In a common emitter amplifier , when a signal of 40 mV is added to the input voltage, the base current changes by `100 muA` and emitter current changes by 2.1mA the trans-conductance is

Options

A. `1/20 Omega^(-1)`
B. `1/50 Omega^(-1)`
C. `50 Omega^(-1)`
D. `15 Omega^(-1)`

Answer

To find the transconductance (gm) of a common emitter amplifier, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Change in input voltage (ΔVbe) = 40 mV = 0.040 V
- Change in emitter current (ΔIe) = 2.1 mA = 0.0021 A

2. **Understand the Formula for Transconductance:**
- The formula for transconductance (gm) is given by:
\[ g_m = \frac{\Delta I_e}{\Delta V_{be}} \]

3. **Substitute the Given Values into the Formula:**
- Substitute ΔIe and ΔVbe into the formula:
\[ g_m = \frac{0.0021 \, \text{A}}{0.040 \, \text{V}} \]

4. **Calculate the Transconductance:**
- Perform the division:
\[ g_m = \frac{0.0021}{0.040} = 0.0525 \, \text{S} \]

5. **Convert to Ohm Inverse:**
- Since transconductance is often expressed in ohm inverse (Ω⁻¹):
\[ g_m = 52.5 \, \text{mS} = 52.5 \, \text{Ω}^{-1} \]

6. **Final Answer:**
- Therefore, the transconductance (gm) is approximately:
\[ g_m \approx 50 \, \text{Ω}^{-1} \]

### Conclusion:
The correct option for transconductance is option 3: 50 Ω⁻¹.