A pnp transistor is used in common-emitter mode in an amplifier circuit. A change of `40 muA` in the base current brings a change of 2 mA in collector current and 0.04 V in base emitter voltage. If a load of `6kOmega` is used, then also find the voltage gain of the amplifier.

Options

A. 1
B. 50
C. 300
D. 900

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Input Resistance (R_in)

The input resistance \( R_{in} \) can be calculated using the formula:

\[ R_{in} = \frac{\Delta V_{in}}{\Delta I_{B}} \]

Where:
- \( \Delta V_{in} = 0.04 \, V \) (change in base-emitter voltage)
- \( \Delta I_{B} = 40 \, \mu A = 40 \times 10^{-6} \, A \) (change in base current)

Substituting the values:

\[ R_{in} = \frac{0.04 \, V}{40 \times 10^{-6} \, A} = \frac{0.04}{0.00004} = 1000 \, \Omega \]

### Step 2: Calculate the Current Gain (β)

The current gain \( \beta \) can be calculated using the formula:

\[ \beta = \frac{\Delta I_{C}}{\Delta I_{B}} \]

Where:
- \( \Delta I_{C} = 2 \, mA = 2 \times 10^{-3} \, A \) (change in collector current)
- \( \Delta I_{B} = 40 \, \mu A = 40 \times 10^{-6} \, A \)

Substituting the values:

\[ \beta = \frac{2 \times 10^{-3} \, A}{40 \times 10^{-6} \, A} = \frac{2 \times 10^{-3}}{40 \times 10^{-6}} = 50 \]

### Step 3: Calculate the Output Voltage (V_out)

The output voltage \( V_{out} \) can be calculated using the formula:

\[ V_{out} = \Delta I_{C} \times R_{L} \]

Where:
- \( R_{L} = 6 \, k\Omega = 6000 \, \Omega \) (load resistance)

Substituting the values:

\[ V_{out} = 2 \times 10^{-3} \, A \times 6000 \, \Omega = 12 \, V \]

### Step 4: Calculate the Voltage Gain (A_V)

The voltage gain \( A_V \) can be calculated using the formula:

\[ A_V = \frac{V_{out}}{V_{in}} \]

Where:
- \( V_{in} = 0.04 \, V \)

Substituting the values:

\[ A_V = \frac{12 \, V}{0.04 \, V} = 300 \]

### Final Answer

The voltage gain of the amplifier is \( 300 \).

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