Photodiode
A photodiode used as a light sensor in an automatic street lamp is deliberately connected in reverse bias rather than forward bias during normal operation. Why is reverse bias preferred for this device?
Select the correct option:
Solution
The fractional change in reverse current with illumination is large and easy to measure
A photodiode detects light because incident photons with energy above the band gap generate extra electron-hole pairs in or near the depletion region, increasing the current. It is operated in reverse bias because, in the dark, only a tiny reverse saturation current flows, set by minority carriers. When light falls on the junction, the additional photo-generated minority carriers cause a relatively large fractional increase in this small reverse current, making the change easy to detect and nearly linear with light intensity. In forward bias the current is already large and dominated by majority carriers, so the small light-induced contribution would be swamped and hard to resolve. The option claiming reverse bias gives a larger total current is wrong; reverse current is actually very small in magnitude. The heat-prevention option is irrelevant to the choice of bias. The option saying forward bias cannot create pairs is false, since light generates pairs regardless of bias. As a check, sensitivity is about resolving fractional change against a low dark background, which reverse bias provides best.
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- photodiode
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
The fractional change in reverse current with illumination is large and easy to measure
A photodiode detects light because incident photons with energy above the band gap generate extra electron-hole pairs in or near the depletion region, increasing the current. It is operated in reverse bias because, in the dark, only a tiny reverse saturation current flows, set by minority carriers. When light falls on the junction, the additional photo-generated minority carriers cause a relatively large fractional increase in this small reverse current, making the change easy to detect and nearly linear with light intensity. In forward bias the current is already large and dominated by majority carriers, so the small light-induced contribution would be swamped and hard to resolve. The option claiming reverse bias gives a larger total current is wrong; reverse current is actually very small in magnitude. The heat-prevention option is irrelevant to the choice of bias. The option saying forward bias cannot create pairs is false, since light generates pairs regardless of bias. As a check, sensitivity is about resolving fractional change against a low dark background, which reverse bias provides best.
This easy difficulty physics question is from the chapter semiconductor electronics, covering the topic of photodiode. It appeared in the 2025 exam.
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