(a) Show that a bubble AND gets is equivalent ot a NOE gate.
(b) The number of electron-hole pairs in an intrinsic semiconductor is `2xx10^(19)m^(-3)` at `27^(@)C` and `E_(g)` is 1eV. Calculate the number of electron-hole pairs at `227^(@)C`. Given that Boltzmann constant is `8.65xx10^(-5)eV//K`
or With the help of neat and labelled diagram, discuss the working of common emitter p-n-p junction transistor amplifier. Obtain expressions for voltage gain and current gain.

Explanation

(a) The bubbled AND gate is a combination of two input NOT gates and an AND gate

The Booklean for output of the gates is `=vecA,vecB=vecA+B`, which is the expression for NOR gate. Hence bubbled AND gate acts as a NOR gate.
(b) As, `n_(t)=Ce^(-Eg//2kT)`
So `n_(27)=2xx10^(19)=Ce^(-E_(g)l(2xxkxx300))`
and `n_(227)=Ce^(-E_(g)l(2kxx500))`
`(n_(227))/(2xx10^(19))=e^((Eg)/(2k)[(1)/(300)-(1)/(500)])=e^((1)/(2xx8.65xx10^(-5))xx(2)/(1500))`
`"log"_(e)(n_(227))/(2xx10^(19))=(2)/(2xx8.65xx10^(-5)xx1500)`
`"log"_(10)(n_(227))/(2xx10^(19))=(2)/(2xx8.65xx10^(-5)xx1500xx2.3026)`
On solving, `n_(227)=4.475xx10^(22)m^(-3)`