JW150F1-T Lineage Power, JW150F1-T Datasheet - Page 15
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JW150F1-T
Manufacturer Part Number
JW150F1-T
Description
CONVERTER DC/DC 3.3V 99W OUT
Manufacturer
Lineage Power
Series
JW150r
Type
Isolated with Remote On/Offr
Datasheet
1.JW050F6.pdf
(20 pages)
Specifications of JW150F1-T
Output
3.3V
Number Of Outputs
1
Power (watts)
99W
Mounting Type
Through Hole
Voltage - Input
36 ~ 75V
Package / Case
9-DIP Module
1st Output
3.3 VDC @ 30A
Size / Dimension
2.40" L x 2.28" W x 0.50" H (61mm x 57.9mm x 12.7mm)
Power (watts) - Rated
99W
Operating Temperature
-40°C ~ 100°C
Efficiency
80%
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:
JW150F1-T
Manufacturer:
LUCENT
Quantity:
27
Company:
Part Number:
JW150F1-T
Manufacturer:
LUCENT
Quantity:
530
April 2008
Thermal Considerations
Heat Transfer with Heat Sinks
Thermal derating with heat sinks is expressed by using
the overall thermal resistance of the module. Total
module thermal resistance (θca) is defined as the max-
imum case temperature rise (ΔT
module power dissipation (P
The location to measure case temperature (T
shown in Figure 26. Case-to-ambient thermal resis-
tance vs. airflow is shown, for various heat sink config-
urations and heights, in Figure 32. These curves were
obtained by experimental testing of heat sinks, which
are offered in the product catalog.
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.
Lineage Power
θ
ca
8
7
6
5
4
3
2
1
0
0
=
Curves; Either Orientation
ΔT
-------------------- -
(100)
0.5
P
C max
,
D
AIR VELOCITY, m/s (ft./min.)
(200)
1.0
=
(
----------------------- -
T
dc-dc Converters; 36 to 75 Vdc Input, 3.3 Vdc Output; 33 W to 99 W
D
(300)
C
1.5
):
P
–
D
C, max
T
A
(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
) divided by the
(continued)
(500)
2.5
C
) is
8-1153 (C)
(600)
3.0
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JW100F
module is operating at V
of 20 A, 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 32 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the JW100F
module is 1.1 m/s (220 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
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 1/2 in.
P
P
O
=
=
=
A
C
I
D
D
= 54 V
= 20 A
= 40 °C
= 85 °C
Sink-to-Ambient
= 15.8 W
2.8 °C/W
∅
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 29:
15.8
C
T
P
C
–
–
D
T
A
)
)
θ
cs
I
= 54 V and an output current
T
S
θ
sa
T
A
8-1304 (C)
15