参数资料
型号: IR3628MPBF
厂商: International Rectifier
文件页数: 16/22页
文件大小: 0K
描述: IC CTLR PWM BUCK SYNC 12-MLPD
标准包装: 122
应用: 控制器,DDR
输入电压: 4.5 V ~ 14 V
输出数: 1
输出电压: 0.6 V ~ 9.94 V
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 12-VFDFN 裸露焊盘
供应商设备封装: 12-MLPD
包装: 管件
IR3628MP b F
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
F P 2 = F o *
Table1- The compensation type and location
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.
1 + Sin Θ
1 ? Sin Θ
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 Ω ; Select : R 3 = 8.06K Ω
C 4 =
; C 4 = 2.46nF, Select : C 4 = 2.2nF
C 3 =
; C 3 = 65 . 8 pF , Select : C 3 = 12 pF
V in =12V
V o =0.9V
V osc =1.25V
V ref =0.6V
g m =1000umoh
L o =0.36uH
C o =6x22uF, ESR=2mOhm
Note: Use 16.5uF instead of 22uF for calculation,
this is due to derating of ceramic capacitor
F s =600kHz
Calculate C 4 , C 3 and C 7 :
1
2 π * F Z1 * R 3
1
2 π * F P 3 * R 3
These result to:
C 7 =
2 π * F o * L o * C o * V osc * 1 . 28
R 3 * V in
; C 7 = 0 . 22 nF ,
F LC =26.6kHz
F ESR =4.8MHz
F s/2 =300kHz
Select crossover frequency:
Select : C 7 = 0 . 22 nF
Calculate R 10 , R 8 and R 9 :
F o < F ESR and F o ≤ ( 1/5 ~ 1/10 ) * F s
Fo=60kHz
Since: F LC <F o <F s/2 <F ESR , typeIII method B is
selected to place the pole and zeros.
R 10 =
R 8 =
1
2 π * C 7 * F P 2
1
2 π * C 7 * F Z 2
; R 10 = 3 . 23 K Ω , Select : R 10 = 3 . 24 K Ω
? R 10 ; R 8 = 41 . 76 K Ω , Select : R 8 = 42 . 20 K Ω
08/10/2007
R 9 =
V ref
V o ? V ref
* R 8 ; R 9 = 84 . 40 K Ω , Select : R 9 = 84 . 50 K Ω
16
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