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Physics Question on Semiconductor electronics: materials, devices and simple circuits

In a transistor amplifier, the two AC current gains α\alpha and β\beta are defined as α=∂IC∂IE\alpha=\frac{\partial I_C}{\partial I_E} and β=∂IC∂IB \beta =\frac{\partial I_C}{\partial I_B} the relation between α\alpha and β\beta is

A

β=1−αα\beta=\frac{1-\alpha}{\alpha}

B

β=α1−α\beta=\frac{\alpha}{1-\alpha}

C

β=α1+α\beta=\frac{\alpha}{1+\alpha}

D

β=1+αα\beta=\frac{1+\alpha}{\alpha}

Answer

β=α1−α\beta=\frac{\alpha}{1-\alpha}

Explanation

Solution

Given, α=∂IC∂IE\alpha=\frac{\partial I_C}{\partial I_E} and β=∂IC∂IB \beta=\frac{\partial I_C}{\partial I_B}
We know that in a transistor amplifier
IE=IC+IB⇒∂IE=∂IC+∂IBI_E =I_C +I_B \Rightarrow \partial I_E =\partial I_C +\partial I_B
Now, β=∂IC∂IB \beta=\frac{\partial I_C}{\partial I_B}
=∂IC∂IE−∂IC=\frac{\partial I_C}{\partial I_E - \partial I_C}
=∂IC/∂IE1−∂IC/∂IE=\frac{\partial I_C / \partial I_E}{1-\partial I_C /\partial I_E}
∴β=α1−α\therefore \beta=\frac{\alpha}{1-\alpha}