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. Hance, 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

Here, at saturation,
`V_(CE)=0 V,V_(BE)=0.8V`
As, `V_(CE)=V_(C C)-I_R_(L)`
`:.I_(C) =(V_(C C))/(R_(C)) =(5.0V)/(1.0k Omega) =5.0 mA`
As, `beta=I_c/I_b`
or `I_b=I_c/beta=(5.0mA)/(250)=20muA`
The input voltage `(V_(i))` in which the transistor will go into the saturation state is given by
`V_(IH)=V(BB)=I_b R_B+V_(BE)`
`=20mu Axx100kOmega +0.8V=2.8V`
We know that in case of Si transistor, as long as input `V_(IL)` is less than `0.6V`, the transistor will be in cut off state and current `I_c` will be zero. Therefore, the value of input voltage below which the transistor remain cut off is `V_(IL)=0.6V`.
It means, the transistor will be in the 'switched off' state between the input voltage `0.0V` to `0.6V` and it will be in 'switch on' state batween the input voltage `2.8 V` to `5.0V`.