ir3088a International Rectifier Corp., ir3088a Datasheet - Page 24
ir3088a
Manufacturer Part Number
ir3088a
Description
Xphase Ic With Fault And Overtemp Detect
Manufacturer
International Rectifier Corp.
Datasheet
1.IR3088A.pdf
(33 pages)
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Part Number:
IR3088A
Manufacturer:
IR
Quantity:
20 000
Part Number:
ir3088aMPBF
Manufacturer:
IR
Quantity:
20 000
Company:
Part Number:
ir3088aMTR
Manufacturer:
ON
Quantity:
225
Part Number:
ir3088aMTR
Manufacturer:
IR
Quantity:
20 000
Part Number:
ir3088aMTRPBF
Manufacturer:
IR
Quantity:
20 000
Bootstrap Capacitor C
Choose C
Decoupling Capacitors for Phase IC and Power Stage
Choose C
VOLTAGE LOOP COMPENSATION
Type II compensation is used for the converter with AL-Polymer output capacitors. Choose the crossover frequency
fc=40kHz, which is 1/10 of the switching frequency per phase, and determine Rcp and C
Choose C
CURRENT SHARE LOOP COMPENSATION
The crossover frequency of the current share loop f
loop f
Choose C
C
C
R
F
CP
MI
CP
SCOMP
Page 24 of 33
=
=
=
C
2 (
10
R
. Choose the crossover frequency of current share loop f
(
V
=
PWMRMP
π
=
V
=
∗
I
∗
O
. 0
BST
VCC
CP1
SCOMP
31
−
. 0
R
*
f
65
L
V
C
65
4 .
CP
E
PWMRMP
=0.1uF
)
=47pF to reduce high frequency noise.
=0.1uF, C
nF
1
*
2
∗
*
*
+
R
∗
16
C
C
=33nF
2 (
PWMRMP
L
E
PWMRMP
E
2 .
π
∗
=
*
*
−
C
10
10
f
V
E
C
DAC
3
VCCL
BST
∗
*
∗
*
*
*
V
R
C
12
I
FB
)
(
f
*
SW
*
220
( *
*
=0.1uF
R
∗
I
105
V
C
O
V
*
I
)
V
RAMP
∗
*
2
−
10
G
PWMRMP
*
V
V
CS
34
−
O
2
DAC
9
=
0 .
∗
_
( *
/
ROOM
2
) 6
∗
2 (
. 1 (
)
π
. 0
10
π
∗
35
∗
47
=
. 1 (
(
3
∗
560
16
f
−
*
40
CI
*
33
10
20
R
2 .
*
∗
LE
−
∗
. 1
*
10
∗
−
(
10
105
12
3
10
10
*
05
3
1 [
−
6
)
−
3
6
−
2
*
* )
*
3
*
+
*
∗
* )
0
10
CI
. 9
10
220
2
8 .
1 [
(
* 1
220
π
6
should be at least one decade lower than that of the voltage
+
)
−
1
*
10
*
. 1
2
+
=
∗
10
f
π
35
CI
2 (
71
10
−
*
4
−
π
)
*
nF
12
)
−
4
( *
9
∗
*
C
*
*
/
CI
40
10
2
12
E
) 6
, Choose C
400
π
=4kHz
( *
*
3
∗
−
∗
10
V
*
(
. 1
4
*
560
O
560
10
*
3
35
*
10
I
3
)
∗
560
,
O
*
*
and calculate C
10
3
)]
10
0
*
*
−
8 .
*
. 1
CP
6
−
F
10
6
05
∗
MI
=
=68nF
*
10
−
. 0
6
10
*
)
*
10
011
∗
*
7
365
6
. 1 (
*
10
33
∗
−
3
0
−
SCOMP
)
CP
8 .
2
105
.
=
2
*
0 .
. 9
,
IR3088A
k
* 1
Ω
10
1/15
−
4
)
/05
105
* ]
. 0
011