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AME5268 参数 Datasheet PDF下载

AME5268图片预览
型号: AME5268
PDF下载: 下载PDF文件 查看货源
内容描述: 3A , 28V , 340KHz同步>整流降压转换器 [3A, 28V, 340KHz Synchronous Rectified Step-Down Converter]
分类和应用: 转换器
文件页数/大小: 20 页 / 354 K
品牌: AME [ ANALOG MICROELECTRONICS ]
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AME5268
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Detailed Description (Contd.)
Compensation Components
AME5268 has current mode control for easy compensa-
tion and fast transient response. The system stability and
transient response are controlled through the C
pin.
OMP
C
OMP
is the output of the internal transconductance error
amplifier. A series capacitor-resistor combination sets a
pole-zero combination to govern the characteristics of the
control system. The DC gain of the voltage feedback loop
is given by:
3A, 28V, 340KHz Synchronous
Rectified Step-Down Converter
In this case, a third pole set by the compensation ca-
pacitor (C6) and the compensation resistor (R3) is used
to compensate the effect of the ESR zero on the loop
gain. This pole is located at:
f
P
3
=
1
2
π
×
C
6
×
R
3
A
VDC
=
R
LOAD
×
G
CS
×
A
EA
×
V
FB
V
OUT
Where V
FB
is the feedback voltage (0.925V), A
VEA
is the
error amplifier voltage gain, G is the current sense
CS
transconductance and R
LOAD
is the load resistor value. The
system has two poles of importance. One is due to the
output capacitor and the load resistor, and the other is due
to the compensation capacitor (C3) and the output resistor
of the error amplifier. These poles are located at:
The goal of compensation design is to shape the con-
verter transfer function to get a desired loop gain. The
system crossover frequency where the feedback loop has
the unity gain is important. Lower crossover frequencies
result in slower line and load transient responses, while
higher crossover frequencies could cause system insta-
bility. A good standard is to set the crossover frequency
below one-tenth of the switching frequency. To optimize
the compensation components, the following procedure
can be used.
1. Choose the compensation resistor (R3) to set
the desired crossover frequency.
Determine R3 by the following equation:
f
P
1
=
f
P
2
=
G
EA
2
π
×
C
3
×
A
VEA
1
2
π
×
C
2
×
R
LOAD
C
R
3
=
2
π
×
C
2
×
f
C
V
OUT
2
π
×
C
2
×
0.1
×
fs V
OUT
×
<
×
G
EA
×
G
CS
V
FB
G
EA
×
G
CS
V
FB
Where G
EA
is the error amplifier transconductance.
The system has one zero of importance, due to the com-
pensation capacitor (C3) and the compensation resistor
e
i
(R3). This zero is located at:
Where f
C
is the desired crossover frequency which is
typically below one tenth of the switching frequency.
2. Choose the compensation capacitor (C3) to achieve
the desired phase margin. For applications with typical
inductor values, setting the compensation zero (fZ1) be-
low one-forth of the crossover frequency provides suffi-
cient phase margin.
Determine C3 by the following equation:
f
Z
1
=
1
2
π
×
C
3
×
R
3
The system may have another zero of importance, if the
output capacitor has a large capacitance and/or a high ESR
value. The zero, due to the ESR and capacitance of the
output capacitor, is located at:
C
3
>
f
ESR
=
1
2
π
×
C
2
×
R
ESR
4
2
π
×
R
3
×
f
C
Where R3 is the compensation resistor.
Rev.B.01
13