JW050R Lineage Power, JW050R Datasheet - Page 17

• Manufacturer Part Number: JW050R
• Manufacturer: Lineage Power

April 2008
Thermal Considerations
Heat Transfer Without Heat Sinks
Figure 40. JW150R 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.-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 35. Case-to-ambient thermal resis-
tance vs. airflow is shown, for various heat sink config-
urations and heights, in Figure 41. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
Lineage Power
θ
ca
16
14
12
10
8
6
4
0.5
=
Output Current at 25 °C
1.0
ΔT
---------------------
P
C max
,
1.5
D
OUTPUT CURRENT, I
V
V
V
I
I
I
2.0
= 36 V
= 48 V
= 75 V
=
(
----------------------- -
2.5
T
D
C
):
P
dc-dc Converters; 36 to 75 Vdc Input, 28 Vdc Output; 50 W to 150
–
D
C, max
3.0
T
A
(continued)
)
O
) divided by the
3.5
(A)
4.0
(continued)
C
) is
5.0
8-1855 (C)
Figure 41. 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 41 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 41 is shown in the following example.
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JW150R1
module is operating at V
of 5.3 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch.
Solution
Given: V
Determine P
8
7
6
5
4
3
2
1
0
I
T
T
Heat sink = 1/2 inch
P
O
0
A
C
I
D
= 48 V
= 5.3 A
= 40 °C
= 85 °C
= 14.8 W
Curves; Either Orientation
D
by using Figure 40:
(100)
0.5
AIR VELOCITY, m/s (ft./min.)
(200)
1.0
I
= 48 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
(500)
2.5
(600)
8-1153 (C)
3.0
17