Question

The input resistance of a common emitter amplifier is $330\Omega $ and the load resistance is $5{\text{k}}\Omega $ . If a change of $15\mu {\text{A}}$ in the base current changes the collector current by $1{\text{mA}}$ , find the voltage gain of the amplifier.
A)\[1000\]
B)\[10001\]
C)\[1010\]
D)\[1100\]

Answer 1

Hint: Voltage gain is a measure of how the output voltage changes corresponding to a change in the input voltage. In a common emitter amplifier, the input voltage is supplied at the base and the corresponding output voltage is obtained at the collector.

Formula used: The voltage gain of a common emitter amplifier is given by, ${A_v} = - \beta \dfrac{{{r_L}}}{{{r_i}}}$ , where $\beta $ is the current gain of the amplifier, ${r_i}$ is the input resistance and ${r_L}$ is the load resistance.
The current gain of the amplifier is given by, $\beta = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}$ , where $\Delta {I_C}$ is the change in the collector current and $\Delta {I_B}$ is the change in the base current of the amplifier.

Complete step by step answer.
Step 1: List the data given in the problem.
The input resistance of the common emitter amplifier is $330\Omega $ .
The load resistance of the common emitter amplifier is $5{\text{k}}\Omega = 5000\Omega $ .
Given, change in the base current is $\Delta {I_B} = 15\mu {\text{A}}$ and the corresponding change in the collector current is $\Delta {I_C} = 1{\text{mA}}$ .
Step 2: Obtain the current gain of the amplifier.
We have, change in base current $\Delta {I_B} = 15\mu {\text{A = 15}} \times {\text{1}}{{\text{0}}^{ - 6}}{\text{A}}$ and change in collector current $\Delta {I_C} = 1{\text{mA = 1}}{{\text{0}}^{ - 3}}{\text{A}}$ .
The current gain is $\beta = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}$ .
Substituting the values $\Delta {I_B} = {\text{15}} \times {\text{1}}{{\text{0}}^{ - 6}}{\text{A}}$ and $\Delta {I_C} = {\text{1}}{{\text{0}}^{ - 3}}{\text{mA}}$ in the above equation, we obtain the current gain $\beta = \dfrac{{{\text{1}}{{\text{0}}^{ - 3}}}}{{{\text{15}} \times {\text{1}}{{\text{0}}^{ - 6}}}} = 66.67$ .
Step 3: Using the value of the current gain, find the voltage gain of the amplifier.
The voltage gain of a common emitter amplifier is given by, ${A_v} = - \beta \dfrac{{{r_L}}}{{{r_i}}}$ ------- (1).
The input resistance ${r_i} = 330\Omega $ and load resistance ${r_L} = 5000\Omega $ of the CE amplifier are known. We have also calculated the current gain, $\beta = 66.67$ .
Substituting these values in equation (1) we obtain the voltage gain as, ${A_v} = 66.67 \times \left( {\dfrac{{5000}}{3}} \right) = - 1010.1$ .
The negative sign implies that the output voltage of the CE amplifier is out of phase with its input voltage.
Therefore the voltage gain of the common emitter amplifier is ${A_v} \cong 1010.1$ .

Note: Instead of finding the current gain first and then finding the voltage gain using its value, we can express the voltage gain as ${A_v} = - \dfrac{{\Delta {I_C} \times {r_L}}}{{\Delta {I_B} \times {r_i}}}$ and then substitute all the values at once. All the quantities are expressed in their respective S. I. units while substitution.