JC050B Datasheet, PDF (13/16 Pages) List of Unclassifed Manufacturers – JC050B

Data Sheet
October 1997
JC050B, JC075B, JC100B Power Modules: dc-dc Converters;
18 Vdc to 36 Vdc Input, 12 Vdc Output; 50 W to 100 W
Thermal considerations (continued)
Heat Transfer with Heat Sinks (continued)
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JC100B
module is operating at nominal line and an output cur-
rent of 8.5 A, maximum ambient air temperature of
40 °C, and the heat sink is 0.5 in.
Solution
Given: VI = 28 V
IO = 8.5 A
TA = 40 °C
TC = 85 °C
Heat sink = 0.5 in.
Determine PD by using Figure 24:
PD = 20 W
Then solve the following equation:
θca = ⁽---T----^C-----^–----T----^A----⁾
PD
θca = ⁽---8----5----^–-----4---0----⁾
20
θca = 2.3 °C/W
Use Figure 25 to determine air velocity for the0.5 inch
heat sink.
The minimum airflow necessary for the JC100B mod-
ule is 1.7 m/s (340 ft./min.).
Custom Heat Sinks
A more detailed model can be used to determine the
required thermal resistance of a heat sink to provide
necessary cooling. The total module resistance can be
separated into a resistance from case-to-sink (θcs) and
sink-to-ambient (θsa) shown below (Figure 26).
TC
TS
TA
PD →
θcs
θsa
8-1304
Figure 26. Resistance from Case-to-Sink and
Sink-to-Ambient
For a managed interface using thermal grease or foils,
a value of θcs = 0.1 °C/W to 0.3 °C/W is typical. The
solution for heat sink resistance is:
θsa = ⁽---T----^C-----^–----T----^A----⁾ – θcs
PD
This equation assumes that all dissipated power must
be shed by the heat sink. Depending on the user-
defined application environment, a more accurate
model, including heat transfer from the sides and bot-
tom of the module, can be used. This equation pro-
vides a conservative estimate for such instances.
Layout Considerations
Copper paths must not be routed beneath the power
module mounting inserts.
Tyco Electronics Corp.
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