参数资料
型号: IR3500MTRPBF
厂商: International Rectifier
英文描述: XPHASE3TM VR11.0 & AMD PVID CONTROL IC
中文描述: XPHASE3TM VR11.0
文件页数: 40/47页
文件大小: 778K
代理商: IR3500MTRPBF
IR3500
Page 40 of 47
June 12, 2007
Calculate constant K
P,
the ratio of inductor peak current over average current in each phase,
1
(
)
(
)
1
=
D
D
f
L
n
I
sw
LIMIT
I
R
)
1
[
_
_
+
+
=
(
)
(
)
126
.
)
108
.
1
108
.
2
800
1
6
/
135
)
108
.
6
1
0
(
)
6
0
108
.
6
)
108
.
1
108
.
12
)
1
2
/
)
=
+
+
=
k
u
D
n
m
n
m
D
n
D
D
V
K
I
P
OCSET
CS
TOFST
CS
P
MAX
L
LIMIT
n
OCSET
I
G
V
K
R
/
]
=
+
=
k
14
)
10
*
40
/(
34
)
10
*
3
126
.
10
*
64
.
6
135
(
6
3
3
Calculate constant K
P,
the ratio of inductor peak current over average current in each phase,
No Load Output Voltage Setting Resistor
R
VSETPT
and Adaptive Voltage Positioning Resistor
R
DRP
From Figure 24, the bias current of VSETPT pin is 40uA with R
OSC
=15k
.
500
10
*
40
VSETPT
I
VCCL Programming Resistor
R
VCCLFB1
and
R
VCCLFB2
Choose VCCL=7V to maximize the converter efficiency. Pre-select R
VCCLFB1
=20k
, and calculate R
VCCLFB2.
3
1
2
=
=
=
10
*
20
6
3
_
NLOFST
O
VSETPT
R
V
=
=
=
k
VCCL
R
R
VCCLFB
VCCLFB
05
.
19
.
7
19
.
*
10
*
20
19
.
19
.
*
VCCL Drive Resistor
R
VCCLDRV
The maximum drive current for the linear regulator is dependent on the type of MosFET used. For this
example, it’s assumed that IR6622/ IRF6691 are used as buck switches.
[
n
n
I
avg
drive
800
)
11
47
(
_
+
=
The minimum input voltage is assumed to be 10.5 V and VCCL is fixed at 6.5V for this design.
=
70
/
350
mA
]
mA
mA
k
350
6
10
=
+
(19)
=
660
5
7
5
10
V
V
V
R
VCCLDRV
(20)
Choose a transistor with
β
(min) of 70. The maximum input voltage is assumed 13.5 V,
V
7
5
13
Thermistor
R
THERM
and Over Temperature Setting Resistors
R
HOTSET1
and
R
HOTSET2
Choose NTC thermistor R
THERM
=2.2k
, which has a constant of B
THERM
=3520, and the NTC thermistor
resistance at the allowed maximum temperature T
MAX
is,
1
1
(
*
[
*
_
_
T
T
ROOM
MAX
L
Select R
HOTSET2
= 931
to linearize the NTC, which has non-linear characteristics in the operational temperature
range. Then calculate R
HOTSET1
corresponding to the allowed maximum temperature TMAX.
mA
mA
10
5
660
5
<
=
(21)
=
+
+
=
=
142
)]
25
273
1
115
273
1
(
*
3520
[
*
10
*
2
)]
3
EXP
B
EXP
R
R
THERM
THERM
TMAX
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