The transfer characteristic curve of a transistor, having input and output resistance 100 Ω and 100 k Ω respectively is shown in the figure. The voltage and power gain, are respectively: (a) 5 × 104, 5 × 105 (b) 5 × 104, 5 × 106 (c) 5 × 104, 2.5 × 106 (d) 2.5 × 104, 2.5 × 106

Answer: (c). From the graph: β = (Δ IC)/(Δ IB) = ((10 - 5) mA)/((200 - 100)) = 50 Voltage gain: Av = β (Ro)/(Ri) = 50 × (100 kΩ)/(100 Ω) = 5 × 10⁴ Power gain: Ap = β Av = 50 × 5 × 10⁴ = 2.5 × 10⁶ So the answer is (c) 5 × 10⁴, 2.5 × 10⁶.

The transfer characteristic curve of a transistor, having input and… | JEE Main 2019 (12 Apr, Shift 1)

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Junction Transistor

The transfer characteristic curve of a transistor, having input and output resistance 100 $\Omega$ and 100 k $\Omega$ respectively is shown in the figure. The voltage and power gain, are respectively:

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Answer: (c)

From the graph: $\beta = \dfrac{\Delta I_C}{\Delta I_B} = \dfrac{(10 - 5)\ \text{mA}}{(200 - 100)\ \mu\text{A}} = 50$
Voltage gain: $A_v = \beta \dfrac{R_o}{R_i} = 50 \times \dfrac{100\ \text{k}\Omega}{100\ \Omega} = 5 \times 10^4$
Power gain: $A_p = \beta A_v = 50 \times 5 \times 10^4 = 2.5 \times 10^6$
So the answer is (c) $5 \times 10^4$ , $2.5 \times 10^6$ .

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