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I built a simple voltage regulator circuit like this:

enter image description here

I'm using an LM723 but any op-amp will do. When I shorted the output, the MOSFET was destroyed instantly but I don't know why.

I have tested many MOSFETs from cheap Chinese ones to a real IRFZ48. They were all destroyed the same way. They were all power MOSFETs able to handle continuous current more than 5A and very high peak current as I have tested. I don't understand how a short circuit can instantly destroy it, even when the 25 V supply instantly changes to 1 A constant current and drops to a very low voltage. When I un-short it, the 3 pins of the MOSFETs are shorted, thus dead.

Some quirks though:

  1. The MOSFETs is only destroyed when I short it while outputting a low voltage like 2 V, they aren't destroyed when I adjust the output to 20 V then short it. Just the slightest touch from 2 V to ground and they are dead, and I can't even see any sparks. After that, the output is stuck at 25 V.

  2. I tried using an IGBT, and it didn't get damaged no matter how I short it. I just don't get why it didn't while the power MOSFETs did. I don't have any power BJTs so I couldn't test those.

  3. I tried using a P-channel one with 25 V to source and controlling it through an NPN transistor, it was also instantly destroyed when shorted, no matter what voltage I short it at.

I don't believe the output capacitor from my SMPS can instantly put out more than 210 A when shorted so it shouldn't exceed any peak current. VGS has a Zener diode so it also shouldn't exceed the maximum rating. I can't seem to find anything that would exceed maximum ratings or break the MOSFETs, so where did it go wrong?

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  • \$\begingroup\$ Your mosfet is not supposed to be used as a linear pass device. You have exceeded it's SOA (safe operating area) which kills it even though you are nowhere near its rating of 210A. \$\endgroup\$
    – Kartman
    Commented Nov 11, 2021 at 10:46

3 Answers 3

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Your MOSFET is a source follower and, when you short it out there is 25 volts between drain and source and several amps flowing (maybe 5 amps). That's a power of roughly 125 watts. Here's the safe operating area for the IRFZ48N from its data sheet: -

enter image description here

On the above graph in red I've drawn a line at 25 volts (drain source voltage) and it crosses the 10 ms duration graph at 3.1 amps. In other words, this device is going to have problems with even a couple of amps and will fail certainly if left shorted out for any reasonable length of time at even very modest currents (less than 1 amp).

The IRFZ48N is intended for switching applications and not linear applications hence why the safe operating area graph doesn't consider time durations greater than 10 ms.

The MOSFETs only gets killed when I short it while outputting a low voltage like 2V, they don't get killed when I adjust the output to 20V

Yes, this is a more likely scenario. When outputting only 2 volts, the gate voltage might be at 4 or 5 volts in order to control the output level to 2 volts. So, as soon as you apply a short, the MOSFET is instantly operating in it's linear region and then, U1 is trying to raise the gate voltage higher but it can't do it quickly enough to avoid the catastrophe of MOSFET failure.

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  • \$\begingroup\$ When you say that it doesn't consider time durations greater than 10ms, do you mean that even with say 20Vdrop and 2A of instant power on the MOSFET that should be within the boundary of the graph, I will still kill it? If U1 can't keep up, wouldn't that mean it would limit the current and thus not kill it? If it did keep up, it would allow more current and wouldn't that kill it instead? How would keeping up avoid the failure? \$\endgroup\$
    – Smh
    Commented Nov 11, 2021 at 11:13
  • \$\begingroup\$ If the gate drive circuit instantly (and I mean in less than 1 microsecond) responded and applied a bigger voltage at the gate, the MOSFET would not be in its linear region and it would survive. With 20 volts dropped at 2 amps the device (I estimate) would survive a few tens of milliseconds. But, it's hard to make that judgement call because this device is not intended for linear apps and the data sheet does not cover those situations. Page 1 has the clue: Fast Switching - this makes it suitable for switching applications and not linear applications. \$\endgroup\$
    – Andy aka
    Commented Nov 11, 2021 at 11:19
  • \$\begingroup\$ Not until the gate-source voltage has risen above about 6 volts (see figure 3) will the temperature coefficient of the device be such that it self-protects. 6 volts is the area on the graph where the two characteristics cross. \$\endgroup\$
    – Andy aka
    Commented Nov 11, 2021 at 11:22
  • \$\begingroup\$ Even if Vgs would rise, Vds will still be kept at more than 20V and have much current flowing through it, isn't that still in the danger zone? Unless responding faster means to have the MOSFET pull 25V into constant current thus lowering Vds faster. Tens of milliseconds is very short, are there any external circuits I can add to detect and prevent the MOSFET from getting damaged in case of accidental shorts? Can you recommend some easy to find, generic power transistors that are suitable for these kinds of applications? (Higher power ones like this IRFZ48N would be better) \$\endgroup\$
    – Smh
    Commented Nov 11, 2021 at 12:09
  • \$\begingroup\$ Yes, but, once Vgs has risen past 6 volts the temperature coefficient of the MOSFET will self-protect it in a lot of circumstances. While ever Vgs is below 6 volts, thermal runaway will happen in microseconds (try searching this site for Spirito effect or try this: electronics.stackexchange.com/questions/472375/…). There are no external circuits that can protect against this. Try IXYS corporation for suitable linear MOSFETs. \$\endgroup\$
    – Andy aka
    Commented Nov 11, 2021 at 12:13
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With lower voltage you drop more voltage on the FET -> more power with the same current.

