参数资料
型号: SC4524BSETRT
厂商: Semtech
文件页数: 13/18页
文件大小: 0K
描述: IC REG BUCK 2A 8SOIC
产品培训模块: Power Supplies 101
标准包装: 1
类型: 降压(降压)
输出数: 1
输入电压: 3 V ~ 18 V
PWM 型: 电流模式
频率 - 开关: 300kHz ~ 1.3MHz
电流 - 输出: 2A
同步整流器:
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 8-SOIC(0.154",3.90mm Width)裸露焊盘
包装: 标准包装
供应商设备封装: 8-SOIC-EP
产品目录页面: 1358 (CN2011-ZH PDF)
其它名称: SC4524BSEDKR
SC4524B
Applications Information (Cont.)
CONTROLLER AND SCHOTTKY DIODE
REF
+
CA
Rs
Io
Including the voltage divider (R 4 and R 6 ), the control to
feedback transfer function is found and plotted in Figure
8 as the converter gain.
FB
-
EA
Vc
Vramp
PWM
MODULATOR
SW
L1
Vo
Since the converter gain has only one dominant pole at
C5
R7
COMP
C8
Co
Resr
R4
R6
low frequency, a simple Type-2 compensation network
is sufficient for voltage loop compensation. As shown in
Figure 8, the voltage compensator has a low frequency
integrator pole, a zero at F Z  , and a high frequency pole
at F P  . The integrator is used to boost the gain at low
frequency. The zero is introduced to compensate the
excessive phase lag at the loop gain crossover due to the
The block = = diagram log ? ? ? Figure 7 ? ? shows the ? ? control loops of a
? ? 20 ? ? log G R
A A C C
?     V FB ?
V V O O ? ? ?
Figure7.Blockdiagramofcontrolloops
20 in ?
? ? G CA R π π F F C C C O O CA
buck converter with the SC4524B. The inner loop (current
integrator pole (-90deg) and the dominant pole (-90deg).
The high frequency pole nulls the ESR zero and attenuates
high frequency noise.
   
   
    . . 0 0 60 ?  5 dB
loop) consists of a current sensing resistor (R s =6. m W )
V FB ? and a current 20 ? ? log ? ? ? (CA) with gain ? ? (G CA =28). The outer ? ? 6 6
? ? 3 . 3 ? ? ? = =  5 . . 9 9 dB
?
A A C C = = ? ? 20 log
? ? loop (voltage loop) consists . .     of ? ?  0 error π π amplifier (EA), ? ?  0
? ? 28 ? ? 6 6
2 2 ? ? 80 ? ?  0 3 ? ? 22 a  0
3 . 3 ? ?
28
 0
?
an
O V O ?
MP
CO
EN
 
 0   . 0 ? 22 k
 0
SA
 5 . 9
TO
  Since 3 the   = = current FB ? loop 3 3 ? is = = internally closed, the remaining
?
RG
20
R R ? 22 ?  0 ? 6 V 3 ? ? . 3 ? ? ?  5 . . 9 3 3 dB
? 22 k AIN
? log π ? ? ? 80 task ? for the 0 0 loop ? ?  0 ?
?  0
2
?
. . 28  0
28 compensation is to design the voltage
7 0 Fp
Fc
OP
R S 2 π F C C O
? G CA compensator (C V , R ? , and C ).
CO
NV GA
ER
    8
TER
C 5 = = 0 . 45 nF
45 nF
5 7 GA
C 5
2 2 π π ? ?  6 ? ?  0   ? ? 22 . .     ? ?  0 3
 6  0
22
? log ? ?
For a converter with switching frequency =  5 SW . , 9 dB
    . 0 ?
F output
?
?
    3 ? 22 ?  0 ? 6 3 . 3 ??
? 6 .   ?  0 =
C 8 =   π ? 80 ?  0 3
? 28 inductance L 2 , output capacitance C = and pF
loading R, the
C 8 O =  2 pF
control (V C ) to output (V O ) transfer function in Figure 7 is
= 0 . 45 nF
= 22 . 3 k V
G PWM (   + s R ESR C O )
V o o =
?  0  2 3 pF
Figure 8. Bode plots for voltage loop design
(8)
=
= (   + s / ω PWM ()   + s / ω ESR Q + O s 2 2 / ω 2 2 )
  V c (   + s / ω p ) (   + s / ω n Q + s / ω n )
= 0 . 45 nF
3 This transfer function has a finite DC gain Therefore, the procedure of the voltage loop design for
6 ?  0 ? 22 .   ?  0
R R
   
   
 
G PWM 3 ≈ = G  2 pF , ω p ≈ R C ,
ω Z Z = = R
, ,
G CA ? R S R ESR C O
00 / ω ?  0 ) 3 ? 22 .   ?  0
R C O
ESR
( ) Plot the converter gain, i.e. control to feedback transfer
amplifier
? 3 80  0 3 22
PWM modulator, and a LC filter. 30 Fz1 Fp1
 5 . 9
20
LO
O IN
IN
3  0 3 -30
Fz Fsw/2
? 3
2 π ? 600 ?  0 3 ? 22 .   ?  0 3 -60
1K 10K 100K 1M 10M
 5 . 9 given by: FREQUENCY (Hz)
20
?
3
the SC4524B can be summarized as:
G PWM ω ω
2 2
n CA S O
an ESR zero F Z at CA A C
  G PWM (   + s R ESR 7 7 C = = O )  0  
R
R / C ω O p ) (   + s / ω n Q + s R 2 g ESR ω C n 2 O )
C C 5 5 = = π F R
a dominant 2 low-frequency pole F P at
2 π F Z Z     R 7 7
ω ≈
ω Z =
,
  ,
,
A ? R S C p = R C O   R ESR C O
V O
R 4 = R 6 ? ?   ?
A C = ? 20 ? log ? ?
? FB ? ?
?     V ?
A C = ? 20 ? log ? ?
?
? 28 ? 6 .   ?  0
2 π ? 80 ?  0 ? 22 ?  0
R    
 0 20 20
R
, ω Z = g m ,
/ m
   
2 2 π π F F P P     R R 7 7
and double poles at half the switching frequency.
?   . 0 V ?
function.
(2) Select the open loop crossover frequency, F C , between
 0% and 20% of the switching frequency. At F C , find the
required compensator gain, A C . In typical applications with
ceramic output capacitors, the ESR zero is neglected and
the required compensator gain at F C can be estimated by
? (9)
? G CA R S 2 π F C C O V O ?
     3
? 3 3
? 6
?
  . 0
3 .
相关PDF资料
PDF描述
SC4524CSETRT IC REG BUCK ADJ 2A 8SOIC
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