The input resistance of a common emitter tran | vedclass

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(A) Given: Output resistance $R_{o} = 500\,k\Omega = 500 	imes 10^3\,\Omega$,current gain $\alpha = 0.98$,and power gain $A_{p} = 6.0625 	imes 10^6$

Vedclass · Accepted Answer

$198$

Vedclass · Answer

(A) Given: Output resistance $R_{o} = 500\,k\Omega = 500 	imes 10^3\,\Omega$,current gain $\alpha = 0.98$,and power gain $A_{p} = 6.0625 	imes 10^6$.First,calculate the current gain $\beta$ for the common emitter configuration:$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.Power gain is defined as the product of voltage gain $(A_{v})$ and current gain $(\beta)$:$A_{p} = A_{v} 	imes \beta$.Substituting the values: $6.0625 	imes 10^6 = A_{v} 	imes 49$.$A_{v} = \frac{6.0625 	imes 10^6}{49} = 1.237245 	imes 10^5$.Voltage gain is also given by $A_{v} = \beta 	imes \frac{R_{o}}{R_{i}}$,where $R_{i}$ is the input resistance.$1.237245 	imes 10^5 = 49 	imes \frac{500 	imes 10^3}{R_{i}}$.$R_{i} = \frac{49 	imes 500 	imes 10^3}{1.237245 	imes 10^5} = \frac{24500 	imes 10^3}{123724.5} \approx 198\,\Omega$.

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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