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What is the constitutive relashionship between current and voltage of a varistor in LTSpice? For example, if I set the control voltage V=1V and Rclamp=2, what is the formula to evaluate the resistance of varistor? I tried to do this simulation (I used a 10V voltage source in serie with R1=1 and a varistor), but the result it's strange for me. The current on R1 is 3A, \$V(R1)=R1\cdot I(R1)=1\cdot3=3V\$, but the resistance of varistor is not Rclamp=2 (is 7/3=2.3). Thank you very much.

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    \$\begingroup\$ Why don't you plot the current when the applied voltage is a ramp - this seems likely to give you what you want. \$\endgroup\$ – Andy aka Nov 8 '16 at 13:36
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If you read in the LTspice manual, you'll see that it says it's breakdown voltage is set by the voltage applied at its control terminals. The voltage can be of any sign, any polarity (it counts as abs(V(ctl))), and the calculation is like this: if your applied voltage is 10V, the series resistance is 1 Ohm, Rclamp=1 and the control voltage is 1V (I presume), then the varistor will breakdown at 1V (the controlling voltage), meaning from the total of 10V you'll have 10-1=9V which, divided by the total resistance of 3 Ohms results in 3A. If your source voltage is 7V, series resistance is 2 Ohms, Rclamp=1.5 and controlling voltage 3V => (7-3)/(2.5+1.5)=4/4=1A.

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