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I'm wanting to use a relay as a SPST switch, and I've noticed many designs incorporate a flyback diode. What instances are these needed and what are potential drawbacks, if any?

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up vote 7 down vote accepted

The flyback diode is used on the coil of the relay, to protect whatever is driving it from being exposed to a high-voltage spike when the current through the coil is suddendly cut off.

The diode allows the current that was flowing through the driver to flow instead back through the coil itself, allowing the stored energy to be dissipated in the diode and the resistance of the coil itself.

The only real drawback to using a flyback diode is that it delays the release of the relay contacts by virtue of the fact that it allows the coil current to decay more slowly. If this is a concern, you can put either a zener diode or an external resistor in series with the flyback diode. This will allow the voltage to rise higher than it would with the diode alone, but will cause the current to decay more rapidly. Select the value of the zener or resistor to be compatible with whatever the driver can withstand.


simulate this circuit – Schematic created using CircuitLab

For a given value of Vpeak, the zener will give the fastest release time.

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Exactly what I needed. Thank you for your advice. – user46703 Jun 18 '14 at 16:16
In this case be sure that the zener diode can appropriately handle the generated voltage and not reach the reverse-breakdown point, otherwise you'll get current conduction and are in the same situation of delayed relay contacts. – sherrellbc Jun 18 '14 at 16:19
@sherrellbc: The whole point of a zener is to have it reach its breakdown voltage! The main concern is whether it can absorb the energy that's stored in the relay coil. The power rating needs to be based on the energy per cycle and how often the relay is switched off. Even if the zener conducts, the relay has the full zener voltage across it, rather than just the Vf of the silicon diode, and this causes the current to decay proportionally faster. – Dave Tweed Jun 18 '14 at 16:23
@DaveTweedIt should decay exponentially faster, right? The current through the inductor is limited by the diodes. I also see your point regarding the tranditional clamping application of the zerner reaching its breakdown, but in this application would it not be more fitting to have a sufficiently large reverse-breakdown voltage such that it is never satisfied? In this way we never have to worry about reverse conduction and the contacts will appropriately instantaneously (~) switch? As a second question, why not just use the zener rather than the series circuit of the silicon diode and zener? – sherrellbc Jun 18 '14 at 16:36
@sherrellbc: Remember, the di/dt of the current is equal to the voltage across the coil. Since we're holding the voltage at a fixed value (the zener voltage), the current decays linearly (not exponentially). And no, if the zener never conducts, that's would be equivalent to having nothing at all across the coil. We need both diodes because we don't want the zener conducting in its forward direction when the relay is energized. The regular diode prevents that. – Dave Tweed Jun 18 '14 at 21:38

It is a protection diode that should prevent inductive kickback from destroying your transistor driver.

You should put it in reverse bias across the relay coil.

This question may have some useful information for you: Where should I put the kickback diode in a transistor switch?

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I have been under the impression that kickback, flyback, freewheel (etc) are all referring to the same diode application. – sherrellbc Jun 18 '14 at 16:21

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