JC050F Lineage Power, JC050F Datasheet - Page 13

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JC050F

Manufacturer Part Number
JC050F
Description
Manufacturer
Lineage Power
Datasheet

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
JC050F
Manufacturer:
LUCENT
Quantity:
27
Thermal Considerations
Heat transfer with Heat Sinks
Figure 27. Case-to-Ambient Thermal Resistance
These measured resistances are from heat transfer
from the sides and bottom of the module as well as the
top side with the attached heat sink; therefore, the
case-to-ambient thermal resistances shown are gener-
ally lower than the resistance of the heat sink by itself.
The module used to collect the data in Figure 27 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 27 is shown in the following example
8
7
6
5
4
3
2
1
0
Data Sheet
March 2008
Lineage Power
0
Curves; Either Orientation
AIR VELOCITY MEASURED IN m/s (ft./min.)
(100)
0.5
(200)
1.0
(300)
1.5
1 1/2 IN HEAT SINK
1 IN HEAT SINK
1/2 IN HEAT SINK
1/4 IN HEAT SINK
NO HEAT SINK
(continued)
(400)
2.0
(continued)
(500)
2.5
JC050F, JC075F, JC100F Power Modules: dc-dc Converters;
(600)
3.0
8-1153
18 Vdc to 36 Vdc Input, 3.3 Vdc Output; 33 W to 66 W
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JC100A
module is operating at nominal line and an output cur-
rent of 20 A, maximum ambient air temperature of
40 °C, and the heat sink is 0.5 in.
Solution
Given: V
Determine P
Then solve the following equation:
Use Figure 27 to determine air velocity for the 0.5 inch
heat sink.
The minimum airflow necessary for the JC100F
module is 1.5 m/s (300 ft./min.).
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 0.5 in.
P
O
=
=
=
A
C
I
D
= 20 A
= 28 V
= 40 °C
= 85 °C
= 18.2 W
2.47
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 26:
C
18.2
P
°C/W
D
T
A
)
)
13

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