In the study of a transistor as an amplifier, | vedclass

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(B) We know that the relationship between the currents in a transistor is given by $I_E = I_C + I_B$.Dividing both sides by $I_C$,we get $\frac{I_E}{I

Vedclass · Accepted Answer

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

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(B) We know that the relationship between the currents in a transistor is given by $I_E = I_C + I_B$.Dividing both sides by $I_C$,we get $\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$.Since $\alpha = \frac{I_C}{I_E}$,it follows that $\frac{1}{\alpha} = \frac{I_E}{I_C}$.Similarly,since $\beta = \frac{I_C}{I_B}$,it follows that $\frac{1}{\beta} = \frac{I_B}{I_C}$.Substituting these into the equation,we get $\frac{1}{\alpha} = 1 + \frac{1}{\beta}$.Rearranging for $\frac{1}{\beta}$,we have $\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1 - \alpha}{\alpha}$.Therefore,$\beta = \frac{\alpha}{1 - \alpha}$.

In the study of a transistor as an amplifier,if $\alpha = \frac{I_C}{I_E}$ and $\beta = \frac{I_C}{I_B}$,where $I_C, I_B$,and $I_E$ are the collector,base,and emitter currents,respectively,then:

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