參數(shù)資料
型號: IR3622AMPBF
廠商: 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: 高頻2相,單或雙輸出同步降壓控制器,輸出跟蹤和排序
文件頁數(shù): 23/33頁
文件大?。?/td> 690K
代理商: IR3622AMPBF
IR3622AMPbF
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.
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
23
OK!
gm
1
0.78k
K
0.33
gm
1
gm
1
R
R
R
Check
K
34
.
6
R
Select
,
K
30
.
6
R
;
R
V
V
V
R
K
87
.
7
R
:
Select
,
K
84
.
7
R
;
R
F
C
2
1
R
K
1
R
Select
,
K
32
.
1
R
F
C
2
1
R
R
and
R
,
R
Calculate
nF
5
.
1
C
Select
,
nF
03
.
1
C
;
V
R
V
C
L
F
2
C
pF
56
C
Select
,
pF
70
C
;
R
F
2
1
C
F
F
2.8nF
C
Select
nF,
90
.
2
C
R
F
*
2
1
C
F
75
.
0
F
:
C
and
C
C
Calculate
6.04K
R
Select
;
K
67
.
0
R
g
2
R
5
6
8
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
s
P3
11
11
7
Z1
11
LC
Z1
10
12
11
7
7
m
7
>
=
>
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
π
π
π
π
π
:
*
*
*
*
*
:
*
*
*
*
*
*
*
*
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
=3000umoh
L
o
=0.4uH, DCR=0.930mOhm
C
o
=4x220uF, ESR= 2.25mOhm
F
s
=375kHz
These result in:
F
LC
=12kHz
(Replace L to L/2 in formula#14 for current share configuration)
F
ESR
=80.38kHz
F
s/2
=185kHz
Select crossover frequency:
F
o
=60kHz
Since:
F
LC
<F
o
<F
ESR
<Fs/2
, typeIII method A is
selected to place the pole and zeros.
(
1/5
)
s
o
ESR
F
o
F
1/10
~
F
and
F
*
<
If this condition is not met, then iteration may be
required by selecting larger R
7
.
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