Question

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 ?
(A) 90
(B) 640
(C) 900
(D) 720

Answer 1

Ans(D)

 \(R_{C}=6 \times 10^{3} \Omega\)
$$
\begin{aligned}
&\Delta \mathrm{I}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}=15 \times 10^{-3} \mathrm{~V} \\
&\Delta \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A} \\
&\Delta \mathrm{I}_{\mathrm{C}}=1.8 \times 10^{-3} \mathrm{~A} \\
&\mathrm{~A}_{\mathrm{V}}=\beta \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\Delta \mathrm{C}}{\Delta_{\mathrm{B}}} \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1.8 \times 10^{-3}}{20 \times 10^{-6}} \times \frac{6 \times 10^{3} \times 20 \times 10^{-6}}{15 \times 10^{-3}}=720
\end{aligned}
$$

Answer 2

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Answer 3

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