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参数资料
| 型号： | IRPP3624-12A |
| 厂商： | International Rectifier |
| 文件页数： | 15/21页 |
| 文件大小： | 0K |
| 描述： | KIT REF DES 12A 1PH SYNC BUCK |
| 产品变化通告： | (EP) Parts Discontinuation 25/May/2012 |
| 标准包装： | 1 |
| 系列： | POWIR+™ |
| 主要目的： | DC/DC，步降 |
| 输出及类型： | 1，非隔离 |
| 输出电压： | 1.8V |
| 电流 - 输出： | 12A |
| 输入电压： | 10.8 ~ 13.2 V |
| 稳压器拓扑结构： | 降压 |
| 频率 - 开关： | 600kHz |
| 板类型： | 完全填充 |
| 已供物品： | 板 |
| 已用 IC / 零件： | IR3624，IRF7823，IRF7832Z |
| 相关产品： | IR3624MPBF-ND - IC REG CTRLR BUCK PWM VM 10-MLPD IR3624MTRPBFCT-ND - IC REG CTRLR BUCK PWM VM 10-MLPD IR3624MTRPBFTR-ND - IC REG CTRLR BUCK PWM VM 10-MLPD IRF7832ZTR-ND - MOSFET N-CH 30V 21A 8-SOIC IRF7832Z-ND - MOSFET N-CH 30V 21A 8-SOIC |
IR3624MPBF
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
- - - (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.
The additional pole is given by:
Gain(dB)
H(s) dB
V REF
2 π * R 3 * 4
F P =
1
C * C POLE
C 4 + C POLE
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
=
For F P <<
C POLE
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, 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.
In such configuration, the transfer function is
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
=
2 π * R 3 ? ? 4
? C * C 3 ?
C 4 + C 3 ? ?
given by:
V e
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
- - - (17)
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:
*
15
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