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  • A common emitter amplifier has a voltage gain of , an input impedance of  <img alt="100 \Omega" src="https://entranceco

A common emitter amplifier has a voltage gain of 50, an input impedance of  100 \Omega and an output impedance of 200 \Omega. The power gain of the amplifier is

 

Option: 1

500


Option: 2

1000


Option: 3

1250


Option: 4

100


Answers (1)

\text { AC power gain }=\frac{\text { Change in output power }}{\text { Change in input power }}

\mathrm{ =\frac{\Delta V_c \times \Delta i_c}{\Delta V_i \times \Delta i_b}=\left(\frac{\Delta V_c}{\Delta V_i}\right) \times\left(\frac{\Delta i_c}{\Delta i_b}\right)=A_V \times \beta_{\mathrm{AC}}}

\mathrm{ where \: A_V \text{is voltage gain and} (\beta)_{\mathrm{AC}}\: is \: \mathrm{AC} \: \text{current gain. }}

\mathrm{Also,}

\mathrm{A_V=\beta_{\mathrm{AC}} \times \text{resistance gain} \left(=\frac{R_o}{R_i}\right)}

\mathrm{Given, A_V=50, R_o=200 \Omega, R_i=100 \Omega}

\mathrm{Hence, 50=\beta_{\mathrm{AC}} \times \frac{200}{100}}

\mathrm{\therefore \beta_{\mathrm{AC}}=25}

\mathrm{Now, \mathrm{AC}\: \text{power gain} =A_V \times \beta_{\mathrm{AC}}=50 \times 25=1250}

Hence option 3 is correct.





 




 

Posted by

Ramraj Saini

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