Q80 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC=5 V, βdc=200, RB=100 kΩ, RC=1 kΩ and VBE=1.0V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

Figure

  • A
    25 μA and 2.8 V
  • B
    20 μA and 2.8 V
  • Correct answer
    C
    25 μA and 3.5 V
  • D
    20 μA and 3.5 V

Solution

When switched on,

VCE=0

VCC-RC iC=0

ic=VCCRC=51×103=5×10-3A

ic=βiB

iB=iCβ=5×10-3200=2.5×10-5A =25 μA

using KVL at input side,

VBB-iB RB-VBE=0

VBB=VBE+iBRB

=1+100×103×25×10-6

=1+2.5

=3.5 V

Hence, the correct answer is option: 'C'.

Relevant Questions