From the output characteristics of common emitter circuit shown in Fig., calculate the value of `beta_(ac)` and `beta_(dc)` of the transistor when `V_(cE)` is `10 V` and `I_c =4.0mA`.

Explanation

Consider any two characteristics for two value of `I_(B)` which are above and below the given value of `I_(C)`, Here `I_(C) = 4.0 mA`.
(Choose characteristics for `I_(B) = 30` and `20 mu A`)
At `V_(CE) = 10 V` we read the two values of `I_(C)` from the graph Then

`Delta I_(B) = (30 - 20) mu A`
`Delta I_(C) =(4.5 - 3.0) mA = 1.5 mA`
Therefore
`beta_(ac)=((Delta i_(c))/(Delta I_(B)))_(V_(CE)) = 1.5 mA//10 mu A = 150`
For determining `beta_(dc)` calculate the two values of `beta_(dc)` for the two characteristics chosen and find their mean. Therefore for
`I_(C) = 4.5 mA` and `I_(B) = 30 mu A`
`beta_(dc) = (I_(C))/(I_(B)) = 4.5 mA//30 mu A = 150`and for
`I_(C) = 3.0 mA` and `I_(B) = 20 mu A`
`beta_(dc) = 3.0 A//20 mu A = 150`
Hence `beta_(dc) = (150 + 150)//2 = 150`.