In a transistor if I_c/I_e=alphaIcIe=α and I_c/I_b=betaIcIb=β, if alphaα varies between (20)/(21)2021 and (100)/(101)100101, then the value of betaβ lied between?

1 Answer
Mar 1, 2017

20 and 10020and100

Explanation:

Representative circuit in Common emitter amplifier configuration is given below
![dnatechindia.com](useruploads.socratic.org)

Given is the ratio of I_c/I_b=betaIcIb=β
And the ratio of I_c/I_e= alphaIcIe=α

From above expressions we have
I_c=beta I_b=alphaI_eIc=βIb=αIe .....(1)

In this configuration, the current flowing out of the transistor must be equal to the currents flowing into the transistor. As such we have

I_e = I_c + I_bIe=Ic+Ib .....(2)

To have a combined the expression for α and β we divide both sides of (2) by I_b and use (1) to get

I_e/I_b = (I_c + I_b)/I_b
=>beta/alpha=beta+1
=>beta=alphabeta+alpha

=>beta=alpha/(1-alpha)

For alpha=20/21

beta=(20/21)/(1-20/21)=20

Again for alpha=100/101
beta=(100/101)/(1-100/101)=100
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Note: value of alpha will always be<1.
and typically, beta values lie between 20 and 200 for most general purpose transistors.