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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.

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    \$\begingroup\$ 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? \$\endgroup\$
    – winny
    Jun 20 at 13:19
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    \$\begingroup\$ 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. \$\endgroup\$
    – Neil_UK
    Jun 20 at 13:25
  • \$\begingroup\$ Actually, diodes have an ON voltage of 1V. So, that's why I am not clear. \$\endgroup\$
    – prashanth
    Jun 20 at 13:26
  • \$\begingroup\$ And what will said 1 V drop cause on your waveform? \$\endgroup\$
    – winny
    Jun 20 at 13:31
  • \$\begingroup\$ 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. \$\endgroup\$
    – prashanth
    Jun 20 at 13:34

1 Answer 1

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Even a "simple" D+C circuit requires the LambertW function in order to determine its analytical formula (see jonk's answer, for example), but here you also have an oscillatory waveform. This is hardly a circuit to analyze using pen and paper. I warmly recommend a simulator.

As for the intuitive part: if the initial voltage is 5 V then, given that there are no resistors in the circuit, the initial oscillations will go well above the diode's forward conduction, which will result in their (most likely permanent) damage. But, since this is a theoretical exercise, the voltage will be strongly attenuated with each cycle, but still continuing due to the resonant LC tank.

As the oscillations return from their peak values, towards zero, the diode's Ron will increase, thus the damping will increase. This cycle will continue until the voltage will get closer to Vfwd, when the oscillations will not cause the diodes to conduct, anymore, thus their resistance will be high, all the time. This will cause the oscillations to stop and whatever residual voltage is across C will slowly discharge through Ron. The lower the voltage will get, the higher the Ron will be, and the more the time will be extended until all the residual will be gone.

You can very easily verify this with any SPICE simulator (I would have used the site's program, but I didn't see how to add initial conditions):

LC + D || ~D

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.

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    \$\begingroup\$ 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? \$\endgroup\$
    – Ste Kulov
    Jun 21 at 0:41
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    \$\begingroup\$ @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. \$\endgroup\$ Jun 21 at 5:57

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