Question

In a common emitter amplifier, when the base to emitter voltage changes by \(20\,\mathrm{mV}\), the change in collector current is \(3\,\mathrm{mA}\). If the output resistance is \(4\,\mathrm{k\Omega}\), then the voltage gain of the amplifier is

• A. \(300\)
• B. \(600\)
• C. \(400\)
• D. \(800\)

Hint:

Voltage gain of a common emitter amplifier is \[ \boxed{ A_v=\frac{\Delta V_o}{\Delta V_i} =\frac{\Delta I_C\,R_o}{\Delta V_{BE}}. } \]

Answer 1

The Correct Option is B

Step 1: Calculate the change in output voltage. The output voltage change is \[ \Delta V_o = \Delta I_C\,R_o. \] Given, \[ \Delta I_C = 3\,\mathrm{mA} = 3\times10^{-3}\,\mathrm{A}, \] \[ R_o = 4\,\mathrm{k\Omega} = 4000\,\Omega. \] Hence, \[ \Delta V_o = 3\times10^{-3}\times4000 = 12\,\mathrm{V}. \]

Step 2: Calculate the voltage gain. Input voltage change, \[ \Delta V_i = 20\,\mathrm{mV} = 0.02\,\mathrm{V}. \] Therefore, \[ A_v = \frac{\Delta V_o}{\Delta V_i} = \frac{12}{0.02} = 600. \] Hence, \[ \boxed{600} \] Therefore, \[ \boxed{(B)} \] is the correct answer.