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I was going through the working of class D commutation and the article said:

As soon as the capacitor completely discharges, its polarities will be reversed but due to the presence of diode the reverse discharge is not possible. 

Why does the polarity of the capacitor reverse as soon as it completely discharges? Is it because it immediately ( at the speed of light) starts to charge?

Moreover, what does reverse discharge mean?

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  • \$\begingroup\$ Short version: the reversal ONLY occurs if the capacitor is connected to an inductor. The inductor-current cannot change rapidly, and this causes the voltage across the capacitor to, rather than just exponentially settling to zero, instead the voltage "overshoots" and becomes reversed. \$\endgroup\$
    – wbeaty
    Apr 16, 2020 at 4:27

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Here's the circuit in your link:-

enter image description here

Initially SCR2 is triggered to charge the capacitor through the load. Once the capacitor has charged up to the supply voltage SCR2 will turn off when current drops below its holding current.

If SCR1 is then triggered to power the load, the capacitor will discharge through the diode and inductor (which is now connected to V+ through SCR1). During this time the inductor produces an opposing voltage that is proportional to the rate of current change (principle of inductance).

The inductor and capacitor form a tuned circuit, so current rises until the capacitor is completely discharged, then starts to drop. Now the current change is negative so the inductor produces opposite voltage, charging the capacitor up in the opposite direction. In a tuned circuit this cycle would normally continue producing a sine wave, but the diode stops the capacitor from discharging again.

Now that the capacitor has 'negative' voltage on it, the '-' side is more positive than the supply. If SCR2 is again triggered it discharges the capacitor into the load, bypassing SCR1 which now turns off because it has negative voltage across it.

To illustrate the waveforms I simulated part of the circuit in LTspice. This graph shows the voltage across the capacitor, so it goes from positive (charged from V+ to V-) to negative ('reverse' charged by the inductor). In the actual circuit it generated a positive voltage of ~9V above the 10V supply.

enter image description here

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  • \$\begingroup\$ so current rises until the capacitor is completely discharged, then starts to drop. How does current increase when a capacitor is discharging? Shouldn't the current decrease from the point when the capacitor has max voltage till the point it is complete discharged (0v)? Moreover what do you mean by it starts to drop? Does the capacitor see an increase and decrease in current during the discharging phase alone? \$\endgroup\$
    – penguin99
    Aug 19, 2019 at 5:51
  • \$\begingroup\$ If the capacitor was discharging into a resistor then the current would start out high and drop as its voltage dropped. However the inductor opposes current change by generating a voltage that matches the capacitor voltage, so current ramps up from zero (at rate dI/dt = V/L). As the capacitor discharges the voltage drops so the current increases more slowly, until at 0V when there is no charge left so the current must drop. At this point the inductor starts generating opposite voltage to keep the current going... \$\endgroup\$ Aug 19, 2019 at 6:51
  • \$\begingroup\$ Overall the current races a half sine wave as it rises and falls, while the voltage is half a cosine wave going from positive to negative peaks (or negative to positive with respect to V+). \$\endgroup\$ Aug 19, 2019 at 6:53

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