The current in the forward bias is known to be more (in `mA`) than the current in the reverse bias (in ` mu A ` ). What is the reason then to operate the photodiodes in reverse bias?

The current in the forward bias is known to be more (in `mA`) than the

1.	The current in the forward bias is known to be more (in `mA`) than the current in the reverse bias (in ` mu A ` ). What is the reason then to operate the photodiodes in reverse bias?
Answer» In case of n-type semiconductor, let n be the majority carrier (i.e., electrons) density and p be the minority carrier (holes) density. Where `n gt p`. On illumination of semiconductor, there will be production of equal number of electrons and holes. Let `Deltan, Deltap` be the increase in majority carrier density and minority carrier density due to illumination of semiconductor, where `Deltan=Deltap`. Hence, fractional change in majority carrier `(=Deltan//n)`, fractional change in minority carrier `(=Deltap//p)`. Since `n gt gt p`, so `(Deltan)/n lt (Deltap)/p` it means, due to photo-effects the fractional change due to minority carriers dominates. As a result of it, the fractional change in the reverse bias current is more easity measurable than the fractional change in the forward bias current. It is due to this reason, photodiodes are preferably used in the reverse bias condition for measuring light intensity.