In the figure, given that V B B supply can vary from 0 to 5.0 V , V C C = 5 V , β d c = 200 , R B = 100 k Ω , R C = 1   k Ω and V B E = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

Step-by-step solution

When switched on, V C E = 0 V C C - R C i C = 0 i c = V C C R C = 5 1 × 10 3 = 5 × 10 - 3 A i c = β i B i B = i C β = 5 × 10 - 3 200 = 2.5 × 10 - 5 A = 25 μ A using KVL at input side, V B B - i B R B - V B E = 0 V B B = V B E + i B R B = 1 + 100 × 10 3 × 25 × 10 - 6 = 1 + 2.5 = 3.5 V