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SC2446 Datasheet(PDF) 23 Page - Semtech Corporation

Part # SC2446
Description  Dual-Phase Single or Two Output Synchronous Step-Down Controllers
PDF  38 Pages
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Manufacturer  SEMTECH [Semtech Corporation]
Direct Link  http://www.semtech.com
Logo SEMTECH - Semtech Corporation

SC2446 Datasheet(HTML) 23 Page - Semtech Corporation

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23
 2004 Semtech Corp.
www.semtech.com
POWER MANAGEMENT
SC2446
Application Information (Cont.)
The transconductance error amplifier (in the SC2446)
has a gain g
m of 260µA/V. The target of the compensation
design is to select the compensation network consisting
of C
2, C3 and R2, along with the feedback resistors Ro1,
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
with
.
V
5
.
0
h
o
=
For the rated output current I
o, the current sensing gain
k is first estimated as
.
1
.
2
I
k
o
=
From Figure 17, the transfer function from the voltage
error amplifier output v
c to the converter output vo is
.
s
s
1
s
s
1
kR
)
s
(
G
:
)
s
(
V
)
s
(
V
1
p
1
z
o
vc
c
o
+
+
=
=
where, the single dominant pole is
,
C
)
R
R
(
1
s
o
oesr
o
1
p
+
=
and the zero due to the output capacitor ESR is
.
C
R
1
s
o
oesr
1
z =
The dominant pole moves as output load varies.
The controller transfer function (from the converter
output v
o to the voltage error amplifier output vc) is
,
s
s
1
s
s
1
)
C
C
(
s
h
g
)
s
(
C
2
p
2
z
3
2
m
+
+
+
=
where
,
C
R
1
s
2
2
2
z
=
and
.
C
C
C
C
R
1
s
3
2
3
2
2
2
p
+
=
The loop transfer function is then
T(s)=G
vc(s)C(s).
To simplify design, we assume that C
3<<C2, Roesr<<Ro,
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.
,
f
2
hkR
g
C
c
o
m
2
π
=
,
C
C
R
R
2
o
o
2 =
and
,
K
R
C
R
C
2
o
oesr
3 =
with a constant K.
For example, if V
o=2.5V, Io=15A, fs=300kHz, Co=1.68mF,
R
oesr=4.67mΩ, one can calculate that
,
m
167
I
V
R
o
o
o
Ω
=
=
,
2
.
0
V
5
.
0
h
o
=
=
and
.
14
.
7
1
.
2
I
k
o
=
=



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