MP Board · Class 12 · Physics · Semiconductor ElectronicsIn a common emitter transistor amplifier, the input resistance is $200 \; \Omega$ and the load resistance is $4 \; \text{k}\Omega$. A change of $15 \; \mu\text{A}$ in the base current produces a change of $3 \; \text{mA}$ in the collector current. Calculate: (i) the current amplification factor ($\beta$), (ii) the voltage gain of the amplifier, and (iii) the power gain of the amplifier.

Given data:

• Input resistance, $R_i = 200 ; \Omega$
• Load resistance, $R_L = 4 ; \text{k}\Omega = 4 \times 10^3 ; \Omega = 4000 ; \Omega$
• Change in base current, $\Delta I_b = 15 ; \mu\text{A} = 15 \times 10^{-6} ; \text{A}$
• Change in collector current, $\Delta I_c = 3 ; \text{mA} = 3 \times 10^{-3} ; \text{A} = 3000 ; \mu\text{A}$

(i) Calculation of Current Amplification Factor ($\beta$):\nThe current amplification factor in a common emitter configuration is given by the formula: $$\beta = \frac{\Delta I_c}{\Delta I_b}$$\nSubstituting the given values: $$\beta = \frac{3 \times 10^{-3} ; \text{A}}{15 \times 10^{-6} ; \text{A}}$$ $$\beta = \frac{3000}{15} = 200$$\nSo, the current amplification factor is 200.

(ii) Calculation of Voltage Gain ($A_v$):\nVoltage gain is the product of the current amplification factor ($\beta$) and the ratio of load resistance to input resistance ($R_L / R_i$): $$A_v = \beta \times \frac{R_L}{R_i}$$\nSubstituting the values: $$A_v = 200 \times \frac{4000 ; \Omega}{200 ; \Omega}$$ $$A_v = 200 \times 20 = 4000$$\nSo, the voltage gain of the amplifier is 4000.

(iii) Calculation of Power Gain ($A_p$):\nPower gain is the product of the current amplification factor and the voltage gain: $$A_p = \beta \times A_v$$\nSubstituting the values: $$A_p = 200 \times 4000 = 800,000 = 8 \times 10^5$$\nSo, the power gain of the amplifier is $8 \times 10^5$.