# Why does the peak value of ripple voltage occurs at T/4 [closed]

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.

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

simulate this circuit – Schematic created using CircuitLab

Figure 1. A half-wave rectifier schematic.

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?