Saved Bookmarks
| 1. |
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 photndiode in reverse bias ? |
|
Answer» Solution :LET us consider an n-type semiconductor which has electrons as majority charge carriers and holes as minority charge carriers such that `n_(e) gt gt n_(h)` . On illumination with appropriate light let excess electrons generated be `Delta n_(e) and Delta n_(p)` such that ` n_(e)' = n_(e) + Delta n_(e) and n_(p)' = n_(e) + Delta n_(e)`.Here `n_(e)' and n_(p)' ` are the electron and hole CONCENTRATIONS at a particular illumination level and `n_(e) and n_(p)`are carrier CONCENTRATION in the absence of illumiation. Due to illumination `Delta n_(e) = Delta n_(p) and n_(e) gt gt n_(p)`.Hence, FRACTIONAL change in majority charge carriers (electrons) `(Delta n_(e))/( n_(e))`would be much less than corresponding change in minority charge carriers (holes)` (Delta n_(p))/(n_(p))`.THUS, fractional change on minority charge carriers is more easily measurable when the semiconductor is illuminated. Due to this reason a photodiode is preferably used in reverse bias only. |
|