PhysicsSemiconductors and Electronic Devices

Which truth table is true for the following circuit?

$Figure for the question$

Answer & explanation

Correct answer: option 3

The correct answer is Option (3)

$Figure for option 3$

Write the output of each gate in terms of A and B. In the figure, B passes through a NOT gate to give $ Y_2 = \bar{B} $, and A passes through a NOT gate to give $ Y_1 = \bar{A} $. One AND gate takes A and $ Y_2 $, the other takes $ Y_1 $ and B, and an OR gate combines them.

$ Y_3 = A\cdot\bar{B},\qquad Y_4 = \bar{A}\cdot B $

$ Y = Y_3 + Y_4 = A\bar{B} + \bar{A}B $

This is the XOR function: the output is 1 only when the inputs differ.

    $ A=0,B=0:\ Y=0 $
    $ A=0,B=1:\ Y=1 $
    $ A=1,B=0:\ Y=1 $
    $ A=1,B=1:\ Y=0 $

The truth table with the outputs 0, 1, 1, 0 is the one shown above.