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. 16 `k Omega`
B. 18 `k Omega`
C. 14 `k Omega`
D. 10 `k Omega`

Answer

C

Explanation

The output AC voltage is 2.0 V. So, the AC collector current, `i_(c)=2.0//2000=1.0mA.`
The signal current throught the base is given by
`i_(B)=(i_(c))/(beta)=(1.0mA)/(100)=0.010 mA`
The DC base current has to be,
`I_(B)=10xx0.0100=0.1mA=10^(-4)A`
`V_(BB)=V_(BE)+I_(B)R_(B)`
`rArr R_(B)=(V_(BB)-V_(BE))//I_(B)`
Assuming, `V_(BE)=0.6V " " `[for Si transistor]
`R_(B)=(2.0-0.6)//10^(-4)=14 k Omega`