Question

Solve this:
Q. In a common emitter transistor circuit, the base current is 40 μA, then VBE is



(1) 2 V
(2) 0.2 V
(3) 0.8 V
(4) Zero

Answer 1

In a common emitter transistor circuit, the base current is 40 μA, then VBE is

(2) 0.2 V

We know,

Input voltage (I.V) = VBE + IB×RB

Where IB is the base current, and,

RB is the resistance offered in the base region.

VBE = I.V - IB×RB

I.V = Vcc = 10V

and IB = 40μA = 40×10⁻⁶ A and,

RB = 245 kΩ = 245×10³Ω   [From the figure]

Putting their respective values, we get:

VBE = 10 - ( 40×10⁻⁶ A)(245×10³Ω)

       = (10 - 9.8) V   = 0.2V

Option (2) is correct.

Answer 2

Answer:

0.2V

Explanation: