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DC/DC DESIGN WITH BOOST TOPLOGY
MODEL:24V 1A DC/DC CONVERTER
WINK.FENG@163.COM
WINK FANG
2007-02-09
D1
Vin
C in
L1
Co
Driver
Q1
Vo set
GND
-------------------------------------------------------------------------------------------------------------------------------------------------------
1. variables explain
1.1 variables unit define
k
:=
10
3
−
6
−
9
−
12
−
5
u
:=
10
n
:=
10
p
:=
10
mil
:=
2.54
⋅
10
m
ounce
:=
0.035mm
-----------------------------------------------------------------------------------------------------------------------------------------------------
μ0
:=
0.4π
⋅
10
−
6
H
m
Ω
⋅
m
The magnetic permeablity of vacuum
The copper resistance coefficient
ρ
:=
1.69
×
10
−
8
-----------------------------------------------------------------------------------------------------------------------------------------------------
1.2 spec define
VINmin
:=
8V
VINnom
:=
12V
VINmax
:=
15V
minmum in put voltage
normal in put voltage
maxmum in put voltage
-----------------------------------------------------------------------------------------------------------------------------------------------------
Voset
:=
24V
Out put voltage
Voripple
:=
0.2V
Out put max ripple voltage
Vo
:=
Voset
+
0.7V
Compensate the out put diode VF
-----------------------------------------------------------------------------------------------------------------------------------------------------
Iomax
:=
1A
Ioocp
:=
1.2A
out put current max
output current ocp limit
-----------------------------------------------------------------------------------------------------------------------------------------------------
Target efficiency
ηset
:=
95%
-----------------------------------------------------------------------------------------------------------------------------------------------------
Note:usually Dmax not recommand <0.7,because the gan is not linear when D>0.7
1.3 design variables define
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1.4 Basic variables calculation
Po( Io)
:=
Voset Io
Po( Iomax)
=
24 W
Pin( Io )
:=
Duty max
0.7
0.6
Dnm( Vin) 0.5
0.4
0.3
8
10
Vin
12
14
Out put power max
In put power max
Vo
⋅
Io
ηset
2. Power component selection
2.1 Boost inductor
VL -voltage wave form
Voltage on inductor
VL( Vin
,
t)
:=
Vin if 0
≤
mod( t
,
Ts)
≤
Dnm( Vin)
⋅
Ts
−
( Vo
−
Vin) if Dnm( Vin)
⋅
Ts
<
mod( t
,
Ts)
≤
Ts
PWM( Vin
,
t)
:=
( 10V) if 0
≤
mod( t
,
Ts)
≤
Dnm( Vin)
⋅
Ts
0 if Dnm( Vin)
⋅
Ts
<
mod( t
,
Ts)
≤
Ts
24.7
12.35
VL( VINmax
,
t)
VL( VINmin
,
t)
−
12.35
−
24.7
t
PWM( VINmax
,
t)
PWM( VINmin
,
t)
24.7
12.35
−
12.35
−
24.7
t
Calculate inductance value
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1.25
×
10
L inductance (H)
1.2
×
10
Lmin( Vin)
1.15
×
10
1.1
×
10
1.05
×
10
−
4
−
4
−
4
−
4
−
4
8
10
Vin
12
14
L1
:=
120μH
ΔIL
( Vin)
:=
Vin
L1
⋅
Dnm( Vin)
⋅
Ts
Choosing a standard value to meet design
actual
Δ
I
0.52
0.5
ΔIL
( Vin) 0.48
0.46
0.44
CURRENT A
8
10
Vin
12
14
Calculate critical current value
ICR( Vin)
:=
ΔIL
( Vin)
2
⋅
( 1
−
Dnm( Vin) )
Critical Io current
ICR( VINmax)
=
0.149 A
0.5
CRITICAL CURRENT (A)
0.375
ICR( Vin) 0.25
0.125
8
10
Vin
12
14
IL wave form
Current through inductor
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Io
Vin
ΔIL
( Vin)
IL( Vin
,
Io
,
t)
:=
⎛
+
⋅
mod( t
,
Ts)
⎞
if 0
≤
mod( t
,
Ts)
≤
Dnm( Vin)
⋅
Ts
⎜
1
−
Dnm( Vin)
−
⎟
L1
2
⎝
⎠
Vo
−
Vin
ΔIL
( Vin)
Io
⎡
+
−
⋅
mod[ ( t
−
Dnm( Vin)
⋅
Ts)
,
Ts]
⎤
if Dnm( Vin)
⋅
Ts
<
mo
⎢
( 1
−
Dnm( Vin) )
⎥
L1
2
⎣
⎦
IL( VINmin
,
1A
,
t)
3
Iomax
⎛
⎞
⋅
t
⎜
⎟
⎝
1
−
Dnm( VINmin)
⎠
t 2
1A
⋅
t
t
1
0
t
3.5
3
ILrms( Vin
,
Iomax) 2.5
2
1.5
8
10
Vin
12
14
ILpeak( Vin
,
Io )
:=
Io
1
−
Dnm( Vin)
+
ΔIL
( Vin)
2
Peak current through inductor
3.5
3
ILpeak( Vin
,
Iomax) 2.5
2
1.5
8
10
Vin
12
14
Choosing a ring-metal powder core to meet the design requirement
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OD1
:=
18mm
ID1
:=
10mm
Ht
:=
5mm
Ae1
:=
2
×
10
−
5 2
m
Ve1
:=
3.6
×
10
AL1
:=
180n
⋅
H
−
7 3
ID
OD
m
Bs1
:=
1.5T
⋅
80%
------------------------------------------------------------------
NL_1
:=
L1
AL1
turns to meet the inductance value
NL_1
=
25.82
Saftey turns to avoid the core saturation
IF the NL_1>NLmin1, the design is acceptable
NL
:=
round( NL_1 )
NL
=
26
-----------------------------------------------------------------------------------------------------------------------------------------------------
Wire_Length
:=
NL
⋅
( OD1
−
ID1
+
2Ht)
⋅
1.1
Wire_Length
=
0.515 m
ΦL1
:=
0.5mm
DCR
:=
Wire_Length
⋅
ρ
ΦL1
⎞
π
⋅
⎛
⎜
2
⎟
⎝
⎠
DCR
=
0.044
Ω
2
Choosing a whole number
The copper wire length
Choosing a wire diameter
the wire dc resistance
conduction loss result from copper wire
0.12
0.115
ΔB
( Vin) 0.11
0.105
0.1
8
10
Vin
12
14
PVe
:=
300
kW
m
3
Pcore_max
:=
PVe
⋅
Ve1
Pcore_max
=
0.108 W
PL1( Vin
,
Io)
:=
PL1cond( Vin
,
Io)
+
Pcore_max
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