InterviewSolution
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InterviewSolution
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The current in the forward bias is known to be more (-mA) than the current in the reverse bias (-µA). What is the reason then, to operate the photodiodes in reverse bias? |
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Answer» In case of n-type semi- conductors, n>p, Where n = majority carrier density p = minority carrier density On illumination of semi-conductor, there will be production of equal number of electrons and holes. If Δn and Δp be the increase in majority carrier density and minority density carrier density due to illumination of semiconductor, then fractional change in majority carrier = \(\frac{Δn}{n}\), and fractional change in minority carrier = \(\frac{Δp}{p}\) Since, n>>p, So, \(\frac{Δn}{n}\) < \(\frac{Δp}{p}\), 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 easily measurable then 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. |
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