The input resistance of a silicon transistor | vedclass

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(A) Given:Input resistance,$R_{i} = 100\,\Omega$Change in base current,$\Delta I_{B} = 40\,\mu A = 40 	imes 10^{-6}\,A$Change in collector current,$\

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$2000$

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(A) Given:Input resistance,$R_{i} = 100\,\Omega$Change in base current,$\Delta I_{B} = 40\,\mu A = 40 	imes 10^{-6}\,A$Change in collector current,$\Delta I_{C} = 2\,mA = 2 	imes 10^{-3}\,A$Load resistance,$R_{L} = 4\,k\Omega = 4000\,\Omega$First,calculate the current gain $(\beta)$:$\beta = \frac{\Delta I_{C}}{\Delta I_{B}} = \frac{2 	imes 10^{-3}}{40 	imes 10^{-6}} = \frac{2000}{40} = 50$The voltage gain $(A_{V})$ of a common emitter amplifier is given by the formula:$A_{V} = \beta 	imes \frac{R_{L}}{R_{i}}$Substituting the values:$A_{V} = 50 	imes \frac{4000}{100}$$A_{V} = 50 	imes 40 = 2000$Thus,the voltage gain of the amplifier is $2000$.

The input resistance of a silicon transistor is $100\,\Omega$. Base current is changed by $40\,\mu A$,which results in a change in collector current by $2\,mA$. This transistor is used as a common emitter amplifier with a load resistance of $4\,k\Omega$. The voltage gain of the amplifier is:

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