Question

With the help of neat circuit diagram obtain an expression for voltage gain of a transistor amplifier in $C-E$ configuration.

Answer 1

The working of an amplifier can be easily understood, if we first assume that $v_i=0$. Then applying Kirchhoff's law to the output loop, we get

$V_{CC}=V_{CE}+I_C{R}_L$ ----------- (1)

Likewise the input loop gives

$V_{BB}=V_{BE}+I_B{R}_B$

When $v_i$ is not zero, we get
$V_{BE}+V_i=V_{BE}+I_B{R}_B+\Delta I_B(R_B+r_i)$

The change in $V_{BE}$ can be related to the input resistance $r_i$ and the change in $I_B$. Hence

$v_i=\Delta I_B(R_B+r_i)=r\Delta I_B$

The change in $I_B$ causes a change in $I_C$. We define a parameter ${\beta }_{ac}=\frac{\Delta I_C}{\Delta I_C}=\frac{i_c}{i_b}$ this is also called current gain.

The change in $I_C$ is due to the change in $I_B$ causes a change in $V_{CE}$ and the voltage drop across the resistor $R_L$ because $V_{CC}$ is fixed.

From equation (1)

$\Delta V_{CC}=\Delta V_{CE}+R_L\delta I_C=0$

$\Delta V_{CE}=-R_L\Delta I_C$

The change in $V_{CE}$ is the output voltage $v_0$

$v_0=\Delta V_{CE}=-{\beta }_{ac}{R}_L\Delta I_B$

Now the voltage gain of the amplifier is

$A_v=\frac{v_0}{v_i}=\frac{\Delta V_{CE}}{r\Delta I_B}=\frac{{\beta }_{ac}{R}_L}{r}$