing variation (jitter) Time (ps) Figure 13-1 Example received eye diagram for a 10-Gb/s 100-bit data sequence. EYE DIAGRAM 551 Bit Pattern 000 Distortion Free Eye Bit Pattern 000 Eye With Distortion 001 010 011 Undistorted Width 001 010 011 Undistorted Width Undistorted Height Distorted Height Undistorted Height 100 101 100 101 Distorted Width 110 110 111 111 Figure 13-2 Eye diagram construction process and the impact of signal distortion. quantitative standpoint, the minimum height and width of the data at the receiver are key metrics for evaluating link performance. The eye must be wide enough to provide adequate time to satisfy the setup and hold requirement of the receiver, and have sufﬁcient height to ensure that the voltage levels meet vih and vil requirements in a system that may possess multiple sources of noise. This allows the receiver to resolve the input signals successfully into digital values. Using the eye diagram properly is critical, as we cannot assess performance simply by comparing the eye diagram to the receiver’s setup and hold window and vih and vil requirements. In addition to the signal distortion induced by the interconnect channel, we must also comprehend the variations in the voltage and timing induced by the clock used to capture data at the receiver and by the transmitter and transmit clock in order to ensure proper operation. We discuss the sources of the timing uncertainties (jitter) and voltage uncertainties (noise), and provide methods for creating jitter and noise budgets in Sections 13.3 and 13.4. A widely used method for evaluating whether or not the eye meets the system timing and noise requirements is to apply an eye mask , as shown in Figure 13-3. The mask represents a forbidden region that the actual eye must not cross, and it includes the receiver setup and hold window and voltage specs, and all jitter and noise terms. With it we can evaluate the performance of a given design by comparing it to worst-case eye obtained using the peak distortion analysis (PDA) method described in Section 13.5. PDA is a deterministic method for ﬁnding the minimum received eye height and width in a signaling system that has signiﬁcant sources of distortion, such as loss, reﬂections, and crosstalk. We can also use the eye to estimate the probability of receiving erroneous bits, which is known as the bit error rate (BER) or bit error ratio. The BER is expressed as the ratio of the number of erroneous received bits to the total number of transmitted bits over a sufﬁciently long interval: BER(ts , vs ) = lim N →∞ Nerr(ts , N vs ) (13-1) 552 MODELING AND BUDGETING OF TIMING JITTER AND NOISE voltage (mV) 50 mV 50 ps 250 200 150 100 50 0 2.5 V/ns Eye Mask −50 −100 −150 −200 −250 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 Time (ps) Figure 13-3 Example eye diagram and spec eye mask. where (ts, vs) represents the relative voltages and times at which the signal is sampled, Nerr is the number of erroneous bits received, and N is the number of bits transmitted over the same time interval. Equation (13-1) suggests that the BER depends on when we sample the data and at what voltage level (i.e., the sampling point position). By varying the sample point, we can generate a diagram that consists of a set of contours at a set of BER values, as shown in Figure 13-4. Ideally, the receiver will sample the data signal in the center of the eye. However, jitter and noise can cause nonideal data sampling, which can result in transmission errors. For example, if the receiver samples in the middle of the data eye, at the point corresponding to 50 mV and 50 ps in the ﬁgure, the error rate is less than 1 bit in 100 trillion (10−14). On the other hand, sampling at the edge of the eye, such as 10 mV and 10 ps, gives a BER in excess of 1 bit per million (10−6). At a transmission rate of 10 Gb/s, a BER of 10−14 will produce one error in approximately 2 hours and 45 minutes, while the 10−6 BER results in 10,000 errors per second! Communicating at high bit error ratios requires that error detection and correction capabilities be designed into the I/O circuits, which adds complexity and consumes power. Errors also degrade performance, as they must be detected and then the data re-transmitted. We discuss BER estimation in Section 13.2. As a ﬁnal thought, we note that the eye diagram applies to both single-ended and differential interfaces. In the case of a differential interface, the eye diagram should plot the differential voltage (vdiff = v − v) at the receiver. 