Question

Derive relation between α, β and γ in a transistor?

Answer 1

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Answer 2

Answer:

γ =β+1 = (1/(1-α))

Explanation:

Common base  current gain,α =(Ic/Ie)

Common emitter current gain,β=(Ic/Ib)

From  transistor current relations, Ie=Ic+Ib

   α=Ic/Ie

    α=Ic/(Ic+Ib)

Dividing numerator and denominator by Ib

α=(Ic/Ib)/((Ic/Ib)+(Ib/Ib))

ie, α=β/β+1

  β=Ic/Ib

  α=Ic/(Ie-Ic)  

  Dividing numerator and denominator by Ie

 β=(Ic/Ie)/((Ie/Ie)-(Ic/Ie))

 ie, β=α/1-α

let consider

Ie = Ic+Ib

divide with Ib

(Ie/Ib)=(Ic/Ib)+(Ib/Ib)

γ = β + 1

hence

γ =β+1 = (1/(1-α))