Peak inverse voltage (PIV) — repaired stem for clarity: For a single-diode half-wave rectifier using an ideal diode and a sinusoidal source with peak amplitude Vp = 3 V, what is the peak inverse voltage (PIV) that the diode must withstand during the nonconducting half-cycle?
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A6 V
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B2.3 V
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C0 V
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D3 V
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E2 * sqrt(2) V
Answer
Correct Answer: 3 V
Explanation
Introduction / Context:PIV (peak inverse voltage) is the maximum reverse voltage that a rectifier diode must withstand when it is reverse-biased. Getting this rating correct is essential to avoid diode breakdown in power supplies. The original item referenced an unspecified “given circuit,” so we minimally repair the stem to a standard, well-defined case: an ideal half-wave rectifier driven by a sinusoidal source of peak Vp = 3 V.
Given Data / Assumptions:
- Topology: single-diode half-wave rectifier, ideal diode.
- Input: sinusoidal source with peak value Vp = 3 V.
- No transformer or additional components; no conduction drop assumed in the ideal model.
- Load is resistive and referenced to ground (typical textbook configuration).
Concept / Approach:In a half-wave rectifier, during the reverse half-cycle the diode is reverse-biased and must block the input's instantaneous voltage. For an ideal source and ideal diode, the maximum reverse stress across the diode equals the peak of the applied sine wave: PIV = Vp. If a transformer and other elements were present, PIV expressions can differ, but in this minimal repaired configuration it equals the source peak.
Step-by-Step Solution:
Identify nonconducting interval: the half-cycle where the source polarity reverse-biases the diode.During that interval, the diode sees the source's instantaneous voltage across it (load is effectively isolated).Find the maximum magnitude of reverse voltage: it occurs at the sine peak.Compute PIV = Vp = 3 V.Verification / Alternative check:Draw the waveform of the input and mark the reverse-biased half. The highest reverse stress across the diode coincides with the negative peak (for the chosen orientation), which equals 3 V in magnitude for this repaired stem.
Why Other Options Are Wrong:
- 6 V: Would apply in some center-tapped full-wave cases (2 * Vp across a nonconducting diode), not in a simple half-wave.
- 2.3 V: No common rectifier formula yields this from Vp = 3 V.
- 0 V: Reverse stress is not zero; the diode must block the peak reverse swing.
- 2 * sqrt(2) V: Not relevant to a half-wave with the given Vp.
Common Pitfalls:Confusing half-wave with full-wave center-tapped PIV requirements; forgetting that quoted Vp is already the peak (not RMS), so no 0.707 factor applies here.
Final Answer:3 V