A flyback diode is used to protect circuits against the reverse voltage generated by removing power from a relay coil.
Like this:

schematic

simulate this circuit – Schematic created using CircuitLab

But what to do if the polarity of my power supply isn't known?
placing two diodes in parallel in opposite won't work (short circuit) Any idea for bi-directional flyback diode?

up vote 3 down vote accepted

You don't need to use a diode; a capacitor will limit the open-circuit energy from the relay coil to a voltage usually not greater than twice 9V.

Lets say your inductance is 1 henry and the coil resistance is normally 750 ohm. The current into the 1 henry is limited to 12 mA and therefore the coil stores an energy of 1 x 0.012 x 0.012 / 2 = 72 uJ.

If all of this is transferred to a 10uF capacitor the voltage increase above 9V on the capacitor is: -

V = \$\sqrt{\dfrac{72uJ \times 2}{10 uF}}\$ = 3.8 volts

i.e. the 9V, when disconnected by the switch, will create a peak voltage of 12.8 volts on the capacitor and it will likely but much less than this because of the series resistance of the coil burning off energy in the transfer of micro joules to the capacitor.

To address Spehro's comment a small values resistor in series with the capacitor can reduce inrush when activating the relay.

  • what is "the 12V" which is disconnected? – AHB Dec 4 '15 at 17:15
  • The problem with just using a capacitor is that it could kill the switch when the switch closes.. so better an R-C snubber. – Spehro Pefhany Dec 4 '15 at 17:16
  • @BlueSky I misread 9V as 12V - i'll fix up my answer and address Spehro's point. – Andy aka Dec 4 '15 at 17:23
  • you adjusted the coil resistance not to change the current through coil? lol. //// thanks. now, Is there any problem if the capacity is more? (or less?) – AHB Dec 4 '15 at 17:29
  • If capacitance reduced to 1uF, the overshoot would be 38 volts on top of 9 volts - still quite practical in many circumstances. More capacitance means less overshoot. You could of course use a resistor - the 12mA into a 1k resistor would be 12V and this isn't much to deal with but can you tolerate the resistor drawing 9mA when the power supply feeds the coil? – Andy aka Dec 4 '15 at 17:38

You are looking for a "bidirectional TVS diode" which is two Zeners back-to-back. You can get them in many different voltages, so you can pick one that is above your supply. That way only the inductor current will break through it.

Alternately you could use a diode bridge to set the polarity of the input no matter what you get fed.

  • Where to place it? provide a schematic. – AHB Dec 4 '15 at 16:31
  • It goes where your diode goes. It works the same. – Daniel Dec 4 '15 at 16:35

Usually the polarity will be known because something like a transistor will not work if the voltage is reversed.

If you have a bidirectional switch such as a mechanical contact or series back-to-back MOSFETs you can use a bidirectional TVS with the TVS voltage higher than the supply voltage. If you put it across the load, and the voltage surges above the breakdown voltage the diode will conduct and the switch and/or TVS could be destroyed, so it's sometimes better to put it across the switch.

Here are some example types. You must make sure that the TVS can safely absorb the energy stored in the coil (less of a problem with the rectifier diode because most of the energy is dissipated in the coil if the operating voltage is much higher than the diode drop). If the inductance is constant, that's simply the operating current squared multiplied by the inductance.

  • Can I place a mini light bulb in series with the TVS diode to avoid dissipation in the TVS? Will the flyback voltage last enough for eye to notice the blink in the light bulb? – AHB Dec 4 '15 at 16:48
  • No. A smallish normal relay coil will not be a problem even with a modest-sized TVS. – Spehro Pefhany Dec 4 '15 at 17:14

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