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The base current of a transistor is 105 muA and collector current is `2.05mA`. Determine the value of `beta, I_(e)` and `alpha` A change of `27muA` in the base current produces a change of 0.65 mA in the collector current. Find `beta_(a.c.)`. |
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Answer» `beta=I_c/I_b=(2.05mA)/(105muA)=(2.05xx10^(-3))/(105xx10^(-6))=19.5` `I_(e)=I_b+I_c=105muA+2.05mA` `=0.105mA+2.05mA=2.155mA` `alpha=I_c/I_(e)=(2.05mA)/(2.155mA)=0.95` `beta_(a.c.)=(DeltaI_c)/(DeltaI_b)=(0.65mA)/(27muA)=(0.65xx10^(-3))/(27xx10^(-6)) =24.07` |
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