Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Photodiodes are preferably operated in reverse bias condition for light intensity measurement.
Reason R : The current in the forward bias is more than the current in the reverse bias for a p-n junction diode.
In the light of the above statements, choose the correct answer from the options given below :

Solution

Physics sir · 5 steps

Reverse bias wins because light changes its tiny current by a big fraction, not because forward is larger.

1. Check R. In forward bias, majority carriers cross the junction and the current is in milliamperes. In reverse bias, only minority carriers flow and the current is in microamperes. So R is true.
2. Check A. Light creates electron-hole pairs near the junction. This mostly adds minority carriers, so it raises the reverse current. Photodiodes are indeed run in reverse bias to measure light intensity, so A is true.
3. Why reverse bias. The extra current from light is about the same size in either bias. Added to a tiny reverse current, it is a large fractional change, which is easy to measure. Added to a large forward current, it is a tiny fraction and gets lost.
4. Does R explain A. R only says the forward current is larger. It says nothing about the fractional change caused by light, which is the real reason for using reverse bias. So R does not explain A.
5. So. Both statements are true, but R is not the correct explanation of A. The answer is (d).

The idea to keep. A detector works best where the signal is a big fraction of the background. For a photodiode that means reverse bias, where the dark current is tiny and light changes it a lot.

Answer from the NTA final answer key. Solution by PrepWiser.