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參數(shù)資料
| 型號： | QW075A1 |
| 英文描述： | Analog IC |
| 中文描述： | 模擬IC |
| 文件頁數(shù)： | 15/20頁 |
| 文件大?。?/td> | 1272K |
| 代理商： | QW075A1 |
Lucent Technologies Inc.
15
Data Sheet
May 2000
dc-dc Converters; 36 to 75 Vdc Input, 5 Vdc Output; 50 W and 75 W
QW050A1 and QW075A1 Power Modules:
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 QW075A1
module is operating at V
I
= 54 V and an output current
of 15 A, maximum ambient air temperature of 40 °C,
and the heat sink is 1/2 inch. The module is oriented in
the transverse direction.
Solution
Given: V
I
= 54 V
I
O
= 15 A
T
A
= 40 °C
T
C
= 85 °C
Heat sink = 1/2 inch
Determine P
D
by using Figure 24:
P
D
= 16 W
Then solve the following equation:
Use Figure 26 to determine air velocity for the 1/2 inch
heat sink.
The minimum airflow necessary for the QW075A1
module is 1.2 m/s (240 ft./min.).
Note:
Pending improvement will lower the power dissi-
pation and reduce the airflow needed.
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) as shown in Figure 32.
8-1304 (C)
Figure 32. 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:
(
)
P
D
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 provides
a conservative estimate for such instances.
EMC Considerations
For assistance with designing for EMC compliance,
please refer to the FLTR100V10 data sheet
(DS99-294EPS).
Layout Considerations
Copper paths must not be routed beneath the power
module mounting inserts. For additional layout guide-
lines, refer to the FLTR100V10 data sheet
(DS99-294EPS).
θ
ca
T
------------------------
T
–
(
)
P
D
=
θ
ca
85
-----------------------
40
–
(
)
16
=
θ
ca
2.8 °C/W
=
P
D
T
C
T
S
T
A
cs
sa
θ
sa
C
-T
T
A
–
θ
cs
–
=
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| QW0806BR | 制造商:NTE Electronics 功能描述:1/4 Watt, Rn 55D, 806 Ohm 1% Metal Film Resistor- 25/Pkg |
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