参数资料
型号: IR3628MPBF
厂商: International Rectifier
文件页数: 15/22页
文件大小: 0K
描述: IC CTLR PWM BUCK SYNC 12-MLPD
标准包装: 122
应用: 控制器,DDR
输入电压: 4.5 V ~ 14 V
输出数: 1
输出电压: 0.6 V ~ 9.94 V
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 12-VFDFN 裸露焊盘
供应商设备封装: 12-MLPD
包装: 管件
IR3628MP b F
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 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, transconductance amplifier has high
impedance (current source) output, therefore,
consider should be taken when loading the error
amplifier output. It may exceed its source/sink
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 typically used compensation network for
voltage-mode controller is shown in figure 15.
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 ? ?
08/10/2007
F o = R 3 * C 7 *
*
15
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