I have a formula here for finding the suitable value of shunt capacitance to be put on an SCR snubber:

$$ C=\frac{1}{2L}\left(\frac{0.546 V}{\frac{dV}{dt}} \right)^2 $$

L is the snubber inductor, V is the DC supply voltage and \$\frac{dV}{dt}\$ the maximum allowable surge voltage. My question is about the derivation of the said equation—if it is a correct one. The formula looks bizarre to me when considering the equation for getting the right inductor, \$V = L\frac{di}{dt}\$, which is more straightforward. Moreover, I'm puzzled why choosing the capacitor value would depend on the series snubber inductor.

  • \$\begingroup\$ L is the load inductance \$\endgroup\$ Commented Aug 24, 2017 at 9:01
  • \$\begingroup\$ Based on my experience I think it's the snubber inductor. \$\endgroup\$
    – DorkOrc
    Commented Aug 24, 2017 at 9:07
  • 2
    \$\begingroup\$ There is no snubber inductor, just RC snubber. The snubber circuit is connected in parallel with SCR. \$\endgroup\$ Commented Aug 24, 2017 at 10:28
  • \$\begingroup\$ Yes, and an inductor would be connected in series to protect the SCR from di/dt. So essentially it is also in series with and becomes part of the load, as you said it is load inductance. \$\endgroup\$
    – DorkOrc
    Commented Aug 24, 2017 at 12:39
  • 1
    \$\begingroup\$ @DorkOrc But even if what you said were true (and I've never heard of a snubber inductor before), this isn't that scenario. Just as you already said, the equation is using dV/dT, not dI/dT. The L here is the load inductance. \$\endgroup\$
    – DKNguyen
    Commented Aug 12, 2019 at 3:26

1 Answer 1


Interesting question.

The inductor stores energy in the current: 0.5 * L * I^2

The snubber capacitor needs to store that same amount of energy, as voltage across the capacitor: 0.5 * C * V^2

Equate the two energy equations, and have fun.


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