This example shows the effect of current chopping in an inductive circuit.
G. Sybille (Hydro-Quebec)
The Ideal Switch is used to switch off and on an inductor (L = 1H, R = 50 ohms) on a 120 Vrms, 60 Hz source. The inductor has a stray capacitance C = 50 nF. The switch resistance is 1e-3 ohm. The switch is initially closed (initial state = 1). It is open at time t = 36ms and then reclosed at time t = 0.15 s.
Start the simulation and observe the inductor voltage U_L and inductor current I_L on Scope1. At the opening time, t = 36 ms, the current in the inductor is I_L= 0.187 A. The current chopping in the inductor produces a high frequency overvoltage (711 Hz) across the inductor. The maximum voltage can be calculated from the following equation:
U_Lmax = I_L*sqrt(L/C) = 0.187*sqrt(1/50e-9) = 836 V
At the reclosing time, t = 0.15 s, the source voltage is zero. Therefore, a full offset is observed in the inductor current. The switch current spike observed on trace 1 of Scope2 at switch reclosing (t = 0.15 s) is due to the capacitor discharge through the source resistance. If you zoom on this current spike you will observe a current step of 30 A (30V capacitor voltage / 1 ohm source resistance), followed by a fast discharge time constant (RC = 1*50e-9 = 50e-9 s).