The input resistance of a common emitter tran | vedclass

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(A) First,calculate the current gain $\beta$ for the common emitter configuration using the relation $\beta = \frac{\alpha}{1 - \alpha}$.Given $\alpha

Vedclass · Accepted Answer

$198$

Vedclass · Answer

(A) First,calculate the current gain $\beta$ for the common emitter configuration using the relation $\beta = \frac{\alpha}{1 - \alpha}$.Given $\alpha = 0.98$,we have $\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.The power gain is given by the formula $A_p = \beta^2 	imes \frac{R_o}{R_i}$,where $R_o$ is the output resistance and $R_i$ is the input resistance.Given $A_p = 6.0625 	imes 10^6$ and $R_o = 500 	imes 10^3 \, \Omega$.Substituting the values: $6.0625 	imes 10^6 = (49)^2 	imes \frac{500 	imes 10^3}{R_i}$.$6.0625 	imes 10^6 = 2401 	imes \frac{500 	imes 10^3}{R_i}$.$R_i = \frac{2401 	imes 500 	imes 10^3}{6.0625 	imes 10^6} = \frac{1200500 	imes 10^3}{6.0625 	imes 10^6} = \frac{1200.5}{6.0625} \approx 198 \, \Omega$.

The input resistance of a common emitter transistor amplifier,if the output resistance is $500 \, k\Omega$,the current gain $\alpha = 0.98$ and power gain is $6.0625 \times 10^6$,is....$\Omega$

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