13.2 BIT ERROR RATE 13.2.1 Worst-Case Analysis Historically, timing analysis for signaling interfaces in PC systems has employed a worst-case methodology. In this approach, all sources of timing uncertainty that Sampling voltage (mV) 0 10 20 30 40 50 60 70 80 90 BIT ERROR RATE BER: <10−14 10−14-10−12 10−12-10−10 10−10-10−8 10−8-10−6 >10−6 553 0 10 20 30 40 50 60 70 80 90 100 Sampling time (ps) Figure 13-4 Example BER contours. would degrade the eye opening are treated as bounded sources that do not exceed a speciﬁed amount. An example of an interface design based on the worst-case approach is the AGP 8X mode interface, which we describe here [Intel, 2002]. The AGP 8X mode is a 533-Mb/s source synchronous interface, which is shown in Figure 13-5a. The AGP source synchronous transmitter sends a clock signal along with a group of 16 data signals. The delay line in the transmitting chip offsets the clock signal by 90◦ from the data signal in order to center it in the data eye. Source synchronous designs achieve high performance by preserving the clock-to-data phase relationship. Doing so requires that the delays of the data signals be matched to those of the clock signal, which is accomplished by using identical transmitters and by matching the interconnect lengths. In theory, the maximum transmission rate of a source synchronous system is limited only by the setup-and-hold window of the receiver, which is described by Dally and Poulton [1998]. In practice, however, variation in the delay of the transmitters, interconnects, and receivers will reduce maximum achievable rate to a substantially lower value. For example, sources of variation in the transmitter delay include differences in the clock distribution path to the various circuits, process variation within the chip, and noise. Timing variation terms for source synchronous interfaces are typically speciﬁed as relative delays between the data and clock signals, so they include variations of each. 554 MODELING AND BUDGETING OF TIMING JITTER AND NOISE Tx Chip x16 Data D Q DQ I/O Delay Clock Line (a) UI 1875 ps Rx Chip Dt TxSu 410 ps Dt chanSu 442.5 ps Dt chanH 267.5 ps Dt TxH 460 ps ideal clock position Dt RxSu 85 ps Dt RxH 210 ps (b) Figure 13-5 AGP 2.0 8X mode source synchronous system timings: (a) system conﬁguration; (b) worst-case timing budget. The worst-case timing equations for the AGP source synchronous link are tmarSu = UI 2 − tmarH = UI 2 − tTxSu − tchanSu − tRxSu tTxH − tchanH − tRxH (13-2) (13-3) where tmarSu and tmarH = timing margins for the setup and hold conditions (ps) UI = unit interval, which is the width of a single bit (ps) tTxSu and tTxH = variation in transmitter delay relative to the clock path for the setup and hold cases (ps) tIntSu and tIntH = variation in interconnect delay relative to the clock path for the setup and hold cases (ps) tRxSu and tRxH = variation in receiver delay relative to the clock path for the setup and cases (ps) At 533 Mb/s, the length of a single bit, known as the unit interval (UI), is 1875 ps. The AGP 8X speciﬁcation at the receiver is 85 ps for the setup case and 210 ps for the hold case. The spec budgets 410 ps for the worst-case variation of the transmitter delay for the data edge that precedes the clock (the BIT ERROR RATE 555 setup case), and 460 ps of worst-case transmitter delay variation for the data edge that follows the clock (hold case). The spec allows for 442.5 ps and 267.5 ps of interconnect delay variation in the setup and hold cases, respectively. The timings are depicted graphically in Figure 13-5b. The total window for setup and hold time at the receiver is 295 ps. This deﬁnes a minimum UI that would give a maximum transfer rate of approximately 3.4 Gb/s if there were no variation in the transmitter and