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The current in the forward bias is known to be more \(\left({\sim}\mathit{mA}\right)\)  than the current in the reverse bias \(\left({\sim}\mu A\right).\)  What is the reason then to operate the photodiodes in reverse bias?

Answer

Consider the case of an \(n\) -type semiconductor. Obviously, the majority carrier density \(\left(n\right)\)  is considerably larger than the minority hole density \(p \;( \text{i.e.}, n\gg p)\) . On illumination, let the excess electrons and holes generated be \(\Delta n\)  and \(\Delta p,\)  respectively:
\(n^\prime =n+\Delta n\)
\(p^\prime =p+\Delta p\)
Here \(n^\prime \)  and \(p^\prime \)  are the electron and hole concentrations at any particular illumination and \(n\)  and \(p\)  are carriers concentration when there is no illumination. Remember \(\Delta n=\Delta p \; \text{and} \;n{\gg}p.\)  Hence, the fractional change in the majority carriers \((\text{i.e.}, \Delta n\text /n )\)  would be much less than that in the minority carriers \(( \text{i.e.}, \Delta p\text /p )\) . In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity.
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