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In the figure, given that V(BB) supply c...


In the figure, given that `V_(BB)` supply can vary from 0 to `5.0V,V_(CC)=5V,beta_(dc)=200,R_(B)=100Komega,R_(C)=1Komega` and `V_(BE)=1.0V`, The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

A

`25muA and 2.8V `

B

`20muA and 3.5V `

C

`20muA and 2.8V `

D

`25muA and 3.5V `

Text Solution

Verified by Experts

The correct Answer is:
D
When switched on,
`V_(CE) = 0`
`V_("CC") -R_Ci_C=0`
`i_C=(V_("CC"))/R_C=5/(1xx10^3)=5xx10^(-3)A`
`i_C=beta i_B`
`i_B=(i_C)/(beta)=(5xx10^(-3))/200 =25 xx10^(-6)A=25muA`
using KVL at input side,
`V_("BB") - i_(B)R_(B) -V_(BE) =0`
`V_(BB) =V_(BE) + i_BR_B`
`=1+100xx10^3 xx 25xx10^(-6)`
`=1+2.5 =3.5 V`
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