interconnect delays. However, the transmitter and interconnect delay variations add a total of 1560 ps to the unit interval, degrading the maximum transfer rate to the 533 Mb/s ﬁnal value. Although the worst case approach treats the sources of timing variation as though they are bounded, in reality this is a faulty assumption. Some sources, for example channel induced jitter due to intersymbol interference, are in fact bounded. Others, however, such as phase-locked loop (PLL) jitter induced by power supply noise, are random in nature. These sources are not bounded, but instead typically ﬁt a Gaussian distribution, in which the timing uncertainty is described by RJ(t ) = √ 1 e−t2/2σR2J 2π σRJ (13-4) where RJ(t) is the probability of having a timing jitter of t ps due to a random source and σRJ is the root-mean-square timing uncertainty (a.k.a. jitter) (ps). With a Gaussian distribution, even very large uncertainties have a nonzero (although extremely small) probability of occurrence, as Figure 13-6 shows. This has the consequence of rendering the notion of worst-case timings meaningless. Instead, we must interpret the timings in terms of the bit error rate, which is really just a probability that the timing uncertainties exceed the width of the unit interval. The signiﬁcance of this discussion is that it implies that prior system designs based on worst-case timings were not really designed to a worst case, since it has no meaning. Why, then, did these designs work? The truth is that worst-case timing-based systems were in reality designed to achieve immeasurably low BER. To illustrate, we examine the mean time between errors for the AGP 8X interface, assuming a BER of 10−18. The AGP data bus is 32 bits wide, running at 533 Mb/s. At the speciﬁed bit error rate, the mean time between errors would be approximately two and one-half years for the entire bus (assuming that errors on different data lines are uncorrelated). The 2.5-year estimate also assumes continuous operation and 100% bus utilization. If we assume that the system is a personal computer, and is only in use half of the time, the mean time between errors increases to ﬁve years. If we assume further that the average trafﬁc on the bus is unlikely to exceed 50%, the mean time between errors increases to 10 years, which far exceeds the expected lifetime of the computer. 13.2.2 Bit Error Rate Analysis As signaling speeds continue to increase, maintaining sufﬁcient margins to guarantee immeasurable bit error rates becomes prohibitive. As a result, high-speed 556 MODELING AND BUDGETING OF TIMING JITTER AND NOISE probability −50 −45 −40 −35 −30 −25 −20 −15 −10 −5 −0 5 10 15 20 25 30 35 40 45 50 0.40 0.35 s = 10 ps 0.30 0.25 0.20 0.15 34.1%34.1% 0.10 0.05 0.1% 2.1% 13.6% 0.00 2.1% 0.1% 13.6% Probability 1.E-02 1.E-06 1.E-10 1.E-14 1.E-18 1.E-22 1.E-26 1.E-30 1.E-34 1.E-38 1.E-42 1.E-46 1.E-50 Timing uncertainty (ps) (a) −150 −100 −50 0 50 100 150 Timing uncertainty (ps) (b) Figure 13-6 Gaussian distribution of timing uncertainties with σ = 10 ps: (a) linear plot; (b) semilog plot. links are moving to a BER-based budgeting approach. Our discussion proceeds with the premise that future high-speed signaling links will be designed to achieve a ﬁnite bit error rate. As such, we need to develop a method for calculating the BER from the timing distributions. We start by formulating an expression for the bit error rate as a function of the distribution of timing jitter: ∞ BER(t) = ρT J (t) dt −∞ (13-5) BIT ERROR RATE 557 where ρT is the transition density, which is the ratio of the number of logic transitions to the total number of bits transmitted (typically, ρT is equal to 0.5), and J (t) is the jitter distribution. The equation is the cumulative distribution function of the timing jitter, which takes into account the probability of actually making a transition on any given bit (since the jitter will be zero if the signal remains at a given level). To calculate the BER we need a model for the jitter distribution. We can use equation (13-4) for the random jitter sources, but it will not be accurate for systems that also include deterministic (bounded) jitter