IR3081 International Rectifier Corp., IR3081 Datasheet - Page 26

• Manufacturer Part Number: IR3081
• Description: Xphasetm Vr 10 Control Ic
• Manufacturer: International Rectifier Corp.

VDAC Slew Rate Programming Capacitor C
From Figure 15, the sink current of VDAC pin corresponding to 400kHz (R
VDAC down-slope slew-rate programming capacitor from the required down-slope slew rate.
Calculate the programming resistor.
From Figure 15, the source current of VDAC pin is 110uA. The VDAC up-slope slew rate is
Over Current Setting Resistor R
The room temperature is 25ºC and the target PCB temperature is 100 ºC. The phase IC die temperature is about 1
ºC higher than that of phase IC, and the inductor temperature is close to PCB temperature.
Calculate Inductor DC resistance at 100 ºC,
The current sense amplifier gain is 34 at 25ºC, and its gain at 101ºC is calculated as,
Set the over current limit at 135A. From Figure 14, the bias current of OCSET pin (I
R
the current sense amplifier input resistor mismatch.
Calculate constant K
No Load Output Voltage Setting Resistor R
From Figure 14, the bias current of FB pin is 41uA with R
C
G
R
SR
R
K
R
=
R
R
L
OSC
OCSET
VDAC
VDAC
DRP
CS
FB
P
(
UP
_
135
MAX
=
Page 26 of 39
_
6
=
MIN
=30.1kΩ. The total current sense amplifier input offset voltage is 0.55mV, which includes the offset created by
=
=
(
V
=
=
∗
R
=
=
R
I
I
. 0
0
C
L
SR
SOURCE
[
FB
=
R
−
5 .
VDAC
61
_
R
I
L
V
G
MAX
SINK
DOWN
LIMIT
∗
+
_
O
n
*
CS
ROOM
R
10
3
)
L
2 .
∗
C
_
_
∗
−
n
=
V
ROOM
VDAC
∗
*
I
MAX
3
V
=
O
∗
∗
LIMIT
110
R
10
∗
33
O
1 [
R
/(
L
2
1
_
O
−
_
5 .
P,
∗
2
+
L
3 .
*
15
NLOFST
I
*
∗
MAX
G
10
3850
∗
76
*
FB
/
+
10
the ratio of inductor peak current over average current in each phase,
1 [
CS
V
n
10
=
. 0
−
−
*
I
−
∗
9
∗
0
_
6
10
−
1470
55
∗
*
5 .
MIN
3
1 (
R
10
=
f
/
−
−
L
*
+
+
10
SW
6
3
−
10
_
V
6
K
(
3 .
*
=
MAX
−
3
33
∗
CS
10
∗
P
−
6
mV
2 .
365
(
3
)
) 2
T
*
OCSET
_
=
)
−
*
+
L
TOFST
10
6
∗
10
_
V
30
/
=
∗
∗
6
30
MAX
uS
CS
−
. 0
∗
−
(
(
4 .
9
12
T
15
2 .
_
61
. 0
)
nF
IC
2
TOFST
−
∗
91
/(
−
*
T
_
=
. 1
41
n
10
, Choose C
ROOM
MAX
*
3
33
∗
10
*
−
5 .
]
FB
R
3
VDAC
10
∗
)
−
Ω
∗
−
∗
O
)]
G
3
30
−
. 1
and Adaptive Voltage Positioning Resistor R
T
CS
=
6
=
ROOM
33
)
2 .
. 0
_
and Resistor R
. 0
=
MIN
47
/(
=
13
61
VDAC
220
. 1
*
)]
.
10
/
*
21
135
k 3
OSC
=
I
10
*
−
OCSET
k
Ω
3
34
10
=33nF
Ω
∗
6 /
−
=30.1kΩ.
3
1 [
∗
−
9
1 [
∗
+
∗
20
3850
−
12
1470
41
*
∗
VDAC
10
*
*
400
10
10
*
−
10
3
−
*
−
6
−
6
10
∗
−
6
(
∗
. 0
100
3
∗
. 0
55
∗
(
OSC
101
61
) 2
−
*
25
*
10
=
=30.1kΩ) is 76uA. Calculate the
−
10
)]
0
25
−
3 .
=
3
−
)]
3
. 0
∗
61
=
6
30
∗
m
Ω
. 0
2 .
91
OCSET
*
DRP
10
10/01
−
) is 41uA with
IR3081
3
=
/04
365
Ω