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IP1201 датащи(PDF) 20 Page - International Rectifier

номер детали IP1201
подробное описание детали  Dual Output Full Function 2 Phase Synchronous Buck Power BlockIntegrated Power Semiconductors, PWM Control & Passives
PDF  29 Pages
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производитель  IRF [International Rectifier]
домашняя страница  http://www.irf.com
Logo IRF - International Rectifier

IP1201 датащи(HTML) 20 Page - International Rectifier

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20
iP1201
(21)
3. Place a zero at 75% of f
LC to cancel one of the LC
filter poles.
o
o
z
C
L
f
×
×
=
π
2
1
75
.
0
(13)
4. Calculate C
9 using equations (11) and (13)
Calculation of compensation components for out-
put1, based on the example above yields:
f
LC = 4.0kHz
f
z = 3.0kHz
f
0 = 20kHz (based on Fsw = 200kHz)
f
esr = 10kHz, per equation (7) using Resr = 11mΩ.
R
5 = 4.5K
C
9 = 12nF
The same steps can be used to determine the val-
ues of the compensation components for output2.
Sometimes, a pole f
p2 is added at half the switching
frequency to filter the switching noise. This is done
by adding a capacitor C
opt in Fig.20 from the output of
the error amplifier (CC pin of iP1201) to ground. This
pole is given by equation (14):
(14)
Select C
opt such that:
10
9
C
C
opt ≤
(15)
Type III
Type III compensation scheme allows the use of
any type of capacitors with esr frequency of any range.
This scheme suggests a double pole double zero
compensation and requires more components
around the error amplifier to achieve the desired
gain and phase margins. Fig. 21 represents the type
III compensation network for iP1201.
The transfer function of the type III compensator is
given by eqaution (16)
)
1
(
)
1
(
)
1
(
)
1
(
1
)
(
21
25
22
26
21
9
9
26
9
9
C
sR
C
sR
C
sR
C
sR
C
sR
s
H
+
×
+
+
×
+
×
≈
(16)
C10
R9
R7
E/A1
Vout1
FB1
VREF
CC1
iP1201
R26
C9
C22
R25
C21
Magnitude(dB)
Frequency
F
Z1
H(s) dB
F
Z2
F
P2
F
P3
Fig. 21: Typical Type III compensation and its gain
plot
The frequencies of the three poles and the two zeros
of the type III compensation scheme are represented
by the following equations:
f
p1= 0
(17)
21
5
2
2
2
1
C
R
f
p
×
×
=
π
(18)
opt
p
C
R
f
×
×
≈
5
2
2
1
π
(20)
(19)
22
6
2
3
2
1
C
R
f
p
×
×
≅
π
9
26
1
2
1
C
R
f
z
×
×
=
π
21
9
2
2
1
C
R
f
z
×
×
=
π



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