The first junction on transistor consisting of two back to back P-N junction was invented in 1951 by William Shockley. Later new Bipolar Junction Transistor (BJT) were invented. Transistor has three doped regions forming two P-N junction between them. There are two types of transistors (i) P-N-P and (ii) N-P-N. Each transistor has three segments (i) Emitter (ii) Base (iii) Collector. The size and level of doping is different of all three segments. For connecting in the circuit three terminals of a transistor are available - emitter, base and collector and it can be connected in three configuration Common Emitter (CE), Common Base (CB) and Common Collector (CC). The transistor can be used as a device application depending on configuration used, the biasing done to the segments and operate in the region namely cut off, Active and Saturation. It can work as an amplifier, a switch and an oscillator too. Based on transistor knowledge give answer.
The \( \alpha \) and \( \beta \) are the current gain in common base and common emitter configuration respectively in the transistor circuit then \( \displaystyle\frac{\beta - \alpha}{\alpha \beta} \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
The two current gains are linked through the current relation \( I_E = I_B + I_C \), with the common-base gain \( \alpha = I_C/I_E \) and the common-emitter gain \( \beta = I_C/I_B \).
\( \displaystyle \frac{1}{\alpha} = \frac{I_E}{I_C} = \frac{I_B + I_C}{I_C} = \frac{1}{\beta} + 1 \)
So \( \displaystyle\frac{1}{\alpha} - \frac{1}{\beta} = 1 \). Now split the given expression:
\( \displaystyle \frac{\beta - \alpha}{\alpha\beta} = \frac{\beta}{\alpha\beta} - \frac{\alpha}{\alpha\beta} = \frac{1}{\alpha} - \frac{1}{\beta} = 1 \)
Shortcut: Put typical values: \( \alpha = 0.99 \) gives \( \beta = \alpha/(1-\alpha) = 99 \), and \( (99 - 0.99)/(0.99 \times 99) = 98.01/98.01 = 1 \).