NEET2018PhysicsSemiconductorsActual
In the circuit shown in the figure, the input voltage V i   is   20   V ,   V B E = 0   and   V C E = 0 . The values of I B ,   I C   and   β are given by
Options
- AI B = 20   μA ,   I C = 5   mA ,   β = 250
- BI B = 25   μA ,   I C = 5   mA ,   β = 200
- CI B = 40   μA ,   I C = 10   mA ,   β = 250
- DI B = 40   μA ,   I C = 5   mA ,   β = 125
Correct answer
D. I B = 40   μA ,   I C = 5   mA ,   β = 125
Step-by-step solution
V i = I B R B + V B E (By Kirchoff's Voltage Law) 20 = I B × 500 × 10 3 + 0 I B = 20 500 × 10 3 = 40  μA V C C = I C R C + V C E   (By Kirchoff's Voltage Law) 20 = I C × 4 × 10 3 + 0 I C = 5 × 10 - 3 = 5   mA β = I C I B = 5 × 10 - 3 40 × 10 - 6 = 125