I'm trying to design a solid state circuit breaker using bi-directional switch modules with IGBTs. (Rating: 3phase: 690V AC, 1000A) The problem here would be the transient voltage spikes across the switch and the solution would be to use some kind of voltage suppressors.

Now, I know that MOVs usually wear out with time. However, the switching events are not going to occur very frequency. The switch is only turned off during fault conditions. Usually how long does an MOV last in this scenario? I mean to ask, Can I go ahead only with an MOV or do I need to include an RC snubber as well keeping in mind the fault levels of current as well?

Assume that the MOV needs to be turned off about 2000 times? (Assuming it's life span to be about 15-20 years.)

  • \$\begingroup\$ MOVs are already used as suppressors with contactors, relays. IMO not a problem if you use it with SSR as well. \$\endgroup\$ Commented Apr 13, 2017 at 10:11
  • \$\begingroup\$ Movs also degrade with use. \$\endgroup\$
    – user16222
    Commented Apr 13, 2017 at 11:17
  • 1
    \$\begingroup\$ @Pujitha MOVs can and do fail short circuit . \$\endgroup\$
    – Autistic
    Commented Apr 13, 2017 at 11:59

1 Answer 1


This question misses the most important point.

You really need to calculate how much energy is stored in the wiring inductance (LI^2) when you open the circuit. At 1000 amps, this could be quite a lot of Joules.

This energy needs to be dissipated into something properly rated, both for the pulse energy, and the pulse power. Enough joules will turn any component into a firecracker.

A capacitor snubber (with a cap of the right value) will store some of this energy in the cap, then slowly dissipate it in the resistor. Pulse power is thus reduced. However, if the cap is too small to absorb the energy stored in the inductance, then voltage will keep rising until something pops.

MOVs tend to be able to dissipate very high peak power, but you will need to check and be absolutely sure that the total energy to dissipate will not exceed the part's capabilities. A wide safety margin will make it last longer, and guard against a use case with higher inductance than you thought...

  • 1
    \$\begingroup\$ Also, I've seen in many places that the leakage inductance for a distribution transformer can be around 0.03-0.1 p.u. Assuming the transformer is about 1.2MVA, (based on my current and votlage ratings: for worst case, 0.1 pu) when calculated, the inductance is very low- around 125uH. ( 1p.u impedance per phase = 400m ohms) Are my calculations correct? If they are, usually how many times this value do I need to consider in my simulations? [Considering all the transformers in the network.] \$\endgroup\$
    – Pujitha
    Commented Apr 13, 2017 at 13:03

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.