The input resistance of a common-emitter amplifier is `2 kOmega` and a.c. Current gain is 20. If the load resistance used is `5 kOmega`, calculate the transconductance of the transistor used

Options

A. `0.01 Omega^(-1)`
B. `0.03 Omega^(-1)`
C. `0.04 Omega^(-1)`
D. `0.07 Omega^(-1)`

Answer

To calculate the transconductance (\( g_m \)) of the transistor used in the common-emitter amplifier, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Input resistance (\( R_{in} \)) = \( 2 \, k\Omega = 2 \times 10^3 \, \Omega \)
- AC current gain (\( \beta \)) = 20
- Load resistance (\( R_L \)) = \( 5 \, k\Omega = 5 \times 10^3 \, \Omega \)

2. **Formula for Transconductance**:
The transconductance (\( g_m \)) can be calculated using the formula:
\[ g_m = \frac{\beta}{R_{in}} \]

3. **Substitute the Values**:
Substitute the values of \( \beta \) and \( R_{in} \) into the formula:
\[ g_m = \frac{20}{2 \times 10^3} \]

4. **Calculate \( g_m \)**:
Now perform the calculation:
\[ g_m = \frac{20}{2000} = 0.01 \, \text{S} \, (\text{Siemens}) \]

5. **Final Result**:
Therefore, the transconductance of the transistor used is:
\[ g_m = 0.01 \, \text{S} \, \text{or} \, 10^{-2} \, \text{S} \]