heat sink calculation for parallel bjt

Question

I'm working on a project and I need to use 3 parallel bjt to give me 20 A current.Total power dissipation is 356 W.I know how to calculate a sink for one transistor,but in this case I'm not sure what is Pdiss in Rja=(Tj-Ta)/Pdiss, is it 356 W or ,as I'm using 3 2SA1302 bjt,356/3 W.Also is Tj (junction temperature) 150 C (for one 2SA1302) or is it 3*150 C ?

Answer 1

The situation is the same as if you had a single transistor with a Tjmax of 150C dissipating 360 watts.

Answer 2

If you want to make a single sink for the three transistors Pdiss will be 356W because it is the power you need to dissipate in the sink. Each BJT is dissipating 356W/3 but the sink has to dissipate the total temperature. With this amount of power I think the sink will be quite huge.

Another thing you can do is designing one sink by BJT, then each sink will have to dissipate 356/3 and then Pdiss will be 356/3.

For Tj it is a caracteristic of the BJT you can not change it. It means at 150°C your BJT will crack. Then even if you have 3 of them, at 150°C they will crack. Which means your sink should be designed to have Tj<150 and not Tj<3*150

And if you want your circuit to live long enough you need to design your sink to have a Tj around at most 125°C cause the hotter it is the earlier it will break.

Answer 3

You are expecting to get rid of 118 watts in each transistor assuming that your proposed 3 transistors share perfectly .Look at the total thermal impedance from the junction to the heatsink because this is the important parameter .Now if I ballpark 1 degree per watt which is reasonable for normal packages and normal metalwork then things start to get bad.So the junction will be 118 degrees hotter than the heatsink .This means that the heatsink will have to be in Ice for good reliability .This means 118 degrees for Tj .So to make mil461 it would have to be on the north or south pole .This is the hot junction ,cold heatsink syndrome that I have seen so often .The sensible thing to do these days is to use more and more transistors until the heatsink becomes reasonable .I use the term " Silicon Aluminium balance " when I have to teach people this stuff .The heatsink allows you to get more out of the transistor but over the decades the heatsink has got dearer and the transistor has got cheaper.

Answer 4

This must be split into two parts: junction to case (package) and case to ambient (heat sink).

Let's assume (since you haven't stated it) that you will use the same heatsink in both cases, but in one case you will use a single transistor, and in the other you will use 3.

In both cases, the heatsink temperature will be the same, since in both cases the total power being dissipated is the same - 356 watts. Just as an arguing number, let's say you have a big heat sink with a thermal resistance of about 0.1 deg/watt. Then the case temperature of the transistors (assuming a real good thermal mounting) will be about 25 + (0.1 x 356), or 61 C.

Now look at the junction to case temperature drop. The data sheet says that the power must be derated from 150 watts by 1.2 watts per deg C at the case. 356/3 is 118 watts, so the maximum case temperature is about 150 - (118/1.2) or about 52 C. So your heatsink will not work - the transistors will get too hot.

You can do the math to find the required thermal impedance of the heat sink required to keep the transistors cool, and it will obviously be less than 0.1 deg/watt. Just off the top of my head, that's a pretty big heat sink, and may well require forced air cooling.