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In the following common emitter configuration an `NPN` transistor with current gain `beta=100` is used. The output voltage of the amlifier will be A. `10mV`B. `0.1V`C. `1.0V`D. `10V` |
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Answer» Correct Answer - C `DeltaI_(B)=(v_(i))/(R_(in))=(10^(-3))/(10^(3))=10^(-6)A` `DeltaI_(c )=betaDeltaI_(B)=100xx10^(-6)=10^(-4)A` `v_(0)=DeltaI-(c )R_(L)=10^(-4)xx10xx10^(3)=1V` |
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