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参数资料
| 型号： | IR3623MPBF |
| 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： | 21/26页 |
| 文件大小： | 523K |
| 代理商： | IR3623MPBF |
IR3623MPbF
www.irf.com
Based on the frequency of the zero generated by
output capacitor and its ESR versus crossover
frequency, the compensation type can be
different.
The
table
compensation types and location of crossover
frequency.
below
shows
the
The following design rules will give a crossover
frequency
approximately
switching frequency. The higher the band width,
the potentially faster the load transient response.
The DC gain will be large enough to provide high
DC-regulation accuracy (typically -5dB to -12dB).
The phase margin should be greater than 45
o
for
overall stability.
Desired Phase Margin:
one-sixth
of
the
Ceramic
F
LC
<F
o
<F
s/2
<F
ESR
TypeIII(PID)
Method B
Tantalum,
ceramic
F
LC
<F
o
<F
ESR
<F
s/2
TypeIII(PID)
Method A
Electrolytic
, Tantalum
F
LC
<F
ESR
<F
o
<F
s/2
TypII(PI)
Output
capacitor
F
ESR
vs. F
o
Compensator
type
21
K
K
K
K
π
2
K
K
π
2
π
2
π
2
π
2
K
Θ
Θ
Θ
Θ
49
.
6
R
Select
,
45
.
6
R
;
R
V
V
V
R
06
.
8
R
:
Select
,
05
.
8
R
;
R
F
C
1
R
68
.
0
R
Select
,
63
.
0
R
F
C
1
R
R
and
R
,
R
Calculate
nF
68
.
0
C
Select
,
nF
67
.
0
C
;
V
R
V
C
L
F
C
pF
47
C
Select
,
pF
53
C
;
R
F
1
C
1.2nF
C
Select
1.19nF,
C
R
F
*
1
C
:
C
and
C
C
Calculate
10K
R
Select
;
72
.
0
R
g
2
R
F
*
0.5
F
and
F
5
.
0
F
:
Select
kHz
21
.
373
F
Sin
1
Sin
1
F
F
kHz
79
.
26
F
Sin
1
Sin
1
F
F
5
5
6
ref
o
ref
5
6
6
8
2
Z
10
6
8
8
2
P
10
8
5
6
8
10
10
in
7
osc
o
o
*
o
10
12
12
7
3
P
12
11
11
7
Z1
11
10
12
11
7
7
m
7
s
P3
2
Z
Z1
2
P
o
2
P
2
Z
o
2
Z
*
*
*
*
*
:
*
*
*
*
*
*
*
*
*
*
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
≥
≥
=
=
=
+
=
=
+
=
3
π
Θ
=
max
Table1- The compensation type and location
of F
ESR
versus F
o
The details of these compensation types are
discussed in application note AN-1043 which can
be downloaded from IR Web-Site.
For this design we have:
V
in
=13.2V
V
o
=1.8V
V
osc
=1.25V
V
ref
=0.8V
g
m
=2800umoh
L
o
=0.34uH, DCR=1.1mOhm
C
o
=15x22uF, ESR= 0.33mOhm
F
s
=600kHz
These result to:
F
LC
=15kHz
F
ESR
=1.46MHz
F
s/2
=300kHz
Select crossover frequency:
F
o
=100kHz
Since:
F
LC
<F
o
<F
s/2
<F
ESR
, typeIII method B is
selected to place the pole and zeros.
(
1/5
)
s
o
ESR
F
o
F
1/10
~
F
and
F
*
≤
<
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