JC050F1 Lineage Power, JC050F1 Datasheet - Page 13

• Manufacturer Part Number: JC050F1
• Description: CONVERTER DC/DC 3.3V 33W OUT
• Manufacturer: Lineage Power
• Series: JC050r
• Type: Isolated with Remote On/Offr

Specifications of JC050F1

• Output: 3.3V
• Number Of Outputs: 1
• Power (watts): 33W
• Mounting Type: Through Hole
• Voltage - Input: 18 ~ 36V
• Package / Case: 9-DIP Module
• 1st Output: 3.3 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: 33W
• Operating Temperature: -40°C ~ 100°C
• Efficiency: 81%
• Approvals: CSA, UL, VDE
• Product: Isolated
• Output Power: 8 W
• Input Voltage Range: 18 V to 36 V
• Input Voltage (nominal): 28 V
• Output Voltage (channel 1): 3.3 V
• Output Current (channel 1): 2.4 A
• Isolation Voltage: 1.5 KV
• Package / Case Size: DIP
• Output Type: Isolated
• Output Voltage: 3.3 V
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• 3rd Output: -
• 2nd Output: -
• 4th Output: -
• Lead Free Status / Rohs Status: No

Thermal Considerations
Heat transfer with Heat Sinks
Figure 27. 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 27 had a
thermal-conductive dry pad between the case and the
heat sink to minimize contact resistance. The use of
Figure 27 is shown in the following example
8
7
6
5
4
3
2
1
0
Data Sheet
March 2008
Lineage Power
0
Curves; Either Orientation
AIR VELOCITY MEASURED IN m/s (ft./min.)
(100)
0.5
(200)
1.0
(300)
1.5
1 1/2 IN HEAT SINK
1 IN HEAT SINK
1/2 IN HEAT SINK
1/4 IN HEAT SINK
NO HEAT SINK
(continued)
(400)
2.0
(continued)
(500)
2.5
JC050F, JC075F, JC100F Power Modules: dc-dc Converters;
(600)
3.0
8-1153
18 Vdc to 36 Vdc Input, 3.3 Vdc Output; 33 W to 66 W
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary? Assume the JC100A
module is operating at nominal line and an output cur-
rent of 20 A, maximum ambient air temperature of
40 °C, and the heat sink is 0.5 in.
Solution
Given: V
Determine P
Then solve the following equation:
Use Figure 27 to determine air velocity for the 0.5 inch
heat sink.
The minimum airflow necessary for the JC100F
module is 1.5 m/s (300 ft./min.).
θ
θ
θ
ca
ca
ca
I
T
T
Heat sink = 0.5 in.
P
O
=
=
=
A
C
I
D
= 20 A
= 28 V
= 40 °C
= 85 °C
= 18.2 W
2.47
D
(
----------------------- -
(
----------------------- -
T
85 40
by using Figure 26:
C
18.2
P
°C/W
–
–
D
T
A
)
)
13