Electronic – Heatsink calculations for a TRIAC

coolingheatsinktriac

It's very rare that I get to design applications in need of cooling so I haven't had the need to study in-depth about that, but now I face an occation where I need cooling for a TO220 and I'm unsure how to calculate what I need.

TRIAC BTA08-600, data sheet: https://www.tme.eu/en/Document/229a1f149ceabe983b1ecaf8c207519b/BTA08-600BRG-STMicroelectronics.pdf

Page 4 tells me the power dissipation relative amps. In my circuit 3A is worst case, but regarding cooling I will calculate for 4A just for the sake of margins. The table tells me 4A => 4W.

Page 3 then tells me Rth(j-a) for TO220 is 60 degrees C per Watt. If I understand this parameter correctly, it would mean that for 4A times 60 degrees, it would go up to 240 degrees C in total. Then, page 2, tells me under "Absolute maximum rating" that Tj is +125 degrees.

If I'm guessing correct here it would mean I would have to use a heatsink that can cool down 240-125 = 115 degrees C.

Question 1: Is this assumption correct?

Question 2: How do I find a TO220-based heatsink that's able to cool 115 degrees? What parameters should I peek on / search for?

Best Answer

If you are using a heatsink the most important figure is \$R_{th(j-c)}\$ and that value is about 2 degC per watt. This means that if you attach your device to an infinite heatsink it will warm 2 degC per watt i.e. it will become 8 degC warmer for a continual dissipation of 4 watts.

However, you won't find such a perfect heatsink but you might find one that is 6 degC/watt (for example) and you add that figure to the 2 degC/watt figure to give you 8 degC/watt and now, your device will warm up by 32 degC in whatever ambient temperature you have.

However, if the heat isn't removed by some moderate air flow, the ambient temperature around the heatsink will rise maybe 10 or 20 degC. You need to be aware of this and take measurements on the final design but, for now I'll assume ambient will warm by 20 degC.

So if ambient around heatsink is (25 + 20) degC then the device's junction temperature will rise to 77 degC.

There is a slight bonus; \$R_{th(j-a)}\$ is a parallel cooling path from junction to ambient so that figure of 60 degC per watt lowers the 8 degC per watt down to 7.05 degC per watt (resistances in parallel).