JW100B Lineage Power, JW100B Datasheet - Page 15

• Manufacturer Part Number: JW100B
• Manufacturer: Lineage Power

Data Sheet
April 2008
Thermal Considerations
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 hole
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 ( ca) is defined as the maximum
case temperature rise ( T
power dissipation (P
The location to measure case temperature (T
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.
Figure 32. Case-to-Ambient Thermal Resistance
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
Lineage Power
ca
8
7
6
5
4
3
2
1
0
0
=
Curves; Either Orientation
-------------------- -
(100)
0.5
T
P
C max
,
D
AIR VELOCITY, m/s (ft./min.)
D
(200)
):
1.0
=
C, max
----------------------- -
T
dc-dc Converters; 36 to 75 Vdc Input, 12 Vdc Output; 50 W to 150 W
(300)
C
1.5
P
–
) divided by the module
D
T
A
(continued)
1 1/2 IN. HEAT SINK
1 IN. HEAT SINK
1/2 IN. HEAT SINK
1/4 IN. HEAT SINK
NO HEAT SINK
(400)
2.0
(500)
2.5
C
JW050B, JW075B, JW100B, JW150B Power Modules:
) is
8-1153 (C)
(600)
3.0
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 JW100B
module is operating at V
of 8.5 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch.
Solution
Given: V
Determine P
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 JW100B
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 ( cs) and
sink-to-ambient ( sa) shown below (Figure 33).
Figure 33. Resistance from Case-to-Sink and
ca
ca
ca
P
I
T
T
Heat sink = 1/2 in.
P
O
D
=
A
=
=
C
I
D
= 54 V
= 8.5 A
= 40 °C
= 85 °C
Sink-to-Ambient
= 15 W
3 °C/W
D
----------------------- -
----------------------- -
T
85 40
by using Figure 30:
T
C
P
15
C
–
–
D
T
A
cs
I
= 54 V and an output current
T
S
sa
T
A
8-1304 (C)
15