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For a CE transistor amplifier, the audio...

For a `CE` transistor amplifier, the audio signal voltage across the collector resistance of `2k Omega` is `2V`. Suppose the current amplification factor of the transistor is `100`. The value of `R_(B)` in series with `V_(BB)` supply of `2V`, if the `DC` base current has to be `10` times the signal current is.

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Here, `DeltaI_c=(2.0V)/(2KOmega)=1.0mA,`
`DeltaI_b=(DeltaI_c)/beta=(1.0mA)/100=0.01mA`
If the dc base current becomes,
`I_b=10xxDeltaI_b=10xx0.01=0.01mA`
As, `V_(BB)=V_(BE)+I_b R_B`,
then, `R_B=(V_(BB)-V_(BE))/I_b=((2.0-0.6))/0.10=14kOmega`
The d.c. collector current,
`I_c=betaI_b=100xx0.1=10mA`
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