IR3080 International Rectifier, IR3080 Datasheet - Page 28

• Manufacturer Part Number: IR3080
• Description: XPHASE VRD10 CONTROL IC WITH VCCVID & OVERTEMP DETECT
• Manufacturer: International Rectifier

The power good delay time is
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
t
C
G
R
SR
R
K
R
=
R
VccPG
L
OSC
OCSET
VDAC
VDAC
FB
CS
P
(
UP
_
135
MAX
=
Page 28 of 41
_
6
=
MIN
=30.1kΩ. The total current sense amplifier input offset voltage is 0.55mV, which includes the offset created by
=
=
(
V
=
=
∗
R
=
=
I
C
I
. 0
0
C
L
SR
SOURCE
[
=
SS
R
−
5 .
_
VDAC
61
I
I
L
V
G
LIMIT
MAX
SINK
/
DOWN
+
_
n
DEL
O
*
CS
ROOM
10
3
)
2 .
∗
C
_
∗
( *
∗
−
=
V
ROOM
VDAC
V
*
I
3
=
I
O
R
. 3
∗
LIMIT
110
O
10
CHG
∗
33
L
1 [
/(
91
2
_
1
_
−
5 .
P,
2
+
L
3 .
I
NLOFST
*
15
MAX
*
∗
FB
−
10
3850
∗
76
*
/
+
10
the ratio of inductor peak current over average current in each phase,
1 [
V
V
n
10
=
∗
. 0
O
−
−
*
I
∗
−
9
0
R
6
10
−
1470
55
−
∗
*
1 (
5 .
3
L
10
. 1
−
=
f
/
+
−
*
_
+
10
SW
V
) 3
6
3
−
10
MAX
K
6
CS
(
3 .
*
−
3
33
∗
P
=
10
∗
−
6
mV
2 .
)
_
(
3
) 2
T
. 0
*
OCSET
+
TOFST
=
)
−
*
L
10
6
∗
* 1
V
10
_
30
/
=
∗
30
CS
MAX
uS
−
10
−
(
(
4 .
9
12
T
15
_
2 .
)
∗
nF
−
IC
TOFST
2
−
6
/(
−
n
T
_
( *
=
70
. 1
41
∗
, Choose C
ROOM
MAX
3
. 3
33
R
*
*
]
5 .
FB
O
91
∗
10
VDAC
10
)
Ω
G
−
∗
)]
=
−
−
−
. 1
and Adaptive Voltage Positioning Resistor R
T
CS
6
=
6
. 1
ROOM
. 0
33
)
. 0
_
33
and Resistor R
=
61
MIN
47
/(
13
−
VDAC
220
*
*
)]
1
10
/
.
10
) 3 .
135
k 3
OSC
I
=
*
OCSET
−
−
Ω
3
34
10
3
=33nF
=
∗
6 /
∗
=30.1kΩ.
1
1 [
∗
−
20
8 .
9
1 [
+
41
∗
ms
3850
−
*
12
10
1470
*
10
∗
VDAC
−
*
400
3
10
−
*
−
6
10
−
∗
. 0
*
6
10
. 0
∗
55
−
6
(
61
100
3
∗
*
∗
(
OSC
10
*
101
) 2
−
10
−
25
3
=
−
=30.1kΩ) is 76uA. Calculate the
−
∗
3
)]
0
25
6
3 .
=
∗
)]
. 0
. 0
61
=
91
30
m
Ω
*
2 .
10
OCSET
−
3
DRP
=
9
/30/04
) is 41uA with
IR3080
365
Ω