JW150C1 Lineage Power, JW150C1 Datasheet - Page 15
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JW150C1
Manufacturer Part Number
JW150C1
Description
CONVERTER DC/DC 15V 150W OUT
Manufacturer
Lineage Power
Series
JW150r
Type
Isolated with Remote On/Offr
Datasheet
1.JW050C1.pdf
(20 pages)
Specifications of JW150C1
Output
15V
Number Of Outputs
1
Power (watts)
150W
Mounting Type
Through Hole
Voltage - Input
36 ~ 75V
Package / Case
9-DIP Module
1st Output
15 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
150W
Operating Temperature
-40°C ~ 100°C
Efficiency
88.5%
Approvals
CE, CSA, UL, VDE
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
3rd Output
-
2nd Output
-
4th Output
-
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
JW150C1
Manufacturer:
LUCENT
Quantity:
27
Part Number:
JW150C1
Manufacturer:
LUCENT
Quantity:
20 000
Part Number:
JW150C1
Quantity:
55
April 2008
Thermal Considerations
Heat Transfer with Heat Sinks
Figure 32. Case-to-Ambient Thermal Resistance
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 32 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 32 is shown in the following example.
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JW100C
module is operating at V
of 6.7 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch.
Solution
Given: V
Determine P
Lineage Power
8
7
6
5
4
3
2
1
0
I
T
T
Heat sink = 1/2 in.
P
O
0
A
C
I
D
= 6.7 A
= 54 V
= 40 °C
= 85 °C
Curves; Either Orientation
= 14 W
D
by using Figure 30:
(100)
0.5
AIR VELOCITY, m/s (ft./min.)
(200)
1.0
I
= 54 V and an output current
dc-dc Converters; 36 to 75 Vdc Input, 15 Vdc Output; 50 W to 150 W
(300)
1.5
(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
(continued)
(500)
2.5
8-1153 (C)
(600)
3.0
Then solve the following equation:
Use Figure 32 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the JW100C
module is 0.91 m/s (180 ft./min.).
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) shown below (Figure 33).
Figure 33. 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.
θ
θ
θ
θ
sa
ca
ca
ca
P
=
D
=
=
=
∅
Sink-to-Ambient
3.2 °C/W
(
------------------------ -
T
(
----------------------- -
(
----------------------- -
T
85 40
C
P
T
C
–
P
14
C
D
–
–
T
D
A
T
)
A
)
)
–
θ
θ
cs
cs
T
S
θ
sa
T
A
8-1304 (C)
15