参数资料
型号: IR3623MTRPBF
厂商: International Rectifier
文件页数: 22/27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
Based on the frequency of the zero generated by
output capacitor and its ESR versus crossover
frequency, the compensation type can be
different. The table below shows the
compensation types and location of crossover
frequency.
The following design rules will give a crossover
frequency approximately one-sixth of the
switching frequency. The higher the band width,
the potentially faster the load transient response.
The DC gain will be large enough to provide high
DC-regulation accuracy (typically -5dB to -12dB).
The phase margin should be greater than 45 o for
F Z 2 ? F o *
Compensator
type
TypII(PI)
TypeIII(PID)
Method A
TypeIII(PID)
Method B
F ESR vs. F o
F LC <F ESR <F o <F s/2
F LC <F o <F ESR <F s/2
F LC <F o <F s/2 <F ESR
Output
capacitor
Electrolytic
, Tantalum
Tantalum,
ceramic
Ceramic
overall stability.
Desired Phase Margin:
1 ? Sin ?
1 ? Sin ?
F Z 2 ? 26 . 79 kHz
1 ? Sin ?
F P 2 ? F o *
1 ? Sin ?
? max ?
?
3
Table1- The compensation type and location
of F ESR versus F o
The details of these compensation types are
F P 2 ? 373 . 21 kHz
Select : F Z1 ? 0 . 5 * F Z 2 and F P3 ? 0.5 * F s
discussed in application note AN-1043 which can
be downloaded from IR Web-Site.
R 7 ?
2
g m
; R 7 ? 0 . 72 K ? ; Select : R 7 ? 10K ?
For this design we have:
Calculate C 11 , C 12 and C 10 :
V in =13.2V
V o =1.8V
V osc =1.25V
V ref =0.8V
g m =2800umoh
L o =0.34uH, DCR=1.1mOhm
C o =15x22uF, ESR= 0.33mOhm
C 11 ?
C 12 ?
1
2 ? * F Z1 * R 7
1
2 ? * F P 3 * R 7
; C 11 ? 1.19nF, Select : C 11 ? 1.2nF
; C 12 ? 53 pF , Select : C 12 ? 47 pF
F s =600kHz
These result to:
C 10 ?
2 ? * F o * L o * C o * V osc
R 7 * V in
; C 10 ? 0 . 67 nF ,
F LC =15kHz
F ESR =1.46MHz
F s/2 =300kHz
Select crossover frequency:
Select : C 10 ? 0 . 68 nF
Calculate R 8 , R 6 and R 5 :
F o ? F ESR and F o ? ? 1/5 ~ 1/10 ? * F s
R 8 ?
1
2 ? * C 10 * F P 2
; R 8 ? 0 . 63 K ? , Select : R 8 ? 0 . 68 K ?
F o =100kHz
Since: F LC <F o <F s/2 <F ESR , typeIII method B is
selected to place the pole and zeros.
R 6 ?
1
2 ? * C 10 * F Z 2
? R 8 ; R 6 ? 8 . 05 K ? , Select : R 6 ? 8 . 06 K ?
www.irf.com
R 5 ?
January 28, 2013
V ref
V o ? V ref
* R 6 ; R 5 ? 6 . 45 K ? , Select : R 5 ? 6 . 49 K ?
22
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