Get Answers to all your Questions

header-bg qa

In a common emitter configuration with suitable bias, it is given that R_{L} is the load resistance and R_{BE} is small signal dynamic
resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by : β is current gain, I_{B} , I_{C} and I_{E} arerespectively base, collector and emitter currents.

  • Option 1)

    \beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{c}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

     

     

     

  • Option 2)

    \beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{E}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

  • Option 3)

    \beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{E}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

  • Option 4)

    \beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

 

Answers (2)

best_answer

As we learned

 

Relation between α and β -

\beta = \frac{\alpha }{1-\alpha }
 

- wherein

\alpha = \frac{I_{C}}{I_{E}}

\beta = \frac{I_{C}}{I_{B}} (current gain )

 

  Current Gain \beta =\frac{\Delta I_{c}}{\Delta I_{B}}

Vol + age Gain = \frac{v_{o}}{v_{1}} = \frac{R_{L}-\Delta I_{c}}{R_{BE}-\Delta I_{B}} = \beta \frac{R_{L}}{R_{BE}}

power gain = (voltage gain) (Current gain)

\beta ^{2}\cdot \frac{R_{L}}{R_{BE}}

 


Option 1)

\beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{c}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

 

 

 

Option 2)

\beta \frac{R_{L}}{R_{BE}}, \frac{\Delta I_{E}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

Option 3)

\beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{E}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

Option 4)

\beta^{2} \frac{R_{L}}{R_{BC}}, \frac{\Delta I_{C}}{\Delta I_{B}}, \beta ^{2 }\frac{R_{L}}{R_{BE}}

Posted by

Avinash

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE