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Lineage Power
15
Data Sheet
April 2008
dc-dc Converters; 36 to 75 Vdc Input, 3.3 Vdc Output; 33 W to 50 W
QW050F1 and QW075F1 Power Modules:
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 QW075F1
module is operating at VI = 54 V and an output current
of 10 A, transverse orientation, maximum ambient air
temperature of 40 °C, and the heat sink is 1/2 inch.
Solution
Given: VI = 54 V
IO = 10 A
TA = 40 °C
TC = 85 °C
Heat sink = 1/2 inch
PD = 8.8 W
Then solve the following equation:
Use
Figure 26 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the QW050F1
module is 0.4 m/s (70 ft./min.).
Note: Pending improvement will lower the power dissi-
pation and reduce the airflow needed.
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 (
8-1304 (C)
Figure 32. 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.
EMC Considerations
For assistance with designing for EMC compliance,
please refer to the FLTR100V10 data sheet
(DS99-294EPS).
Layout Considerations
Copper paths must not be routed beneath the power
module mounting inserts. For additional layout guide-
lines, refer to the FLTR100V10 data sheet
(DS99-294EPS).
θca
TC TA
–
()
PD
------------------------
=
θca
85 40
–
()
8.8
------------------------
=
θca
5.1 °C/W
=
PD
TC
TS
TA
θcs
θsa
→
θsa
TC TA
–
()
PD
------------------------
θcs
–
=