QHW075Y1 Lineage Power, QHW075Y1 Datasheet - Page 16

• Manufacturer Part Number: QHW075Y1
• Description: CONVERTER DC/DC 1.8V 27W OUT
• Manufacturer: Lineage Power
• Series: QHWr
• Type: Isolated with Remote On/Offr

Specifications of QHW075Y1

• Output: 1.8V
• Number Of Outputs: 1
• Power (watts): 27W
• Mounting Type: Through Hole
• Voltage - Input: 36 ~ 75V
• Package / Case: 8-DIP Module
• 1st Output: 1.8 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: 27W
• Operating Temperature: -40°C ~ 100°C
• Efficiency: 82%
• 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, 1.8 Vdc Output; 18 W to 36 W
Thermal Considerations
Heat Transfer with Heat Sinks
Figure 35. Heat Sink Power Derating Curves;
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 30 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 30 is shown in the following example.
16
16
20
15
10
5
0
0
1/2 IN. HEAT SINK
1/4 IN. HEAT SINK
1.0 m/s (200 lfm); Longitudinal
Orientation
10
1 IN. HEAT SINK
LOCAL AMBIENT TEMPERATURE, T
NO HEAT SINK
20
30
40
50
60
(continued)
70
(continued)
80
A
(°C)
90 100
8-2892(F)
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the QHW100Y1
module is operating at V
of 20 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 30 to determine air velocity for the
1/2 inch heat sink.
The minimum airflow necessary for this module is
0.4 m/s (80 ft./min.).
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 1/2 inch
P
O
=
=
=
A
C
I
D
= 20 A
= 48 V
= 40 °C
= 85 °C
= 8.0 W
5.63 °C/W
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 29:
C
8.0
P
–
–
D
T
A
)
)
I
= 48 V and an output current
Lineage Power
April 2008