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The current in theforward bias is known to be more ( ~ mA) than the current in the reverse bias ( ~muA) .What is the reasonthen to operate the photodiodes in reverse bias ? |
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Answer» SOLUTION :Consider the CASE of an n-type semconductor . Obviously, the majority carrier density ( n ) is considerably larger than the minority hole density p( i.e., `n gt gt p ) `. On illumination, let the excess electrons and holes generated be `Delta n ` and `Delta p` , respectively. ` n' = n + Delta n ` `p' = p +Delta p` Here n' and p' are the electron and hole concentrations at any particular illumination and n and p are carriers concentrationwhen there is no illumination. Remember `Delta n = Delta p ` and `n gt gt p` . HENCE,the fractional change in the majoritycarriers (i.e., `Delta n //n ) ` WOULD be much less than that in the minoritycarriers ( i.e., `Delta p //p ` ) . In general, we can state thatthe fractional change due to the photo-effects on the minority carrier dominated REVERSE bias current is more easily measurable than the fractional changein the forward bias current. Hence, photodiodes are preferablyused in the reverse bias condition for measuring light intensity . |
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