Question

Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, Av, of the amplifier is given by $A_v=\frac{-{\beta }_{ac}{R}_L}{r_i}$, where ${\beta }_{ac}$ is the current gain; $R_L$ is the load resistance and $r_i$ is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?

Answer 1

In the circuit the emitter is forward biased and collector is reversed biased. this makes input resistance $\left(R_i\right)$ very low and out resistance $\left(R_{out}\right)$ very high.during positive half cycle of input AC decreases the forward bias hence emitter current $\left(I_E\right)$ and by transistor collector current increases. this tend to increase the collector voltage.

which is given by

$V_0=V_{CE}=V_{CC}-I_C{R}_L$

The high value of $R_L$ produces large change in $V_0$ corresponding to low change in $V_i$ . thus amplified pulse obtained at collector.

Voltage gain : $voltagegain\left(A_V\right)=\frac{\Delta V_{CE}}{\Delta V_{BE}}=\frac{\left(\Delta I_C\right)R_L}{\left(\Delta I_B\right)R_i}$

Where $R_L$ and $R_i$ are output and input resistance respectively.

$\left(A_V\right)=\left(\frac{\left(\Delta I_C\right)}{\left(\Delta I_B\right)}\right)\frac{R_L}{R_i}$

$Since{\beta }_{AC}=\frac{\left(\Delta I_C\right)}{\left(\Delta I_B\right)}$

$\left(A_V\right)={\beta }_{AC}\frac{R_L}{R_i}$

The output voltage of CE amplifier differ in phase from input voltage by 180 degree or in opposite phase is represented by

Therefore voltage gain

$\left(A_V\right)={-\beta }_{AC}\frac{R_L}{R_i}$