参数资料
型号: IR3622AMPBF
厂商: International Rectifier
文件页数: 22/33页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 32MLPQ
标准包装: 73
PWM 型: 电压模式
输出数: 2
频率 - 最大: 600kHz
占空比: 84%
电源电压: 4.5 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: 管件
配用: IRDC3622S-ND - BOARD EVALUATION W/IR3622MPBF
IRDC3622D-ND - BOARD EVAL W/IR3622MPBF DUAL OUT
IR3622AMPbF
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 π L o * C o
- - - -(20)
C 10
R 8
R 6
R 7
C 11
Z f
Using equations (18) and (20) to calculate C9.
C 9 =
1
2 π * 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
F P =
2 π * R 4 * 9
C POLE =
?
π * R 4 * F s
π * R 4 * F s ?
F Z 1
F Z 2
F P 2
F P 3
1
C * C POLE
C 9 + C POLE
The pole sets to one half of switching frequency
which results in the capacitor C POLE :
1 1
1
C 9
Frequency
Fig. 21: Compensation network with local
feedback and its asymptotic gain plot
As known, transconductance amplifier has high
impedance (current source) output, which needs
to be considered when loading the E/A output. If
the source/sink output current capability is
For F P <<
F s
2
exceeded the amplifier will not be able to swing
its output voltage over the necessary range.
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 (typeIII).
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 7 ? ?
? ?
F z 1 =
The typically used compensation network for
voltage-mode controller is shown in figure 21.
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 8 * C 10
1
? C 11 * C 12 ?
? C 11 + C 12 ?
1
2 π * R 7 * C 11
?
1
2 π * R 7 * C 12
g m * Z f >> 1 and g m * Z in >> 1
- - - -(21)
F z 2 =
1
2 π * C 10 * ( R 6 + R 8 )
?
1
2 π * C 10 * R 6
By replacing Z in and Z f according to figure 15, the
transformer function can be expressed as:
Cross over frequency is expressed as:
1 ( 1 + sR 7 C 11 ) * 1 + sC 10 ( R 6 + R 8 ) ]
? 1 + sR 7 ? ?
? ? ? * ( 1 + sR 8 C 10 )
V in 1
V osc 2 π * L o * C o
H ( s ) =
*
sR 6 ( C 11 + C 12 ) ? ? C 11 * C 12 ? ?
? ? C 11 + C 12 ? ?
F o = R 7 * C 10 *
*
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