Question

A transistor is used as a common emitter amplifier with a load resistance of 2 k\(\Omega\). The input resistance is 150 \(\Omega\). Base current is changed by 20 \(\mu A\), which results in a change in collector current by 1.5 mA. The amplified voltage gain of the amplifier is

• A. 500
• B. 1000
• C. 750
• D. 1250

Hint:

The voltage gain of a transistor amplifier is given by the product of the current gain and the ratio of load resistance to input resistance.

Answer 1

The Correct Option is B

Step 1: Calculate the current gain.
The current gain \( \beta \) (also called the current amplification factor) is the ratio of the change in collector current (\(\Delta I_C\)) to the change in base current (\(\Delta I_B\)): \[ \beta = \frac{\Delta I_C}{\Delta I_B} \] Given that \( \Delta I_C = 1.5 \, \text{mA} = 1.5 \times 10^{-3} \, \text{A} \) and \( \Delta I_B = 20 \, \mu A = 20 \times 10^{-6} \, \text{A} \), we have: \[ \beta = \frac{1.5 \times 10^{-3}}{20 \times 10^{-6}} = 75 \] Step 2: Calculate the voltage gain.
The voltage gain \( A_v \) of the amplifier is the product of the current gain and the ratio of load resistance \( R_L \) to the input resistance \( R_{\text{in}} \): \[ A_v = \beta \times \frac{R_L}{R_{\text{in}}} \] Given that \( R_L = 2 \, \text{k}\Omega = 2000 \, \Omega \) and \( R_{\text{in}} = 150 \, \Omega \), we have: \[ A_v = 75 \times \frac{2000}{150} = 75 \times 13.33 = 1000 \] Step 3: Conclusion.
Thus, the amplified voltage gain of the amplifier is 1000, which corresponds to option (B).