ir3500 International Rectifier Corp., ir3500 Datasheet - Page 40

• Manufacturer Part Number: ir3500
• Description: Xphase3 Vr11.0 & Amd Pvid Control Ic
• Manufacturer: International Rectifier Corp.

Calculate constant K
Calculate constant K
No Load Output Voltage Setting Resistor R
From Figure 24, the bias current of VSETPT pin is 40uA with R
VCCL Programming Resistor R
Choose VCCL=7V to maximize the converter efficiency. Pre-select R
VCCL Drive Resistor R
example, it’s assumed that IR6622/ IRF6691 are used as buck switches.
The minimum input voltage is assumed to be 10.5 V and VCCL is fixed at 6.5V for this design.
Choose a transistor with β(min) of 70. The maximum input voltage is assumed 13.5 V,
Thermistor R
Choose NTC thermistor R
resistance at the allowed maximum temperature T
Select R
range. Then calculate R
K
R
=
R
R
The maximum drive current for the linear regulator is dependent on the type of MosFET used. For this
R
TMAX
P
OCSET
VSETPT
VCCLFB
(
135
=
6



V
=
I
∗
=
R
Page 40 of 47
2
⋅
HOTSET2
=
THERM
D
. 0
(
[
=
I
I
V
⋅
LIMIT
64
R
1 (
LIMIT
O
VCCL
I
n
VCCLFB
−
VSETPT
_
THERM
*
*
/
NLOFST
D
n
10
EXP
)
)
∗
= 931 to linearize the NTC, which has non-linear characteristics in the operational temperature
⋅
⋅
−
−
L
1
n
R
3
[
*
. 1
⋅
⋅
L
B
P,
P,
∗
and Over Temperature Setting Resistors R
(
. 1
f
_
19
THERM
D
sw
. 1
=
MAX
19
the ratio of inductor peak current over average current in each phase,
the ratio of inductor peak current over average current in each phase,
−
HOTSET1
⋅
126
VCCLDRV
2
20
40
I
m
=
n
drive
⋅
∗
D
( *
)
*
20
*
R
THERM
⋅
1 (
+
⋅
(
_
T
10
10
VCCLDRV
1 (
m
avg
*
L
7
+
0
10
_
n
−
+
−
3 .
1
−
−
MAX
K
corresponding to the allowed maximum temperature TMAX.
3
6
D
. 1
1
VCCLFB1
=
13
3
*
=2.2k , which has a constant of B
−
)
P
*
19
=
[
5 .
10
(
. 1
)
=
D
−
47
V
500
+
19
)
T
10
660
−



−
V
n
_
3
ROOM
=
)
5 .
0
CS
=
1
+
Ω
Ω
7 .
∗
V
and R



350
. 4
11
_
12
−
34
TOFST
05
−
n
6
)]
⋅
0
mA
VSETPT
5 .
. 0
)
k
/(
7 .
=
Ω
⋅
108
(
40
135
VCCLFB2
=
800
V
2
]
/
2 .
9
∗
MAX
70
−
5 .
⋅
*
*
6 /
G
1 (
k
6
10
mA
10
5 .
CS
and Adaptive Voltage Positioning Resistor R
)
−
+
3
⋅
is,
V
. 0
0
−
*
10
<
/
1 .
6
108
EXP
I
=
10
)
u
OCSET
mA
=
⋅
660
mA
800
)
[
14
⋅
3520
]
6
⋅
Ω
OSC
⋅
k
k
6
(
Ω
⋅
. 0
=
2
( *
108
HOTSET1
⋅
=15k .
350
. 0
273
VCCLFB1
108
−
mA
1
+
0
6
115
⋅
)
1 (
THERM
⋅
(
−
0
and R
−
. 0
+
6
=20k , and calculate R
(21)
273
1
108
(20)
−
=3520, and the NTC thermistor
1
+
. 0
)
HOTSET2
25
108
)]
)
=



142
=
. 0
(19)
Ω
June 12, 2007
126
IR3500
VCCLFB2.
DRP