參數(shù)資料
型號: JC100B
廠商: Electronic Theatre Controls, Inc.
英文描述: JC050B
中文描述: JC050B
文件頁數(shù): 13/16頁
文件大?。?/td> 711K
代理商: JC100B
Tyco Electronics Corp.
13
Data Sheet
October 1997
18 Vdc to 36 Vdc Input, 12 Vdc Output; 50 W to 100 W
JC050B, JC075B, JC100B Power Modules: dc-dc Converters;
Thermal considerations
(continued)
Heat Transfer with Heat Sinks
(continued)
Example
If an 85 °C case temperature is desired, what is the
minimum airflow necessary Assume the JC100B
module is operating at nominal line and an output cur-
rent of 8.5 A, maximum ambient air temperature of
40 °C, and the heat sink is 0.5 in.
Solution
Given: V
I
= 28 V
I
O
= 8.5 A
T
A
= 40 °C
T
C
= 85 °C
Heat sink = 0.5 in.
Determine P
D
by using Figure 24:
P
D
= 20 W
Then solve the following equation:
Use Figure 25 to determine air velocity for the0.5 inch
heat sink.
The minimum airflow necessary for the JC100B mod-
ule is 1.7 m/s (340 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 (
θ
cs) and
sink-to-ambient (
θ
sa) shown below (Figure 26).
8-1304
Figure 26. Resistance from Case-to-Sink and
Sink-to-Ambient
For a managed interface using thermal grease or foils,
a value of
θ
cs = 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.
Layout Considerations
Copper paths must not be routed beneath the power
module mounting inserts.
θ
ca
C
-T
T
A
(
)
P
D
=
θ
ca
-85
40
20
(
)
=
θ
ca
2.3
°C/W
=
P
D
T
C
T
S
T
A
θ
cs
θ
sa
θ
sa
C
-T
T
A
(
)
P
D
θ
cs
=
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