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Electric welding. How to calculate the radiator. Encyclopedia of radio electronics and electrical engineering

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During operation of a semiconductor device, power is released in its crystal, which leads to heating of the latter. If more heat is released than is dissipated in the surrounding space, then the temperature of the crystal will rise and may exceed the maximum allowable. In this case, its structure will be irreversibly destroyed.

Therefore, the reliability of semiconductor devices is largely determined by cooling efficiency. The most effective is the convective cooling mechanism, in which heat is carried away by the flow of a gaseous or liquid coolant, washing the cooled surface.

The larger the cooled surface, the more efficient the cooling, and therefore powerful semiconductor devices must be installed on metal radiators with a developed cooled surface. Ambient air is usually used as the heat transfer medium.

According to the method of moving the coolant, they distinguish:

  • natural ventilation;
  • forced ventilation.

In the case of natural ventilation, the movement of the coolant is carried out due to the draft that occurs near the heated radiator. In the case of forced ventilation, the coolant is moved by means of a fan. In the second case, it is possible to obtain higher flow rates and, accordingly, better cooling conditions.

Thermal calculations can be greatly simplified if we use the thermal cooling model (Fig. 18.26) Here the difference between the crystal temperature TJ and ambient temperature TA causes a heat flow moving from the crystal to the environment through thermal resistances RJC (crystal - housing), RCS (housing - radiator) and RSA (radiator - environment).

How to calculate the radiator
Figure 18.26. Thermal cooling model

Thermal resistance has the unit of °С/W. Total maximum thermal resistance RJA on the site of the crystal - the environment can be found by the formula:

where PPP - power dissipated on the crystal of a semiconductor device, W.

Thermal resistance RJC and RCS indicated in the reference data for semiconductor devices. For example, according to the reference data, for the IRFP250N transistor, its thermal resistance in the crystal-radiator section is RJC + RCS \u0,7d 0,24 + 0,94 \uXNUMXd XNUMX ° C / W.

This means that if a power of 10 W is dissipated on the chip, then its temperature will be 9,4 °C higher than the temperature of the heatsink.

Heat sink thermal resistance can be found by the formula:

The following methodology is based on AAVID THERMALLOY's Max Clip System™ Aluminum Radiator Selection Guidelines.

On fig. 18.27 shows a graphical relationship between the perimeter of the section of an aluminum radiator and its thermal resistance for natural (red line) and forced (blue line) airflow cooling.

By default, it is assumed that:

  • the radiator has a length of 150 mm;
  • difference between heatsink temperature TS and ambient temperature Tа is ;
  • forced cooling flow velocity is 2 m/s.

If the cooling conditions differ from those accepted by default, then the necessary correction can be made using the graphs in Fig. 18.28 - fig. 18.30.

How to calculate the radiator
Rice. 18.27. Relationships between the cross section of an aluminum radiator and its thermal resistance

How to calculate the radiator
Rice. 18.28. Correction factor for the difference between the temperature of the radiator and the environment

How to calculate the radiator
Rice. 18.29. Correction factor for airflow rate

How to calculate the radiator
Rice. 18.30. Correction factor for radiator length

For example, let's calculate a radiator that provides cooling for an ERST transistor, consisting of 20 IRFP250N type transistors. The calculation of the radiator can be carried out for one transistor, and then the resulting size can be increased by a factor of 20.

Since the total power dissipated on the key transistor is 528 W, then the power dissipated on each IRFP250N transistor is 528/20 = 26,4 W. The heatsink must ensure the maximum temperature of the transistor crystal is not more than +110 °C at a maximum ambient temperature of +40 °C.

Will find thermal resistance RJA for one transistor IRFP250N:

Now find heat sink thermal resistance:

Knowing the maximum temperature of the crystal and the thermal resistance in the section of the crystal-radiator, we determine the maximum temperature of the radiator:

According to the graph (Fig. 18.28), we determine the correction factor Kt for the temperature difference between the radiator and the environment:

Used to cool the radiator fan type 1,25EV-2,8-6-3270U4, having a capacity of 280 m3/h. To calculate the flow rate, you need to divide the performance by the cross section of the duct blown by the fan.

If the duct has a cross-sectional area:

then the airflow rate will be:

According to the graph (Fig. 18.29), we determine the correction factor Kv to the actual airflow speed:

Let's assume that we have a large number of ready-made radiators with a section perimeter of 1050 mm and a length of 80 mm. According to the graph (Fig. 18.30), we determine the correction factor KL for the length of the radiator:

To find the total correction, we multiply all the correction factors:

Subject to amendments, the radiator should provide thermal resistance:

Using the graph (Fig. 18.27), we find that one transistor requires a radiator with a cross-sectional perimeter of 200 mm. For a group of 20 IRFP250N transistors, the radiator must have a cross-sectional perimeter of at least 4000 mm. Since the available radiators have a perimeter of 1050 mm, 4 radiators will have to be combined.

Less power is dissipated on the ERST diode, but for design reasons, a similar radiator can be used for it.

Often, cooler manufacturers indicate the surface area of ​​\uXNUMXb\uXNUMXbthe radiator, and not the perimeter and length.

To obtain the radiator area from the proposed method, it is enough to multiply the radiator length by its perimeter SP = 400 x 8 = 3200 cm2.

Author: Koryakin-Chernyak S.L.

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Comments on the article:

a guest
But what if the thermal resistance Rsa is negative?

Zufar
Hello. Thanks a lot for the helpful article. A small question: what does the data on the radiator mean, for example 13 inches * degrees / Watt (for a HS 107-100 radiator)? zufarakhmetvaliev@gmail.com

Sergeij
Class!

Mikola
Available - the basics of calculation and the most important parameters. [roll]


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