MP Board · Class 12 · Physics · Semiconductor ElectronicsIn a common-emitter amplifier circuit using an n-p-n transistor, the current gain $\beta$ is 50. If the load resistance $RL$ is $4 \text{ k}\Omega$ and the input resistance $ri$ is $1 \text{ k}\Omega$, calculate the voltage gain and the power gain of the amplifier.

Given Data:

• Current gain ($\beta$) = $50$
• Load resistance ($R_L$) = $4 \text{ k}\Omega = 4 \times 10^3 \Omega$
• Input resistance ($r_i$) = $1 \text{ k}\Omega = 1 \times 10^3 \Omega$

Formulae:

1. Voltage Gain ($A_v$): $$A_v = \beta \times \frac{R_L}{r_i}$$ Alternatively, $A_v = \text{Current Gain} \times \text{Resistance Gain}$

2. Power Gain ($A_p$): $$A_p = \beta^2 \times \frac{R_L}{r_i}$$ Alternatively, $A_p = \text{Voltage Gain} \times \text{Current Gain}$

Step-by-Step Calculation:

Step 1: Calculate Voltage Gain ($A_v$)\nSubstitute the given values into the voltage gain formula: $$A_v = 50 \times \left( \frac{4 \times 10^3 \Omega}{1 \times 10^3 \Omega} \right)$$ $$A_v = 50 \times 4$$ $$A_v = 200$$

Step 2: Calculate Power Gain ($A_p$)\nSubstitute the calculated voltage gain and given current gain into the power gain formula: $$A_p = A_v \times \beta$$ $$A_p = 200 \times 50$$ $$A_p = 10,000$$ \nAlternatively, using the direct power gain formula: $$A_p = (50)^2 \times \left( \frac{4 \times 10^3}{1 \times 10^3} \right)$$ $$A_p = 2500 \times 4 = 10,000$$

Final Answer:

• The voltage gain of the amplifier is 200.
• The power gain of the amplifier is 10,000.