In the circuit of CE amplifier, a silicon transistor is used. The value of `V_("CC")=+20V, R_(L)=3kOmega`, collector voltage `=5v, beta=100`. Then the base resistance `R_(B)` should be `……xx10^(5)Omega`
(Take input voltage drop across Si transistor = 0.7 V)

Answer

3.86

Explanation

`V_("CC") = V_(C ) + I_(C ) R_(L)`
`:.` 20 = 5 + 3 ` xx 10^(3) I_(C )`
`:. I_(C ) = (15)/(3xx 10^(3))`
`= 5 xx 10^(-3) A = 5 ` mA ,
`beta = (I_(c))/(I_(B))`
`:. I_(B) = (I_(C))/(I_(beta)) = (5xx 10^(-3))/(100)`
`= 5 xx 10^(-5)` A
`V_("CC" ) = I_(B) R_(B) +P.D `
across input junction
As silicon transistor is in forward bias, the P. D. across input junction is 0. 7 volt.
`:. 20 = 5xx 10^(-5) R_(B) + 0.7 `
`:. R_(B) = (19.3)/(5 xx 10^(-5)) `
`:. R_(B) = 3.86 xx 10^(5) Omega`