参数资料
型号: AEH20G48-8
元件分类: 电源模块
英文描述: 1-OUTPUT 50 W DC-DC REG PWR SUPPLY MODULE
封装: 2.400 X 2.280 INCH, 0.500 INCH HEIGHT, HALF BRICK PACKAGE-9
文件页数: 25/35页
文件大小: 409K
代理商: AEH20G48-8
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-31-
USA
Europe
Asia
TEL:
1-760-930-4600
44-(0)1384-842-211
852-2437-9662
FAX:
1-760-930-0698
44-(0)1384-843-355
852-2402-4426
www.astec.com
Basic Thermal Model
There is another approach to analyze module
thermal performance, to model the overall ther-
mal resistance of the module. This presentation
method is especially useful when considering
heat sinks. The following equation can be used
to calculate the total thermal resistance .
RCA =
TC, max / PD
Where RCA is the module thermal resistance,
TC, max is the maximum case temperature
rise,
PD is the module power dissipation.
In this model, PD,
TC, max, and RCA are equals
to current flow, voltage drop, and electrical
resistance, respectively, in Ohm's law, as
shown in Figure 34. Also,
TC, max is defined as
the difference between the module case tem-
perature (TC) and the inlet ambient temperature
(TA).
TC, max = TC TA
Where TC is the module case temperature;
TA is the inlet ambient temperature.
For AEH Series 50W to 150W 2.5V output con-
verters, the module's thermal resistance values
versus air velocity have been determined
experimentally and shown in figure 35. The
highest values on each curve represents the
point of natural convection.
Figure 35 is used for determining thermal per-
formance under various conditions of airflow
and heat sink configurations.
Example 4. How to determine the allowable
minimum airflow to heat sink combinations
necessary for a module under a desired Tc
and a certain condition?
Although the maximum case temperature for
the AEH Series converters is 100° C, you can
improve module reliability by limiting Tc,max to
a lower value. How to decide? For example,
what is the allowable minimum airflow for AEH
150W heat sink combinations at desired Tc of
80 °C?
The working condition is as following:
Vin = 48 V, IO = 30 A, TA = 40 °C
Determine PD ( Fig.26 )
PD = 13.5 W
Then solve RCA
::
RCA =
TC, max / PD
RCA =
(TC TA)
/ PD
RCA =
(80 40)
/ 13.5 = 3°C/W
determine air velocity from figure 35:
If no heat sink:
v = 2.6 m/s (520 ft./min.)
If 1/4 in. heat sink:
v = 1.8 m/s (360 ft./min.)
If 1/2 in. heat sink:
v = 1.1 m/s (220 ft./min.)
If 1 in. heat sink:
v = 0.4 m/s (80 ft./min.)
BMPM
PD
= BMPM
THERMAL
RESISTANCE
Fig.34 Basic Thermal Resistance Model
0
0.5
(100)
1.0
(200)
1.5
(300)
2.0
(400)
2.5
(500)
3.0
(600)
0
1
5
6
7
8
Air Velocity m/s (ft./min.)
4
3
2
Case-Ambient
Thermal
Resistance
R
CA
(°C/W)
1 in. HEAT SINK
1/2 in. HEAT SINK
1/4 in. HEAT SINK
NO HEAT SINK
Fig.35. Case-to-Ambient Thermal
Resistance Curves; Either Orientation
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