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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.

A

`4 kOmega`

B

`14 kOmega`

C

`28 kOmega`

D

`54 kOmega`

Text Solution

Verified by Experts

The correct Answer is:
B
The output AC voltage is 2V. So the AC collector current `i_(e)=(2)/(2000)=1mA`
The signal current through the base is therefore, given by
`i_(B)=(i_(c))/(beta)`
The D.C base current has to be `10xxi_(B)`
`:.R_(B)=((V_(BB)-V_(BE)))/(l_(B)),V_(BE)=0.6V`
`:.R_(B)=((2-0.6))/(0.10)=14kOmega`
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