Concept:A photodiode operates in reverse bias because the wide depletion region and low dark current allow incident photons to generate detectable electron-hole pairs via the photoelectric effect.
Formula:$$I_{\text{total}} = I_0 + I_{\text{photo}} \quad \text{where } I_{\text{photo}} \propto \Phi \text{ (optical flux)}$$
Solution:- In reverse bias, the dark current (background current) is extremely small (order of \( \text{nA} \)).
- When light with \( h f > E_g \) strikes the depletion region, newly generated electron-hole pairs are separated by the internal electric field.
- This produces a measurable reverse photocurrent proportional to light intensity.
Why other options are incorrect:- Option A: In forward bias, large majority carrier diffusion current would obscure small photo-induced current changes.
- Option B: Thermal generation produces noise without the field-assisted carrier separation of reverse bias.
- Option D: Alternating bias causes periodic switching transients that interfere with steady optical detection.
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