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I'm trying to design some sort of arc suppression for a relay under short circuit when tripped. So i'm trying to understand how these arcs form. From my understanding, at the instant of tripping, the relay contacts instantly changes its voltage to the supply voltage. Since the switch takes some time to fully open, E = V/d, and we have a very small distance (at instant of tripping) with a very large voltage (230 V AC rms in my case). The electric field is so high that it causes breakdown of the air and a conduction arc forms.

Correct ?

So my main goal is to reduce the dv/dt across the switch which will reduce the electric field and prevent breakdown ?

How can I prevent or suppress the arc from forming ?, will a snubber across the contacts work ?

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When the contacts close, if a short circuit exists in the load, the current will increase very quickly (based on the supply voltage and the impedance of the source and load and wiring (a loop). The contact does not close instantaneously, it may bounce a few times, and the pressure starts out very low, so a great deal of heat is produced right at the small contact surface. That heat will cause a rapid increase in temperature, vaporizing some of the contact material and it will often cause the contacts to weld together.

The only way to prevent this from happening is to limit the current, and/or prevent it from rising so rapidly. Unfortunately, the obvious way to do the latter (a huge choke) will cause arcing when the contacts open, if they are carrying significant current.

It is possible to make an electronic circuit breaker capable of opening into a short circuited load at the peak mains voltage. That would protect the contact but the seniconductor has to be rated for the service and there are other issues.

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  • \$\begingroup\$ So if I connect an inductor in series to restrict the current rise and then put a snubber across the contacts to prevent arcing from the inductor when the relay trips, would that help ?, would the inductor interfere with any load I connect though ? \$\endgroup\$
    – Deadshot
    Oct 7, 2017 at 8:06
  • \$\begingroup\$ Depends on the load. An inductor large enough to make much of a difference might be physically large and expensive. It will have some resistance which will cause a voltage drop and power loss, and a phase shift if it is AC. For example a Hammond 10mH 50A inductor weighs 36kg and costs hundreds of dollars. And you might need bigger than that though some optimizations would be possible- lower flux and thinner wire to have higher saturation current and but lower RMS current rating. \$\endgroup\$
    – Spehro Pefhany
    Oct 7, 2017 at 19:44

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