Concept:In a center-tapped full-wave rectifier, during the half-cycle when one diode conducts (acting as a closed switch with nearly zero drop), the non-conducting diode is subjected to the full voltage of the entire transformer secondary winding, which equals \(2V_m\).
Formula:$$\text{PIV} = V_{\text{upper}} - (-V_{\text{lower}}) = V_m - (-V_m) = 2 V_m$$
Solution:- Let the upper terminal be at \(+V_m\) and the lower terminal at \(-V_m\) relative to the center tap (\(0\text{ V}\)).
- Diode \(D_1\) conducts, placing its cathode at approximately \(+V_m\).
- Diode \(D_2\) has its anode connected to the lower terminal at \(-V_m\).
- The reverse voltage across \(D_2\) 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 diode, not a center-tapped rectifier diode.
- Option C: \(0.5 V_m\) is mathematically incorrect and underestimates the reverse stress.
- Option D: \(\sqrt{2} V_m\) confuses RMS-peak relationships with circuit loop analysis.
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