Question

(a) Draw the circuit diagram of an n-p-n transistor amplifer in common emitter configuration.
(b) Derive an experssion for voltage gain of the amplifier and hence show that the output voltage is in opposite phase with the input voltage.

Answer 1

(a)

(b) Voltage gain of the amplifier :
If $V_i=0$
$V_{CC}=V_{CE}+I_C{R}_L$
Similarly, $V_{BB}=V_{BE}+I_B{R}_B$
If $V_i$ is not zero,
$V_{BB}+V_i=VBE+I_B{R}_B+\Delta I_B(R_B+R_i)$
$\Rightarrow V_i=\Delta I_B(R_B+R_i)$
$\Rightarrow V_i=r\Delta I_B$
Change in $I_B$ causes a change in $I_C$.
${\beta }_{ac}=\frac{\Delta I_C}{\Delta I_B}=\frac{I_C}{I_B}$
As $\Delta V_{CC}=\Delta V_{CE}+R_L\Delta I_C=0$
$\Rightarrow \Delta V_{CE}=-R_L\Delta I_C$
The change in $V_{CE}$ is the output voltage $V_0$,
$\Rightarrow V_0=R_L\Delta I_C$
$=-{\beta }_{ac}\Delta I_B{R}_L$
$\therefore \text{Voltage gain},A_V=\frac{V_o}{V_i}=-\frac{{\beta }_{ac}\Delta I_B{R}_L}{r\Delta I_B}AV=-{\beta }_{ac}\frac{R_L}{r}$
The negative sign shows that there is a phase difference of radians between input and output.