The current in the forward bias is known to be more (~ mA) than the current in the reverse bias (~ μA), thwn why photon diodes is to operate in reverse bias. Rajiv did not know its cause. So he requested his friend Sanjiv for its answer. Sanjiv explained him that in case of n-type semiconductor, the majority carrier (electron) density a is considerably larger than the minority hole density p, i.e., n > > p. On illumination, if the excess electrons and holes generated be Δn and Δp respectively, then
= n + Δn and p' = p + Δp
Here, An : Δp and n > > p. Hence the fractional change Δn/n would be much less than Δp/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. So, photo diodes are preferably used in the reverse bias condition for measuring light intensity.
(a) What values are noticed in Sanjiv ?
(b) For a CE transistor amplifier the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Current amplification factor of the transistor is 100. Find the input signal voltage and the base current, if its resistance is 1 kΩ.