JW150H Lineage Power, JW150H Datasheet - Page 16
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JW150H
Manufacturer Part Number
JW150H
Description
Manufacturer
Lineage Power
Datasheet
1.JW150H.pdf
(20 pages)
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dc-dc Converters; 36 to 75 Vdc Input, 24 Vdc Output; 50 W to 175 W
Thermal Considerations
Heat Transfer Without Heat Sinks
Figure 35. JW175H Power Dissipation vs.
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/lbs.).
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 29. Case-to-ambient thermal resis-
tance vs. airflow is shown, for various heat sink config-
urations and heights, in Figure 36. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
16
16
θ
30
25
20
15
10
ca
5
0
0
=
Output Current
ΔT
-------------------- -
1
P
C max
,
D
OUTPUT CURRENT, I
2
=
3
(
----------------------- -
T
D
C
V
V
V
):
P
I
I
I
–
= 72 V
= 54 V
= 36 V
D
C, max
T
4
A
(continued)
)
O
) divided by the
(A)
5
(continued)
6
C
) is
8-2655 (C)
7
Figure 36. 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 36 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 36 is shown in the following example.
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JW100H
module is operating at V
of 4.2 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 in.
Solution
Given: V
Determine P
8
7
6
5
4
3
2
1
0
I
T
T
Heat sink = 1/2 in.
P
O
0
A
C
I
D
= 4.2 A
= 54 V
= 40 °C
= 85 °C
Curves; Either Orientation
= 12 W
D
AIR VELOCITY MEASURED IN m/s (ft./min.)
by using Figure 33:
(100)
0.5
(200)
1.0
I
= 54 V and an output current
(300)
1.5
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
Lineage Power
April 2008
(500)
2.5
8-1153 (C)
(600)
3.0