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I am building a Spot Welder with a big powerful transformer. I have to switch a big inductive load of 70mH (input of a transformer) The switching will be done with an SSR (Opto Triac Zero crossing SSR-90DA).

When the transformer is under load the input impedance is almost resistive. My problem is this transformer is also open circuit output sometime. This is equivalent to switching a 120VAC into a 70mH inductance. The zero crossing feature is great but with at purely inductive load it become a problem. Switching ON is OK. Switching OFF is not OK because the inductive load will shoot a big reverse pulse much higher than the MAX voltage of the SSR. I want to put an bipolar AC capacitor in parallel with the inductor, reducing the pulse to acceptable voltage. I am missing math involved necessary to calculate the Cap value. What is the energy stored in the coil at Switch OFF ? What voltage this energy will charge the capacitor at Switch OFF ?

The Questions:
How can I calculate the pulse voltage .vs. Capacitor value (at switch OFF time) ?
How can I calculate the necessary value of this capacitor ?

Fred Cailloux
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    A zero-cross approach may actually be the very worst possible way to switch a highly inductive load, as surge currents can be an order of magnitude (and more) than steady state current. Doing it at peak voltage may be better because it can mean little or no surge. – jonk Jan 05 '19 at 20:50
  • My quick calculations indicate a polycarbonate around 100nF will help with 1kOhm 1W reduce dV/dt – Tony Stewart EE75 Jan 05 '19 at 22:53
  • zero (current) crossing is the best time to turn off the transformer. any other time you have energy that needs to be dissipated to dissipated. use a series RC snubber designed to keep dv/dt within acceptable limits. – Jasen Слава Україні Jan 06 '19 at 00:48
  • Tony EE rocketscientist - I just tried your suggestion. And I instantly fried my 50Amps SSR. My scope was showing some thousands of volts kickback. Conclusion: that is not the solution. Beside, I am more looking for the Math involved here than a physical solution. Thank you for your participation anyway. – Fred Cailloux Jan 06 '19 at 17:56

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