参数资料
型号: IRDC3629
厂商: International Rectifier
文件页数: 15/23页
文件大小: 0K
描述: BOARD EVAL SYNC BUCK CONTROLLER
标准包装: 1
系列: *
IR3629/IR3629A MPbF
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
Z IN
V O UT
C 3
F z = 0 . 75 *
F z = 75 % F LC
1
2 π L o * C o
- - - (16)
C 7
R 10
R 8
R 3
C 4
Z f
Using equations (15) and (16) to calculate C9.
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 the switching
frequency which results in the capacitor C POLE :
Fig.15: Compensation network with local
feedback and its asymptotic gain plot
C POLE =
1
π * R 3 * F s ?
1
C 4
?
1
π * R 3 * F s
As known, the transconductance amplifier has a
high impedance (current source) output,
therefore, consideration should be taken when
loading the error amplifier output. It may exceed
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.
its 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:
V e 1 ? g m Z f
F P 2 =
F P 3 =
2 π * R 3 ? ? 4
? C * C 3 ?
C 4 + C 3 ? ?
F z 1 =
In such configuration, the transfer function is
given by:
=
V o 1 + g m Z IN
The error amplifier gain is independent of the
transconductance under the following condition:
F P 1 = 0
1
2 π * R 10 * C 7
1
?
1
2 π * R 3 * C 4
?
?
1
2 π * R 3 * C 3
g m * Z f >> 1 and g m * Z in >> 1
- - - (17)
F z 2 =
1
2 π * C 7 * ( R 8 + R 10 )
?
1
2 π * C 7 * R 8
Cross over frequency is expressed as:
V in 1
V osc 2 π * L o * C o
1 ( 1 + sR 3 C 4 ) * 1 + sC 7 ( R 8 + R 10 ) ]
? ? ? * ( 1 + sR 10 C 7 )
? 1 + sR 3 ? ?
By replacing Z in and Z f according to figure 15, the
transfer function can be expressed as:
H ( s ) =
*
sR 8 ( C 4 + C 3 ) ? ? C 4 * C 3 ? ?
? ? C 4 + C 3 ? ?
11/29/2007
F o = R 3 * C 7 *
*
15
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