IR3629 International Rectifier Corp., IR3629 Datasheet - Page 16

• Manufacturer Part Number: IR3629
• Description: High Frequency Synchronous Pwm Buck Controller With Power Good Output
• Manufacturer: International Rectifier Corp.

Based on the frequency of the zero generated by
the output capacitor and its ESR versus
crossover frequency, the compensation type can
be different. The table below shows the
compensation
crossover frequency.
Table1- The compensation type and location
The details of these compensation types are
discussed in application note AN-1043 which can
be downloaded from IR’s website at www.irf.com.
For this design we have:
V
V
V
V
g
L
C
These result to:
F
F
F
Select crossover frequency:
Fo=60kHz
Since: F
selected to place the pole and zeros.
11/29/2007
Compensator
type
m
o
in
o
osc
ref
LC
ESR
s/2
o
Type III(PID)
Type III(PID)
=0.6uH
=1.8V
=1000umoh
=2x330uF, ESR=6mOhm/each
=12V
Type II(PI)
Method A
Method B
F
=0.6V
=8kHz
=150kHz
=1.25V
o
=80.38kHz
<
F
ESR
LC
<F
and
of F
o
<F
F
types
ESR
o
F
F
F
ESR
≤
LC
LC
LC
(
1/5
F
versus F
<F
<F
<F
<F
ESR
ESR
o
o
s/2
~
<F
<F
and
1/10
, Type III Method A is
vs. F
<F
ESR
s/2
o
<F
o
)
<F
<F
*
o
location
ESR
F
s
s/2
s/2
Output
capacitor
Electrolytic
, Tantalum
Tantalum,
Ceramic
ceramic
of
the
The following design rules will give a crossover
frequency
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
overall stability.
Detailed calculation of compensation Type III
Method A:
Calculate
C
C
C
Select
Calculate
F
F
Select
R
R
R
R
10
Z
P
8
9
4
3
7
3
2
2
=
=
=
=
=
≥
=
=
=
V
2
2
2
g
2
2
F
F
π
:
2
π
π
C :
o
π
IR3629/IR3629A MPbF
m
π
V
LC
ESR
F
*
*
*
*
ref
;
*
7
Z1
V
C
F
F
F
=
R
1
C
C
1
=
R
1
=
ref
Z1
P
o
1
7
=
8
3
7
4
3
. 0
10
80
*
*
,
kHz
*
*
. 0
≥
*
*
approximately
C
L
,
F
56
R
F
.
R
R
R
R
75
2K
o
Z
38
3
P
3
3
3
2
8
*
nF
8
and
2
;
;
*
;
*
*
kHz
C
Ω
and
;
R
C
C
V
F
R
o
R
;
9
4
in
3
Z
10
*
C
10
2
=
=
=
V
Select
R
;
7
=
15
osc
=
R
. 0
39
:
9
. 3
8
:
6kHz
99
8 .
*
.
=
74
. 1
54
K
nF,
R :
31
28
Ω
pF
K
.
Ω
,
3
97
;
and
,
Select
=
C
,
Select
K
7
Select
6 2
Select
one-fifth
Ω
=
F
7 .
,
. 0
P3
:
K
58
R :
Select
=
C
Ω
C :
R :
nF
0.5
4
9
3
=
=
10
=
,
1
*
15
:
nF
=
39
F
R
8 .
s
. 3
8
of
=
pF
K
57
=
150
Ω
31
K
o
Ω
.
kHz
60
the
for
K
16
Ω