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Consider the above circuit. The N-channel MOSFET has been used as a low side switch to provide reverse polarity protection. A capacitor charges to the 12V supply(battery) voltage and subsequently the supply terminals are reversed. Since the capacitor will hold its charge for a certain period of time, it is therefore able to keep the MOSFET Vgs well above the minimum threshold voltage of 2.8V. As such on reversal, the capacitor discharges and settles its voltage to 2.8V after which the MOSFET turns off completely thereby disconnecting the circuit.

Well the above case seems to be impractical, but consider a situation where a startup delay of few seconds is given, the user might unknowingly reverse the polarity. As circuits contain innumerable capacitor and magnetic components that will be able to keep the mosfet on till their energy collapses in the biasing and other circuit loads, for that period the circuit seems to be in conduction mode even under reverse polarity.

Can anyone explain what will be the effect on vulnerable components that are a part of such a circuit? Also is there any way to mitigate this effect?

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  • \$\begingroup\$ If the user applies, say, -12 V to the 12 V terminal, their source will have to discharge the capacitor before it can actually supply -12 V to the protected circuit. There might be a few ns delay for the carriers in the channel to clear out after the gate voltage is reversed (and also reverse-recovery time of the body diode to consider). \$\endgroup\$
    – The Photon
    Apr 4, 2016 at 16:55
  • \$\begingroup\$ Can you explain in detail. Is there any current actually flowing or its that the capacitor is getting sudden discharge because the positive charge that collected previously on its plate now face the negative charge, as in the case where a capacitor discharges when we short its terminals. I'm totally lost. \$\endgroup\$
    – Akash
    Apr 4, 2016 at 17:09
  • \$\begingroup\$ The positive terminal of the capacitor is connected to the gate of the FET. So you can't change the voltage of the FET gate without changing the voltage of the capacitor at the same time. Or vice versa. \$\endgroup\$
    – The Photon
    Apr 4, 2016 at 17:13

1 Answer 1

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Yes, it's possible the surge could damage something (perhaps blow a fuse). The current could be quite high if the MOSFET is large and the source impedance low, perhaps hundreds of A, but not likely with only 1000uF. An ordinary 1000uF cap might have an ESR of a few ohms, so the surge would be maybe 10A for a few ms, no big deal most likely. Typically your load will bleed the voltage off quickly anyway. Bad layout might allow some transient negative voltage into your circuit, so don't do that.

Also, it is unlikely (but not impossible) that a user would reverse from a working to a non-working configuration.

Never underestimate the ability of users to come up with innovative ways to break things.

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  • \$\begingroup\$ hahahaha....especially when you are dealing with the customers of rural market.... They can do anyyyything... \$\endgroup\$
    – Akash
    Apr 4, 2016 at 17:16
  • \$\begingroup\$ 1000uf was just to demonstrate the problem..... In my circuit it takes 1.7s before voltage reduces to the Mosfet's off state. I thought that its a short period, but when I tried it myself, I could actually do that in 700ms with ease. I dont want to use a schotkky, can you suggest me any other method or should I take the risk. \$\endgroup\$
    – Akash
    Apr 4, 2016 at 17:23
  • \$\begingroup\$ You can estimate the surge and confirm it by measurement. You could degrade the cap ESR with a resistor or use a smaller MOSFET, or add a current limiter (BJT + sense resistor). All the add-on things add a bit of cost and reduce reliability a bit so it's a trade-off. \$\endgroup\$
    – Spehro Pefhany
    Apr 4, 2016 at 17:32
  • \$\begingroup\$ I'm more concerned about the chips that I'm using in the circuit for performing other secondary functions. None of them have reverse polarity protection. Maybe I should find a way to quickly discharge the MOSFET charge. Also it becomes important because there is one more possibility if the input leads are put to a state of short rather reverse polarity. \$\endgroup\$
    – Akash
    Apr 4, 2016 at 17:44
  • \$\begingroup\$ Reverse polarity shouldn't happen unless your layout is bad. \$\endgroup\$
    – Spehro Pefhany
    Apr 4, 2016 at 17:58

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