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參數(shù)資料
| 型號(hào)： | IRU3138CS |
| 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)： | 10/18頁(yè) |
| 文件大小： | 281K |
| 代理商： | IRU3138CS |
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10
Rev. 1.0
01/29/04
IRU3138
www.irf.com
The IRU3138’s error amplifier is a differential-input
transconductance amplifier. The output is available for
DC gain control or AC phase compensation.
The E/A can be compensated with or without the use of
local feedback. When operated without local feedback,
the transconductance properties of the E/A become evi-
dent and can be used to cancel one of the output filter
poles. This will be accomplished with a series RC circuit
from Comp pin to ground as shown in Figure 9.
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:
F
ESR
= 1
2
π×
ESR
×
Co
V
OUT
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:
|H(s)| is the gain at zero cross frequency.
First select the desired zero-crossover frequency (Fo):
Use the following equation to calculate R
4
:
Fo
×
F
V
IN
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
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
Using equations (17) and (19) 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 suppress the switching noise. The additional
pole is given by:
1
C
9
×
C
POLE
C
9
+ C
POLE
The pole sets to one half of switching frequency which
results in the capacitor C
POLE:
1
1
C
9
For F
P
<< f
S
/2
R
4
=17.8K and F
S
=400KHz will result to C
POLE
=44pF.
Choose C
POLE
=47pF.
C
9
2.4nF; Choose C
9
=2.2nF
H(s) = g
m
× ×
---(15)
R
5
1 + sR
C
sC
9
R
4
= V
× × ×
g
OSC
1
m
Fo > F
ESR
and F
O
≤
(1/5 ~ 1/10)
×
f
S
For:
Lo = 1.1
μ
H
Co = 990
μ
F
F
Z
75%F
LC
F
Z
0.75
×
1
2
π
L
O
×
C
O
---(19)
F
Z
= 3.6KHz
R
4
= 17.8K
F
P
=
2
π×
R
4
×
F
Z
= 1
2
π×
R
4
×
C
9
|H(s=j
×
2
π×
F
O
)| = g
m
×
×
R
4
---(16)
R
5
R
6
×
R
5
Vp=V
REF
R
5
R
6
R
4
C
9
Ve
E/A
F
Z
H(s) dB
Frequency
Gain(dB)
Fb
Comp
Optional
F
LC
= 4.82KHz
R
5
= 1K
R
6
= 1K
g
m
= 600
μ
mho
For:
V
IN
= 5V
V
OSC
= 1.25V
Fo = 40KHz
F
ESR
= 12KHz
This results to R
4
=17.32K
Choose R
4
=17.8K
C
POLE
=
π×
R
4
×
f
S
-
1
π×
R
4
×
f
S
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