JFW050A1 Lineage Power, JFW050A1 Datasheet - Page 18

• Manufacturer Part Number: JFW050A1
• Description: CONVERTER DC/DC 5V 50W OUT
• Manufacturer: Lineage Power
• Series: JFWr
• Type: Isolatedr

Specifications of JFW050A1

• Output: 5V
• Number Of Outputs: 1
• Power (watts): 50W
• Mounting Type: Through Hole
• Voltage - Input: 36 ~ 75V
• Package / Case: Module
• 1st Output: 5 VDC @ 10A
• Size / Dimension: 2.40" L x 2.28" W x 0.50" H (61mm x 57.9mm x 12.7mm)
• Power (watts) - Rated: 50W
• Operating Temperature: -40°C ~ 100°C
• Efficiency: 84%
• Approvals: CE, CSA, UL, VDE
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• 3rd Output: -
• 2nd Output: -
• 4th Output: -

dc-dc Converters; 36 to 75 Vdc Input, 5 Vdc Output; 50 W to 150 W
Thermal Considerations
Heat Transfer with Heat Sinks
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 27. Case-to-ambient thermal resis-
tance vs. airflow for various heat sink configurations is
shown in Figure 33. 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-
ally lower than the resistance of the heat sink by itself.
The module used to collect the data in Figure 33 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 33 is shown in the following example.
Figure 33. Case-to-Ambient Thermal Resistance
18
18
θ
ca
8
7
6
5
4
3
2
1
0
0
=
Curves; Either Orientation
ΔT
-------------------- -
(100)
0.5
P
C max
,
D
AIR VELOCITY, IN m/s (ft./min.)
(200)
1.0
=
(
----------------------- -
T
D
(300)
C
):
P
1.5
–
D
C, max
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
) divided by the
(continued)
(500)
2.5
C
) is
8-1153 (C).a
(600)
3.0
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JFW150A
module is operating at V
of 20 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:
Use Figure 33 to determine air velocity for the 1/2 inch
heat sink. The minimum airflow necessary for the
JFW150A module is about 1.3 m/s (260 ft./min.).
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 1/2 inch
P
O
=
=
=
A
C
I
D
= 20 A
= 54 V
= 40 °C
= 85 °C
= 17 W
2.6 °C/W
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 32:
C
P
17
–
–
D
T
A
)
)
I
= 54 V and an output current
April 2008
Lineage Power