From the output characteristics shown in Figure , calculate the values of `beta_(ac) ` and `beta_(dc)` of the transistor when `V_(CE)` is 10 V and `I_(C) = 4.0` mA .

Explanation

`beta_(ac) = ((Delta I_(C))/(DeltaI_(B)))_(V_(CE)) , beta = ((I_(C))/(I_(B)))`
For determining `beta_(ac)` and `beta_(dc)` at the stated values of `V_(CE)` and `I_(C)` consider any two charateristics for two values of `I_(B)` which lie 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 = 10 mu A , Delta I_(C) = (4.5 -3.0) mA = 1.5 mA`
Therefore , `beta_(ac) = (1.5 mA)/(10 muA) = 150`
For determining `beta_(ac)` and `beta_(dc)` at the stated values of `V_(CE)` and `I_(C)` consider any two characteristics for two values of `I_(B)` which lie above and below the given value of `I_(C)`. Here `I_(C) = 4.0` mA . (Choose characteristic for `I_(B) = 30` aove `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 = 10 mu A , Delta I_(C) = (4.5 - 3.0) mA = 1.5 mA`
Therefore , `beta_(ac) = (1.5 mA)/(10 mu A) = 150`
For determining `beta_(dc)` . either estimate the value of `I_(B)` corresponding to `I_(C) = 4.0 mA` at `V_(CE)= 10 V` or 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) = (4.5 mA)/(30 mu A) = 150` and for `I_(C) = 3.0` mA and `I_(B) = 20 mu A`
`beta_(dc) = (3.0 mA)/(20 mu A) = 150 `
Hence , `beta_(dc) = ((150 + 150))/(2) = 150`