1
\$\begingroup\$

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?

\$\endgroup\$
  • \$\begingroup\$ 4A times 60 degrees you mean: 4W times 60 degrees \$\endgroup\$ – Oldfart Jan 26 '18 at 14:14
  • \$\begingroup\$ If you're adding a heatsink anyway, design conservatively. The device may survive 125C but it'll have much higher reliability at 60-80C. (I notice Andy's doing that anyway, just wanted to mention why) \$\endgroup\$ – Brian Drummond Jan 26 '18 at 14:23
1
\$\begingroup\$

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).

\$\endgroup\$
  • \$\begingroup\$ "There is a slight bonus; Rth(j−a)Rth(j−a) is a parallel cooling path from junction to ambient " Andy you need to be a little careful with that one, that value assumes an unobstructed package in free air not with a heatsink attached. Also if you are that close to the line it would not be good to rely on this. \$\endgroup\$ – RoyC Jan 26 '18 at 14:29
  • \$\begingroup\$ @RoyC playing safe I wouldn't rely on it for sure. \$\endgroup\$ – Andy aka Jan 26 '18 at 14:34
0
\$\begingroup\$

Rth(j-a) is the thermal resistance you need to take only if you expect the triac to dissipate all the thermal power directly to air. Since with a heatsink the heatsink will be mostly in contact with air instead of your triac, this is not what you should be looking at.

First the triac's junction needs to transfer heat to the case, sometimes datasheets call this the mounting base. So you look at Rth(j-c) or Rth(j-mb). Then this needs to be transferred over to the heatsink, this is represented by Rth(c-hs). This will be a parameter used by your thermal grease, for example. Then the last thermal resistance value is the Rth(hs-a) (heatsink-to-ambient), which is a property of the heatsink.

So instead of Rth(j-a), your total thermal resistance from the triac to fresh air is Rth(j-c)+Rth(c-hs)+Rth(hs-a), which, multiplied by the thermal power you need to dissipate, must not be more than 125degrees minus the ambient temperature.

\$125 >= (R_{th(j-c)}+R_{th(c-hs)}+R_{th(hs-a)})*P + T_a\$

Now you solve the equation for Rth(hs-a) and you know what heatsink to choose :)

EDIT: some additional info:

  • note that Rth(j-c) is much smaller than Rth(j-a), so it is not as impossible to find a heatsink as it looks (unless ofc you have too much power to dissipate)

  • Rth(hs-a) is a property of the heatsink, but it is not constant. Most importantly it depends on air movement and less obviously, also on the temperature difference between the heatsink's surface and air. So Rth(hs-a) is larger if the surrounding air is hot. For this reason, good heatsink datasheets specify Rth(hs-a) not with a single number, but with one or more curves.

  • For general info on estimating thermal characteristics of a TRIAC, have look at application note AN10384 by Philips Semiconductors.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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