Concept:In reverse bias, the applied voltage reinforces the built-in potential barrier, widening the depletion region and preventing majority carrier flow, which results in a very high junction resistance (several \( \text{M}\Omega \)).
Formula:$$R_{\text{reverse}} = \frac{\Delta V_R}{\Delta I_R} \to \infty \quad (\text{Ideal Diode } I_R \approx 0)$$
Solution:- Connecting the P-side to the negative terminal and the N-side to the positive terminal widens the depletion layer.
- Because no majority carrier current can cross the barrier, the diode allows only a negligible minority leakage current (\( \mu\text{A} \) to \( \text{nA} \)).
- Therefore, it acts effectively as an OFF or open switch.
Why other options are incorrect:- Option A: A diode does not store magnetic energy like an inductor.
- Option B: A forward-biased diode behaves as an ON or closed switch.
- Option C: A diode in normal reverse bias is not a constant current power supply; it only conducts tiny leakage current.
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