Concept:In a center-tapped rectifier with a capacitor filter, the capacitor charges to \( +V_m \). During the opposite peak, the non-conducting diode experiences the sum of the capacitor voltage (\( +V_m \)) and the opposite half-winding peak voltage (\( -V_m \)), resulting in \( \text{PIV} = 2V_m \).
Formula:$$\text{PIV}_{\text{center-tapped with filter}} = V_C - (-V_m) = V_m + V_m = 2 V_m$$
Solution:- The filter capacitor holds the cathode of the non-conducting diode at \( +V_m \).
- The anode of the non-conducting diode reaches \( -V_m \) at the negative peak of its half-winding.
- The total peak reverse voltage across the diode is \( V_{\text{cathode}} - V_{\text{anode}} = V_m - (-V_m) = 2V_m \).
Why other options are incorrect:- Option A: \( V_m \) is the PIV for a bridge rectifier, but is insufficient for a center-tapped design.
- Option B: \( V_m / 2 \) would cause reverse breakdown.
- Option D: \( 4V_m \) would apply only across a full end-to-end doubler configuration.
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