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In CE transistor amplifier, the audio signal voltage across the collector resistance of $2 \mathrm{k} \Omega$ is $2 \mathrm{~V}$. If the base resistance is $1 \mathrm{k} \Omega$ and the current amplification of the transistor is $100,$ the input signal voltage is
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The correct answer is:
$10 \mathrm{mV}$
Given : Voltage across the collector $\mathrm{V}_{0}=2$ $\mathrm{V} ;$ collector resistance, $\mathrm{R}_{\mathrm{c}}=2 \times 10^{3} \Omega ;$ Base resistance $\mathrm{R}_{\mathrm{B}}=1 \times 10^{3} \Omega ;$ Input signal voltage, $\mathrm{V}_{\mathrm{i}}=?$
$$
\begin{array}{l}
\mathrm{V}_{0}=\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}=2 \\
\Rightarrow \quad \mathrm{I}_{\mathrm{C}}=\frac{2}{2 \times 10^{3}}=10^{-3} \mathrm{~A} \\
\text { Current gain } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=100 \\
\Rightarrow \quad \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\
\quad \mathrm{~V}_{\mathrm{i}}=\mathrm{R}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{V}_{\mathrm{i}}=1 \times 10^{3} \times 10^{-5} \\
\Rightarrow \quad \mathrm{V}_{\mathrm{i}}=10^{-2} \mathrm{~V} \Rightarrow \mathrm{V}_{\mathrm{i}}=10 \mathrm{mV}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{V}_{0}=\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}=2 \\
\Rightarrow \quad \mathrm{I}_{\mathrm{C}}=\frac{2}{2 \times 10^{3}}=10^{-3} \mathrm{~A} \\
\text { Current gain } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=100 \\
\Rightarrow \quad \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\
\quad \mathrm{~V}_{\mathrm{i}}=\mathrm{R}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{V}_{\mathrm{i}}=1 \times 10^{3} \times 10^{-5} \\
\Rightarrow \quad \mathrm{V}_{\mathrm{i}}=10^{-2} \mathrm{~V} \Rightarrow \mathrm{V}_{\mathrm{i}}=10 \mathrm{mV}
\end{array}
$$
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