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參數(shù)資料
| 型號(hào)： | IRU3046CF |
| 廠商： | International Rectifier |
| 元件分類： | 基準(zhǔn)電壓源/電流源 |
| 英文描述： | DUAL SYNCHRONOUS PWM CONTROLLER WITH CURRENT SHARING CIRCUITRY AND LDO CONTROLLER |
| 中文描述： | 雙同步PWM控制器，帶有均流電路和LDO控制器 |
| 文件頁(yè)數(shù)： | 9/20頁(yè) |
| 文件大?。?/td> | 163K |
| 代理商： | IRU3046CF |
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IRU3046
9
Rev. 1.9
09/27/02
www.irf.com
The master error amplifier is a differential-input transcon-
ductance 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 Comp1 pin to ground as shown in Figure 6.
The ESR zero of the LC filter expressed as follows:
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:
The gain is determined by the voltage divider and E/A's
transconductance gain.
First select the desired zero-crossover frequency (Fo):
Use the following equation to calculate R
4
:
Where:
V
IN(MASTER)
= Maximum Input Voltage
V
OSC
= Oscillator Ramp Voltage
F
O1
= Crossover Frequency for the master E/A
F
ESR
= Zero Frequency of the Output Capacitor
F
LC(MASTER)
= Resonant Frequency of Output Filter
g
m = Error Amplifier Transconductor
R
5
and R
6
= Resistor Dividers for Output Voltage
Programming
For:
V
IN(MASTER)
= 5V
V
OSC
= 1.25V
F
O1
= 30KHz
F
ESR
= 25.26KHz
F
LC(MASTER)
= 3.57KHz
R
5
= 1K
R
6
= 200
g
m = 600
μ
mho
This results to: R
4
=29.7K
. Choose: R
4
=29.4K
To cancel one of the LC filter poles, place the zero be-
fore the LC filter resonant frequency pole:
Using equations (12) and (14) 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
F
O1
> F
ESR
and F
O1
≤
(1/5 ~ 1/10)
×
f
S
F
P
=
2
π
×
R
4
×
C
9
×
C
POLE
C
9
+ C
POLE
C
9
= 2003pF
Choose: C
9
= 2200pF
V
OUT
V
REF
R
5
R
6
R
4
C
9
Ve
E/A1
F
Z
H(s) dB
Frequency
Gain(dB)
Fb1
Comp1
F
ESR
= 1
2
π×
ESR
×
Co
|H(s)| =
g
m
× ×
R
4
---(11)
×
5
F
Z
= 1
2
π
×
R
4
×
C
9
F
Z
75%F
LC(MASTER)
F
Z
0.75
×
1
2
π
L
O
×
C
O
---(14)
For:
Lo = 2.2
μ
H
Co = 900
μ
F
Fz = 2.67KHz
R
4
= 24.9K
R
4
= V
---(13)
V
IN(MASTER)
F
LC2
×
×
OSC
F
O1
×
F
ESR
R
5
+ R
6
R
5
1
g
m
H(s) = g
m
×
---(10)
×
R
5
1 + sR
C
sC
9
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