JAW050A Lineage Power, JAW050A Datasheet - Page 13

• Manufacturer Part Number: JAW050A
• Description: CONVERTER DC/DC 5V 50W OUT
• Manufacturer: Lineage Power
• Series: JAWr
• Type: Isolated with Remote On/Offr

Specifications of JAW050A

• Output: 5V
• Number Of Outputs: 1
• Power (watts): 50W
• Mounting Type: Through Hole
• Voltage - Input: 36 ~ 75V
• Package / Case: 9-DIP 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
• Product: Isolated
• Output Power: 50 W
• Input Voltage Range: 36 V to 75 V
• Input Voltage (nominal): 48 V
• Output Voltage (channel 1): 5 V
• Output Current (channel 1): 10 A
• Isolation Voltage: 1.5 KV
• Package / Case Size: DIP
• Output Type: Isolated
• Output Voltage: 5 V
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• 3rd Output: -
• 2nd Output: -
• 4th Output: -
• Lead Free Status / Rohs Status: No

Data Sheet
April 2008
Thermal Considerations
Heat Transfer with Heat Sinks
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 Figures 23 and
24 had a thermal-conductive dry pad between the case
and the heat sink to minimize contact resistance. The
use of Figure 23 is shown in the following example.
Example
If an 82 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JAW075A
module is operating at V
15 A, longitudinal orientation, maximum ambient air
temperature of 40 °C, and the heat sink is 1/4 inch.
Solution
Given: V
Determine P
Then solve the following equation:
Use Figure 23 to determine air velocity for the 1/4 inch
heat sink.
The minimum airflow necessary for this module is
1.1 m/s (220 ft./min.).
Lineage Power
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 1/4 inch.
P
O
=
=
=
A
C
I
D
= 15 A
= 55 V
= 40 °C
= 82 °C
= 14 W
3.0 °C/W
D
(
----------------------- -
(
----------------------- -
T
82 40
by using Figure 20:
C
P
14
–
–
D
T
A
)
)
I
= 55 V, an output current of
(continued)
JAW050A and JAW075A Power Modules; dc-dc Converters:
(continued)
36 Vdc to 75 Vdc Input, 5 Vdc Output; 50 W to 75 W
Custom Heat Sinks
A more detailed model can be used to determine the
required thermal resistance of a heat sink to provide
necessary cooling. The total module resistance can be
separated into a resistance from case-to-sink (θcs) and
sink-to-ambient (θsa) as shown in Figure 25.
Figure 25. Resistance from Case-to-Sink and
For a managed interface using thermal grease or foils,
a value of θcs = 0.1 °C/W to 0.3 °C/W is typical. The
solution for heat sink resistance is:
This equation assumes that all dissipated power must
be shed by the heat sink. Depending on the user-
defined application environment, a more accurate
model, including heat transfer from the sides and bot-
tom of the module, can be used. This equation pro-
vides a conservative estimate for such instances.
EMC Considerations
For assistance with designing for EMC compliance,
refer to the FLTR100V10 Filter Module Data Sheet
(DS99-294EPS).
Layout Considerations
Copper paths must not be routed beneath the power
module standoffs. For additional layout guidelines,
refer to the FLTR100V10 Filter Module Data Sheet
(DS99-294EPS).
θ
sa
=
P
D
Sink-to-Ambient
(
------------------------ -
T
C
T
P
–
C
D
T
A
)
–
θ
θ
cs
cs
T
S
θ
sa
T
A
8-1304(F).e
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