Difficulty: Easy
Correct Answer: the peak value of the secondary voltage
Explanation:
Introduction / Context:
Peak inverse voltage (PIV) dictates the minimum reverse-voltage rating of rectifier diodes. Choosing an underrated diode risks breakdown and failure. The PIV differs between bridge and center-tapped rectifiers, so recognizing the correct relationship prevents costly mistakes.
Given Data / Assumptions:
Concept / Approach:
In a bridge, two diodes conduct each half-cycle, delivering the rectified output; the other two diodes are reverse-biased. The nonconducting diode sees at most the transformer secondary peak voltage across it because the bridge configuration prevents the full end-to-end peak from appearing across a single reverse-biased device, unlike a center-tapped rectifier where PIV ≈ 2 * V_secondary_peak.
Step-by-Step Solution:
Let Vp be the secondary peak voltage from the winding.Analyze a positive half-cycle: two diodes conduct, two are reverse-biased.The reverse-biased diode sees approximately Vp across its junction.Therefore, PIV per diode ≈ Vp in a bridge rectifier.
Verification / Alternative check:
Compare with center-tapped full-wave rectifier: there, each nonconducting diode must withstand roughly twice the half-winding peak (≈ Vp_end-to-center), effectively ≈ 2 * Vp_half. The bridge requires a lower PIV rating per diode.
Why Other Options Are Wrong:
Half or four times Vp: inconsistent with bridge voltage distribution.
Twice Vp: true for center-tapped, not for a bridge.
Zero: a reverse-biased diode absolutely sees reverse voltage; otherwise blocking would not occur.
Common Pitfalls:
Interchanging bridge and center-tap PIV values. Always sketch the conducting path and the reverse-biased diode terminals relative to the secondary to determine PIV correctly.
Final Answer:
the peak value of the secondary voltage
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