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参数资料
| 型号： | IRDC3822A |
| 厂商： | International Rectifier |
| 文件页数： | 14/21页 |
| 文件大小： | 0K |
| 描述： | BOARD EVAL SYNC BUCK CONVERTER |
| 产品变化通告： | (EP) Parts Discontinuation 25/May/2012 |
| 标准包装： | 1 |
| 系列： | * |
PD-60345
IR3822AMPbF
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
Z IN
V OUT
C 3
F z = 0 . 75 *
F z = 75 % F LC
1
2 π L o * C o
- - - (17)
C 7
R 10
R 8
R 3
C 4
Z f
Use equations (15) and (16) to calculate C4.
One more capacitor is sometimes added in
parallel with C4 and R3. This introduces one
R 9
Fb
E/A
Comp
Ve
more pole which is mainly used to suppress the
switching noise.
Gain(dB)
V REF
F P =
2 π * R 3 * 4
The additional pole is given by:
1
C * C POLE
C 4 + C POLE
H(s) dB
F Z 1
F Z 2
F P 2
F P 3
Frequency
The pole sets to one half of switching frequency
which results in the capacitor C POLE :
Fig.15: Compensation network with local
=
C POLE
For F P <<
1
π * R 3 * F s ?
F s
2
1
C 4
?
1
π * R 3 * F s
feedback and its asymptotic gain plot
As known, a transconductance amplifier has high
impedance (current source) output, therefore,
consideration should be taken when loading the
error amplifier output. It may exceed its
For a general solution for unconditional stability
for any type of output capacitors, in a wide range
of ESR values we should implement local
feedback with a compensation network (type III).
The typically used compensation network for
voltage-mode controller is shown in figure 15.
In such a configuration, the transfer function is
source/sink output current capability, so that the
amplifier will not be able to swing its output
voltage over the necessary range.
The compensation network has three poles and
two zeros and they are expressed as follows:
F P 1 = 0
1 ? g m Z f
V e
2 π * R 3 ? ? 4
? C * C 3 ?
C 4 + C 3 ? ?
given by:
=
V o 1 + g m Z IN
The error amplifier gain is independent of the
transconductance under the following condition:
g m * Z f >> 1 and g m * Z in >> 1
- - - (18)
F P 2 =
F P 3 =
F z 1 =
1
2 π * R 10 * C 7
1
?
1
2 π * R 3 * C 4
?
?
1
2 π * R 3 * C 3
By replacing Z in and Z f according to figure 15, the
transformer function can be expressed as:
F z 2 =
1
2 π * C 7 * ( R 8 + R 10 )
?
1
2 π * C 7 * R 8
1 ( 1 + sR 3 C 4 ) * 1 + sC 7 ( R 8 + R 10 ) ]
sR 8 ( C 4 + C 3 ) ?
? C 4 * C 3 ? ?
? 1 + sR 3 ? ? ? * ( 1 + sR 10 C 7 )
F o = R 3 * C 7 *
V in 1
V osc 2 π * L o * C o
H ( s ) =
*
? ? C 4 + C 3 ? ?
Cross over frequency is expressed as:
*
11/04/08
14
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