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.

Options

A. `4 k Omega`
B. `14 k Omega`
C. `28 k Omega`
D. `54 k Omega`

Answer

B

Explanation

The output `AC` voltage is `2V`. So the `AC` collector current `i_(c) = (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 `10 xx i_(B)`
`:. R_(B)=((V_(BB)-V_(BE)))/(l_(B)),V_(BE)=0.6 V`
`:. R_(B) =((2-0.6))/(0.10)=14 kOmega`.