sources such as ISI, as Figure 13-7 demonstrates. In particular, the jitter histograms extracted from the zero crossing points of the eye show bimodal distributions, which suggests that the jitter is comprised of a combination of deterministic and Gaussian jitter sources. So we need a model for deterministic jitter that we can combine with the Gaussian random jitter model. Deterministic jitter sources can ﬁt a variety of distributions, examples of which are shown in Figure 13-11 and described in the next section. In creating system-level jitter budgets, we employ the dual Dirac model in equation (13-6) to express the probability density function (PDF) of the deterministic jitter, DJ(t): DJ(t) = δ (t − DJδδ/2) + δ (t + DJδδ/2) 2 2 (13-6) where DJδδ is the dual Dirac deterministic jitter (ps) and δ(t) is the Dirac delta function, δ(t) = 0 t = 0 1 t=0 As the equation shows, the dual Dirac model, which is widely used in industry, treats the deterministic jitter as though it is equally distributed at extreme values. The deterministic jitter in real systems does not ﬁt a dual delta distribution function. The usefulness of the model is that it allows us easily to combine the deterministic and random jitter distributions. The reason that the dual Dirac model works is that we really only need to be accurate in estimating the jitter at low BER, which is governed by the random jitter. As a result, we can use the dual Dirac distribution model to shift the “tails” of the jitter distributions (the RJ) to their proper locations. Conceptually, the dual Dirac model gives us a Gaussian approximation to the outer edges of the jitter distribution when displaced by DJδδ. For a more in-depth treatment of jitter distributions and the dual Dirac model, we refer the reader to a report by Stephens [2004] and a book by Li [2008]. We provide an example that demonstrates the application of the dual Dirac model in Section 13.3. The total jitter PDF, JT(t), is created by convolving the DJ and RJ models: JT(t) = RJ(t) ∗ DJ(t) = ∞ √ 1 e−t2/2σR2J −∞ 2π σRJ t − DJδδ + t + DJδδ dt 2 2 558 0.01 0.005 High Signal Amplitude Low Signal Amplitude 0 −200 −150 −100 −50 0 50 100 −100 −50 0 50 100 150 200 voltage (mV) Leading Edge Jitter 0.06 Trailing Edge Jitter 0.06 0.03 0.03 0 −20 −15 −10 −5 0 5 10 15 20 250 0 −20 −15 −10 −5 0 5 10 15 20 10 Gb/s Received Eye Diagram 200 150 100 50 0 −50 −100 −150 −200 −250 75 85 95 105 115 125 135 145 155 165 175 185 195 205 215 225 Time (ps) Figure 13-7 Example eye diagram with jitter and noise histograms. BIT ERROR RATE 559 We can use the convolution property of the delta function, ∞ −∞ f (t )δ (t − a) dt = f (a), to simplify the total jitter model: JT(t) = √ 1 e−(t−DJδδ/2)/2σR2J + e−(t+DJδδ/2)/2σR2J 2 2π σRJ (13-7) The resulting jitter distribution is bimodal and is composed of a pair of Gaussian distributions with equal variance (σR2J) whose mean values are separated by DJδδ. As an example, we return to the timings for the AGP 8X interface. To illustrate the distributions, we assume that DJδδ = 1000 ps and σRJ = 50 ps, which includes the jitter contributions of the transmitter, receiver, and interconnect. We allocated the jitter distributions so that the sum of the DJ plus 17.5 × σRJ is equal to the unit interval, which will result in a BER of approximately 10−19. The dual Dirac model jitter distributions for the leading edge of the data eye are shown in Figure 13-8. Using the probability density function for the total system jitter, we calculate the bit error rate as the area under the PDF. The expression for the bit error rates Random Jitter PDF 0.01 0.005 sRJ = 50 ps Random Jitter Probability −01000 −500 0 Deterministic Jitter (dual Dirac) PDF 0.6 Deterministic Jitter Probability 0.4 0.2 0 −1000 −500 0 Total Jitter PDF 0.004 500 1000 1500 Time (ps) 2000 2500 3000 DJ = 1000 ps 500 1000 1500 Time (ps) 2000 2500 3000 0.002 DJ = 1000 ps,sRJ = 50 ps Total Jitter Probability 0 −1000 −500 0 500 1000 1500 Time (ps) 2000 2500 3000 Figure 13-8 Hypothetical jitter PDF models for AGP 8X interface (UI = 1.875 ns). 