In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be:

  • Option 1)

    \frac{5}{4}\text{G}

  • Option 2)

    \frac{2}{3}\text{G}

  • Option 3)

    1.5 G

  • Option 4)

    \frac{1}{3}\text{G}

 

Answers (1)

As we learnt in

Amplifier -

Transistor is used as an amplifier

- wherein

in such circuit output is amplified as compared to input .

 

 

 

As  A_{V}=\beta \frac{R_{L}}{R_{i}}\:\:\:\: or\:\:\:\: G = \left( \frac{\beta}{R_{i}} \right )R_{L}

\Rightarrow G=g_{m}R_{L}

\Rightarrow G \propto g_{m}\left[ \because g_{m}= \frac{\Delta I_{C}}{\Delta I_{B}R_{1}} \right ]

\therefore \frac{G_{2}}{G_{1}}= \frac{gm_{2}}{gm_{1}}=\frac{0.02}{0.03}=\frac{2}{3}

\therefore voltage\:\: gain=\frac{2}{3}G

Correct option is 2.


Option 1)

\frac{5}{4}\text{G}

This option is incorrect.

Option 2)

\frac{2}{3}\text{G}

This option is correct.

Option 3)

1.5 G

This option is incorrect.

Option 4)

\frac{1}{3}\text{G}

This option is incorrect.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Exams
Articles
Questions