A common-emitter amplifier is designed with an n-p-n transistor `(alpha=0.99)`.The input impedance is `1kOmega`and the load is `10kOmega`.The voltage gain will be

Options

A. 9.9
B. 99
C. 990
D. 9900

Answer

To solve the problem of finding the voltage gain of a common-emitter amplifier with the given parameters, we will follow these steps:

### Step 1: Understand the parameters given
- **Alpha (α)** of the transistor = 0.99
- **Input impedance (Ri)** = 1 kΩ
- **Load impedance (RL)** = 10 kΩ

### Step 2: Calculate the current gain (β)
The current gain (β) for a common-emitter amplifier is given by the formula:
\[ \beta = \frac{\alpha}{1 - \alpha} \]
Substituting the value of α:
\[ \beta = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99 \]

### Step 3: Calculate the transconductance (K)
Transconductance (K) is defined as the ratio of the change in output current to the change in input voltage. For our case, it can be calculated as:
\[ K = \frac{\beta}{R_i} \]
Substituting the values:
\[ K = \frac{99}{1 \text{ kΩ}} = \frac{99}{1000} \text{ A/V} = 0.099 \text{ A/V} \]

### Step 4: Calculate the voltage gain (Av)
The voltage gain (Av) of the common-emitter amplifier can be calculated using the formula:
\[ A_v = K \times R_L \]
Substituting the values:
\[ A_v = 0.099 \text{ A/V} \times 10 \text{ kΩ} = 0.099 \times 10000 = 990 \]

### Conclusion
The voltage gain of the common-emitter amplifier is **990**.