It makes sense to me that the PIV occurs at the bottom of the negative half cycle for the rectifier without the filter. What doesn't make sense to me is why the PIV for the rectifier with the filter occurs at the top of the positive half cycle. It seems like the reverse voltage is equal to the capacitor voltage minus the supply voltage, and at the top of the positive half cycle that would be zero. Am I missing something here?
This image from the above link is what's confusing me. The above answers state that PIV is 2Vp because the voltage from the capacitor and the maximum negative voltage from the source are equal in magnitude and across the diode at the same time. However, this diagram shows the capacitor voltage is only equal in magnitude to Vp at the same peaks as the waveform without the capacitor (the top of the positive half cycle).