# LC circuit with back to back diodes [closed]

I know how an LC circuit works, but in this circuit there are 2 diodes and I am not able to intuitively get the waveforms of current and voltage across the capacitor. Could you please help understand the basic concept from scratch and how to deal with this kind of circuits?

Initially, the capacitor is charged to 5V and ON voltage of diodes is 1V, so D1 is ON and the capacitor discharges till it becomes 0V in a normal LC circuit, but here I am not clear.

• Start by drawing the waveform with the diodes shorted. What influence would a diode have on that? What influence would two anti-parallel diodes have on that? Jun 20 at 13:19
• A simple diode model is all that's needed to get the educational intent from this circuit. The model is that a diode can conduct any current through it when the voltage is above a threshold, and no current below. A pair of anti-parallel diodes look like a 'loss' of the threshold voltage when a current is flowing through them. As winny says, plot the waveform with a diode threshold voltage of 0 (aka 'shorted'). Now increase the diode threshold to 0.3 V (schottky) or 0.7 V (silicon), a minor change. Jun 20 at 13:25
• Actually, diodes have an ON voltage of 1V. So, that's why I am not clear. Jun 20 at 13:26
• And what will said 1 V drop cause on your waveform? Jun 20 at 13:31
• I am not clear whether the voltage across capacitor discharges completely or not. In simulation, its not discharging till 0V. Also the oscillations are damped. Jun 20 at 13:34

The middle plot shows the voltage across the capacitor and the current through D1 for the whole timespan, showing the damped oscillations, and how the diodes would have been smoking after the first cycles. Above, the voltage across the capacitor but only the part after the oscillations have stopped. You can see how the last cycle is snuffed and what remains is the voltage that seems to decay faster initially, then slower, and slower, and ever slower due to the increasing Ron -- which is the bottom plot.
• What happens if you run it with the default diode model D like in the question? Does it oscillate forever without an Rs parameter defined? Or is the 1milliohm (default) in the inductor enough to make it decay eventually? Jun 21 at 0:41
• @SteKulov The much lower Ron will cause much less damping so the oscillations should go much higher, but they will pass through zero, eventually, which is where the maximum damping occurs with Roff. However, given its very large value (1/Gmin) the steadily decaying part will take a very long time to decay. Funny thing, the help says default Vfwd=0, but it seems to not be the case; if it were, then the only damping would be Ron. With two really ideal diode (Ron=0, Roff=inf, Vfwd=0) they will count as a short. Jun 21 at 5:57