Concept:With a shunt filter capacitor, the capacitor charges to \( +V_m \) and holds its charge. During the negative half-cycle, the AC secondary voltage reaches \( -V_m \), subjecting the reverse-biased diode to a total reverse voltage of \( 2V_m \).
Formula:$$\text{PIV}_{\text{half-wave with capacitor filter}} = V_{\text{capacitor}} - V_{\text{secondary,min}} = (+V_m) - (-V_m) = 2 V_m$$
Solution:- The capacitor charges to \( +V_m \) at the diode cathode.
- At the negative peak, the transformer secondary applies \( -V_m \) to the diode anode.
- The maximum reverse voltage across the diode is \( V_{\text{cathode}} - V_{\text{anode}} = V_m - (-V_m) = 2V_m \).
Why other options are incorrect:- Option B: \( V_m \) is the PIV for an unfiltered half-wave rectifier with a purely resistive load.
- Option C: \( V_m / 2 \) would cause reverse breakdown.
- Option D: The diode experiences maximum reverse voltage at the negative peak.
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