In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductor 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will

Options

A. `(2)/(3)G`
B. `1.5G`
C. `(1)/(3)G`
D. `(5)/(4)G`

Answer

To solve the problem, we need to find the new voltage gain \( G' \) when the transistor in a common emitter amplifier is replaced. We start with the given parameters and use the relationship between transconductance, current gain, and voltage gain.

### Step-by-Step Solution:

1. **Understand the Voltage Gain Formula**:
The voltage gain \( A_v \) of a common emitter amplifier can be expressed as:
\[ A_v = \beta \times \frac{R_o}{R_i} \]
where \( \beta \) is the current gain, \( R_o \) is the output resistance, and \( R_i \) is the input resistance.

2. **Calculate Input Resistance for the First Transistor**:
For the first transistor, we have:
- Transconductance \( g_m = 0.03 \, \text{mho} \)
- Current gain \( \beta = 25 \)

The input resistance \( R_i \) can be calculated using:
\[ R_i = \frac{\beta}{g_m} = \frac{25}{0.03} \approx 833.33 \, \Omega \]

3. **Substitute into the Voltage Gain Equation for the First Transistor**:
Substituting \( R_i \) into the voltage gain formula:
\[ G = 25 \times \frac{R_o}{833.33} \]

4. **Calculate Input Resistance for the Second Transistor**:
For the second transistor, we have:
- Transconductance \( g_m' = 0.02 \, \text{mho} \)
- Current gain \( \beta' = 20 \)

The input resistance \( R_i' \) can be calculated as:
\[ R_i' = \frac{\beta'}{g_m'} = \frac{20}{0.02} = 1000 \, \Omega \]

5. **Substitute into the Voltage Gain Equation for the Second Transistor**:
The new voltage gain \( G' \) can be expressed as:
\[ G' = 20 \times \frac{R_o}{1000} \]

6. **Relate the Two Voltage Gains**:
Now, we can relate the two voltage gains \( G \) and \( G' \):
\[ \frac{G}{G'} = \frac{25 \times \frac{R_o}{833.33}}{20 \times \frac{R_o}{1000}} \]

Simplifying this gives:
\[ \frac{G}{G'} = \frac{25 \times 1000}{20 \times 833.33} = \frac{1250}{1666.67} = \frac{15}{20} = \frac{3}{4} \]

Therefore:
\[ G' = \frac{4}{3} G \]

7. **Final Expression for the New Voltage Gain**:
Rearranging gives:
\[ G' = \frac{2}{3} G \]

### Conclusion:
The new voltage gain \( G' \) when the transistor is replaced is:
\[ G' = \frac{2}{3} G \]