Question

Which of the following the correct relationship between two current gains \(\alpha\) and \(\beta\) in a transistor? (A) \(\beta=[(1+\alpha) / \alpha]\) (B) \(\beta=[\alpha /(1+\alpha)]\) (C) \(\alpha=[(1+\beta) / \beta]\) (D) \(\alpha=[\beta /(1+\beta)]\)

Short answer

The correct relationship between α and β in a transistor is given by option (D): \( \alpha = \frac{\beta}{1 + \beta} \).

Step-by-step solution

Identify the definitions of alpha and beta

In a transistor, α (alpha) is the common base current gain and is defined as the ratio of output current (Ic) in the collector to the input current (Ie) in the emitter, which can be mathematically represented as: \( \alpha= \frac{I_c}{I_e} \) β (beta) is the common emitter current gain, and it is defined as the ratio of output current (Ic) in the collector to the input current (Ib) in the base, which can be mathematically represented as: \( \beta= \frac{I_c}{I_b} \)

Find the relationship between Ic, Ie, and Ib

We know that in a transistor, the total current entering the transistor is equal to the total current leaving the transistor, so: \( I_e = I_c + I_b \)

Substitute the value of Ic in the equation for alpha

Plug in the value of \( I_c \) from the definition of β (beta) into the equation from step 2, which results in: \( I_e = (\beta I_b) + I_b \)

Simplify and solve for alpha

Simplifying the above equation we get, \( I_e = I_b (1 + \beta) \) Now we can solve for α (alpha) by plugging in the relationship between Ie and Ib, which we got from step 2 as, \( \alpha = \frac{I_c}{I_e} \) Substituting Ic and Ie with their expressions in terms of Ib we get, \( \alpha = \frac{\beta I_b}{I_b (1 + \beta)} \)

Derive the relationship between alpha and beta

We can simplify the above expression to derive the relationship between α and β, \( \alpha = \frac{\beta}{1 + \beta} \) Comparing this derived relationship with the given options, we see that it matches option (D), which is: \( \alpha = \frac{\beta}{1 + \beta} \) So, the correct relationship between α and β in a transistor is given by option (D).