QW075A1 Lineage Power, QW075A1 Datasheet - Page 14

• Manufacturer Part Number: QW075A1
• Description: CONVERTER DC/DC 5V 75W OUT
• Manufacturer: Lineage Power
• Series: QWr
• Type: Isolated with Remote On/Offr

Specifications of QW075A1

• Output: 5V
• Number Of Outputs: 1
• Power (watts): 75W
• Mounting Type: Through Hole
• Voltage - Input: 36 ~ 75V
• Package / Case: 8-DIP Module
• 1st Output: 5 VDC @ 15A
• Size / Dimension: 2.28" L x 1.45" W x 0.50" H (57.9mm x 36.8mm x 12.7mm)
• Power (watts) - Rated: 75W
• 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: -

QW050B1 and QW075B1 Power Modules; dc-dc Converters:
36 Vdc to 75 Vdc Input, 12 Vdc Output; 50 W to 75 W
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 Figure 23 and
Figure 24 had a thermal-conductive dry pad between
the case and the heat sink to minimize contact resis-
tance. The use of Figure 23 and Figure 24 are shown
in the following example.
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the QW075B1
module is operating at V
of 5.5 A, transverse orientation, maximum ambient air
temperature of 40 °C, and the heat sink is 1/2 inch.
Solution
Given: V
Determine P
Then solve the following equation:
Use Figure 23 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for this module is
1.25 m/s (250 ft./min.).
14
14
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 1/2 inch
P
O
=
=
=
A
C
I
D
= 75 V
= 5.5 A
= 40 °C
= 85 °C
= 16 W
2.8 °C/W
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 21:
C
P
16
–
–
D
T
A
)
)
I
= 75 V and an output current
(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 29.
Figure 29. 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 Filter Module 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 Filter Module Data
Sheet (DS99-294EPS).
θsa
P
=
D
Sink-to-Ambient
(
----------------------- -
T
T
C
P
C
–
D
T
A
)
θ
–
cs
θcs
T
S
θ
sa
Lineage Power
Data Sheet
April 2008
T
A
8-1304 (F).e