560 MODELING AND BUDGETING OF TIMING JITTER AND NOISE of the leading and trailing edges of the eye are given by BERlead(t) = 0.5 erfc BERtrail(t) = 0.5 erfc t −√DJδδ/2 + erfc t +√DJδδ/2 2 σRJ 2 σRJ (13-8) UI −√t − DJδδ/2 + erfc UI −√t + DJδδ/2 2 σRJ 2 σRJ (13-9) where erfc(t) is the complementary error function, which is deﬁned as erfc(t) = √2 ∞ e−x2 dx πt Note that in developing the equation for the leading-edge BER, we calculate the cumulative density function (CDF) moving from right to left (toward decreasing time values), while we move in the positive time direction when calculating the trailing-edge CDF. This explains the difference in the sign of the time argument in the two equations. Using the equations, we can plot the bit error rate as a function of the horizontal position in the data eye. We call this plot a BER “bathtub plot” and show an example in Figure 13-9 using the AGP 8X timing distributions that we just developed. The plot has ﬂat regions near the edges of the eye in which the BER is dominated by deterministic jitter and sections with steep slope near the center of the eye that are dominated by random jitter. We can use the plot to assess the ability of a given design to meet the BER requirements by ﬁnding the rate at which the two curves intersect. For the example in the ﬁgure, the crossing point is 10−18, which represents the minimum BER that the design can meet. 13.3 JITTER SOURCES AND BUDGETS In this section we expand our discussion of jitter to provide a fuller understanding of the various sources of jitter and how they propagate through a signaling system so that we can budget for their effects. To start with, we deﬁne jitter as the deviation of a signal timing event from its ideal position. This is what causes the “smearing” of a data eye along the time axis. Any component that transmits, propagates, or receives a signal can contribute to jitter. As such, we ﬁrst categorize the types of jitter and their origins, followed by an analysis of how the different components combine to produce the jitter signature for the entire system. JITTER SOURCES AND BUDGETS 561 1000 Voltage (mV) 500 BER Deterministic Jitter Random Jitter Random Jitter Deterministic Jitter 0 0 125 250 375 500 625 750 875 1000 1125 1250 1375 1500 1625 1750 1875 Time (ps) BER Bathtub Plot 1 UI = 1875 ps, DImax = 1000 ps, sRJ = 50 ps 01 0.01 1-10−3 1-10−4 1-10−5 1-10−6 1-10−7 1-10−8 1-10−9 1-10−10 1-10−11 1-10−12 1-10−13 1-10−14 1-10−15 1-10−16 1-10−17 1-10−18 1-10−19 0 200 400 600 800 1000 1200 1400 1600 1800 Time (ns) Figure 13-9 Hypothetical BER bathtub plot for AGP 8X interface. 13.3.1 Jitter Types and Sources We have already introduced the notions of deterministic jitter (DJ) and random jitter (RJ) in Section 13.2. They constitute the two major categories, and all the types of jitter that relate to speciﬁc causes that we present in this section are subcategories of either DJ or RJ. The key characteristics of deterministic and random jitter sources are described in Figure 13-10 and are discussed in detail below. As we have shown, RJ is expressed as a Gaussian distribution which is described by a mean value, µ (typically equal to zero), and a standard deviation, σRMS. As a result, RJ is not bounded, although the probability of having 562 MODELING AND BUDGETING OF TIMING JITTER AND NOISE Total Jitter Random Jitter (RJ) Characteristics • Unbounded, Gaussian distributed • Key parameters: m= 0, sRMS • Sources: Device noise (shot, flicker, thermal) Deterministic Jitter (DJ) Sinusoidal Jitter (SJ) Characteristics • Bounded, peak-to-peak • Key parameters: Maximum pk-pk jitter • Sources: Losses, reflections, tr/tf mismatch, spread spectrum clocking, crosstalk Data Dependent Jitter (DDJ) Intersymbol Interference (ISI) Duty Cycle Distortion (DCD) Bounded Uncorrelated Jitter (Crosstalk) Figure 13-10 Summary of jitter types and their characteristics. random jitter values at several standard deviations from the mean is extremely low. Random jitter is caused by device effects such as thermal noise and shot noise (see Section 13.4.1). Random jitter shows up as the “tails” of the jitter distribution, and we use it to budget peak-to-peak jitter as a function of BER, as we describe in the material that follows. As