JAW075B1 Lineage Power, JAW075B1 Datasheet - Page 12

• Manufacturer Part Number: JAW075B1
• Manufacturer: Lineage Power

36 Vdc to 75 Vdc Input, 12 Vdc Output; 50 W to 75 W
Thermal Considerations
Heat Transfer Without Heat Sinks
Figure 21. Forced Convection Power Derating with
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.).
Thermal derating with heat sinks is expressed by using
the overall thermal resistance of the module. Total
module thermal resistance (θca) is defined as the max-
imum case temperature rise (ΔT
module power dissipation (P
The location to measure case temperature (T
shown in Figure 18. Case-to-ambient thermal resis-
tance vs. airflow for various heat sink configurations
and heights is shown in Figure 22. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
12
12
θ
ca
14
12
10
8
6
4
2
0
0
=
No Heat Sink; Either Orientation
10
LOCAL AMBIENT TEMPERATURE, T
ΔT
-------------------- -
(200 ft./min.)
(400 ft./min.)
(600 ft./min.)
(20 ft./min.)
0.1 m/s
1.0 m/s
2.0 m/s
3.0 m/s
P
C max
20
,
D
30
=
40
(
----------------------- -
T
D
C
50
):
P
–
D
C, max
T
60
A
(continued)
)
) divided by the
70
80
A
(continued)
(°C)
C
90 100
) is
8-3231 (F)
Figure 22. 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 22 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 22 is shown in the following example.
Example
If an 82 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JAW075B1
module is operating at V
6.2 A, longitudinal orientation, maximum ambient air
temperature of 40 °C, and the heat sink is 1/4 inch.
Solution
Given: V
Determine P
9
8
7
6
5
4
3
2
1
0
I
T
T
Heat sink = 1/4 inch
P
O
0
A
C
I
D
= 6.2 A
= 55 V
= 40 °C
= 82 °C
Curves; Either Orientation
= 11 W
D
(100)
by using Figure 20:
0.5
AIR VELOCITY, m/s (ft./min.)
(200)
1.0
I
= 55 V, an output current of
(300)
1.5
NO HEAT SINK
1/4 IN. HEAT SINK
1/2 IN. HEAT SINK
1 IN. HEAT SINK
1 1/2 IN. HEAT SINK
(400)
2.0
Lineage Power
April 2008
(500) (600)
2.5
8-3233 (F)
3.0