Lucent Technologies Inc.
15
Data Sheet
May 2000
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 V
I
= 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: V
I
= 54 V
I
O
= 10 A
T
A
= 40 °C
T
C
= 85 °C
Heat sink = 1/2 inch
Determine P
D
by using Figure 24:
P
D
= 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 (
θ
cs) and
sink-to-ambient (
θ
sa) as shown in Figure 32.
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
θ
cs = 0.1 °C/W to 0.3 °C/W is typical. The
solution for heat sink resistance is:
(
)
P
D
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 provides
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).
θ
ca
T
------------------------
T
–
(
)
P
D
=
θ
ca
85
-----------------------
40
–
(
)
8.8
=
θ
ca
5.1 °C/W
=
P
D
T
C
T
S
T
A
cs
sa
θ
sa
C
-T
T
A
–
θ
cs
–
=