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)

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

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