参数资料
型号: SC4525ASETRT
厂商: Semtech
文件页数: 13/19页
文件大小: 0K
描述: IC REG BUCK 3A 8SOIC
标准包装: 1
类型: 降压(降压)
输出数: 1
输入电压: 8 V ~ 28 V
PWM 型: 电流模式
频率 - 开关: 300kHz ~ 1.3MHz
电流 - 输出: 3A
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: SC4525ASEDKR
SC4525A
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
? 1 1 V FB ?
V V O O ? ? ?
? G CA R S S 2 2 π π F F C C C C O O
? the SC4525A. The inner loop (current
buck converter with
Figure7.Blockdiagramofcontrolloops
20 in ?
CA
integrator pole (-90deg) and the dominant pole (-90deg).
The high frequency pole nulls the ESR zero and attenuates
high frequency noise.
1 1
1 1
1 1 . . 0 0 60 ? 15 dB
loop) consists of a current sensing resistor (R s =4. m W )
V FB ? and a current 20 ? ? log ? ? ? (CA) with gain ? ? (G CA =28). The outer ? ? 6 6
? ? 3 . 3 ? ? ? = = 15 . . 9 9 dB
?
A A C C = = ? ? 20 log
80 10 3 22
? ? loop (voltage loop) consists . . 1 1 of ? ? 10 error π π amplifier (EA), ? ? 10
? ? 28 ? ? 6 6
2 2 ? ? 80 ? ? 10 3 ? ? 22 a 10
3 . 3 ? ?
28
10
?
an
O V O ?
MP
CO
EN
1
10 1 . 0 ? 22 k
10
SA
15 . 9
TO
1 Since 3 the 1 = = current FB ? loop 3 3 ? is = = internally closed, the remaining
?
RG
20
R R 22 ? 10 ? 6 V 3 ? ? . 3 ? ? ? 15 . . 9 3 3 dB
22 k
? log π ? ? ? 80 task ? for the 0 0 loop ? ? 10
? 10 ? 7
? 7 AIN
2
?
. . 28 10
28 compensation is to design the voltage
?
Fc
OP
R S 2 π F C C O
? G CA compensator (C V , R ? , and C ).
CO
NV GA
ER
O
1 1 8
TER
C 5 = = 0 . 45 nF
45 nF
5 7 GA
C 5
10 3
2 2 π π ? ? 16 ? ? 10 3 1 ? ? 22 . . 1 1 ? ? 10 3
16 10
22
-30
? log ? ?
For a converter with switching frequency = 15 SW . , 9 dB
1 1 . 0 ?
F output
?
?
1 3 ? 22 ? 10 ? 6 3 . 3 ??
? 6 . 1 ? 10 =
C 8 =   π ? 80 ? 10 1 3
? 28 inductance L 2 , output capacitance C = and pF
loading R, the
12 pF
10 3
2 2 π π ? ? 600 ? ? 10 3 ? ? 22 . . 1 1 ? ? 10 3
C 8 O = 12
600 10 22
control (V C ) to output (V O ) transfer function in Figure 7 is
= 0 . 45 nF
= 22 . 3 k V
V o = G PWM ( 1 + s R ESR C O )
? 10 12 3 pF
Figure 8. Bode plots for voltage loop design
(8)
=
= ( 1 + s / ω PWM () 1 + s / ω ESR Q + O s 2 2 / ω 2 2 )
V V c c
( 1 + s / ω p p ) ( 1 + s / ω n n Q + s / ω n n )
1
16 ? 10 3 ? This . 1 transfer function has a finite DC gain
= 0 . 45 nF
R R
1 1
1 1
1
G PWM 3 ≈ = G 12 pF , ω p ≈ R C ,
ω Z Z = = R
, ,
G CA ? R S R ESR C O
600 / ω ? 10 ) 3 ? 22 . 1 ? 10
R C O O
CA
ESR
( ) Plot the converter gain, i.e. control to feedback transfer
amplifier
? ? 3 3
PWM modulator, and a LC filter. 30 Fz1 Fp1
15 . 9
20
0 Fp
LO
IN
IN
3
Fz Fsw/2
? 3
-60
1K 10K 100K 1M 10M
15 . 9 given by: FREQUENCY (Hz)
0 20
?
3
22 ? 10 3 Therefore, the procedure of the voltage loop design for
the SC4525A can be summarized as:
G PWM ω ω
2 2
n
an ESR zero F Z at CA A C
1 G PWM ( 1 + s R ESR 7 7 C = = O ) 10 1
R
R / C ω O p ) ( 1 + 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 1 1 R 7 7
R 1 1
C 8 p = = R C O 1
2 2 π π F F P P 1 1 R R 7 7
V O
R 4 = R 6 ? ? 1 ?
A C = ? 20 ? log ? ?
? FB ? ?
? 1 1 V ?
A C = ? 20 ? log ? ?
1 1  3
?
? 28 ? 6 . 1 ? 10
2 π ? 80 ? 10 ? 22 ? 10
A ? R S C 8 R ESR C O
10 20 20
R
, ω Z = g m ,
/ m
1 1
,
, ω ≈ 1 ω Z = ,
and double poles at half the switching frequency.
? 1 . 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
? 6
?
1 . 0
3 .
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