Question

Why is a photodiode used in reverse bias?
(A). Photocurrent flows only in reverse bias.
(B). Thermally generated current is large in reverse bias as compared to the photocurrent.
(C). Photocurrent is large in forward bias as compared to the forward bias current.
(D). Photocurrent is significant in reverse bias as compared to the reverse bias current

Answer 1

Hint: The incident light creates electron-hole pairs in the depletion region. Greater the area of depletion region, more are the electron-hole pairs that can be created by the incident light which is directly proportional to the amount of current produced.

Complete step-by-step answer:
A photodiode is a semiconductor electronic device which is used to convert light energy into electrical energy. When light is made to fall on the depletion layer of the photodiode, the photons get absorbed and their energy is transferred to the creation of electron-hole pairs which constitute a current flowing through the circuit.
A photodiode is basically a p-n junction. When a photon of sufficient energy called the threshold energy falls on the diode, it leads to the creation of an electron-hole pair in the depletion region. These newly created electron-hole pairs move towards the opposite electrode which constitutes what we call photocurrent. This field that sweeps charge carriers towards opposite electrodes comes from the built-in electric field of the depletion region.
Now when diode is forward biased, the current is mostly due to the majority charge carriers whereas in case of reverse bias, the current is due to the minority charge carriers. When a diode is forward biased, the width of the depletion layer decreases whereas the width increases in case of reverse biasing.
Now if the width of the depletion layer is large like in the case of reverse biasing then photons will create more electron-hole pairs, that is, more is the photocurrent as compared to when width is smaller like in the case of forward biasing.
Therefore the photocurrent produced in reverse bias is greater in comparison to the reverse bias current which is smaller since it is to minority charge carriers.
Hence, the correct answer to our question is option D.

Note: So that student is not confused by other options, following points need to be kept in mind.
The photocurrent produced in case of forward is as such small compared to the forward current which is large since its being produced by majority charge carriers. So, option C is wrong.
Option A and B are obviously wrong as photocurrent flows both in forward and reverse bias, just their magnitude is different due to which depletion layer. And as discussed, the photocurrent is much larger than reverse bias current.