zener
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?
The current from the relay opening doesn't go into the Vcc rail at all. It follows the path shown here:
The stored energy is dissipated in the diode drop and the coil resistance of the relay.
In the Zener diode configuration, the stored energy is dissipated in the full Zener voltage of the diode. V*I is a lot higher power, so the current will fall faster and the relay might open a little faster:
MOVs are different than Zeners, but fulfill a similar circuit function: They absorb energy when the voltage exceeds a certain level. They are used for overvoltage protection, not for precision things like voltage regulators.
First, why will the zener fry? You may have a 12V supply powering your regulator, but your supply must have a fuse or circuit breaker somewhere to be safe. If this fuse or circuit breaker has a current rating which the zener can tolerate, then your PCB should be protected by the zener while the fuse shuts off, and no harm will come to the zener. If the current rating is higher than the zener can tolerate, well, you've made a mistake in choosing a current limiter.
What is the function of the 6V zener? If its function is to protect the 5V rail of your circuit, then I suggest that it's not going to do a very good job. Many 5V components have a maximum input voltage of around 5.5V or 6V, so a 6.2V zener won't help much. I'm not sure what your environment is like, but it's usually better to just shut everything down if your regulator fails than to try to let the zener run everything.
One common use of a diode for protection from error conditions is to use a reverse-biased rectifier diode across the input terminals. That way, if anyone connects the source backwards, the power will be dissipated in your designated diode. Make sure that this diode's forward voltage is less than the maximum reverse voltage on the protected components. In this configuration, when the source is plugged in backwards, nothing will work: The supply voltage will be at -0.7V. Presumably, you'd notice that the power LED was not on or that the circuit was not functioning, and correct your error.
Second, no, you can't say whether the result will be a short or open circuit. You've operated the device outside of the specifications, so anything could happen.
Best Answer
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.