参数资料
型号: IRPP3624-12A
厂商: International Rectifier
文件页数: 16/21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:
已用 IC / 零件: IR3624,IRF7823,IRF7832Z
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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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