The input resistance of a silicon transistor is `665Omega`. Its base current is changed by `15muA` which results in the change in collector current by `2mA`. This transistor is used as a common emitter amplifier with a load resistance of `5kOmega`. What is the voltage gain of the amplifier.

Options

A. 100
B. 1000
C. 500
D. 2000

Answer

To find the voltage gain of the common emitter amplifier, we will follow these steps:

### Step 1: Calculate the current gain (β) of the transistor.
The current gain (β) is defined as the ratio of the change in collector current (ΔI_C) to the change in base current (ΔI_B).

\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]

Given:
- ΔI_C = 2 mA = 2 × 10^-3 A
- ΔI_B = 15 μA = 15 × 10^-6 A

Substituting the values:

\[ \beta = \frac{2 \times 10^{-3}}{15 \times 10^{-6}} = \frac{2 \times 10^{3}}{15} \approx 133.33 \]

### Step 2: Calculate the voltage gain (A_v) of the amplifier.
The voltage gain of a common emitter amplifier can be calculated using the formula:

\[ A_v = \beta \times \frac{R_L}{R_{in}} \]

Where:
- \( R_L \) is the load resistance
- \( R_{in} \) is the input resistance of the transistor

Given:
- \( R_L = 5 k\Omega = 5 \times 10^{3} \Omega \)
- \( R_{in} = 665 \Omega \)

Substituting the values into the formula:

\[ A_v = 133.33 \times \frac{5 \times 10^{3}}{665} \]

Calculating \( \frac{5 \times 10^{3}}{665} \):

\[ \frac{5 \times 10^{3}}{665} \approx 7.51 \]

Now substituting back into the voltage gain equation:

\[ A_v \approx 133.33 \times 7.51 \approx 1000 \]

### Conclusion
The voltage gain of the amplifier is approximately **1000**.