Lineage Power 13
Data Sheet
March 2008 18 Vdc to 36 Vdc Input, 15 Vdc Output; 50 W to 100 W
JC050C, JC075C, JC100C Power Modules: dc-dc Converters;
Thermal considerations (continued)
Heat Transfer with Heat Sinks (continued)
Solution
Given: V
I = 28 V
I
O = 6 A
T
A = 40 °C
T
C = 85 °C
Heat sink = 0.5 in.
Determine P
D by using Figure 23:
P
D = 14.0 W
Then solve the following equation:
Use Figure 24 to determine air velocity for the0.5 inch
heat sink.
The minimum airflow necessary for the JC100C
module is 1.0 m/s (200 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 25).
8-1304
Figure 25. 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:
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.
θca
TC TA–()
P
D
------------------------
=
θca
85 40–()
14.0
------------------------
=
θca 3.2 °C/W=
PD
TC TS TA
θcs θsa
→
θsa
T
C TA–()
P
D
-------------------------
θcs–=