we shall see, the amount of RJ (in terms of the number of sigma) that we must take into account increases as we decrease the target BER. Deterministic jitter traces to speciﬁc causes and is bounded, meaning that the probability of exceeding the peak-to-peak maximum value is equal to zero. We can categorize DJ in terms of sinusoidal or periodic jitter (PJ) and sources of data-dependent jitter (DDJ). Data-dependent jitter, as its name suggests, depends on the data pattern that is being transmitted. Prominent types of DDJ include duty-cycle distortion (DCD), intersymbol interference (ISI), and crosstalk. We describe each source in the paragraphs that follow. Periodic Jitter PJ repeats at a ﬁxed frequency and is caused by modulating effects, such as spread-spectrum clocking. For a system with multiple periodic sources, we model the total periodic jitter as P J (t) = Ai cos(ωit + θi) i (13-10) JITTER SOURCES AND BUDGETS 563 where Ai is the amplitude of source i, ωi the frequency of source i, and θi the phase of source i. The probability density function for an individual periodic jitter source is given by (see Figure 13-11a): PDFPJ(t ) = π √1 A2 0 − t 2 A > |t| A ≤ |t| (13-11) Probability 0.05 0.04 0.03 0.02 0.01 0 −40 0.6 0.5 0.4 0.3 0.2 0.1 0.0 −40 −20 0 20 Jitter (ps) (a) −20 0 20 Jitter (ps) (b) 40 40 Probability Probability 0.05 0.04 0.03 0.02 0.01 0 −40 −20 0 20 40 Jitter (ps) (c) 0.2 0.16 0.12 0.08 0.04 0 −40 −20 0 20 40 Jitter (ps) (d) Probability Figure 13-11 Jitter distributions for Example 13-1: (a) periodic jitter (20 ps amplitude); (b) duty cycle distortion jitter ω/αDCD = 10%); (c) ISI jitter with no equalization; (d) ISI jitter with equalization; (e) random jitter (σ RJ = 2.5 ps); (f) system deterministic jitter; (g) system total jitter. (Continued) 564 MODELING AND BUDGETING OF TIMING JITTER AND NOISE Probability Probability 0.1 0.08 0.06 0.04 0.02 0 −40 −20 0 20 40 Jitter (ps) (e) 0.02 0.016 0.012 0.008 0.004 0 −40 −20 0 20 40 Jitter (ps) (f) 0.015 0.012 0.009 0.006 0.003 0 −40 −20 0 20 40 Jitter (ps) (g) Figure 13-11 (Continued ) Probability where A is the amplitude of the periodic jitter and t is the time relative to the ideal edge position. Duty Cycle Distortion (DCD) Duty cycle distortion results from variation in the amount of time that the signal spends in the logic states, such as rise and fall time mismatch. The probability density function for jitter caused by duty cycle distortion is the sum of two delta functions: 1 PDFDCD(t) = 2 δ t − αDCD 2 +δ t + αDCD 2 (13-12) In equation (13-12), αDCD is the peak-to-peak duty cycle distortion and t is the time relative to the ideal edge position. The PDF for duty cycle distortion jitter is shown in Figure 13-11b. JITTER SOURCES AND BUDGETS 565 Intersymbol Interference As we have seen, ISI is caused by losses, dispersion, and reﬂections on the transmission lines that make up the interconnect channel. We show examples of probability density functions for channels with and without equalization in Figure 13-11c and d. In general, the ISI PDF for a nonequalized channel has multiple peaks, whereas that for an equalization channel looks like a truncated Gaussian distribution. Bounded Uncorrelated Jitter BUJ is deterministic jitter that is not aligned in time to the data stream. The most common source of BUJ is crosstalk. Recall that we have shown that crosstalk affects signal delay and amplitude through coupling of signals between neighboring transmission lines. As a result, crosstalk can be a signiﬁcant source of jitter in high-speed systems. Crosstalk-induced jitter is correlated to the data on neighboring signal, but is not correlated to the data pattern on the signal itself. Hence, it is uncorrelated. The key characteristics of deterministic and random jitter types are described in Figure 13-10. Example 13-1 Jitter Distributions for a 10-Gb/s Interface We now present an example running at 10 Gb/s to demonstrate multiple aspects of system jitter and bit error rate calculations. We start by showing how the various jitter sources combine to give the total system jitter. We then develop the dual Dirac model parameters and show how the