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# Problem with Formulae for Heatsink with Rectifier

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#### Wylbur

##### Newbie level 1
First thing: I have been searching trying to find what I am doing wrong. And I have been out of electronics for 20+ years where I had to do math on a daily basis (I use to work on all tube type equipment, hardly ever solid state).

I have built a test circuit and my results (in the area of heat) are just not matching my calculations (2A draw doesn't raise the temperature enough for a heat probe to detect).

I have a spec sheet on a 40A 50VDC diode/rectifier (NTE5980) which is the diode I am working with.

What I am working on is a circuit for an automotive application where I must assume a max of 15VDC and the alternator has a max output of 63A (rated).

W=I**2xR tells me that I must dissipate 11200 watts if the circuit is drawing 40A.

This just failed the reasonableness test. I say that because, 15*40 (V*A) gives 600 watts. Something wrong with conservation of energy from where I sit.

So I have to have something wrong in the print of the spec sheet (the junction resistance is supposed to be 7 ohms). Or, more likely, I'm just missing something here.

I also seem to have a problem with the spec sheet because I can't seem to be able to determine the junction temperature (Tj) delta so I know how much heat I have to dissipate at max current.

And I say this because I have formulae for these things (all nicely built into a spread sheet), but I can't seem to determine how much heat I actually have to dissipate (even with a test circuit to see what heat I'm getting). So I can't compute the right size for a heatsink.

Could someone give me a pointer or two on what I'm missing or doing wrong?

Regards,
Steve.T

You speek in a very confusing fashion. Are you trying to say you have a junction diode that has a 7 ohm series resistance that is passing 63 amps of DC current? Your DC generator would have to have over 441 volts DC to achieve that.

Maybe you are confusing load power with the diode dissipation. It is also wrong to specify the resistance of a diode, it's a dynamic characteristic and in any case 7Ω seems massively too high for a power rectifier.

The diode dissipates power according to the voltage dropped across it and the current through it. In most cases this will be I x Vf where Vf is the forward voltage drop. Your diode is specified to drop 1.3V at 40A, it will be less voltage if the current is lower. So if you actually passed a constant 40A, it would dissipate 52W.

Incidentally, the resistance at 40A works out to be 0.0325Ω not 7Ω !

Brian.

Wylbur said:
So I have to have something wrong in the print of the spec sheet (the junction resistance is supposed to be 7 ohms).

The only reference to junction resistance in the NTE5980 datasheet available here, mentions the maximum forward slope resistance as being 3.79 milliohms, which is a long long way away from the 7 ohms mentioned in whichever version of the datasheet you're looking at.

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