For a common emitter amplifier, the audio frequency voltage across the collector resistance `2kOmega` is 2V. If the current amplication factor of the transistor is 200, and the base resistance is `1.5 kOmega`, the input signal voltage and base current are

Options

A. `0.1 V "and" 1 mu A`
B. `0.15 V "and" 10 mu A`
C. `0.015 V "and" 1 A`
D. `0.0075 V "and" 5mA`

Explanation

`V_(0)=2V, R_(C)=2KOmega, beta=200`
`R_(B)=1.5kOmega`
`A_(V)=(V_(0))/(V_(i))=beta(R_(C))/(R_(B))`
`impliesV_(i)=(V_(0)R_(B))/(betaR_(C))=(2xx1.5xx10^(3))/(200xx2xx10^(3))=0.0075V`
`I_(C)=(V_(0))/(R_(C))=(2)/(2xx10^(3))=1mA`
`beta=(I_(C))/(I_(B))impliesI_(B)=(10^(-3))/(200)=5muA`