IR3086AMTRPBF International Rectifier, IR3086AMTRPBF Datasheet - Page 24
![IC CTLR XPHASE 28-MLPQ](/photos/7/48/74806/20_mlpq_sml.jpg)
IR3086AMTRPBF
Manufacturer Part Number
IR3086AMTRPBF
Description
IC CTLR XPHASE 28-MLPQ
Manufacturer
International Rectifier
Series
XPhase™r
Datasheets
1.IR3086AMTRPBF.pdf
(33 pages)
2.IR3086AMTRPBF.pdf
(33 pages)
3.IR3086AMTRPBF.pdf
(33 pages)
Specifications of IR3086AMTRPBF
Package / Case
20-MLPQ
Mounting Type
Surface Mount
Current - Supply
10mA
Voltage - Supply
8.4 V ~ 14 V
Operating Temperature
0°C ~ 125°C
Applications
Processor
Package
20-Lead MLPQ
Circuit
X-Phase Phase IC
Iout (a)
2.5A Gate Driver
Pbf
PbF Option Available
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
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
=
=
=
C
2 (
10
R
. Choose the crossover frequency of current share loop f
(
V
=
PWMRMP
π
Page 24 of 33
=
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 .
IR3086A
. 9
,
k
* 1
Ω
10
1/31
−
4
)
/05
105
* ]
. 0
011