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I'm switching a solenoid with a relay. The relay has failed after just a few hundred operations, I suspect because of the flyback from the solenoid.

I have a P6KE20A TVS diode that I want to use to protect the relay contacts.

Now I wonder, does it matter whether I place it parallel to the solenoid or parallel to the relay contacts? Does the fact that it's a TVS diode have any influence on placement? (Unfortunately it is unidirectional I believe.)

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

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So technically its likely that the problem was back EMF from your inductor, rather than "transient voltage" (although the back EMF is sort of transient, and it does cause a voltage, so maybe that counts as a transient voltage). The solution to the back EMF is to put a diode in parallel with L1, as you have done in the first image.

Using a TVS diode as you have in the first image will also work, but the transient voltage suppression part isn't really doing anything, any diode (suitably rated for current) will work.

Interestingly, although your second arrangement wouldn't work with a normal diode, it probably sort of maybe does work with a TVS diode. I wouldn't recommend doing this, because it could do some weird stuff to your power supply.

In summary, feel free to use the arrangement you have in the first picture, and you can use a TVS which will protect you a tiny bit more from random static or whatever on the inductor contacts (although I doubt it will really be different to a normal diode).

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  • \$\begingroup\$ "your second arrangement wouldn't work with a normal diode", of course, that's why no one is using that arrangement. That's what I didn't realize. \$\endgroup\$ – AndreKR Aug 27 '20 at 7:12
  • \$\begingroup\$ Depending on the amount of current going through the inductor, you might have to limit the current going through diode at turn-off (looking at the first arrangement). This could be done with a resistor in series with the diode. Otherwise you might blow your diode and then blow the relay again, thinking that the solution didn't work ;) \$\endgroup\$ – Bonnevie Aug 27 '20 at 10:04
  • \$\begingroup\$ @Bonnevie This is true, but a resistor will not limit the current (although it will decrease the decay time for the current). AndreKR you should instead choose a larger diode if necessary. The maximum current through the diode will be equal to the maximum current that you allow through your inductor. \$\endgroup\$ – BeB00 Aug 27 '20 at 10:27
  • \$\begingroup\$ Ah yes I see. So if it only switches once in a while it is really just about if your diode package can handle the energy stored in the inductor in the time frame. \$\endgroup\$ – Bonnevie Aug 27 '20 at 10:53
  • \$\begingroup\$ If the solenoid can tolerate it and still work, adding a series resistance in the loop may help to knock down the 'spike' a bit and save your relay. Adding a small capacitor across the solenoid is also a valid 'spike-knock-down' approach. GIve the energy somewhere to go besides arcing across the relay. Make sure it's a high-voltage cap of course! I might start with something like 1uF 100V ceramic. It's pretty easy to use an oscilloscope to see if either attempt is having the effect you want. \$\endgroup\$ – Kyle B Aug 27 '20 at 19:32

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