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`.
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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`
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