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Question: In an n – p – n transistor \(10^{10}\) electron enter the emitter in \(10^{- 6}\) s. if 2% of the el...

In an n – p – n transistor 101010^{10} electron enter the emitter in 10−610^{- 6} s. if 2% of the electrons are lost in the base, the ratio of current transfer ratio and the current amplification factor is.

A

0.020.02

B

7

C

33

D

4.94.9

Answer

0.020.02

Explanation

Solution

: As, IE=nE×etI_{E} = \frac{n_{E} \times e}{t}

And IC=nC×et=(98/100)nE×et=98100×IEI_{C} = \frac{n_{C} \times e}{t} = \frac{(98/100)n_{E} \times e}{t} = \frac{98}{100} \times I_{E}

Current transfer ratio,

α=ICIlE=98100=0.98\alpha = \frac{I_{C}}{IlE} = \frac{98}{100} = 0.98

Current amplification factor,

β=α1−α=0.981−0.98=49∴αβ=0.9849=0.02\beta = \frac{\alpha}{1 - \alpha} = \frac{0.98}{1 - 0.98} = 49\therefore\frac{\alpha}{\beta} = \frac{0.98}{49} = 0.02