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Hi There!

I will be using a high power TRIAC BTA41 to turn on/off a 1 HP motor.  This TRIAC is capable of handling the load up to 40 amps.  As per my calculation load current will be around 5 amps and watts will be 1100 (5amps * 220VAC).  What I would like to know is, do I need a heat sink attached to this TRIAC to handle this load?  If yes, how to calculate the dimensions of the heat sink.  Datasheet of BTA41 is attached. Thanks in advance!


* BTA41.pdf (102.95 KB - downloaded 30 times.)
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For power calculations see this:-
http://www.thebox.myzen.co.uk/Tutorial/Power.html
However you need to know how much power the Triac will dissipate when it is on. Form a quick look at the data sheet I can't see the voltage between the two anodes when it is on. You need to multiply this voltage by the 5A current to get the amount of heat you will generate. The thermal resistance of the case is given so it should be plane sailing after that.

OK figure 1 will give you that power information, for 5A it looks like you are generating 5W of heat.
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Thanks for your quick response! In the data sheet Junction to case thermal resistance mentioned as 0.9 degrees C/W.  Considering its generating 5W, can I assume the heat it will produce is 0.9 * 5,  i.e 4.5 degrees centigrade?   is it correct? or I have to multiply with junction to ambient thermal resistance i.e, 50 degrees C/W? in this case the heat its going to produce is 250 degrees centigrade.
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The equation you need is:

Tj = Ta + (P * (Rjc + Rch + Rha))

where Tj is the junction temperature, Ta is the ambient temperature, P is the power dissipated in the device (which according to Mike is 5W), Rjc is the thermal resistance junction to case (0.9 degC/W from your triac datasheet), Rca is the thermal resistance case to heatsink, and Rha is the thermal resistance heatsink to ambient. The datasheet for a commercial heatsink will give you the value of Rha. You need to choose a heatsink with a low enough Rha such that Tj is below the device limit. See http://www.jaycar.com.au/images_uploaded/heatsink.pdf for more.
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The data sheet says junction to ambient is 50 degrees C per watt, so yes if the ambient was at 0 degrees C then the junction would be at 250C, which is too hot.
Therefore you need to reduce this figure by adding an external heat sink.
That link shows you how to do this as does the equation that dc42 posted. You can think of thermal resistances like electrical resistances and the thermal gradient (difference in temperature between the junction and the ambient) as the voltage across the resistance.
So to have a junction temperature of 100C with an ambient of 30C you have a 70C thermal gradient. So the total thermal resistance needs to be less than 70C / 5W = 14 degrees C per watt. You already have an inbuilt thermal resistance of 0.9 (lets call it 1), so your heat sink needs to be 14 -1 = 13 degrees C per watt.
So look for a heat sink with that resistance or smaller.
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Thank you dc42 and Mike for your detailed explanation.  It's a great learning.  I have chosen a 75°C @ 15W (natural convection) heat sink for my application. I believe its bigger than what is needed, but I feel its no harm.
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