In Fig (a) , the `V_(BB)` supply can be varied from 0 V to 5.0 V . The Si transistor has `beta_(dc) = 250` and `R_(B) = 100 kOmega , R_(C) = 1 k Omega , V_(C C) = 5.0 V` . Assume that when the transistor is saturated , `V_(CE) = 0V ` and `V_(B E) = 0.8 V ` . Calculate (a) the minimum base current , for which the transistor will reach saturation .

Explanation

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