model provides a suitable approximation of the system jitter at low error rates. Our system contains sources for periodic jitter, duty cycle distortion (DCD) jitter, ISI jitter, and random jitter. The amplitude of the periodic jitter is 20 ps, and it has the PDF in Figure 13-11a. The DCD jitter is caused by a 10% duty cycle variation (αDCD = 0.1), and the resulting PDF is shown in Figure 13-11b. The ISI jitter in our example is obtained from a signaling system that is based on the design that is presented in Section 14.2. This particular case is an 84differential pair that is driven by a 5-mA current source with 40- termination at each end. The system uses a single tap equalizer with a coefﬁcient value of −0.27. The resulting ISI PDFs for the nonequalized and equalized cases are shown in Figure 13-11c and d, respectively. For this analysis we focus on the equalized case, which has a distribution that lies between ±6 ps. The random jitter has an RMS value of 2.5 ps, resulting in the Gaussian jitter PDF shown in Figure 13-11e. We calculate the PDFs for the system-level DJ and total jitter via convolution using equations (13-13) and (13-14). The resulting distributions are shown in Figure 13-11f and g. PDFDJ(t) = PDFPJ(t) ∗ PDFDCD(t) ∗ PDFISI(t) PDFTJ(t) = PDFDJ(t) ∗ PDFRJ(t) (13-13) (13-14) Our next step is to use the distribution in Figure 13-11f to ﬁnd the maximum (peak-to-peak) deterministic jitter. By plotting the PDF on a log scale, as shown in Figure 13-12, we see that distribution shows a steep slope at the lower and upper jitter extremes. This indicates that we have reached the bounds of the 566 MODELING AND BUDGETING OF TIMING JITTER AND NOISE Probability 1 0.1 0.01 1⋅10−3 1⋅10−4 1⋅10−5 1⋅10−6 1⋅10−7 1⋅10−8 DJpp = 62 ps 1⋅10−9 1⋅10−10 1⋅10−11 1⋅10−12 1⋅10−13 1⋅10−14 1⋅10−15 1⋅10−16 1⋅10−−1750 −40 −30 −20 −10 0 10 20 30 40 50 Jitter (ps) Figure 13-12 Extraction of system-level deterministic jitter for Example 13-1. DJ distribution, which allows us to estimate that the peak-to-peak deterministic jitter is 62 ps. We can check the accuracy of the system DJ distribution by comparing the peak-to-peak DJ against the sum of the peak-to-peak jitter for the individual DJ components. The sum of PJ (±20 ps), ISI (±6 ps), and DCD (±5 ps) components also equals 62 ps, indicating that our system DJ distribution is accurate. To estimate the DJδδ terms we must look at the PDF for the total system jitter. Recall that the dual Dirac model treats the DJ as being distributed as a pair of delta functions. The model works by setting DJδδ such that the tails of the total jitter distribution model accurately reﬂect the tails of the actual distribution. We start by converting the probability values from the total jitter PDF to Q-scale values. The Q scale, denoted as QBER, speciﬁes the amount of eye closure due to random jitter that we must account for at a given BER and is described by QBER(BER) = √ 2 erf−1 1 − BER ρT (13-15) Values of QBER over a wide range of error rates are listed in Table 13-1. Figure 13-13 shows the QBER versus jitter plot. The utility of the plot is that it is approximately linear at low error rates, with a slope equal to ±σR−J1. We obtain the deterministic jitter term for the dual Dirac model by extrapolating the linear slope starting at very small BER. The DJδδ is simply the jitter value obtained via the linear extrapolation to a BER value of 1. From the ﬁgure we calculate that the values are −28.5 ps and 27.7 ps for the left- and right-hand side of the plots, so that DJδδ = 56.2 ps. JITTER SOURCES AND BUDGETS TABLE 13-1. BER 1 × 10−3 1 × 10−4 1 × 10−5 1 × 10−6 1 × 10−7 1 × 10−8 1 × 10−9 QBER as a Function of the Bit Error Rate QBER BER QBER BER 6.180 7.438 8.530 9.507 10.399 11.224 11.996 1 × 10−10 1 × 10−11 1 × 10−12 1 × 10−13 1 × 10−14 1 × 10−15 1 × 10−16 12.723 13.412 14.069 14.698 15.301 15.882 16.444 1 × 10−17 1 × 10−18 1 × 10−19 1 × 10−20 1 × 10−21 1 × 10−22 7.7 × 10−24 567 QBER 16.987 17.514 18.026 18.524 19.010 19.484 20.000 −DJpp/2 −DJδδ/2 0 1 Slope = 1/σRJ 2 −DJδδ/2 −DJpp/2 Slope = 1/σRJ 3 Q 4 DJδδ = 57.2 ps 5 DJpp = 62 ps 6 ½QBER 7 8 −50 −40 −30 −20 −10 0 10 20 30 40 50 ½TJ (BER = 10−12) Jitter (ps) ½TJ (BER = 10−12) Figure 13-13 Extraction of DJδ − δ for Example 13-1. As a ﬁnal step, we create a total jitter PDF for the dual Dirac model by applying our DJδδ and σRJ to equation (13-7) and then compare it against the actual total jitter PDF obtained via convolution of the individual PDFs at low error rates. If we are truly operating in the region of the plot that is dominated by RJ, we expect exact agreement. Figure 13-14 shows good, though not perfect agreement between the model and the actual system distribution. The error in the comparison is attributed to numerical errors in our frequency-domain convolution algorithm. We have thus illustrated the application of the dual Dirac model for estimating the jitter of high-speed signaling systems. Finally, we can use the model to estimate the total jitter as a function of bit error rate. For example, in Figure 13-13 we show that the total jitter is approximately 91 ps for a BER of 10−12( 1 2 QBER = 7.034), leaving us with 9 ps of margin in our design. 568 MODELING AND BUDGETING OF TIMING JITTER AND NOISE 1⋅10−8 1⋅10−9 1⋅10−10 1⋅10−11 1⋅10−12 1⋅10−13 Probability 1⋅10−14 1⋅10−15 1⋅10−16 1⋅10−17 1⋅10−18 1⋅10−19 1⋅10−20 1⋅10−2−1 50 −49.5 −49 −48.5 −48 −47.5 −47 −46.5 −46 −45.5 −45 −44.5 −44 −43.5 −43 −42.5 −42 Actual PDF Dual Dirac PDF Jitter (ps) Figure 13-14 Comparison of the dual Dirac and actual system TJ PDFs at the low BER tail for Example 13-1. 13.3.2 System Jitter Budgets At the system level we allocate the deterministic and random jitter for a given BER using equation (13-16): UI = DJδδ(sys) + QBERσRMS(sys) (13-16) Applying the dual Dirac model, we can approximate the convolution of the individual jitter distribution by adding the deterministic jitter components linearly, so that the total DJ for the system is DJδδ(sys) = DJδδ(i) i (13-17) The convolution of random jitter components results in a root-sum-of-squares relationship, leading to the following equation for the system RJ: σRMS(sys) = σR2MS(i) i (13-18) As mentioned, any component that transmits, propagates, or receives signals will add jitter to the system. Although speciﬁc implementations will vary, all signaling JITTER SOURCES AND BUDGETS 569 systems include data transmitters and receivers, interconnect channels, and clock sources. In designing a system, we must account for the DJ and RJ from each. Example 13-2 2.5-Gb/s PCI Express System Jitter Budget To illustrate, we use the system jitter budget for the ﬁrst-generation PCI Express (PCIe) interface. Figure 13-15a depicts the main components of a PCIe link. PCI Express systems do not distribute a global high-frequency I/O clock, but instead, use I/O clocks that are locally generated from a low-frequency reference clock, often using phase-locked loops (PLLs). In addition, a clock-and-data recovery (CDR) circuit at the receiver handles synchronization of the data signals [Martwick, 2005]. Figure 13-15b contains a block diagram that describes the system model for the generation and propagation of jitter in a PCI Express link. The 100-MHz reference clock generates jitter, TJrefclk, which enters the phase-locked loops for both the transmitter and receiver. The PLLs act as high-pass ﬁlters that allow high-frequency jitter to pass through them. In addition, each will also act as a source of additional jitter (TJTx,gen and TJRx,gen) caused by noise introduced both locally (e.g., thermal noise) and by the system (e.g., supply noise). The channel adds jitter (TJchan) due to ISI and crosstalk, but is assumed not to ﬁlter the incoming jitter noticeably. The comparator function reﬂects the fact that both PLLs track the jitter, so that the jitter propagation due to the transmitter and receiver is a function of the difference in their transfer functions. Intuitively, if both PLLs have the same impulse response, they will pass identical jitter, given the same input, TJrefclk. As a result, the receiver will tend to track the jitter passed by the transmitter, and the total jitter from the reference clock that gets passed by the PLLs is determined by the difference in their transfer functions. Finally, the CDR circuit