FW300F1 Lineage Power, FW300F1 Datasheet - Page 16

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FW300F1

Manufacturer Part Number
FW300F1
Description
CONVERTER DC/DC 3.3V 198W OUT
Manufacturer
Lineage Power
Series
FW300r
Type
Isolated with Remote On/Offr
Datasheet

Specifications of FW300F1

Output
3.3V
Number Of Outputs
1
Power (watts)
198W
Mounting Type
Through Hole
Voltage - Input
36 ~ 75V
Package / Case
Module
1st Output
3.3 VDC @ 60A
Size / Dimension
4.60" L x 2.40" W x 0.53" H (116.8mm x 61mm x 13.5mm)
Power (watts) - Rated
198W
Operating Temperature
-40°C ~ 100°C
Efficiency
79%
Approvals
CE, CSA, UL, VDE
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
3rd Output
-
2nd Output
-
4th Output
-

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dc-dc Converters; 36 to 75 Vdc Input, 3.3 Vdc Output; 165 W to 198 W
Thermal Considerations
Heat Transfer with Heat Sinks
Figure 28. 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 Figures 27 and
28 had a thermal-conductive dry pad between the case
and the heat sink to minimize contact resistance.
To choose a heat sink, determine the power dissipated
as heat by the unit for the particular application.
Figure 29 shows typical heat dissipation for a range of
output currents and three voltages for the FW250F1
and FW300F1.
16
16
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0
Curves; Longitudinal Orientation
(100)
0.5
AIR VELOCITY, m/s (ft./min.)
(200)
1.0
(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
8-1320 (C)
(600)
3.0
Figure 29. FW250F1 and FW300F1 Power
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the FW300F1
module is operating at nominal line and an output cur-
rent of 50 A, maximum ambient air temperature of
40 °C, and the heat sink is 0.5 inch.
Solution
Given: V
Determine P
60
50
40
30
20
10
0
I
T
T
Heat sink = 0.5 inch.
P
0
O
A
I
C
D
= 50 A
= 54 V
= 40 °C
= 85 °C
Dissipation vs. Output Current
= 42 W
D
by using Figure 29:
1 0
O UTPUT CURRENT, I
2 0
30
V
V
V
Tyco Electronics Corp..
I
I
I
= 72 V
= 54 V
= 36 V
40
O
(A)
July 2002
50
8-1737 (C)
60

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