As I get to know more about the MOV operation, there are more things needed to understand.

My MOV (V420La20AP) has Varistor voltage of 680 Vrms (10%) at 1 mA, and maximum clamping voltage of 1120 V at 50 A. Now my system (where this MOV is installed) generates a surge of 4 kV with timing 1.2/50 us. It has a source impedance of 2 ohm, so it has a short circuit current of 2 kA. I have this info, but I am not sure what would be the clamping voltage of MOV when this pulse is applied or what voltage it will output to the load. I found a formula for mov non-linear region which is

I = K * V_power(a).

Can this formula be used to find out about the clamping voltage at given current flowing throw mov. Then again what will be values of "K" and "a". They are dependent on MOV but manufacturer doesn't list these values.

Actually, I want to find energy of the surge handled by mov. According to SEMTECH,

E = E1+E2 = (0.5 * Vc * Ip * t_front_time) + (1.4 * Vc * Ip * t_50%time)

For this I need Vc (Clamping voltage at 2 kA peak current passing through MOV).

To know this, I need "K" and "a" , can I find these by simple mathematics applying 680 V (1 mA) and 1120 V (50), this way I will get two equation with two variables, and possibly I can find both of these values.

Am I doing it right? I mean this varistor equation (I=kV_power(a)) can be used this way to find any varistor voltage given current or vice versa?

  • 2
    \$\begingroup\$ Please stop sprinking underscores into your text. Capital I for I. k for kilo. \$\endgroup\$
    – winny
    Commented Aug 28, 2018 at 7:56

1 Answer 1


I used Webplot digitizer https://automeris.io/WebPlotDigitizer/ on a screenshot of figure 12 from the datasheet to see that the clamping voltage for that varistor at 2kA is 1700V.


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