Practical examples: If you'd set the output to 25 V, there would be hardly any power dissipation on the FET, as ideally the FET would then be completely open and the power dissipation would be only I x R_dson -> 1 A x 0.014 ohm = 0.014 W.

With 20 V output the power would be I x voltage drop over the FET -> 1 A x (25-20) V = 5 W.

Edit, case added: With 2 V output the power would be I x voltage drop over the FET -> 1 A x (25-2) V = 23 W.

Junction to ambient thermal resistance on the FET is 62 degrees per celcius. So even 5 W is too much without a heat sink.

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  • \$\begingroup\$ It didn't break at 20V, but at 2V, and I only lightly swiped it across the ground, not even 0.5 seconds of contact I bet. \$\endgroup\$
    – Smh
    Commented Nov 11, 2021 at 10:59
  • \$\begingroup\$ I added a case for 2 V. With lower voltage the effect is even more drastic. If you would hold it in short circuit the chip would reach 1400 C, if nothing would fail. In practice it will fail far before reaching the end temperature. It doesn't take a long to heat the mass of the IC. \$\endgroup\$
    – Ralph
    Commented Nov 11, 2021 at 11:19
  • \$\begingroup\$ If the output voltage actually drops under 2 V or whatever it's adjusted to, then surely the voltage drop over the FET should drop and therefore decrease power dissipation in it. The datasheet didn't completely describe usage of C2 in your circuit. Maybe that causes intentional delay in the feedback loop. \$\endgroup\$
    – Ralph
    Commented Nov 11, 2021 at 11:31
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When you output 2V that means 23V (25 - 2 = 23) are dropped on the MOSFET itself. Running 1A through the MOSFET would mean you're dissipating 23 watts of heat through the MOSFET's body.

When you output 20V that means 5V (25 - 20 = 5) are dropped on the MOSFET itself. Running 1A through the MOSFET would mean you're dissipating 5 watts of heat through the MOSFET's body.

You are effectively using the MOSFET as a big variable resistor to dissipate the excess power as heat.

In both cases the MOSFET can only survive if the dissipated heat is removed fast enough to keep its temperature in the safe operating range. The MOSFET can survive without a heatsink if it has a junction-to-case thermal-resistance low enough to not exceed the maximum operating temperature.

This applies to any type of series pass transistor being used; BJT, FET or IGBT.

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  • \$\begingroup\$ This doesn't make sense as just a slight touch and I mean literally removing the contact to ground as fast as I can by just swiping it across the leads, still rendered the MOSFET dead, it was instant. And it doesn't even feel hot, I don't think there is a "removed fast enough" for this. \$\endgroup\$
    – Smh
    Commented Nov 11, 2021 at 16:26
  • \$\begingroup\$ Then I would suggest learning to walk before we can run. How about you build an adjustable constant current load like this link and see how your test devices react in a steady state at various load currents and output voltages, then we can move one to testing transients? \$\endgroup\$ Commented Nov 11, 2021 at 16:41
  • \$\begingroup\$ See this answer \$\endgroup\$ Commented Nov 11, 2021 at 16:54

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