参数资料
型号: IR3622AMTRPbF
厂商: International Rectifier
英文描述: HIGH FREQUENCY 2-PHASE, SINGLE OR DUAL OUTPUT SYNCHRONOUS STEP DOWN CONTROLLER WITH OUTPUT TRACKING AND SEQUENCING
中文描述: 高频2相,单或双输出同步降压控制器,输出跟踪和排序
文件页数: 15/33页
文件大小: 690K
代理商: IR3622AMTRPBF
IR3622AMPbF
www.irf.com
To cancel one of the LC filter poles, place the
zero before the LC filter resonant frequency pole:
Using equations (18) and (20) to calculate C9.
One more capacitor is sometimes added in
parallel with C
9 and R4. This introduces one more
pole which is mainly used to suppress the
switching noise.
The additional pole is given by:
The pole sets to one half of switching frequency
which results in the capacitor C
POLE:
For a general solution for unconditional stability
for any type of output capacitors in a wide range
of ESR values, we should implement local
feedback with a compensation network (typeIII).
The typically used compensation network for
voltage-mode controller is shown in figure 21.
In such configuration, the transfer function is
given by:
The error amplifier gain is independent of the
transconductance under the following condition:
By replacing Z
in and Zf according to figure 15, the
transformer function can be expressed as:
-(20)
-
C
L
2
1
75
0
F
75
F
o
z
LC
z
*
.
%
π
=
As known, transconductance amplifier has high
impedance (current source) output, which needs
to be considered when loading the E/A output. If
the source/sink output current capability is
exceeded the amplifier will not be able to swing
its output voltage over the necessary range.
The compensation network has three poles and
two zeros and they are expressed as follows:
Cross over frequency is expressed as:
C
R
2
1
F
POLE
9
POLE
9
4
P
+
=
*
π
2
F
For
F
R
*
1
C
1
F
R
1
C
s
P
s
4
9
s
4
POLE
<<
=
*
π
VOUT
VREF
R5
R6
R8
C10
C12
C11
R7
Ve
FZ1
FZ2
FP2
FP3
E/A
Zf
ZIN
Frequency
Gain(dB)
H(s) dB
Fb
Comp
Fig. 21: Compensation network with local
feedback and its asymptotic gain plot
IN
m
f
m
o
e
Z
g
1
Z
g
1
V
+
=
22
(
)
[]
)
(
*
)
(
*
)
(
)
(
10
8
12
11
12
11
7
8
6
10
11
7
12
11
6
C
sR
1
C
sR
1
R
sC
1
C
sR
1
C
sR
1
s
H
+
+
=
6
10
8
6
10
2
z
11
7
1
z
12
7
12
11
12
11
7
3
P
10
8
2
P
1
P
R
C
2
1
R
C
2
1
F
C
R
2
1
F
C
R
2
1
C
R
2
1
F
C
R
2
1
F
0
F
*
)
(
*
π
+
=
+
=
o
osc
in
10
7
o
C
L
2
1
V
C
R
F
*
π
=
-(21)
-
1
Z
*
g
and
1
Z
*
g
in
m
f
m
>>
F
R
2
1
C
z
4
9
*
π
=
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