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While reading about how a half wave rectified output is filtered using a capacitor, I came across the following waveforms:

enter image description here

corresponding to the circuit: enter image description here

In the interval from 0 to piOverTwo, the diode is forward biased, and hence, conducts. This causes the capacitor to be charged up to the peak voltage. In the interval from pi to twoPi, the diode is reverse biased, and therefore, does not conduct. So the capacitor discharges through the load resistance at a rate longer than that of the input.

That much is clear to me. However, I am not exactly sure about what happens in the interval from piOverTwo to pi. According to me, the input voltage is simply decreasing here, so the diode should remain forward biased, and hence conduct. But the above waveform shows otherwise. So, is the capacitor discharging in this interval ? If yes, then is it different from the discharge in the interval from pi to twoPi ?

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... in the interval from \$ \frac {\pi} {2}\$ to \$ \pi \$. According to me, the input voltage is simply decreasing here, so the diode should remain forward biased, and hence conduct. But the above waveform shows otherwise.

The diode is only forward biased when the \$ V_{AC} \$ > \$ V_C \$. Once the capacitor charges to peak voltage and the AC falls faster than \$ V_C \$ then the diode is reverse biased.

enter image description here

Figure 1. OP's graph with diode current superimposed.

Figure 1 shows a typical current pulse in this situation. All the charge is passed in the short duration when \$ V_{AC} \$ > \$ V_C \$.

So, is the capacitor discharging in this interval?

Yes. \$ V_C \$ > \$ V_{AC} \$ so the diode is reverse biased.

If yes, then is it different from the discharge in the interval from \$ \pi \$ to \$ 2\pi \$?

No, it is a continuation of the same discharge because the diode is reverse biased in both cases.

Note: The original diagram might be a little clearer if the red line was 0.7 V below the blue waveform as would be the case with a real diode due to its forward voltage drop.

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  • \$\begingroup\$ Green rectifier current will always exceed Iavg load current by a ratio function dependent on R_load/ESR(tot) , RC/T values such that the area under each I(t) curves are equal for diode and load R \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 9 '17 at 15:21
  • \$\begingroup\$ It wasn't to scale. Better now? \$\endgroup\$ – Transistor Sep 9 '17 at 15:28
  • \$\begingroup\$ Much better. Now the $64k question is for ripple pp <20% what is approx. equation for diode Ripple Current vs Load Regulation or R/ESR where ESR combines cap, diode and voltage source. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 9 '17 at 15:53

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