With the help of circuit diagram, explain the working of transistor as a common emitter amplifier.

Explanation

The Amplifier circuit using p-n-p transistor is shown in the following Figure. Figure shows that the base-emitted circuit is forward biased. Therefore the resistance in the input circuit in less. The collector-emitter circuit is output circuit and reverse biased. Therefore resistance in the output circuit is more. When no a.c input signal is applied, then the transistor works normally and `I_(e)= I_(b) + I_(c )`. When `I_(c)` flows through load resistance `R_(L)`, then potential drop equal to `I_(C )R_(L)` will occur. Therefore output voltage is

`V_(o)= V_(CE)- I_(C )R_(L)` ..(i)
When input a.c signal is applied, it will change emitter current `I_(e)` and hence collector current `I_(c)`. Therefore, from eqn (i), output voltage `(V_(o))` will appear as amplifier voltage.
During positive half cycle of a.c., the forward voltage of emitter -base junction decreases. Due to this, emitter current `(I_(e))` and hence collector current `(I_(c))` also decrease. Therefore, voltage drop across `R_(L)` i.e., `I_(C )R_(L)` will also decrease. Therefore, from eqn (i) output voltage increases. As the collector is connected to negative terminal of the battery, so increase in output voltage means the collector voltage becomes more negative i.e., the output voltage `V_(o)` will vary in negative sense and corresponding to positive half cycle of a.c. input, negative output half cycle will be obtained. During negative half-cycle of a.c. the forward voltage of emitter -base junction increases. Due to this, emitter current `(I_(e))` and hence collector current `(I_(o))` also increase. Therefore, voltage drop across `R_(L)` i.e., `I_(C )R_(L)` increases. Hence from eqn (i), the output voltage decreases. As the collector is connected to negative terminal of battery `V_(CE)`, so decrease in output voltage means the output voltage becomes less negative i.e output voltage `V_(o)` will vary in positive sense and corresponding to negative half cycle of a.c input, positive output half cycle will be obtained. Hence, both input and output voltage are out of phase.
(i) A.C. current gain `(beta)` It is the ratio of change in output current `(Delta I_(c))` to the change in input current `(Delta I_(b))` i.e., `beta= (Delta I_(c ))/(Delta I_(b))`
(ii) A.C. voltage gain `(A_(V))` It is the ratio of change in output voltage `(Delta V_(0))` to change in input voltage `(Delta V_(i))`
`A_(V)= (Delta V_(o))/(Delta V_(i))`. Here `Delta V_(o)= Delta I_(c) R_(o) and Delta V_(i) = Delta I_(b) R_(i)`
`A_(V)= (Delta I_(c)R_(o))/(Delta I_(b) R_(i))= beta xx` resistance gain
(iii) AC power gain: It is defined as the ratio of change in output power to change in input power. AC power gain `("Change in output power")/("Change in input power") = ((Delta I_(c))^(2)R_(o))/((Delta I_(b))^(2) R_(i))= beta_(ac)^(2) xx` resistane gain.