An NPN transistor is used in common emitter c | vedclass

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(A) Given:Load resistance $R_L = 1\,k\Omega = 1000\,\Omega$Input signal voltage $\Delta V_{in} = 10\,mV = 10 	imes 10^{-3}\,V$Change in collector cur

Vedclass · Accepted Answer

$0.67\,k\Omega, 300$

Vedclass · Answer

(A) Given:Load resistance $R_L = 1\,k\Omega = 1000\,\Omega$Input signal voltage $\Delta V_{in} = 10\,mV = 10 	imes 10^{-3}\,V$Change in collector current $\Delta I_C = 3\,mA = 3 	imes 10^{-3}\,A$Change in base current $\Delta I_B = 15\,\mu A = 15 	imes 10^{-6}\,A$$1$. Input resistance $(r_{in})$:The input resistance is given by the ratio of input voltage change to base current change:$r_{in} = \frac{\Delta V_{in}}{\Delta I_B} = \frac{10 	imes 10^{-3}}{15 	imes 10^{-6}} = \frac{10000}{15} \approx 666.67\,\Omega = 0.67\,k\Omega$$2$. Voltage gain $(A_v)$:The voltage gain is the ratio of output voltage change to input voltage change:Output voltage change $\Delta V_{out} = \Delta I_C 	imes R_L = (3 	imes 10^{-3}\,A) 	imes (1000\,\Omega) = 3\,V$$A_v = \frac{\Delta V_{out}}{\Delta V_{in}} = \frac{3\,V}{10 	imes 10^{-3}\,V} = \frac{3}{0.01} = 300$Thus,the input resistance is $0.67\,k\Omega$ and the voltage gain is $300$.

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