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(D) The current gain $A_i$ for a common-emitter transistor amplifier is given by the formula:$A_i = - \left( \frac{h_{fe}}{1 + h_{oe} R_L} \right)$Giv

Vedclass · Accepted Answer

$-48.78$

Vedclass · Answer

(D) The current gain $A_i$ for a common-emitter transistor amplifier is given by the formula:$A_i = - \left( \frac{h_{fe}}{1 + h_{oe} R_L} ight)$Given values are $h_{fe} = 50$,$h_{oe} = 25 \ \mu A/V = 25 	imes 10^{-6} \ S$,and $R_L = 1 \ k\Omega = 10^3 \ \Omega$.Substituting these values into the formula:$A_i = - \left( \frac{50}{1 + (25 	imes 10^{-6} 	imes 10^3)} ight)$$A_i = - \left( \frac{50}{1 + 0.025} ight)$$A_i = - \left( \frac{50}{1.025} ight)$$A_i \approx -48.78$

For a transistor amplifier in a common-emitter configuration,given the load resistance $R_L = 1 \ k\Omega$,$h_{fe} = 50$,and $h_{oe} = 25 \ \mu A/V$,calculate the current gain $(A_i)$.

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