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AN533 bảng dữ liệu(PDF) 7 Page - STMicroelectronics

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Giải thích chi tiết về linh kiện  The behavior of a semiconductor device depends
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AN533 bảng dữ liệu(HTML) 7 Page - STMicroelectronics

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AN533
Through-hole packages
7/22
Figure 4.
Rth(h-a) versus the length of a flat square heatsink
1.1.7
Forced cooling
For high power or very high power applications, a forced-air or liquid cooling heatsink may
be required. Heatsink manufacturers give a coefficient depending on the air or liquid flow.
However, in some applications like vacuum cleaners, dissipated power is only a few watts
and air flow cooling is available. This allows a very small heatsink to be used, very often a
flat aluminium heatsink. In this case it is necessary to measure the case temperature in the
worst case scenario and to check the following formula:
Tc < Tjmax - P . Rth(j-c)
1.2
Thermal impedance
In steady state, a thermal equivalent circuit can be made only with thermal resistances.
However, for pulse operation it can be useful to consider the thermal impedance, especially
when the component is on during a time lower than the time to reach the thermal resistance.
The thermal impedance value versus pulse duration is given in the datasheets (see an
example in Figure 5), in the form of the relationship Zth/Rth plotted against pulse duration.
For example, BTA08-600SW is able to dissipate
≈ 27 W without heatsink during 1 s:
Zth(j-a) can be obtained from the datasheet by reading the value of the ratio Zth/Rth from the
curve (in the case of this product the ratio is 0.06 as seen in Figure 5) and multiplying the
ratio by the value of Rth(j-a) from the datasheet. For this example Rth(j-a) is 60 °C/W
Rth(h-a)
100
50
30
20
10
1
2
3
5
STEEL
Thickness of
the plate in mm
0
2
8
14
4
10
16
6
12
18
20
0.5
1
2
5
Rth(h-a)
100
50
30
20
10
1
2
3
5
(°C/W)
COPPER
Thickness of
the plate in mm
(cm)
0
2
8
14
4
10
16
6
12
18
20
Length
0.5
1
2
5
Rth(h-a)
100
50
30
20
10
1
2
3
5
(°C/W)
ALUMINIUM
Thickness of
the plate in mm
(cm)
0
2
8
14
4
10
16
6
12
18
20
Length
0.5
1
2
5
Length
(°C/W)
(cm)
- Semiconductor device
in the center
- Bare convector (no ventilation)
- Vertical position
Thermal model for calculation
based on square heatsink
L
e
P =
Tjmax -Tamax
Zth(j-a) (1 s)
P =
125 - 25
60 x 0.06
= 27.5 W



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