21 January 2020

Relation between current amplification factor α and current amplification factor β



                      α = IC / IE……….. ( 1 )

                      β = IC / IB………….( 2 )
Relation between α and β
As IE = IC + IB
IE = IC + IB
Therefore ∆IB = IE – ∆IC………. ( 3 )
From equation ( 2 ) and ( 3 )
β = IC / IE – ∆IC
Divide numerator and denominator by IE
β = [ IC / IE ]  / [ IE – ∆IC ] / IE
   = α / ( 1 α )

β = α / ( 1 α )


      OR
β ( 1 α ) = α
β βα = α
β = α + βα
β = ( 1 + β ) α

α = β / ( 1 + β )


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