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(C) The voltage gain $A_v$ is defined as the ratio of output signal voltage to input signal voltage:$A_v = \frac{V_{out}}{V_{in}} = \frac{2.5 \ V}{0.0

Vedclass · Accepted Answer

$75$

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(C) The voltage gain $A_v$ is defined as the ratio of output signal voltage to input signal voltage:$A_v = \frac{V_{out}}{V_{in}} = \frac{2.5 \ V}{0.02 \ V} = 125$.We also know that the voltage gain for a $CE$ amplifier is given by $A_v = \beta 	imes \frac{R_c}{R_b}$,where $\beta$ is the current amplification factor,$R_c$ is the collector resistance,and $R_b$ is the base resistance.Substituting the given values:$125 = \beta 	imes \frac{2.5 \ k\Omega}{1.5 \ k\Omega}$.$125 = \beta 	imes \frac{5}{3}$.$\beta = 125 	imes \frac{3}{5} = 25 	imes 3 = 75$.Thus,the current amplification factor is $75$.

In a $CE$ transistor amplifier,the signal voltage across the collector resistance is $2.5 \ V$. The input signal voltage is $0.02 \ V$. If the base and collector resistances are $1.5 \ k\Omega$ and $2.5 \ k\Omega$,then the current amplification factor is

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