A transistor having `alpha = 0.99` is used in a common base amplifier. If the load resistance is `4.5 kOmega` and the dynamic resistance of the emitter junction is `50Omega` the voltage gain of the amplifier will be

Options

A. `79.1`
B. `8910`
C. `78.2`
D. `450`

Answer

To find the voltage gain of a common base amplifier with the given parameters, we can follow these steps:

### Step 1: Understand the relationship between alpha and beta
The relationship between alpha (α) and beta (β) in a transistor is given by the formula:
\[ \beta = \frac{\alpha}{1 - \alpha} \]
Given that α = 0.99, we can substitute this value into the formula.

### Step 2: Calculate beta (β)
Substituting α into the formula:
\[ \beta = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99 \]
So, β = 99.

### Step 3: Identify the load resistance (RL) and dynamic resistance (re)
From the question, we have:
- Load resistance, \( R_L = 4.5 \, k\Omega = 4500 \, \Omega \)
- Dynamic resistance of the emitter junction, \( r_e = 50 \, \Omega \)

### Step 4: Use the formula for voltage gain (Av)
The voltage gain (Av) for a common base amplifier is given by the formula:
\[ A_v = \frac{\beta \cdot R_L}{r_e} \]
Substituting the values we have:
\[ A_v = \frac{99 \cdot 4500}{50} \]

### Step 5: Calculate the voltage gain (Av)
Now, performing the calculations:
\[ A_v = \frac{445500}{50} = 8910 \]

### Final Answer
The voltage gain of the amplifier is \( A_v = 8910 \).

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