In a CE transistor amplifier,the audio signal | vedclass

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(D) Given: Collector resistance $R_{C} = 2 \, k\Omega = 2 	imes 10^{3} \, \Omega$.Output voltage $V_{o} = 2 \, V$.Base resistance $R_{B} = 1 \, k\Ome

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$10 \, mV$

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(D) Given: Collector resistance $R_{C} = 2 \, k\Omega = 2 	imes 10^{3} \, \Omega$.Output voltage $V_{o} = 2 \, V$.Base resistance $R_{B} = 1 \, k\Omega = 1 	imes 10^{3} \, \Omega$.Current amplification factor $\beta = 100$.The output voltage is given by $V_{o} = I_{C} R_{C}$.Therefore,the collector current is $I_{C} = \frac{V_{o}}{R_{C}} = \frac{2 \, V}{2 	imes 10^{3} \, \Omega} = 10^{-3} \, A = 1 \, mA$.Since $\beta = \frac{I_{C}}{I_{B}}$,the base current is $I_{B} = \frac{I_{C}}{\beta} = \frac{10^{-3} \, A}{100} = 10^{-5} \, A$.The input signal voltage $V_{i}$ is given by $V_{i} = I_{B} R_{B}$.Substituting the values: $V_{i} = (10^{-5} \, A) 	imes (1 	imes 10^{3} \, \Omega) = 10^{-2} \, V = 10 \, mV$.

In a $CE$ transistor amplifier,the audio signal voltage across the collector resistance of $2 \, k\Omega$ is $2 \, V$. If the base resistance is $1 \, k\Omega$ and the current amplification of the transistor is $100$,the input signal voltage is

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