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参数资料
| 型号： | IP1206TRPBF |
| 厂商： | International Rectifier |
| 文件页数： | 21/30页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK SYNC ADJ 30A/15A LGA |
| 产品变化通告： | (EP) Parts Discontinuation 25/May/2012 |
| 标准包装： | 750 |
| 系列： | iPOWIR™ |
| 类型： | 降压（降压） |
| 输出类型： | 可调式 |
| 输出数： | 1 或 2 |
| 输出电压： | 0.8 V ~ 5.5 V |
| 输入电压： | 7.5 V ~ 14.5 V |
| PWM 型： | 电压模式 |
| 频率 - 开关： | 200kHz ~ 600kHz |
| 电流 - 输出： | 30A，15A |
| 同步整流器： | 是 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 36-BFLGA |
| 包装： | 带卷 (TR) |
| 供应商设备封装： | 36-LGA（9.25x15.5） |
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i P1206PbF
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
- - - -(18)
C 10
R 8
R 6
R 7
C 11
Z f
Using equations (16) and (18) 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
2 π * 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
C POLE =
?
π * R 4 * F s
π * R 4 s ?
* F
For F P <<
The pole sets to one half of switching frequency
which results in the capacitor C POLE :
1 1
1
C 9
F s
2
For a general solution for unconditionally stability
and 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 Fig. 22.
Fig. 22: Compensation network with local
feedback and its asymptotic gain plot
As known, transconductance amplifiers have
high impedance (current source) outputs,
therefore, care should be taken when loading the
E/A output. It may exceed 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:
F P 1 = 0
1 ? g m f
V e
Z
2 π * R 7 ? ? 11
? C * C 12 ?
C 11 + C 12 ? ?
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:
g m * Z f >> 1 and g m * Z in >> 1 - - - -(19)
By replacing Z in and Z f according to figure 15, the
F P 2 =
F P 3 =
F z 1 =
F z 2 =
1
2 π * R 8 * C 10
1
?
?
1
2 π * R 7 * C 11
1
2 π * C 10 * ( R 6 + R 8 )
?
?
1
2 π * R 7 * C 12
1
2 π * C 10 * R 6
transformer function can be expressed as:
1 ( 1 + sR 7 C 11 ) * 1 + sC 10 ( R 6 + R 8 ) ]
sR 6 11 + C 12 ) ?
?
? 1 + sR 7 ? ?
? ? ? * ( 1 + sR 8 10 )
( C ? C 11 * C 12 ?
? C 11 + C 12 ? ?
F o = R 7 * C 10 *
V in 1
V osc 2 π * L o * C o
H ( s ) =
*
?
C
Cross over frequency is expressed as:
*
www.irf.com
2/26/2008
21
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