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

Answer

A

Explanation

Let us take an example of P - type semiconductor .
Without illumination
number of holes `(n_h ) gtgt` number of electrons ` (n_e)`
This is because holes are the majority charge carriers in p - type semiconductor .
On illumination, let ` Delta n_e` and ` Delta n_h ` are the excess electrons and holes generated .
` Delta n_e = Delta n_h`
From Eqs. (i) and (ii) , we can see that
` (Delta n_e) / n_e gtgt (Delta n_h )/ n_h`
From here, we can say that the fractional change due to illumination on the minority carrier
dominated reverse bias current is more easily measurable than the fractional change in the
forward bias current.