Concept:In a bridge rectifier, the two non-conducting reverse-biased diodes are connected in parallel with the load and the two conducting forward-biased diodes. The maximum reverse voltage appearing across any single diode is limited to the peak secondary voltage \(V_m\).
Formula:$$\text{PIV}_{\text{Bridge}} = V_m$$
Solution:- During the peak of a half-cycle, two diodes conduct (forward drop \(\approx 0\text{ V}\)).
- Applying Kirchhoff's Voltage Law around the loop shows that each reverse-biased diode experiences a peak voltage of \(V_m\).
- This is a major advantage of the bridge rectifier over the center-tapped topology (which requires diodes rated for \(2V_m\)).
Why other options are incorrect:- Option A: \(2 V_m\) is the PIV required for diodes in a center-tapped full-wave rectifier.
- Option C: \(V_m / 2\) underestimates the peak reverse voltage applied across the bridge arm.
- Option D: \(4 V_m\) is an arbitrary multiple.
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