State the reason, why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode, has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of wavelength of 400nm incident on it

Explanation

Because photodiodes conduct in reverse biased condition only whe a light of suitable frequency falls on it is used to detect the variation of intensity of light. During reverse bias, the current is mainly due to the diffusion of minority carries through the junctions. The variation in intensity of light affects the minority carrier-based reverse current much more than the majority carrier-based forward current. So, a photodiode is used in reverse-bias.
For operating a photodiode, reverse bias is kept below the breakdown voltage. Conductivity of a photodiode changes with the absorption of light around depletion layer.
For the relation, `E = hv (" where " v = .^(c)//_(lamda))`
`rArr E = (hc)/(lamda) = (6.63 xx 10^(-34) xx 10^(-34) xx 3 xx 10^(8))/(1.6 xx 10^(-19) xx 400 xx 10^(-19)) rArr E = 3.105 eV`
As the energy gap is `E_(g) = 1.2 eV` (for photodiode,) so light can be detected (`(hc)/(lamda)` energy photon)