In the following circuit how does the zener diode D1 provide reverse polarity protection?

enter image description here

Lets say V+=12V and V-=-12V and the zener is 36V zener.

It means for correct polarity zener will not pass any current since it sees 24V across itself.

But in case of reverse polarity, the zener will short the opamp maybe protecting it, but wouldn't that also mean it will short the power supply? Is this a good reverse polarity protection?

  • 1
    \$\begingroup\$ You might want to read about crowbar circuits. If the zener blows up, you are left with no protection, if it shorts, you are left with a short. Either way not ideal. \$\endgroup\$
    – PlasmaHH
    Jan 29, 2018 at 12:59
  • \$\begingroup\$ Without context this circuit cannot be described as good or bad. \$\endgroup\$
    – Andy aka
    Jan 29, 2018 at 13:13

1 Answer 1


Protection relies on two things:

  • the voltage must not exceed a certain value

  • the current must not exceed a certain value

This is valid for all components including the protection devices themselves !

In your circuit the Zener diode will go into forward mode when the supply is reversed. For a BZX55 Zener diode I just looked up the forward voltage, it is 1.5 V at 200 mA. But the opamp is an IC and it is bound to have ESD protection diodes on each pin, it is a circuit like this:

enter image description here

Those diodes (present on each pin!) will also go into forward mode. Since they're Silicon diodes they have a forward voltage of around 1 V to 1.5 V (for two in series). So the ESD diodes might even start to conduct before the Zener kicks in! So it might be that the Zener doesn't to anything, all the reverse supply current goes through the opamp! Ouch!

Also, if you do not limit that current to a sane value, like 500 mA, preferably a lot less, you will destroy the opamp.


  • the zener might not do anything because of the ESD protection diodes.

  • Without current limiting on the reversed supply voltage, something will blow eventually. My guess: the opamp goes first, if it fails open then the zener will be destroyed.

  • \$\begingroup\$ Does that also mean if a signal is applied to the opamp's input pin when the power supply is off, those diodes will be an issue? \$\endgroup\$
    – user16307
    Jan 29, 2018 at 14:12
  • \$\begingroup\$ Yes it does / it can be. You can supply power to the opamp through the signal pins via the ESD diodes! See this EEVBlog video for an explanation: youtube.com/watch?v=2yFh7Vv0Paw&t=6s there Dave demonstrates the effect using a uC but it will work the same for an opamp or any IC really. \$\endgroup\$ Jan 29, 2018 at 14:19
  • \$\begingroup\$ Omg so better to turn of the function generator for the input signal before the power supply. I never did that until now. \$\endgroup\$
    – user16307
    Jan 29, 2018 at 14:23
  • \$\begingroup\$ It does not have to be an issue / cause any damage as long as you limit the current somehow. Function gens. have 50 ohm output impedance so that helps a little. To be 100% safe use a 1 kohm series resistor on the signal inputs if possible (if noise is not an issue and the input is high impedance). I think only a really large signal from the generator could damage the opamp. \$\endgroup\$ Jan 29, 2018 at 14:30

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