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AN533 数据表(PDF) 7 Page - STMicroelectronics |
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AN533 数据表(HTML) 7 Page - STMicroelectronics |
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7 / 22 page ![]() 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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