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 parameters
We are given:
- Alpha (α) = 0.99
- Load resistance (Ro) = 4.5 kΩ = 4500 Ω
- Dynamic resistance of the emitter junction (Ri) = 50 Ω

### Step 2: Write the formula for voltage gain
The voltage gain (Av) of a common base amplifier can be expressed as:
\[ A_v = \frac{V_{out}}{V_{in}} = \frac{I_C \cdot R_o}{I_E \cdot R_i} \]
where:
- \(I_C\) = Collector current
- \(I_E\) = Emitter current
- \(R_o\) = Load resistance
- \(R_i\) = Dynamic resistance of the emitter junction

### Step 3: Relate collector current to emitter current
From the definition of alpha, we have:
\[ \alpha = \frac{I_C}{I_E} \]
This can be rearranged to express \(I_C\) in terms of \(I_E\):
\[ I_C = \alpha \cdot I_E \]

### Step 4: Substitute \(I_C\) in the voltage gain formula
Substituting \(I_C\) in the voltage gain formula gives:
\[ A_v = \frac{\alpha \cdot I_E \cdot R_o}{I_E \cdot R_i} \]

### Step 5: Simplify the expression
Since \(I_E\) appears in both the numerator and the denominator, we can cancel it out:
\[ A_v = \frac{\alpha \cdot R_o}{R_i} \]

### Step 6: Substitute the known values
Now, substituting the known values into the equation:
\[ A_v = \frac{0.99 \cdot 4500}{50} \]

### Step 7: Calculate the voltage gain
Calculating the above expression:
\[ A_v = \frac{4455}{50} = 89.1 \]

### Final Answer
Thus, the voltage gain of the amplifier is approximately:
\[ A_v \approx 89.1 \]