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