Concept:In a half-wave rectifier, an applied negative half-cycle places a negative potential on the anode and a positive potential on the cathode, reverse-biasing the diode.
Formula:$$V_{\text{out}} = I_{\text{load}} R_L \approx 0\text{ V} \quad (\text{when } D \text{ is OFF})$$
Solution:- Reverse biasing widens the depletion layer, increasing diode resistance to near-infinite levels.
- The diode acts as an open switch (non-conducting state), dropping the full supply voltage across itself and delivering zero voltage to the load resistor.
Why other options are incorrect:- Option A: Low dynamic resistance occurs exclusively during the positive half-cycle when the diode is forward-biased.
- Option B: Avalanche breakdown occurs only if the reverse voltage exceeds the Peak Inverse Voltage (\( \text{PIV} \)), which causes diode failure in standard rectifiers.
- Option C: Frequency doubling and polarity inversion occur in full-wave rectifiers, not in half-wave rectifiers during the negative half-cycle.
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