In the common-base configuration,a transistor | vedclass

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

Question

(D) Given,current amplification factor in common-base configuration,$\alpha = 0.95$.We know that the relationship between $\alpha$ and $\beta$ (curren

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(D) Given,current amplification factor in common-base configuration,$\alpha = 0.95$.We know that the relationship between $\alpha$ and $\beta$ (current amplification factor in common-emitter configuration) is given by $\beta = \frac{\alpha}{1 - \alpha}$.Substituting the value of $\alpha$:$\beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} = 19$.In common-emitter configuration,the current gain $\beta$ is defined as the ratio of change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$:$\beta = \frac{\Delta I_C}{\Delta I_B}$.Given $\Delta I_B = 2 \mu A$,we can calculate $\Delta I_C$:$\Delta I_C = \beta 	imes \Delta I_B = 19 	imes 2 \mu A = 38 \mu A$.

In the common-base configuration,a transistor has a current amplification factor of $0.95$. If the transistor is used in a common-emitter configuration and the base current changes by $2 \mu A$,then the change in the collector current is: (in $\mu A$)

Similar Questions

An $n-p-n$ transistor has three leads $A, B$ and $C$. Connecting $B$ and $C$ by moist fingers,$A$ to the positive lead of an ammeter,and $C$ to the negative lead of the ammeter,one finds large deflection. Then,$A, B$ and $C$ refer respectively to

The part of a transistor which is heavily doped to produce a large number of majority carriers is

In an $n-p-n$ transistor amplifier,the collector current is $8 \ mA$. If $80 \%$ of the electrons from the emitter reach the collector,then ($\alpha$ and $\beta$ are current ratios):

For a transistor to act as a switch,it must be operated in

For a $CE$ amplifier,the current gain is $60$. If the load resistance is $5 \times 10^3 \ \Omega$ and the input resistance is $5 \times 10^2 \ \Omega$,then the voltage gain is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module