The collector supply in a common emitter amplifier is 8 V and the voltage drop across the load of 800 `Omega` is 0.4 V. If the current gain for common base be `alpha`=0.96 , then the base current

Options

A. `15 muA`
B. `21 muA`
C. `25 muA`
D. `30 muA`

Answer

B

Explanation

`V_("cc")=8V, R_(l)=800 Omega, V_(0)=0.4 V, alpha=0.96, I_(b)=? `
`V_(0)=I_(c)R_(l)`
`therefore I_(c)=(V_(0))/(R_(l))=(0.4)/(800)=5xx10^(-4)A`
`I_(c)=500 muA`
Now `alpha=(I_(c))/(I_(e))=(I_(c))/(I_(c)+I_(b))`
`therefore I_(c)=alphaI_(c)+ alpha I_(b)`
`I_(b)=((1-alpha)/(alpha))I_(c)`
`=((1-0.96)/(0.96))500`
`=21 muA`