Concept:A series diode conducts whenever its anode potential exceeds its cathode potential by at least the forward cut-in drop: \( V_{\text{in}} > V_K + V_D \).
Formula:$$V_{\text{out}} = V_{\text{in}} - V_D \quad (\text{when } V_{\text{in}} \ge V_K + V_D)$$
$$V_{\text{out}} = 0\text{ V} \quad (\text{when } V_{\text{in}} < V_K + V_D, \text{ diode OFF})$$
Solution:- When \( V_{\text{in}} \ge V_{\text{ref}} + V_D \), the diode becomes forward-biased.
- Current flows through the load, producing an output voltage of \( V_{\text{out}} = V_{\text{in}} - V_D \).
Why other options are incorrect:- Option A: The output is zero only when the diode is reverse-biased (\( V_{\text{in}} < V_{\text{ref}} + V_D \)).
- Option B: The diode blocks negative voltages rather than inverting them.
- Option C: Passive circuits cannot produce infinite voltages.
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