Tyco Electronics Corp
15
Data Sheet
July 1999
dc-dc Converters; 36 to 75 Vdc Input, 5 Vdc Output; 50 W to 150 W
JW050A, JW075A, JW100A, JW150A Power Modules:
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 mod-
ule thermal resistance (
θ
ca) is defined as the maximum
case temperature rise (
T
C, max
) divided by the module
power dissipation (P
D
):
The location to measure case temperature (T
C
) is
shown in Figure 26. Case-to-ambient thermal resis-
tance vs. airflow is shown, for various heat sink config-
urations and heights, in Figure 32. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
8-1153 (C)
Figure 32. Case-to-Ambient Thermal Resistance
Curves; Either Orientation
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 32 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 32 is shown in the following example.
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary Assume the JW100A
module is operating at V
I
= 54 V and an output current
of 20 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch.
Solution
Given: V
I
= 54 V
I
O
= 20 A
T
A
= 40 °C
T
C
= 85 °C
Heat sink = 1/2 in.
Determine P
D
by using Figure 30:
P
D
= 17 W
Then solve the following equation:
Use Figure 32 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the JW100A mod-
ule is 1.3 m/s (260 ft./min.).
θ
ca
T
C max
,
D
-------P
T
C
-------P
T
A
–
(
)
D
=
=
0
0.5
(100)
1.0
(200)
1.5
(300)
2.0
(400)
2.5
(500)
3.0
(600)
0
1
5
6
7
8
AIR VELOCITY, m/s (ft./min.)
4
3
2
1 1/2 IN. HEAT SINK
1 IN. HEAT SINK
1/2 IN. HEAT SINK
1/4 IN. HEAT SINK
NO HEAT SINK
C
R
θ
C
°
C
θ
ca
T
C
-------P
T
A
–
(
)
D
=
θ
ca
85
-----------------------
40
–
(
)
17
=
θ
ca
2.6 °C/W
=