Find out relations between α, β and γ as current amplification factors for CE, CB and CC configurations.

Mumbai University > Information Technology > Sem3 > Analog and Digital Circuits

Marks: 4M

Year: May 14

We know that, $α = I_C/I_E$

I.e.

$α = I_C/ (I_B+I_C)$

Divide both numerator and denominator of RHS by $I_C$, we get

$α = I_C/ [I_B/I_C+1]$

Since $(I_C / I_B = β)$

$α=\frac{β}{(β+1)}$

Also we have

α (l + β) = β

α + α β = β

α = β - αβ

α = β (1- α)

$β=\frac{α}{(1-α)}$

We know that, $γ = I_E/I_B$

I.e. $γ = (I_B+I_C)/I_B$

Divide both Numerator and denominator by $I_B$, we get

$γ = [1+ I_C / I_B]/1$

Since $(I_C / I_B = β)$

γ = 1+ β

$γ=1+\frac{α}{(1-α)}$

i.e $γ=\frac{1}{(1-α)}$