I have just been cogitating on the tutorial at http://www.electronics-tutorials.ws/io/io_5.html, and in the discussion of flywheel diodes it includes this sentence without further elaboration:
As well as using flywheel Diodes for protection of semiconductor components, other devices used for protection include RC Snubber Networks, Metal Oxide Varistors or MOV and Zener Diodes.
I can kind of see how an RC network might be needed if it is a large device and therefore the coil could be kicking back more current than you want to dissipate through a single diode. (Please correct me if that's not the reason.)
I don't have a clue what an MOV is so for the moment I'll ignore that one. :-)
I have read a bit about Zener diodes, but I don't understand why their lower reverse breakdown voltage might be desirable here?
Edit: I'm also puzzled by the following diagram from the tutorial above:
Wouldn't this take any flyback voltage and dump it into the Vcc net? Would it not be a better idea to have the relay coil be between TR1 and ground, and the diode dissipating the flyback voltage to ground?
Answer
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.
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