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| 型号： | IP1206TRPBF |
| 厂商： | International Rectifier |
| 文件页数： | 22/30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
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 Θ
1 + Sin Θ
F Z 2 = 10 . 72 kHz
1 + Sin Θ
F P 2 = F o *
1 ? Sin Θ
Θ max =
π
3
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
F P 2 = 81 . 2 kHz
Select : F Z1 = 0 . 5 * F Z 2 and F P3 = 0.5 * F s
be downloaded from the IR Web-Site.
For this design we have:
R 7 ≥
2
g m
; R 7 ≥ 0 . 72 K Ω ; Select : R 7 = 6 . 81 K Ω
V in =12V
Calculate C 11 , C 12 and C 10 :
V o =1.2V
V osc =1.25V
V ref =0.8V
g m =2800umoh
L o =1uH, DCR=2.4mOhm
C o =15x22uF, ESR= 0.33mOhm
F s =300kHz
C 11 =
C 12 =
1
2 π * F Z1 * R 7
1
2 π * F P 3 * R 7
; C 11 = 4 . 36 nF, Select : C 11 = 5.6nF
; C 12 = 155 pF , Select : C 12 = 100 pF
These result to:
F LC =10.73kHz
C 10 =
2 π * F o * L o * C o * V osc
R 7 * V in
; C 10 = 0 . 85 nF ,
(Replace L to L/2 in formula#14 for current share configuration)
F ESR =1.46MHz
F s/2 =150kHz
Select : C 10 = 1 nF
Calculate R 8 , R 6 and R 5 :
Select crossover frequency:
F o < F ESR and F o ≤ ( 1/5 ~ 1/10 ) * F s
F o =40kHz
R 8 =
R 6 =
1
2 π * C 10 * F P 2
1
2 π * C 10 * F Z 2
; R 8 = 1 . 96 K Ω , Select : R 8 = 2 K Ω
? R 8 ; R 6 = 13 . 9 K Ω , Select : R 6 = 16 . 9 K Ω
Since: F LC <F o <F s/2 <F ESR , typeIII method B is
selected to place the pole and zeros.
www.irf.com
2/26/2008
R 5 =
V ref
V o ? V ref
* R 6 ; R 5 = 39 . 2 K Ω , Select : R 5 = 39 . 2 K Ω
22
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