参数资料
型号: SC2446ITETRT
厂商: Semtech
文件页数: 23/38页
文件大小: 0K
描述: IC SYNC STEP DOWN CTRLR 28-TSSOP
标准包装: 1
应用: 控制器,DDR
输入电压: 4.7 V ~ 16 V
输出数: 2
输出电压: 可调至 0.5V
工作温度: -40°C ~ 85°C
安装类型: 表面贴装
封装/外壳: 28-SOIC(0.173",4.40mm 宽)裸露焊盘
供应商设备封装: 28-TSSOP 裸露焊盘
包装: 标准包装
产品目录页面: 1358 (CN2011-ZH PDF)
其它名称: SC2446ITEDKR
SC2446
POWER MANAGEMENT
Application Information (Cont.)
s z 2 =
,
s p 2 =
C 2 C 3
C 2 + C 3
The transconductance error amplifier (in the SC2446)
has a gain g m of 260 μ A/V. The target of the compensation
design is to select the compensation network consisting
of C 2 , C 3 and R 2 , along with the feedback resistors R o1 ,
R o2 and the current sensing gain, such that the converter
output voltage is regulated with satisfactory dynamic
performance.
With the output voltage V o known, the feedback gain h
and the feedback resistor values are determined using
the equations given in the “Output Voltage Setting” section
1
R 2 C 2
and
1
R 2
The loop transfer function is then
.
h =
.
k =
I o
with
0 . 5
V o
For the rated output current I o , the current sensing gain
k is first estimated as
.
2 . 1
T(s)=G vc (s)C(s).
To simplify design, we assume that C 3 <<C 2 , R oesr <<R o,
selects S p1 =S z2 and specifies the loop crossover
frequency f c . It is noted that the crossover frequency
determines the converter dynamic bandwidth. With these
assumptions, the controller parameters are determined
as following.
From Figure 17, the transfer function from the voltage
error amplifier output v c to the converter output v o is
C 2 =
g m hkR o
2 π f c
,
: = G vc ( s ) = kR o
R 2 =
C 3 =
V o ( s )
V c ( s )
where, the single dominant pole is
1 +
1 +
s
s z 1
s
s p 1
.
and
R o C o
C 2
R oesr C o
R 2
,
K ,
s p 1 =
,
1
( R o + R oesr ) C o
and the zero due to the output capacitor ESR is
with a constant K.
For example, if V o =2.5V, I o =15A, f s =300kHz, C o =1.68mF,
R oesr =4.67m ? , one can calculate that
s z 1 =
1
R oesr C o
.
R o =
V o
I o
= 167 m ? ,
The dominant pole moves as output load varies.
The controller transfer function (from the converter
h =
0 .5
V o
= 0 . 2 ,
output v o to the voltage error amplifier output v c ) is
C ( s ) =
g m h
s ( C 2 + C 3 )
1 +
1 +
s
s z 2
s
s p 2
,
and
k =
I o
2 . 1
= 7 . 14 .
where
? 2004 Semtech Corp.
23
www.semtech.com
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