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 find the voltage gain of a common-emitter amplifier designed with an NPN transistor, we will follow these steps:

### Step 1: Understand the parameters given
- Alpha (α) = 0.99
- Input impedance (R_in) = 1 kΩ = 1000 Ω
- Load resistance (R_L) = 10 kΩ = 10000 Ω

### Step 2: Calculate Beta (β)
Beta (β) can be calculated using 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 resistance gain (R_g)
The resistance gain (R_g) is given by the ratio of the load resistance (R_L) to the input impedance (R_in):
\[ R_g = \frac{R_L}{R_{in}} = \frac{10 \times 10^3}{1 \times 10^3} = 10 \]

### Step 4: Calculate the voltage gain (V_g)
The voltage gain (V_g) can be calculated using the formula:
\[ V_g = \beta \times R_g \]
Substituting the values of β and R_g:
\[ V_g = 99 \times 10 = 990 \]

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

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