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I'm using several MOSFETs to control various loads, e.g. pumps, lighting, ventilation etc.

In the implementation, I have several driver circuits on one board (simple emitter follower, which gives good switching speed with a gate voltage of ~5V; using pull-ups were too slow/inefficient), with jumper cables to a board with several MOSFET circuits. Each circuit is like the schematic.

While testing, I touched the signal cable and it caused the MOSFET to turn on, but it got extremely hot in just half a second of on-time.

Now I'm worried that a small failure in the driver circuit, e.g. if MCU is disconnected and the signal wire somehow picks up a small voltage, it will be amplified into the MOSFET gate, causing a meltdown in just a short period of time. Could this start a fire? Or will the device just fail closed?

Ideally the circuit would not even allow the MOSFET to operate in high resistance part of the linear mode. Can I improve my circuit to avoid gate voltages less than, for example, 4V? Is there a way that I can detect/prevent overheating (besides a complicated setup of sensors).

schematic

simulate this circuit – Schematic created using CircuitLab

UPDATE: Pull-down on signal Adding a 10k pull-down on the base of the BJTs caused the noise to no longer turn on the MOSFET.

QUESTION: However, it didn't really answer the core of my question, which is more that I'm worried about situations where a MOSFET may turn on ever so slightly, yet able to conduct, causing extreme heat and consequently potentially starting a fire. I can't use a fuse, since the problem is not overcurrent. Perhaps a thermal fuse mounted to the MOSFET, but that seems rather unorthodox. Is this a non-problem that I'm overthinking?

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  • \$\begingroup\$ If it is off, the power dissipated in the MOSFET is about zero. If it is fully on, the power dissipation in the MOSFET is about zero. When the MOSFET is half-on, in linear mode, the power dissipation is at least 25 % pu and this is why it will blow the switch. (During switching the power might be up to 100 % pu or more, but only for a short time.) \$\endgroup\$ – skvery Feb 22 '17 at 19:52
  • \$\begingroup\$ On your follow-up: if heat/fire safety is of paramount importance, then look into fault-protected "intelligent" MOSFETS like the NID5004N (of course, a lot of other options from major manufacturers exist). \$\endgroup\$ – anrieff Feb 22 '17 at 23:25
  • \$\begingroup\$ It is very simple. When the gate-to-source voltage of a MOSFET is zero, it will be off. So just make sure something is pulling it low during those times you need it to be off. Your drive circuit was not doing that, apparently. But the pulldown fixed it. \$\endgroup\$ – mkeith Feb 23 '17 at 16:35
  • \$\begingroup\$ @mkeith, The problem was that signal noise was being amplified just enough to turn on the NMOS just enough to conduct but with such high resistance that it was causing extreme dissipation. \$\endgroup\$ – user95482301 Feb 23 '17 at 19:33
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A simple, modular, and robust way to get better behavior from a circuit when the input is in an invalid state is to place a Schmitt trigger between the exposed input and the driver. A Schmitt trigger is (almost) guaranteed to produce a logic high or logic low even if the input is at an invalid intermediate level, weak, or floating.

  • Note that the integrated MOSFET driver that anrieff's answer mentioned has an internal Schmitt trigger in its input logic (denoted by the ⎎ symbol).

  • If your existing circuit is adequate at the actual job of of driving the MOSFET when the input is 0 V or 5 V as intended, then you can add a Schmitt trigger in the form of an inverter, buffer, or other logic gate that has a Schmitt trigger input, preceding your driving circuit.

  • As a further example of their use, Schmitt triggers are also commonly used in microcontroller inputs to protect the internal logic — ordinary logic gates can misbehave, and even draw excessive current and overheat, when the input is invalid, whereas a Schmitt trigger turns an invalid signal into some valid signal.

Another hazard specifically for driving MOSFETs (or any power transistor) that is not solved just by using a Schmitt trigger is if the input is is commanding the driver to switch on and off rapidly (whether this is deliberate, as in PWM drive, or accidental noise). Then the MOSFET will spend more of its time in the middle of switching, and so dissipate more power. To reduce this you must either avoid such frequent transitions reaching the driver (probably the better option, if needed at all, in your application since your loads might also be unhappy) or ensure that the driver switches very fast (low rise/fall time).

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  • \$\begingroup\$ Great answer, thanks. This also justifies better the use of a dedicated driver circuit. I was feeling that they were a bit expensive when I could achieve 'the same' with a simple emitter follower. \$\endgroup\$ – user95482301 Feb 22 '17 at 22:55
  • \$\begingroup\$ @user95482301 Glad it was helpful. I've reorganized and expanded my answer a bit — please let me know if I lost anything that you liked about the original. \$\endgroup\$ – Kevin Reid Feb 23 '17 at 15:28
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You can use dedicated MOSFET drivers like the TC4424. You can use those instead of your Q1/Q2 pair. There are MOSFETs with integrated temperature protection as well.

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I am not sure your FET went in linear mode, it could also have been oscillating. This can easily happen when the gate drive has inductance (ie, a 10cm wire).

I would advise a 50 ohm gate resistor to avoid this. For your next board, it is better to put the drivers right next to the MOSFETs.

For low frequency switching, high gate drive current is not necessary as switching losses will be very low anyway. Slow switching (1µs, not 10ns) is better, as it will radiate less EMI.

If your FETs are logic level, you can use 74HC logic gates as drivers also.

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I believe something as simple a pull-down resistor at the BJT will resolve this issue by providing a well define voltage level even when the MCU pin will go to high impedance or the cable goes open or disconnected. I suggest trying like a 10K pull-down resistor and trying it out before attempting any more complex solutions.

To answer your question is that keeping the MOSFET from operating in the linear region is the job of your driving circuitry. As an engineer to must design and test your MOSFET driving circuitry to operate the MOSFET out of the linear region as much as possible while juggling all other trade offs. In your case you found a fault with driving circuitry because under some circumstances its input was not in a well defined state. If you were looking into making your device safer then you can start looking at thermal limit switches, temp sensor, more sofisticaed devices that have inherit thermal protection built-in etc. Maybe even something as simple as proper heating sinking using the PCB copper is good enough to keep the device cool even in the linear region. All depends on your application and its requirements

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  • \$\begingroup\$ The 10k pull-down solved the problem :) \$\endgroup\$ – user95482301 Feb 22 '17 at 19:12

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