In common emitter transistor as shown in Fig., the `V_(BB)` supply can be varied from 0 V to 5.0V. The Si. Transistor has `beta_(ac)=250` and `R_(B)=100kOmega, R_(c)=1k Omega,V_(C C)=5.0V`. Assume that when the transistor is saturated, `V_(CE)=0V` and `V_(BE)=0.8V`. Calculate the minimum base current, for which the transistor will reach saturation. Hence, determine `V_(i)` when the transistor is 'switched on' find ranges of `V_(i)` for which the transistor is switched off and switched on.

Explanation

Given a saturation `V_(CE) = 0V, V_(BE) = 0.8 V`
`V_(CE) = V_(C C) -I_(C) R_(C) rArr I_(C) =V_(C C)//R_(C)`
=`5.0V//1.0 K Omega = 5.0 mA`
Therefore
`I_(B) =I_(C)//beta 5.0 mA//250 = 20 mu A`
The input voltage at which the transistor will go into saturation is given by
`V_(IH) =V_(BB) =I_(B) R_(B) + V_(BC)`
`= 20 mu A xx 100 K Omega + 0.8 V = 2.8 V`
The value of input voltage below which the transistor remains cut off is given by `V_(IL) = 0.6 V,V_(IH) = 2.8 V`.
Between `0.0V` and `0.6 V`, the transistor will be in the 'switched off state. Between `2.8 V` and `5.0V`, it will be in 'switched on' state
Note that the transistor is in active state when `I_(B)` varies from `0.0mA` to `0.20 mA`
In this range, `I_(C) = beta I_(B)` is valied. In the
In this range, `I_(C) = beta I_(B)` is valid.
in the saturation range, `I_(C) le beta I_(B)`.