参数资料
型号: IR3624MPBF
厂商: International Rectifier
文件页数: 16/21页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 10-MLPD
标准包装: 100
PWM 型: 电压模式
输出数: 1
频率 - 最大: 660kHz
占空比: 71%
电源电压: 4.5 V ~ 14 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: 10-VFDFN 裸露焊盘
包装: 管件
配用: IRPP3624-12A-ND - KIT REF DES 12A 1PH SYNC BUCK
IR3624MPBF
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-tenth 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
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 Θ
F Z 2 = F o *
1 + Sin Θ
F Z 2 = 16 kHz
Θ max =
π
3
Table1- The compensation type and location
F P 2 = F o *
1 + Sin Θ
1 ? Sin Θ
of F ESR versus F o
The details of these compensation types are
discussed in application note AN-1043 which can
be downloaded from IR Web-Site.
F P 2 = 224 kHz
Select : F Z1 = 0 . 5 * F Z 2 and F P3 = 0.5 * F s
For this design we have:
R 3 ≥
2
g m
; R 3 ≥ 2K ? ; Select : R 3 = 5K ?
V in =13.2V
V o =1.8V
V osc =1.25V
Calculate C 4 , C 3 and C 7 :
V ref =0.6V
g m =1000umoh
L o =0.82uH
C 4 =
1
2 π * F Z1 * R 3
; C 4 = 3.96nF, Select : C 4 = 3.9nF
C o =2x22uF, ESR=1.5mOhm
F s =600kHz
These result to:
C 3 =
1
2 π * F P 3 * R 3
; C 3 = 106 pF , Select : C 3 = 100 pF
F LC =26.5kHz
F ESR =2.4MHz
C 7 =
2 π * F o * L o * C o * V osc
R 3 * V in
; C 10 = 0 . 26 nF ,
F s/2 =300kHz
Select crossover frequency:
F o < F ESR and F o ≤ ( 1/5 ~ 1/10 ) * F s
Select : C 7 = 0 . 33 nF
Calculate R 10 , R 8 and R 9 :
Fo=60kHz
Since: F LC <F o <F s/2 <F ESR , typeIII method B is
R 10 =
1
2 π * C 7 * F P 2
; R 10 = 2 . 1 K ? , Select : R 10 = 2 K ?
selected to place the poles and zeros.
R 8 =
1
2 π * C 7 * F Z 2
? R 10 ; R 8 = 28 K ? , Select : R 8 = 28 K ?
R 9 =
V ref
V o ? V ref
* R 8 ; R 9 = 14 K ? , Select : R 9 = 14 K ?
16
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