Concept:Incident photons with energies greater than the bandgap (\( hf \ge E_g \)) are absorbed in or near the depletion region, generating electron-hole pairs. The reverse electric field sweeps these carriers across the junction to produce a detectable photocurrent.
Formula:$$I_{\text{photo}} = q \eta_{\text{quantum}} \left(\frac{P_{\text{opt}}}{h f}\right) \propto \Phi \quad (\text{Optical Power})$$
Solution:- Incoming photons with \( hf \ge E_g \) excite valence electrons into the conduction band within the depletion layer.
- The built-in reverse electric field rapidly separates the resulting electron-hole pairs before they can recombine.
- Electrons are swept to the N-side and holes to the P-side, producing a reverse photocurrent proportional to light intensity.
Why other options are incorrect:- Option B: Photodiodes are solid-state semiconductor devices, not vacuum tubes.
- Option C: Electroluminescence occurs in forward-biased LEDs, not reverse-biased photodetectors.
- Option D: Photodiodes operate entirely via internal solid-state carrier generation.
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