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In a `CE` transistor amplifier, the audio signal voltage across the collector resistance of `2 k Omega` is `2 V`. If the base resistance is `1 k Omega` and the current amplification of the transistor is `100`, the input signal voltage isA. `10 mV`B. `0.1 V`C. `1.0 V`D. `1mV` |
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Answer» Correct Answer - A Current amplification factor `beta=(DeltaI_(C))/(DeltaI_(B))` Collector resistance `DeltaI_(C)=(2V)/(2xx10^(3) Omega)=1xx10^(-3) A` Base current `Delta I_(B)=(V_(B))/(R_(B))=(V_(B))/(R_(B))=(V_(B))/(1xx10^(3))=V_(B)xx10^(-3)` Given, `beta=100` Now, `100=(10^(-3))/(V_(B)xx10^(-3))` `V_(B)=1/100 V=10 mV` |
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