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If you use transistors for an amplifier scheme, you want to make sure that when you parallel devices, you don't start to conduct more current if your transistors heat up.

I know that when using MOSFETs you always have negative feedback. If one takes more current, it will heat up more, and therefore the resistance will become bigger, and you will conduct less.

When using a BJT the device that dissipates the most current, will heat up most and will start conducting even more. This is unsafe and to solve this I always place a small resistance in series with the emitter.

But what is the physical explanation that in a MOSFET you have this positive temperature coefficient and in a BJT you have this negative coefficient?

Peter Mortensen
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J. Joly
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  • Old style power mosfets were safe; the automotive industry and the satellite industry(JPL) discovered the new-style mosfets are not safe. NASA has some papers out on this. – analogsystemsrf Dec 28 '18 at 16:46

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I do not know the answer to your question about physic behind this.

But your statement about MOSFET temperature coefficient is only true when MOSFET work as a switch.

In linear region (saturation) most MOSFET will have as a negative temperature coefficient as BJT have.

And in general terms, the MOSFET can have negative, positive and zero temperature coefficient. And we can see this on Transfer characteristic Id=f(Vgs) in the datasheet.

enter image description here

More here

https://www.infineon.com/dgdl/Infineon-ApplicationNote_Linear_Mode_Operation_Safe_Operation_Diagram_MOSFETs-AN-v01_00-EN.pdf?fileId=db3a30433e30e4bf013e3646e9381200

https://www.onsemi.com/pub/Collateral/AND8199-D.PDF

And here I found a specially designed MOSFET for audio amplifiers application.

http://www.exicon.info/PDFs/ecw20n20-z.pdf

enter image description here

As you can see for this lateral MOSFET if you set the quiescent current at Id > 0.2A you are in the safe zone (zero or positive temperature coefficient) And no temperature compensation network is needed.

G36
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  • okay good sidenote that I didn't know. Thanks, I'll keep this in mind! I was talking about an amplifier circuit, so as you mentioned I was using the MOSFET as a switch. But still have to keep your comment in mind! Thanks! – J. Joly Dec 28 '18 at 09:17
  • Related: https://electronics.stackexchange.com/a/357641/17387 – try-catch-finally Dec 28 '18 at 09:25
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    @J.Joly Are you building a class D amplifier? – G36 Dec 28 '18 at 10:13
  • I'm not building it, I'm trying to analyse it. [MOSFET power amplifier](http://amplifiercircuit.net/mosfet-power-amplifier-5200w-irfp250.html) – J. Joly Dec 28 '18 at 10:25
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    That schematic shows a linear (not switching) amplifier so you are NOT using the MOSFET as a switch. –  Dec 28 '18 at 11:12
  • Okay so for this amplifier it is not applicable, as I've seen they put small resistors at the source, probably for the same reason they do this with BJT, as I mentioned above? Or is this for another reason? As said before, I didn't know the MOSFET had a varying temp coefficient, this schematic is pointed out in my slides when we discuss the facts about temperature coefficient. But Thank you for pointing out that it is a linear amplifier. – J. Joly Dec 28 '18 at 12:49
  • The schematic you have given will do not work. So do not even try to analyze it. – G36 Dec 28 '18 at 13:34
  • As for the IRP250 we can see http://www.vishay.com/docs/91212/91212.pdf (fig.3) that we have the negative temperature coefficient for current in the range from 0A to 12A. So, definitely, you will need a "big" resistance in the source. And a temperature compensation circuit to prevent thermal runaway to occur (vbe multiplier will do the job). – G36 Dec 28 '18 at 13:41
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    Also good read about this you can find here http://sound.whsites.net/articles/hexfet.htm#s51 – G36 Dec 28 '18 at 14:03