I am envisaging a relay which switches an inductive load consisting of a couple of switched-mode power supplies (standard ATX computer power supplies) in addition to several wall-wart style power supplies powering other equipment. The total power consumption is about 600 W at 220 VAC. The exact nature of the load may vary over time as different equipment is substituted, but what I have described is the typical size and character of the load.

I've been reading about snubber designs and it seems that some combination of a RC snubber and/or a bidirectional TVS (Transorb/Tranzorb) diode in parallel with the load would be required. I'm not comfortable with the alternative design of snubbing in parallel with the contacts because I want zero current flow when the relay contacts are open. Something like the diagram in this application note for Driving and Relay and Contact Snubbing [PDF] is what I imagine.

First of all, what combination of snubber components is appropriate for my situation - RC network, TVS diode or maybe a MOV? Secondly how can I conservatively estimate the values and ratings of the required components to ensure that my relay contacts don't erode or weld?


1 Answer 1


The RC network's purpose is to provide a dampened path for the charge stored in the inductor. The MOV and TVS diode purposes' is to limit the high voltage that may appear. So they're complementary. Use both a RC network plus either the MOV or TVS.

I used 100 nF + 100 ohm for my loads for years and never had any problems so that's a couple of starting values.

  • \$\begingroup\$ Thank you. I'd noticed that 100 nF + 100 ohm appeared to be very common and a sort of default value for such RC networks, but I didn't want to jump the gun without being better informed. \$\endgroup\$ Commented Aug 17, 2011 at 16:07
  • \$\begingroup\$ Can you provide any guidance on TVS selection? The 1.5KE series seems widely used. Given my 220 V RMS potential (which is 311 V peak) it looks like the 1.5KE400CA bidirectional TVS might work. It has a 342 V reverse stand-off voltage a ~400 V breakdown voltage, but its maximum peak pulse current is (only?) 2.8 A. Is that likely to be enough? Is there a simple way to determine this? \$\endgroup\$ Commented Aug 17, 2011 at 16:35
  • \$\begingroup\$ @RobSmallshire Just to answer your question (for anyone else who may have it), you want to pay attention to three characteristics. A) TVS Peak pulse power rating > Max power of the relay's load B) TVS Max reverse standoff voltage > Relay's switching voltage (not Vrms, make sure to use PEAK voltage, ie (240Vrms = 340Vp) C) TVS Max clamping voltage < relay contact breakdown voltage \$\endgroup\$ Commented Sep 28, 2015 at 0:40
  • 2
    \$\begingroup\$ A TVS diode works to dampen the inductive energy with or without a RC snubber. The only difference is that the snubber is active when ANY AC voltage is across it, whereas a properly spec'd TVS only activates during an inductive spike event. Bottom line: The TVS by itself is OK for protecting the relay contacts. In fact, your solid state relays often have a sole TVS built-in. Just make sure it's sized properly. \$\endgroup\$ Commented Sep 28, 2015 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.