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A common emitter transistor amplifier is connected with a load resistance of $6 \mathrm{k} \Omega$. When a small a.c. signal of 15 $\mathrm{mV}$ is added to the base emitter voltage, the alternating base current is $20 \mu \mathrm{A}$ and the alternating collector current is $1.8 \mathrm{~mA}$. What is the voltage gain of the amplifier?
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The correct answer is:
720
Hint:
$R_{C}=6 \times 10^{3} \Omega$
$\Delta \mathrm{l}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}=15 \times 10^{-3} \mathrm{~V}$
$\Delta \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}$
$\Delta \mathrm{l}_{\mathrm{C}}=1.8 \times 10^{-3} \mathrm{~A}$
$\mathrm{~A}_{\mathrm{v}}=\beta \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\Delta \mathrm{l}_{\mathrm{C}}}{\Delta_{\mathrm{B}}} \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1.8 \times 10^{-3}}{20 \times 10^{-6}} \times \frac{6 \times 10^{3} \times 20 \times 10^{-6}}{15 \times 10^{-3}}=720$
$R_{C}=6 \times 10^{3} \Omega$
$\Delta \mathrm{l}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}=15 \times 10^{-3} \mathrm{~V}$
$\Delta \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}$
$\Delta \mathrm{l}_{\mathrm{C}}=1.8 \times 10^{-3} \mathrm{~A}$
$\mathrm{~A}_{\mathrm{v}}=\beta \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\Delta \mathrm{l}_{\mathrm{C}}}{\Delta_{\mathrm{B}}} \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1.8 \times 10^{-3}}{20 \times 10^{-6}} \times \frac{6 \times 10^{3} \times 20 \times 10^{-6}}{15 \times 10^{-3}}=720$
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