Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
Reason R : For a p-n junction diode, at applied voltage V the current in the forward bias is more than the current in the reverse bias for |V_z| > ± v ≥ |v₀| where v₀ is the threshold voltage and V_z is the breakdown voltage.
In the light of the above statements, choose the correct answer from the options given below

Solution

Physics sir · 4 steps

A photodiode works in reverse bias, where light makes a big relative change in a tiny current.

1. Check Assertion A. In forward bias the current is large and comes from majority carriers. The few extra carriers made by light hardly change it, so the fractional change is too small to measure well.
2. Why reverse bias. In reverse bias the current is a tiny minority-carrier current. Light creates electron-hole pairs, which add directly to it, so the fractional change is large and easy to detect. Photodiodes are therefore used in reverse bias, and A is false.
3. Check Reason R. For a voltage magnitude between the threshold v₀ and the breakdown voltage |V_z|, the forward current is in milliamperes. The reverse current is only microamperes, because breakdown has not yet been reached. So the forward current is larger, and R is true.
4. So. A is false and R is true, which is option (d).

The idea to keep. Photodiodes are used in reverse bias because light changes the tiny reverse current by a large fraction. That makes intensity easy to read. In forward bias the same change is lost in a big majority-carrier current.

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