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參數(shù)資料
| 型號： | IR3637STRPBF |
| 廠商： | International Rectifier |
| 英文描述： | 1% ACCURATE SYNCHRONOUS PWM CONTROLLER |
| 中文描述： | 1％準確的同步PWM控制器 |
| 文件頁數(shù)： | 9/21頁 |
| 文件大?。?/td> | 655K |
| 代理商： | IR3637STRPBF |
IR3637SPBF
9
Rev. 1.1
06/16/05
www.irf.com
Note that this method requires that the output capacitor
should have enough ESR to satisfy stability requirements.
In general the output capacitor’s ESR generates a zero
typically at 5kHz to 50kHz which is essential for an ac-
ceptable phase margin.
The ESR zero of the output capacitor expressed as fol-
lows:
F
ESR
= 1
2
π
×
ESR
×
Co
Figure 9 - Compensation network without local
feedback and its asymptotic gain plot.
The transfer function (Ve / V
OUT
) is given by:
(
)
R
6
+ R
5
The (s) indicates that the transfer function varies as a
function of frequency. This configuration introduces a gain
and zero, expressed by:
R
5
R
6
×
R
5
The gain is determined by the voltage divider and E/A's
transconductance gain.
First select the desired zero-crossover frequency (Fo):
Fo > F
ESR
and F
O
≤
(1/5 ~ 1/10)
×
f
S
Use the following equation to calculate R
4
:
Where:
V
IN
= Maximum Input Voltage
V
OSC
= Oscillator Ramp Voltage
Fo = Crossover Frequency
F
ESR
= Zero Frequency of the Output Capacitor
F
LC
= Resonant Frequency of the Output Filter
R
5
and R
6
= Resistor Dividers for Output Voltage
Programming
g
m
= Error Amplifier Transconductance
This results to R
4
=16.06K
. Choose R
4
=16K
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
Using equations (11) and (13) to calculate C
9
, we get:
One more capacitor is sometimes added in parallel with
C
9
and R
4
. This introduces one more pole which is mainly
used to supress the switching noise. The additional pole
is given by:
1
The pole sets to one half of switching frequency which
results in the capacitor C
POLE:
For:
V
IN
= 5.5V
V
OSC
= 1.25V
Fo = 40kHz
F
ESR
= 26.5kHz
F
LC
= 7.50kHz
R
5
= 1K
R
6
= 1.25K
g
m
= 600
μ
mho
C
9
= 1.77nF
Choose C
9
= 1.8nF
F
P
=
2
π
×
R
4
×
C
9
×
C
POLE
C
9
+ C
POLE
Ve
V
OUT
V
REF
R
5
R
6
R
4
C
9
E/A
F
Z
H(s) dB
Frequency
Gain(dB)
Fb
Comp
C
POLE
F
Z
75%F
LC
F
Z
0.75
×
---(13)
2
π
L
O
×
C
O
For:
Lo = 1.5
μ
H
Co = 300
μ
F
F
Z
= 5.6kHz
R
4
= 16K
H(s) = g
m
× ×
---(9)
R
5
1 + sR
4
C
9
sC
9
F
Z
= 1
2
π×
R
4
×
C
9
|H(s)| = g
m
×
×
R
4
---(10)
R
4
= V
---(12)
V
F
LC
2
OSC
Fo
×
F
ESR
R
5
+ R
6
R
5
1
g
m
×
×
×
C
POLE
=
π×
R
4
×
f
S
-
1
for F
P
<<f
S
2
1
C
9
1
π×
R
4
×
f
S
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