For a common emitter configuration,if alpha a | vedclass

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

Question

(C) We know that the current gain in common base configuration is $\alpha = \frac{I_C}{I_E}$ and in common emitter configuration is $\beta = \frac{I_C

Vedclass · Accepted Answer

$\alpha = \frac{\beta}{1 - \beta}$

Vedclass · Answer

(C) We know that the current gain in common base configuration is $\alpha = \frac{I_C}{I_E}$ and in common emitter configuration is $\beta = \frac{I_C}{I_B}$.Since $I_E = I_B + I_C$,we can write $\alpha = \frac{I_C}{I_B + I_C}$.Dividing numerator and denominator by $I_B$,we get $\alpha = \frac{I_C/I_B}{1 + I_C/I_B} = \frac{\beta}{1 + \beta}$.From this,we can derive $\frac{1}{\alpha} = \frac{1 + \beta}{\beta} = \frac{1}{\beta} + 1$,which implies $\frac{1}{\alpha} - \frac{1}{\beta} = 1$ or $\frac{1}{\alpha} = \frac{1}{\beta} + 1$.Comparing the given options,option $C$ states $\alpha = \frac{\beta}{1 - \beta}$,which is incorrect because the correct relation is $\alpha = \frac{\beta}{1 + \beta}$.

For a common emitter configuration,if $\alpha$ and $\beta$ have their usual meanings,the incorrect relationship between $\alpha$ and $\beta$ is:

Similar Questions

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

Which region is the least doped in a transistor?

In a transistor, if $\frac{I_C}{I_E} = \alpha$ and $\frac{I_C}{I_B} = \beta$. If $\alpha$ varies between $\frac{20}{21}$ and $\frac{100}{101}$, then the value of $\beta$ lies between:

For $I_b = 20 \ \mu A$ and $\beta = 100$,the value of the emitter current is ........ $mA$.

For a given transistor,if $\alpha = I_C / I_E = 0.96$,then the current gain $\beta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module