Question

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`.

Answer 1

Consider any two characteristic for two values of `I_b` which lie above and below the given value of `I_c`. Here, `I_c =4.0mA`, therefore we select the two characteristics for `I_b=20muA` and `30 muA`.
From the graph, at `V_(CE)=10V`,
`DeltaI_(b)=(30-20)muA=10muA,`
`DeltaI_(c)=(4.5-3.0)mA=1.5mA`
Therefore, `beta_(ac)=(DeltaI_c)/(DeltaI_b)=(1.5mA)/(10muA)=150`
At `V_(CE)=10V`,
either (i) `I_b=30muA` and `I_c=4.5mA`
or (ii) `I_b=20muA` and `I_c=3mA`. Therefore
For (i), `beta_(dc)=I_c/I_b=(4.5mA)/(30muA)=150`,
For (ii), `beta_(dc)=I_c/I_b=(3mA)/(20muA)=150`