JW050A, JW075A, JW100A, JW150A Power Modules:
dc-dc Converters; 36 to 75 Vdc Input, 5 Vdc Output; 50 W to 150 W
Data Sheet
July 1999
These measured resistances are from heat transfer
Thermal Considerations (continued)
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.
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.).
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JW100A
module is operating at VI = 54 V and an output current
of 20 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch.
Thermal derating with heat sinks is expressed by using
the overall thermal resistance of the module.Total mod-
ule thermal resistance (θca) is defined as the maximum
case temperature rise (∆TC, max) divided by the module
power dissipation (PD):
Solution
Given: VI = 54 V
IO = 20 A
(TC – TA)
∆TC, max
TA = 40 °C
TC = 85 °C
θca =
=
------------------------
--------------------
PD
PD
Heat sink = 1/2 in.
The location to measure case temperature (TC) 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.
Determine PD by using Figure 30:
PD = 17 W
Then solve the following equation:
(TC – TA)
θca =
θca =
------------------------
8
PD
1 1/2 IN. HEAT SINK
1 IN. HEAT SINK
1/2 IN. HEAT SINK
1/4 IN. HEAT SINK
7
6
5
4
3
2
1
0
(85 – 40)
-----------------------
17
NO HEAT SINK
θca = 2.6 °C/W
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.).
0
0.5
(100)
1.0
(200)
1.5
(300)
2.0
(400)
2.5
3.0
(500) (600)
AIR VELOCITY, m/s (ft./min.)
8-1153 (C)
Figure 32. Case-to-Ambient Thermal Resistance
Curves; Either Orientation
Tyco Electronics Corp
15
TE [ TE CONNECTIVITY ]
