Concept:The built-in barrier potential \( V_0 \) arises from uncompensated ionized dopants in the depletion layer and depends on the semiconductor bandgap and doping levels.
Formula:$$V_{0,\text{Silicon}} \approx 0.7\text{ V} \quad \text{and} \quad V_{0,\text{Germanium}} \approx 0.3\text{ V} \quad (\text{at } T = 300\text{ K})$$
Solution:- For silicon, the built-in potential barrier at room temperature is approximately \( 0.7\text{ V} \).
- Silicon diodes require a forward bias of \( \ge 0.7\text{ V} \) to overcome this barrier and conduct substantial forward current.
Why other options are incorrect:- Option A: 0.1 V is too low for silicon or germanium junctions.
- Option B: 0.3 V is the characteristic barrier potential of Germanium (Ge), not Silicon (Si).
- Option D: 1.5 V is typical for wide-bandgap LEDs (e.g., GaAs, GaP), not standard silicon diodes.
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