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The current transfer ratio `beta` of a transistor is 50. The input resistance of the transistor when used in common emitter mode is 1 kilo ohm. The peak value of the collector alternating current for an input peak voltage of 0.01 volt isA. `0.01 muA`B. `0.25 muA`C. `100 muA`D. `500 muA` |
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Answer» Correct Answer - D Here `R_(i)=10^(3) Omega, beta=50, V_(i)=0.01 V`, `I_(b)=V_(i)/R_(i)=0.01/10^(3)=10^(-5)A` As `I_(C)=betaxxI_(b)` `=50xx10^(-5)=500xx10^(-6)A=500 muA` |
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