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For a half wave rectifier, why does the diode turn off-in other words peak value of ripple voltage occurs- when AC voltage whose period is T starts to decrease from its peak value? I have an intuiton that it must happen a bit after AC signal starts to decrease from its peak value, not at the same time that it has its max value.

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    \$\begingroup\$ turn off where? what signal? what is T? \$\endgroup\$
    – PlasmaHH
    Commented Oct 29, 2017 at 21:25
  • \$\begingroup\$ provide some context, please \$\endgroup\$
    – user28910
    Commented Oct 29, 2017 at 21:27
  • \$\begingroup\$ Tnx for comments, I edited it accordingly. \$\endgroup\$ Commented Oct 29, 2017 at 21:30

1 Answer 1

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A half-wave rectifier schematic.

enter image description here

Figure 2. Simulation waveforms for Figure 1.

Notes:

  • There is capacitance in the circuit.
  • With the arrangement shown the capacitance maintains voltage while the source sine falls away. i.e., The supply voltage is falling faster than the load voltage.
  • At (1) the capacitor is fully charged. Note that the current (lower trace) turns off very slightly after \$ V_p \$ probably as the diode continues to conduct a little as its forward voltage drops from 0.7 to 0.5 or so. (Not that the source is still > \$ V_C \$ for some time after the peak.
  • The first cycle current is very high as it has to provide inital charge to C1.
  • At (2) we can see that the current starts to flow again when \$ V_S > V_C \$.
  • At (3) the current stops again at the same point as (1).

I have an intuition that it must happen a bit after AC signal starts to decrease from its peak value, not at the same time that it has its max value.

How close was your intuition?

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