Question:
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:
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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Find the current gain \(\beta = \Delta I_c/\Delta I_b\) from the graph; then \(A_v = \beta R_{out}/R_{in}\) and power gain \(= \beta A_v\).
Updated On: Nov 5, 2026
- \(5 \times 10^{4},\ 5 \times 10^{5}\)
- \(5 \times 10^{4},\ 5 \times 10^{6}\)
- \(5 \times 10^{4},\ 2.5 \times 10^{6}\)
- \(2.5 \times 10^{4},\ 2.5 \times 10^{6}\)
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The Correct Option is C
Solution and Explanation
From the graph, points (100 \(\mu\)A, 5 mA), (200 \(\mu\)A, 10 mA), etc.: \(\beta = \dfrac{\Delta I_c}{\Delta I_b} = \dfrac{5\ \text{mA}}{100\ \mu\text{A}} = \dfrac{5\times10^{-3}}{10^{-4}} = 50\).
Voltage gain: \(A_v = \beta\dfrac{R_{out}}{R_{in}} = 50\times\dfrac{100\times10^{3}}{100} = 5\times10^{4}\).
Power gain: \(A_p = \beta\times A_v = 50\times5\times10^{4} = 2.5\times10^{6}\).
Answer: (3) \(5\times10^{4},\ 2.5\times10^{6}\)
Voltage gain: \(A_v = \beta\dfrac{R_{out}}{R_{in}} = 50\times\dfrac{100\times10^{3}}{100} = 5\times10^{4}\).
Power gain: \(A_p = \beta\times A_v = 50\times5\times10^{4} = 2.5\times10^{6}\).
Answer: (3) \(5\times10^{4},\ 2.5\times10^{6}\)
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