JFW100G1 Lineage Power, JFW100G1 Datasheet - Page 16

• Manufacturer Part Number: JFW100G1
• Manufacturer: Lineage Power

dc-dc Converters; 36 to 75 Vdc Input; 2.5 Vdc Output; 25 W to 75 W
Thermal Considerations
Heat Transfer Without Heat Sinks
Figure 24. JFW150G Power Dissipation vs. Output
Heat Transfer with Heat Sinks
The power module has 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 21. Case-to-ambient thermal resis-
tance vs. airflow for various heat sink configurations is
shown in Figure 25. These curves were obtained by
experimental testing of heat sinks, which are offered in
the product catalog.
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-
16
16
θ
30
25
20
15
10
ca
5
0
0
=
3
Current at 80 °C
ΔT
---------------------
P
C max
,
OUTPUT CURRENT, I
D
V
V
V
8
I
I
I
= 75 V
= 54 V
= 36 V
=
13
(
----------------------- -
T
D
C
):
18
P
–
D
C, max
T
O
A
(continued)
)
(A)
23
) divided by the
28
(continued)
C
30
) is
8-2462 (C)
33
ally lower than the resistance of the heat sink by itself.
The module used to collect the data in Figure 25 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 25 is shown in the following example.
Figure 25. Case-to-Ambient Thermal Resistance
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JFW150G
module is operating at V
of 30 A, maximum ambient air temperature of 40 °C,
and heat sink of 1/2 inch.
Solution
Given: V
Determine P
Then solve the following equation:
θ
θ
θ
ca
ca
ca
8
7
6
5
4
3
2
1
0
0
I
T
T
Heat sink = 1/2 inch
P
O
=
=
=
A
C
I
D
= 30 A
= 54 V
= 40 °C
= 85 °C
Curves; Either Orientation
= 24 W
1.9 °C/W
D
(
----------------------- -
(
----------------------- -
(100)
T
85 40
0.5
by using Figure 24:
C
P
24
–
–
D
AIR VELOCITY, IN m/s (ft./min.)
T
A
(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).a
(600)
3.0