it is a typical diagram of reverse voltage protection using N channel MOSFET. what one should do is to connect the anode to the source of the MOSFET. But here instead of that I have connected to the negative or drain of the mosfet. The mosfet should trigger since still the gate is positive (V_z) with respect to the source, but also we can see that the zener voltage is appearing at the gate-drain of the mosfet. will that cause any harm or any functional difference? Because since, the drain of the mosfet is isolated from the body, so does it even matter if we apply a voltage between GATE and drain?
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\$\begingroup\$ Why don't you use a p-channel mosfet? Source connected to the positive terminal and gate to GND? \$\endgroup\$– FranMartinApr 12, 2021 at 14:58
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\$\begingroup\$ @mkeith sry that was normal operation, \$\endgroup\$– SayanApr 12, 2021 at 16:37
2 Answers
You have your zener diode connected incorrectly (reason below): -
That's how it should be connected to prevent gate-source over-voltage.
Reason
If you connect it as you originally had it then, when trying to protect against a reverse voltage of 30 volts, the zener would conduct (forward biased) and apply 29.3 volts to the gate. Given the the source would be connected to the negative end of the reversed supply via the load impedance, you could certainly expect damage to the gate-source region. Typical gate-source maximum voltages are around 20 volts.
Your zener diode is only 5 V, so there is a possibility of 25 V between the gate and source while the gate-source capacitance is charging up. That is dangerous.
Also, once the FET is "on" (enhanced), Vgs is only 5 V because of the zener. The 530 is not a logic-level MOSFET, and this type of part usually is driven with at least 10 V to assure minimum Rdson.
Move the diode as in Andy's answer, increase the zener voltage to something in the 9 V - 12 V range, and increase R3 to something in the 1K - 10K range.
For a more rapid and clean turn-off, add a resistor in parallel with the diode.