A transistor is used in common-emitter mode i | vedclass

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(B) The input resistance $r_i$ is calculated as the ratio of the change in base-emitter voltage to the change in base current:$r_i = \frac{\Delta V_{B

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(B) The input resistance $r_i$ is calculated as the ratio of the change in base-emitter voltage to the change in base current:$r_i = \frac{\Delta V_{BE}}{\Delta I_B} = \frac{10 \, mV}{10 \, \mu A} = \frac{10 	imes 10^{-3} \, V}{10 	imes 10^{-6} \, A} = 1000 \, \Omega = 1 \, k\Omega$.The current gain $\beta$ is the ratio of the change in collector current to the change in base current:$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{1.5 \, mA}{10 \, \mu A} = \frac{1.5 	imes 10^{-3} \, A}{10 	imes 10^{-6} \, A} = 150$.The voltage gain $A_V$ for a common-emitter amplifier is given by the product of the current gain and the ratio of load resistance $R_L$ to input resistance $r_i$:$A_V = \beta 	imes \frac{R_L}{r_i} = 150 	imes \frac{5000 \, \Omega}{1000 \, \Omega} = 150 	imes 5 = 750$.

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