Concept:A diode conducts substantial current only when the applied forward voltage equals or exceeds its built-in potential barrier (the knee voltage). If the applied voltage is less than the barrier voltage, the remaining internal field continues to prevent majority carrier diffusion.
Formula:$$V_{\text{net}} = V_0 - V_{\text{applied}} = 0.3\text{ V} - 0.2\text{ V} = 0.1\text{ V} > 0$$
Solution:- The applied forward voltage (\(0.2\text{ V}\)) is lower than the barrier potential of Germanium (\(0.3\text{ V}\)).
- A net barrier of \(0.1\text{ V}\) remains across the depletion region.
- Consequently, majority carriers cannot cross the junction in large numbers, and only negligible current flows.
Why other options are incorrect:- Option A: Large conduction occurs only when \(V_{\text{applied}} \ge 0.3\text{ V}\).
- Option B: Breakdown occurs only under large reverse voltages, not small forward bias.
- Option C: Forward bias narrows the depletion layer rather than widening it.
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