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A p-n junction in REVERSE BIAS is irradiated with light of photon energy slightly GREATER than the band gap of the semiconductor. What happens to the reverse current?
AIt drops to zero because the depletion region is destroyed
BIt remains unchanged — light has no effect on diodes
CIt increases — photons generate extra electron-hole pairs
DIt oscillates with the frequency of the incident light
Answer & Solution
Correct answer: C. It increases — photons generate extra electron-hole pairs
1. When photons with energy $h\nu > E_g$ are absorbed in the depletion region (or near it), they create extra electron-hole PAIRS by exciting valence electrons across the band gap.
2. The strong reverse-bias internal field IMMEDIATELY separates these pairs and sweeps them across the junction: electrons to the n side, holes to the p side.
3. This generates an additional component of reverse current — the PHOTOCURRENT — that adds to the small thermally-generated $I_0$.
4. This is the principle of the PHOTODIODE: a reverse-biased p-n junction whose reverse current measures the incident light intensity. Solar cells operate on the same principle (but in open-circuit / load mode).
5. The depletion region is not destroyed (A), light DOES affect a reverse-biased junction (B), and the current is steady, not oscillating at the optical frequency (D).
_Source: NCERT Class 12 Physics Part 2, Ch 14, §14.6.2 (Reverse bias and special diode applications — photodiode, solar cell), p. 13–14._
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