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| 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： | 16/33页 |
| 文件大小： | 690K |
| 代理商： | IR3622AMTRPBF |
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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
below
shows
the
compensation types and location of crossover
frequency.
The following design rules will give a crossover
frequency
approximately
one-sixth
of
the
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 45o for
overall stability.
Ceramic
F
LC<Fo<Fs/2<FESR
TypeIII(PID)
Method B
Tantalum,
ceramic
F
LC<Fo<FESR<Fs/2
TypeIII(PID)
Method A
Electrolytic
, Tantalum
F
LC<FESR<Fo<Fs/2
TypII(PI)
Output
capacitor
F
ESR vs. Fo
Compensator
type
23
OK!
gm
1
0.78k
K
0.33
gm
1
gm
1
R
Check
K
34
6
R
:
Select
,
K
30
6
R
V
R
K
87
7
R
:
Select
K
84
7
R
F
C
2
1
R
K
1
R
:
Select
K
32
1
R
;
F
C
2
1
R
and
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
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
6
ref
o
ref
5
6
8
2
Z
10
6
8
2
P
10
8
5
6
8
10
in
7
osc
o
10
12
7
3
P
12
s
P3
11
7
Z1
11
LC
Z1
10
12
11
7
m
7
>
=
>
=
=
=
≥
π
π
:
.
;
*
.
,
.
;
*
,
.
*
:
,
.
,
.
;
*
,
;
*
.
*
.
Table1- The compensation type and location
of F
ESR versus Fo
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:
FLC<Fo<FESR<Fs/2, typeIII method A is
selected to place the pole and zeros.
(
)
s
o
ESR
o
F
1/10
~
1/5
F
and
F
*
≤
<
If this condition is not met, then iteration may be
required by selecting larger R
7.
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