参数资料
型号: IR3623MTRPBF
厂商: International Rectifier
文件页数: 21/27页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 32MLPQ
标准包装: 1
PWM 型: 电压模式
输出数: 2
频率 - 最大: 1.2MHz
占空比: 85%
电源电压: 8.5 V ~ 14.5 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: -40°C ~ 125°C
封装/外壳: 32-VFQFN 裸露焊盘
包装: 标准包装
其它名称: IR3623MTRPBFDKR
IR3623MPbF
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
F z ? 75 % F LC
Z IN
V OUT
C 12
F z ? 0 . 75 *
1
2 ? L o * C o
- - - -(21)
C 10
R 8
R 6
R 7
C 11
Z f
Using equations (19) and (21) to calculate C9.
C 9 ?
1
2 ? * R 4 * F z
R 5
Fb
E/A
Comp
Ve
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:
Gain(dB)
H(s) dB
V REF
2 ? * R 4 * 9
F P ?
1
C * C POLE
C 9 ? C POLE
F Z 1
F Z 2
F P 2
F P 3
Frequency
Fig. 20: Compensation network with local
The pole sets to one half of switching frequency
which results in the capacitor C POLE :
feedback and its asymptotic gain plot
C POLE ?
1
? * R 4 * F s ?
1
C 9
?
1
? * R 4 * F s
As known, transconductance amplifier has high
impedance (current source) output, therefore,
consider should be taken when loading the E/A
output. It may exceed its source/sink output
For F P ??
F s
2
current capability, so that the amplifier will not be
able to swing its output voltage over the
necessary range.
For a general solution for unconditionally stability
for any type of output capacitors, in a wide range
of ESR values we should implement local
feedback with a compensation network (typeIII).
The compensation network has three poles and
two zeros and they are expressed as follows:
1 ? g m Z f
V e
F P 2 ?
F P 3 ?
2 ? * R 7 ? ?
? ?
F z 1 ?
The typically used compensation network for
voltage-mode controller is shown in figure 20.
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 ? * R 8 * C 10
1
? C 11 * C 12 ?
? C 11 ? C 12 ?
1
2 ? * R 7 * C 11
?
1
2 ? * R 7 * C 12
g m * Z f ?? 1 and g m * Z in ?? 1
- - - -(22)
F z 2 ?
1
2 ? * C 10 * ( R 6 ? R 8 )
?
1
2 ? * C 10 * R 6
By replacing Z in and Z f according to figure 15, the
1 ( 1 ? sR 7 C 11 ) * 1 ? sC 10 ? R 6 ? R 8 ? ?
? 1 ? sR 7 ? ?
? ? ? * ( 1 ? sR 8 C 10 )
F o ? R 7 * C 10 *
V in 1
V osc 2 ? * L o * C o
transformer function can be expressed as:
H ( s ) ? *
sR 6 ( C 11 ? C 12 ) ? ? C 11 * C 12 ? ?
? ? C 11 ? C 12 ? ?
Cross over frequency is expressed as:
*
www.irf.com
January 28, 2013
21
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