Concept:The knee (cut-in or threshold) voltage \( V_k \) is the forward voltage at which the applied field neutralizes the built-in potential barrier, allowing majority carriers to diffuse freely across the junction.
Formula:$$V_F \ge V_k \approx V_0 \implies I_F = I_0 \left( e^{\frac{q V_F}{\eta k T}} - 1 \right) \quad (\text{exponential current rise})$$
Solution:- Below \( V_k \) (\( \approx 0.7\text{ V} \) for Si, \( \approx 0.3\text{ V} \) for Ge), forward current is very small.
- Above \( V_k \), the barrier is overcome, and forward current increases exponentially with small increases in voltage.
Why other options are incorrect:- Option B: Destructive breakdown occurs under excessive reverse voltage, not forward bias.
- Option C: Forward current increases rapidly above the knee voltage; it does not drop to zero.
- Option D: The depletion layer narrows under forward bias.
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