In a common base circuit of a transistor,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the current amplification factor $\alpha = 0.95$ and the emitter current $I_{E} = 2\,mA$.The collector current $I_{C}$ is given by the relat

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) Given the current amplification factor $\alpha = 0.95$ and the emitter current $I_{E} = 2\,mA$.The collector current $I_{C}$ is given by the relation $I_{C} = \alpha 	imes I_{E}$.Substituting the values,we get $I_{C} = 0.95 	imes 2\,mA = 1.9\,mA$.In a transistor,the emitter current is the sum of the collector current and the base current: $I_{E} = I_{C} + I_{B}$.Therefore,the base current $I_{B} = I_{E} - I_{C}$.Substituting the values,$I_{B} = 2\,mA - 1.9\,mA = 0.1\,mA$.

In a common base circuit of a transistor,the amplification factor is $0.95.$ The base current when the emitter current is $2\,mA,$ is ....$mA$

Similar Questions

$A$ transistor is used as a common emitter amplifier with a load resistance $2 \ k\Omega$. The input resistance is $150 \ \Omega$. The base current is changed by $20 \ \mu A$,which results in a change in collector current by $1.5 \ mA$. The voltage gain of the amplifier is

$A$ transistor is used in common emitter mode as an amplifier. Then:

In a $CE$ amplifier,the input $ac$ signal to be amplified is applied across

The correct relation between $\alpha$ (ratio of collector current to emitter current) and $\beta$ (ratio of collector current to base current) of a transistor is :

In a $NPN$ transistor,$10^8$ electrons enter the emitter in $10^{-8} \ s$. If $1\%$ of electrons are lost in the base,the fraction of current that enters the collector and the current amplification factor $\beta$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module