参数资料
型号: IR3622AMPBF
厂商: International Rectifier
文件页数: 24/33页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 32MLPQ
标准包装: 73
PWM 型: 电压模式
输出数: 2
频率 - 最大: 600kHz
占空比: 84%
电源电压: 4.5 V ~ 14.5 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: 32-VFQFN 裸露焊盘
包装: 管件
配用: IRDC3622S-ND - BOARD EVALUATION W/IR3622MPBF
IRDC3622D-ND - BOARD EVAL W/IR3622MPBF DUAL OUT
IR3622AMPbF
Compensation for Current Loop
(slave channel)
The slave error amplifier is differential
transconductance amplifier, in 2-phase
configuration the main goal for the slave channel
feedback loop is to control the inductor current to
match the master channel inductor current as
Select a zero frequency for current loop (F o2 ) 1.2
times larger than zero cross frequency for
voltage loop (F o1 ).
F O 2 ? 1 . 25 % * F O 1
well provides highest bandwidth and adequate
phase margin for overall stability. The following
analysis is valid for both using external current
H ( F O 2 ) = g m * R s 1 * R 2 *
V in
2 π * F O 2 * L 2 * V osc
= 1
- - - - ( 24 )
sense resistors and using DCR of the inductor.
The transfer function of power stage is
expressed by:
From (24), R2 can be expressed as:
G ( s ) =
I L 2 ( s )
V e
=
V in
sL 2 * V osc
- - - - ( 22 )
R 2 =
1
g m * R s 1
*
2 π * F O 2 * L 2 * V osc
V in
- - - - ( 25 )
Where:
V in =Input voltage
L 2 =Output inductor
V osc =Oscillator Peak Voltage
As shown the G(s) is a function of inductor
current. The transfer function for compensation
network is given by equation (23), when using a
series RC circuit as shown in figure22.
I L2
L 2
V in =13.2V
V osc =1.25V
g m =3000umoh
L 2 =0.4uH
R s1 =DCR=0.930mOhm
F o2 =72kHz
This results to : R 2 =6.14K
Select R 2 =6.09K
The power stage of current loop has a dominant
pole (Fp) at frequency expressed by:
F P =
R S2
R S1
Fb2
Vp2
E/A2
Comp2
R 2
Ve
R eq
2 π * L 2
R eq = R ds ( on 1 ) * D + R ds ( on 2 ) * ( 1 ? D ) + R L
L 1
C 2
I L1
Fig. 22: The Compensation network for current loop
Where R ds(on1) is the on-resistance of control
FET, R ds(on2) is the on-resistance of synchronous
FET, R L is the DCR of output inductance and D
is the duty cycle
R eq =3.7mOhm
= ? ? g m * s 1 ? ? * ? ?
? ?
T ( s ) =
V e ( s ) ? R ? ? 1 + sC 2 R 2 ?
R s 2 ? R s 2 ? ? sC 2 ?
- - - - ( 23 )
Set the zero of compensator at 10 times the
dominant pole frequency F P , the compensator
capacitor, C2 can be expressed as:
The loop gain function is:
F z = 10 * F P
? ? * ? ?
H ( s ) = R s 2 * ? ? g m * s 1 ? ? * ? ?
?
C 2 =
www.irf.com
H ( s ) = [ G ( s ) * T ( s ) * R s 2 ]
? R ? ? 1 + sR 2 C 2 ? ?
? R s 2 ? ? sC 2 ? ?
V in ?
sL 2 * V osc ? ?
1
2 π * R 2 * F z
C 2 =1.8nF
All design should be tested for stability
to verify the calculated values.
24
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