JC050A1 Datasheet, PDF (13/16 Pages) Lineage Power Corporation – Small size: 61.0 mm x 57.9 mm x 12.7 mm (2.40 in. x 2.28 in. x 0.50 in.)

Data Sheet
March 2008
JC050A, JC075A, JC100A Power Modules: dc-dc Converters;
18 Vdc to 36 Vdc Input, 5 Vdc Output; 50 W to 100 W
Thermal Considerations (continued)
Heat Transfer with Heat Sinks
The power modules have through-threaded, M3 x 0.5
mounting holes, which enable heat sinks or cold plates
to attach to the module. The mounting torque must not
exceed 0.56 N–m (5 in.–lb.). For a screw attachment
from the pin side, the recommended hole size on the
customer’s PWB around the mounting holes is
0.130 ± 0.005 inches. If a larger hole is used, the
mounting torque from the pin side must not exceed
0.25 N–m (2.2 in.–lb.).
Thermal derating with heat sinks is expressed by using
the overall thermal resistance of the module. Total
module thermal resistance (θca) is defined as the max-
imum case temperature rise (ΔTC, max) divided by the
module power dissipation (PD):
θca = ^Δ----^T---P-^C---^,-_D-^m----^a--^x- = -⁽--T----^C--P--^–---_D--T----^A---⁾-
The location to measure case temperature (TC) is
shown in Figure 23. Case-to-ambient thermal resis-
tance vs. airflow is shown, for various heat sink config-
urations and heights, in Figure 28. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
8
7
1 1/2 IN HEAT SINK
6
1 IN HEAT SINK
1/2 IN HEAT SINK
5
1/4 IN HEAT SINK
NO HEAT SINK
4
3
2
1
0
0
0.5
1.0
1.5
2.0
2.5 3.0
(100) (200) (300) (400) (500) (600)
AIR VELOCITY MEASURED IN m/s (ft./min.)
8-1153
Figure 28. Case-to-Ambient Thermal Resistance
Curves; Either Orientation
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 28 is shown in the following example
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: VI = 28 V
IO = 20 A
TA = 40 °C
TC = 85 °C
Heat sink = 0.5 in.
Determine PD by using Figure 27:
PD = 22.8 W
Then solve the following equation:
θca = -⁽--T----^C--P--^–---_D--T----^A---⁾-
θca = -⁽--8---5-----^–-----4---0----⁾
17
θca = 1.97 °C/W
Use Figure 28 to determine air velocity for the 0.5 inch
heat sink.
The minimum airflow necessary for the JC100A mod-
ule is 2.0 m/s (400 ft./min.).
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 28 had a
Lineage Power
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