参数资料
型号: APW7066RC-TRL
厂商: ANPEC ELECTRONICS CORP
元件分类: 稳压器
英文描述: Dual Synchronous Buck PWM Controllers and One Linear Controller
中文描述: SWITCHING CONTROLLER, PDSO24
封装: LEAD FREE, MO-153ADT, TSSOP-24
文件页数: 18/33页
文件大小: 338K
代理商: APW7066RC-TRL
Copyright
ANPEC Electronics Corp.
Rev. A.7 - Mar., 2008
APW7066
25
Application Information (Cont.)
The PWM modulator is shown in Figure. 11. The input
is the output of the error amplifier and the output is the
PHASE node. The transfer function of the PWM
modulator is given by:
GAINPWM =
OSC
IN
V
PWM Compensation (Cont.)
VOSC
PWM
Comparator
Driver
Output of
Error
Amplifier
VIN
PHASE
Figure 11. The PWM Modulator
The compensation circuit is shown in Figure 12. It
provide a close loop transfer function with the highest
zero crossover frequency and sufficient phase margin.
The transfer function of error amplifier is given by:
GAINAMP =
+
+
sC3
1
R3
//
R1
sC2
1
R2
//
sC1
1
=
OUT
COMP
V
()
×
+
×
×
+
×
+
×
×
+
×
+
C3
R3
1
s
C2
C1
R2
C2
C1
s
C3
R3
R1
1
s
C2
R2
1
s
C1
R3
R1
R3
R1
The poles and zeros of the transfer function are:
FP2
=
C3
R3
2
1
×
π
×
FZ1
=
C2
R2
2
1
×
π
×
FZ2
=
() C3
R3
R1
2
1
×
+
×
π
×
Figure 12. Compensation Network
VCOMP
C2
VOUT
R3
VREF
C1
FB
-
+
C3
R2
R1
The closed loop gain of the converter can be written
as:
GAINLC x GAINPWM x GAINAMP
Figure 13. shows the asymptotic plot of the closed
loop converter gain and the following guidelines will
help to design the compensation network. Using the
below guidelines should give a compensation similar
to the curve plotted. A stable closed loop has a -20dB/
decade slope and a phase margin greater than 45
degree.
1.Choose a value for R1, usually between 1K and
5K.
2.Select the desired zero crossover frequency FO:
(1/5 ~ 1/10) x FS >FO>FESR
Use the following equation to calculate R2:
3.Place the first zero FZ1 before the output LC filter
double pole frequency FLC.
FZ1 = 0.75 x FLC
Calculate the C2 by the equation:
4.Set the pole at the ESR zero frequency FESR:
FP1 = FESR
Calculate the C1 by the equation:
1
R
F
V
R2
LC
O
IN
OSC
×
=
0.75
F
R2
2
1
C2
LC
×
π
×
=
1
F
C2
R2
C2
C1
ESR
×
π
×
=
2
FP1
=
+
×
π
×
C2
C1
C2
C1
R2
2
1
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