In a single state transistor amplifier, when the signal changes by `0.02 V` the base current by `10 mu A` and collector current by `1 mA`. If collector load `R_(C) = 2k Omega` and `R_(L) = 10 k Omega`, Calculate :
(i) Current Gain
(ii) Input Impedance,
(iii) Effective `AC` load,
(iv) Voltage gain and
(v) Power gain.

Explanation

(i) Current Gain `beta = (Delta i_(c))/(Delta i_(b)) = (1mA)/(10 muA) = 100`
(ii) Input impedance
`R_(i)=(Delta V_(BE))/(Delta i_(b)) = (0.02)/(10 muA) = 2000 Omega = 2k Omega`
(iii) Effective `(a,c)` load
`R_(AC) =R_(C)||R_(L) :. R_(AC) = (2 xx 10)/(2+10) =1.66 k Omega`
(iv) Voltage gain `A_(v) = beta xx (R_(AC))/(R_(i n)) = (100 xx 1.66)/(2) = 83`
(v) Power gain, `A_(P)` = Current gain xx Voltage gain `= 100 xx 83 = 8300`.