QW050A1 Lineage Power, QW050A1 Datasheet - Page 15

• Manufacturer Part Number: QW050A1
• Description: CONVERTER DC/DC 5V 50W OUT
• Manufacturer: Lineage Power
• Series: QWr
• Type: Isolated with Remote On/Offr

Specifications of QW050A1

• Output: 5V
• Number Of Outputs: 1
• Power (watts): 50W
• Mounting Type: Through Hole
• Voltage - Input: 36 ~ 75V
• Package / Case: 8-DIP Module
• 1st Output: 5 VDC @ 10A
• Size / Dimension: 2.28" L x 1.45" W x 0.50" H (57.9mm x 36.8mm x 12.7mm)
• Power (watts) - Rated: 50W
• Operating Temperature: -40°C ~ 100°C
• Efficiency: 84%
• Approvals: CE, CSA, UL, VDE
• Output Current: 10A
• Input Voltage: 48V
• Output Voltage: 5V
• Output Power: 50
• Screening Level: Industrial
• Operating Temperature Min Deg. C: -40C
• Operating Temperature Max Deg. C: 100C
• Product Length (mm): 36.8mm
• Product Depth (mm): 57.9mm
• Product Height (mm): 12.7mm
• Mounting Style: Through Hole
• Pin Count: 8
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• 3rd Output: -
• 2nd Output: -
• 4th Output: -
• Lead Free Status / Rohs Status: RoHS Not Applicable

April 2008
Thermal Considerations
Heat Transfer with Heat Sinks
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the QW075A1
module is operating at V
of 15 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch. The module is oriented in
the transverse direction.
Solution
Given: V
Determine P
Then solve the following equation:
Use Figure 26 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the QW075A1
module is 1.2 m/s (240 ft./min.).
Note: Pending improvement will lower the power dissi-
Lineage Power
θ
θ
θ
ca
ca
ca
pation and reduce the airflow needed.
I
T
T
Heat sink = 1/2 inch
P
O
=
=
=
A
C
I
D
= 15 A
= 54 V
= 40 °C
= 85 °C
= 16 W
2.8 °C/W
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 24:
C
P
16
–
–
D
T
A
)
)
I
= 54 V and an output current
dc-dc Converters; 36 to 75 Vdc Input, 5 Vdc Output; 50 W and 75 W
(continued)
(continued)
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 32.
Figure 32. 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,
please refer to the FLTR100V10 data sheet
(DS99-294EPS).
Layout Considerations
Copper paths must not be routed beneath the power
module mounting inserts. For additional layout guide-
lines, refer to the FLTR100V10 data sheet
(DS99-294EPS).
θsa
P
D
=
→
Sink-to-Ambient
(
----------------------- -
T
T
C
P
C
–
D
T
A
)
θcs
–
θcs
T
S
θsa
T
A
8-1304 (C)
15