参数资料
型号: IRDC3629
厂商: International Rectifier
文件页数: 16/23页
文件大小: 0K
描述: BOARD EVAL SYNC BUCK CONTROLLER
标准包装: 1
系列: *
IR3629/IR3629A MPbF
Based on the frequency of the zero generated by
the output capacitor and its ESR versus
crossover frequency, the compensation type can
be different. The table below shows the
compensation types and location of the
crossover frequency.
The following design rules will give a crossover
frequency approximately one-fifth 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
Type II(PI)
Type III(PID)
Method A
Type III(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.
Detailed calculation of compensation Type III
Method A:
F Z 2 = F LC = 8 kHz
F P 2 = F ESR = 80 . 38 kHz
Table1- The compensation type and location
Select : F Z1 = 0 . 75 * F Z 2 = 6kHz and F P3 = 0.5 * F s = 150 kHz
of F ESR versus F o
The details of these compensation types are
R 3 ≥
2
g m
; R 3 ≥ 2K Ω ; Select : R 3 = 2 6 . 7 K Ω
discussed in application note AN-1043 which can
be downloaded from IR’s website at www.irf.com.
Calculate C 4 , C 3 and C 7 :
; C 4 = 0 . 99 nF, Select : C 4 = 1 nF
; C 3 = 39 . 74 pF , Select : C 3 = 39 pF
2 π * F o * L o * C o * V osc * 1 . 28
For this design we have:
V in =12V
V o =1.8V
V osc =1.25V
V ref =0.6V
g m =1000umoh
L o =0.6uH
C o =2x330uF, ESR=6mOhm/each
C 4 =
C 3 =
C 7 =
1
2 π * F Z1 * R 3
1
2 π * F P 3 * R 3
R 3 * V in
; C 7 = 0 . 58 nF ,
These result to:
F LC =8kHz
F ESR =80.38kHz
Select : C 7 = 0 . 56 nF
Calculate R 10 , R 8 and R 9 :
F s/2 =150kHz
Select crossover frequency:
F o < F ESR and F o ≤ ( 1/5 ~ 1/10 ) * F s
R 10 =
R 8 =
1
2 π * C 7 * F P 2
1
2 π * C 7 * F Z 2
; R 10 = 3 . 54 K Ω , Select : R 10 = 3 . 57 K Ω
* R 10 ; R 8 = 31 . 97 K Ω , Select : R 8 = 31 . 60 K Ω
Fo=60kHz
Since: F LC <F o <F ESR <F s/2 , Type III Method A is
selected to place the pole and zeros.
11/29/2007
R 9 =
V ref
V o V ref
* R 8 ; R 9 = 15 . 8 K Ω , Select : R 9 = 15 . 8 K Ω
16
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