0
\$\begingroup\$

For a voltage regulator without a heatsink, the temperature change ΔT is calculated by using the electrical power P by the regulator which is basically roughly P=Iin*(Vin-Vout) and the total thermal resistance θ_total. For example a data-sheet for a linear regulator mentions:

Thermal resistance of the TO-220 package (T) is typically 4°C/W junction to case and 50°C/W case to ambient.

So one can write down the following:

θ_JC = 4

θ_CA = 50

θ_total_no_heatsink = 54

If I'm not wrong, as you see above the total thermal resistance θ_total_no_heatsink is easily found from the data-sheet.

But what happens to the total thermal resistance when the regulator is used with a heatsink with a thermal resistance θ_H°C/W? Would the total thermal resistance be θ_total = θ_JC+θ_H or only θ_H? And why? There are examples they take into account resistance from the case to the heatsink ect. So I'm not sure if in the case of heatsink what should be added to the thermal resistance of the heatsink itself.

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

You should add some value to all of the following thermal resistances:

  • junction to case
  • case to heatsink
  • heatsink to ambient

The total thermal resistance is the sum of the stack.

The first is easily found from the component datasheet. The 3rd is found from heatsink's datasheet. The second depends on two different products and the used assembly method. It's presented in application notes or found by making measurements. An example of application note:

https://www.fairchildsemi.com/application-notes/AN/AN-4166.pdf

Thermal resistance from case to heatsink is increased if there's some insulation between the parts. The insulators usually have their own specified thermal resistance, which should be added to the stack.

Rth from case to heatsink can be substantially reduced with thermal conducting grease. There are also thermal contact plates which do the job of the grease and are not intended to be insulators. See the linked application note!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.