ir3500a International Rectifier Corp., ir3500a Datasheet - Page 40

• Manufacturer Part Number: ir3500a
• 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
OCSET
VSETPT
P
VCCLFB
(
135
=
6



V
=
I
∗
=
R
Page 40 of 47
2
HOTSET2
=
⋅
(
THERM
. 0
D
[
=
I
I
V
LIMIT
64
⋅
R
LIMIT
1 (
O
VCCL
I
n
VCCLFB
VSETPT
−
_
THERM
*
*
/
NLOFST
D
10
n
EXP
∗
)
)
= 931 to linearize the NTC, which has non-linear characteristics in the operational temperature
⋅
⋅
−
−
1
n
L
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
20
40
I
2
m
n
=
drive
⋅
∗
)
( *
D
*
20
*
R
⋅
THERM
1 (
(
+
⋅
_
T
10
10
VCCLDRV
m
1 (
avg
*
L
7
+
0
n
10
_
−
+
−
3 .
1
−
−
MAX
K
corresponding to the allowed maximum temperature TMAX.
1
D
3
6
. 1
VCCLFB1
=
13
3
−
*
)
=2.2k , which has a constant of B
P
*
19
=
[
5 .
10
(
D
. 1
)
=
−
47
V
500
+
)
19
T
10



660
−
−
V
n
_
3
=
ROOM
)
5 .
0
CS
=
1
+
Ω
Ω
7 .



∗
V
and R
12
350
. 4
11
_
−
34
TOFST
05
−
⋅
n
6
)]
. 0
0
mA
VSETPT
5 .
)
k
/(
7 .
=
(
108
135
Ω
⋅
40
VCCLFB2
=
800
V
2
]
/
2 .
9
∗
MAX
70
⋅
−
6 /
5 .
1 (
*
*
G
k
6
10
mA
10
−
)
5 .
CS
and Adaptive Voltage Positioning Resistor R
⋅
+
. 0
3
0
is,
V
−
1 .
*
10
<
108
/
6
EXP
u
I
=
10
)
OCSET
mA
⋅
=
)
800
660
mA
⋅
[
14
6
3520
]
⋅
k
⋅
Ω
(
OSC
k
6
. 0
⋅
2
Ω
108
=
( *
⋅
. 0
HOTSET1
=15k .
350
273
108
−
VCCLFB1
0
6
mA
1
+
⋅
)
1 (
115
⋅
(
THERM
−
0
. 0
6
and R
+
−
1
=20k , and calculate R
(21)
108
273
−
(20)
. 0
=3520, and the NTC thermistor
)
1
+
108
HOTSET2
25
)]
)



=
=
142
. 0
126
(19)
Ω
July 3, 2008
IR3500A
VCCLFB2.
DRP