Electronic – Zener + MOSFET overvoltage protection

mosfetprotectionzener

I need a simple circuit to protect my device from overvoltage (more than 13V). Typical working DC input voltage is 7-12V. Can I use this circuit:

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Will it work?

UPD. OK, it will not. I was too stupid. Yet another try:

enter image description here

(Sorry, I forget to change default names. My MOSFET is IRLML6402, transistors is BC817, zener 13V.)

When input < 13V, Q2 closed, Q5 open, MOSFET open. Else Q2 open, Q5 close, MOSFET close. OK?

Best Answer

Your circuits (whether they work as you intend or not) seem to be based on the idea that when an over-voltage condition occurs, you will isolate the load with a MOSFET.

I think this is not a great approach. Consider, why did over-voltage occur anyway? You could say that it's because the impedance of the load is too high. Were the load impedance lower, then more current would flow, providing a way to sink whatever excess energy there is without the voltage becoming too high.

So, if you try to isolate the load with a MOSFET, that just effectively increases the load impedance. For a great many things that might have led to your over-voltage fault in the first place, they will simply respond by increasing voltage even more. This is problematic, because MOSFETs have relatively low maximum voltages. When a source-drain voltage is applied in excess of this, they go into breakdown and conduct (not unlike reverse breakdown of a diode). Consequently your load gets toasted anyway.

A different approach is a crowbar circuit. Instead of disconnecting the load, you short it out (usually with a TRIAC or SCR). If you put a short across the load, the voltage will be reduced (to 0V, for an ideal short). This provides very effective protection to the load from over-voltage.

Of course this also results in huge current drawn from the power supply. That blows the fuse, which isolates the power supply from the circuit, and does so much more effectively than a MOSFET can. Furthermore, the crowbar continues to conduct until there is no more energy which can result in a voltage across the load, which might take a little time, depending on just what caused the fault, and what reactive components are in the circuit.