91Ó°ÊÓ

The relationship between common-base current gain (α) and common-emitter current gain (β) is given as: \(β = \frac{\text{Output current (C E)}}{\text{Input current (C E)}}\) \(α = \frac{\text{Output current (C B)}}{\text{Input current (C B)}}\) Output current in both configurations is the same (i.e., collector current \(I_C\)), while input current in C B configuration is base current (\(I_B\)), and input current in C E configuration is emitter current (\(I_E\)).

Knowing the relationship between α, β, and the input currents, we can express the collector current (\(I_C\)) and emitter current (\(I_E\)) in terms of α and β as follows: \(I_C = αI_B\) (for C B configuration) \(I_C = βI_E - βI_B\) (for C E configuration, since \(I_E = I_B + I_C\))

Now that we have the collector current in terms of α and β, we will equate them to find the relationship between α and β. \(αI_B = βI_E - βI_B\) Now, solving for β in terms of α: \(β = \frac{αI_B + βI_B}{I_E}\) \(β = \frac{I_B(α + β)}{I_E}\) As \(I_E = I_B + I_C\) and \(I_C = αI_B\), we get: \(β = \frac{I_B(α + β)}{I_B + αI_B}\) Dividing both sides by \(I_B\): \(β = \frac{α + β}{α+1}\) Now, solving for \(\frac{\beta - α}{αβ}\):

\(\frac{\beta - α}{αβ} = \frac{(\frac{α + β}{α+1}) - α}{\frac{α(α + β)}{α+1}}\) Now, multiplying both numerator and denominator by \((α+1)\): \(\frac{\beta - α}{αβ} = \frac{(α + β) - α(α+1)}{α(α+β)}\) Simplifying the expression further: \(\frac{\beta - α}{αβ} = \frac{-α^2+1}{α(α+β)}\) Since \(α\) is the current gain, it is always greater than zero. For this expression to be equal to the given options, the value of \((-α^2 + 1)\) should be equal to \(α(α+\beta)\). Let's check with the given options: If we compare the expression with option (A) - 1: \(-α^2 + 1 = α(α+β) = 1\) We know that since \(α\) is always greater than zero, this is not possible, so option (A) is ruled out. Now let's check with option (D) - Zero: \(-α^2 + 1 = 0\) In this situation, the value of \(α\) would be 1 or -1. As the current gain is always positive, it should be 1. So, α = 1. Now let's consider the situation where α = 1. If we plug α = 1 into the equation relating α and β: \(β = \frac{α + β}{α+1}\) \(β = \frac{1 + β}{1+1}\) Plugging α = 1 in the equation: \(β = \frac{1 + β}{2}\) Multiplying both sides by 2: \(2β = 1 + β\) \(β = 1\) So, this situation is possible. It means that for certain transistor configurations, the given expression can be equal to zero.

According to the steps mentioned above, we can conclude that the expression \(\frac{\beta - \alpha}{\alpha \beta}\) is equal to Zero (Option D).