Design with Operational Amplifiers and Analog Integrated Circuits –3rd EditionDesign with Operational Amplifiers and Analog Integrated Circuits – 3rd EditionSergio Franco - McGraw-Hill, 2002 - ISBN 0-07-232084-2°First-Printing Errata (Updated May 20, 2007)Page 47, Problem 1.3, 1st line: change Aoc to ArPage 104, Fig. P2.54: swap the resistance labels R and )1(δ+RPage 113, Eq. (3.11a): change H1−H2 to H1/H2Page 114, Eq. (3.13c): change 1/H1dB to 1/HdBPage 137, Example 3.11, Solution (b): change 136.69° to −136.69° (twice), and 46.69°to 133.31° (twice)Page 145, denominator of the right-hand side of the equation following Eq. (3.78): replace + R4(1 + with + R4R7C2(1 +Page 162, Fig. 4.2b: change arrow as shown on the r ight:age 162, text preceding Eq. (4.2): change b2 to b1, and replace b3 with b2 = b3PPage 205, Problem 4.8, 2nd line from the end: change 11.080 Hz to 11.080 kHzPage 285, Eq. (6.38), last denominator term: change –(f/ft)4 to +(f/ft)4Page 295, 2nd line: change –10 dec/dec to –1 dec/decPage 351, last line: change (5 – 1) to (5 – 2) Page 352, right edge of Fig. 8 (b): chan 45° t .4 ge o –45°R1/R2)/2πR2CfπR2Cf ≅ 210 kHz;Page 3 Problem 8.48; Example 8.16,Page 3 a10(e text 5 lines below: change exp[(t – t0)Page 5 line: change VBE3(on)/(R3 + R4) to VBE3(on)/R4; 4th line: changePage 5 to tOFF/2 with to tON + tOFF/2; Eq. (11.48):Page 358, Fig. 8.9: swap + and − inside the op ampPage 362, Solution: change fx = 107/( to fx = 2×107/(Page 363, after Eq. (8.20a): change 1/2πR2Cf to (1 +Page 364, Solution (b): change 1/2πR2Cf ≅ 140 kHz to (1 + R1/R2)/2in the denominator of A(jf), change 140 to 21085, line after Eq. (8.36): change Problem 8.46 tosecond line: change 99 k to 99 kΩ86, 4th line: change a0 = a01( to a0 =Page 451, expression after Eq. (10.2), and in thto exp[–(t – t0)22, Solution, 3rd160 Ω to 210 Ω, and 540 Ω to 700 Ω41, 3rd line before Eq. (11.46): replacechange numerator from IO(1 – VI/VO) to IO/(1 – VI/VO)
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