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參數(shù)資料
| 型號(hào)： | IRU3038CS |
| 廠商： | International Rectifier |
| 英文描述： | SYNCHRONOUS PWM CONTROLLER FOR TERMINATION POWER SUPPLY APPLICATIONS |
| 中文描述： | 同步PWM控制器終止電源應(yīng)用 |
| 文件頁(yè)數(shù)： | 8/18頁(yè) |
| 文件大?。?/td> | 139K |
| 代理商： | IRU3038CS |
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8
Rev. 2.0
09/12/02
IRU3038
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
acceptable phase margin.
The ESR zero of the output capacitor expressed as fol-
lows:
1
2
π×
ESR
×
Co
Figure 6 - 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
1
2
π
×
R
4
×
C
9
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
=26.52K
. Choose R
4
=26.1K
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:
C
9
1800pF
One more capacitor is sometimes added in parallel with
C
9
and R
4
. This introduces one more pole which is mainly
used to suppress 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
= 5V
V
OSC
= 1.25V
Fo = 30KHz
F
ESR
= 26.5KHz
F
LC
= 5KHz
R
5
= 1K
R
6
= 1K
g
m = 600
μ
mho
F
P
=
2
π
×
R
4
×
C
9
×
C
POLE
C
9
+ C
POLE
|H(s)| =
g
m
×
×
R
4
---(10)
F
Z
= ---(11)
V
OUT
Vp=V
REF
R
5
R
6
R
4
C
9
Ve
E/A
F
Z
H(s) dB
Frequency
Gain(dB)
Fb
Comp
F
ESR
= ---(8)
H(s) = g
m
× ×
R
5
---(9)
4
C
9
sC
9
R
4
= V
× × ×
---(12)
V
IN
Fo
×
F
OSC
R
+ R
1
m
C
POLE
= 1
π ×
R
4
×
f
S
-
1
C
9
1
π ×
R
4
×
f
S
for F
P
<<f
S
2
F
Z
75%F
LC
F
Z
0.75
×
1
2
π
L
O
×
C
O
For:
Lo = 10
μ
H
Co = 300
μ
F
F
Z
= 3.8KHz
R
4
= 26.1K
---(13)
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