IR3081AMPBF International Rectifier, IR3081AMPBF Datasheet - Page 33
![IC CTRLR XPHASE VR10.0 28MLPQ](/photos/7/48/74875/ir3081ampbf_sml.jpg)
IR3081AMPBF
Manufacturer Part Number
IR3081AMPBF
Description
IC CTRLR XPHASE VR10.0 28MLPQ
Manufacturer
International Rectifier
Series
XPhase™r
Datasheet
1.IR3081AMPBF.pdf
(39 pages)
Specifications of IR3081AMPBF
Applications
Processor
Current - Supply
11mA
Voltage - Supply
9.5 V ~ 14 V
Operating Temperature
0°C ~ 100°C
Mounting Type
Surface Mount
Package / Case
28-MLPQ
Package
28-Lead MLPQ
Circuit
X-Phase Controller IC
Switch Freq (khz)
150kHz to 1.0MHz
Pbf
PbF Option Available
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
R
Bootstrap Capacitor C
Choose C
Decoupling Capacitors for Phase IC and Power Stage
Choose C
VOLTAGE LOOP COMPENSATION
Type III compensation is used for the converter with only ceramic output capacitors. The crossover frequency and
phase margin of the voltage loop can be estimated as follows.
Choose
Choose the desired crossover frequency fc (=140kHz) around fc1 estimated above, and calculate
Choose C
CURRENT SHARE LOOP COMPENSATION
The crossover frequency of the current share loop f
loop f
θ
C
C
C
C
R
F
f
C
PHASE41
CP
MI
C
CP
SCOMP
FB
DRP
1
1
=
Page 33 of 39
=
=
=
=
=
C
90
=
10
2
4
2 (
R
. Choose the crossover frequency of current share loop f
(
π
π
(
V
=
PWMRMP
−
π
=
R
=
R
∗
I
∗
∗
=10kΩ, R
. 0
FB
BST
VCC
A
CP1
FB
∗
20
−
. 0
C
R
f
65
tan(
L
V
1
C
f
65
1
CP
6 .
+
C
E
E
PWMRMP
=0.1uF
=
=47pF to reduce high frequency noise.
∗
=0.1uF, C
R
*
nF
)
R
∗
*
∗
2
R
DRP
*
. 0
R
2
3
FB
18
C
∗
G
FB
C
PWMRMP
R
) 5
∗
1
E
L
PWMRMP
CS
2 .
)
1
DRP
R
E
∗
PHASE42
∗
=
FB
=
*
∗
V
180
C
−
∗
10
π
O
10
C
4
V
FB
R
=
π
E
DAC
3
VCCL
BST
*
FB
∗
=
2
3
∗
∗
*
=
*
V
12
140
63
∗
R
I
)
∗
(
f
(
=768Ω, R
162
100
162
FB
SW
*
( *
*
=0.1uF
R
°
I
*
105
V
1
O
∗
LE
*
10
I
*
V
V
=
+
*
10
−
RAMP
110
G
110
3
PWMRMP
*
=
V
V
. 1
∗
CS
−
34
O
DAC
576
2
110
9
65
)
Ω
π
∗
_
/
PHASE43
*
∗
=
) 6
ROOM
2
∗
0 (
∗
5
)
π
10
=
6 .
2 (
∗
(
. 1 (
5 .
∗
62
=
5
(
3
*
π
*
22
33
2 .
18
f
10
*
10
CI
∗
∗
nF
R
140
*
2 .
−
=2.80kΩ
−
22
*
−
9
10
LE
105
3
. 1
*
, choose C
=
*
10
−
*
(
*
05
6
12
6
10
10
2
1 [
* )
*
*
3
7 .
*
. 9
*
−
+
3
62
−
1 [
10
CI
nF
)
100
6
* 1
. 0
2
2
+
)
)
π
6
∗
should be at least one decade lower than that of the voltage
75
∗
10
2
=
576
*
(
*
π
100
34
−
27
10
f
−
FB
*
CI
4
. 1
∗
3500
−
nF
)
*
) 3
=5.6nF
12
1
*
162
∗
10
3 .
C
2
*
( *
CI
π
E
−
−
800
12
*
9
∗
=3.5kHz
( *
∗
20
22
/
0 (
3500
V
) 6
−
*
*
O
*
5 .
. 1
10
10
∗
10
) 3
*
(
I
3
*
22
−
O
10
−
*
3
. 1
6
)]
,
. 0
*
*
05
and calculate C
−
*
10
75
3
62
F
*
/
MI
−
10
) 6
*
=
6
. 1 (
∗
. 0
6
=
62
011
33
146
)
∗
−
162
105
kHz
*
*
SCOMP
. 0
. 9
75
* 1
10
=
. 1
,
−
4
65
IR3081A
1/31
)
k
105
Ω
/05
* ]
. 0
011