Question 171: Semiconductors and Logic Gates

Question

The truth table given in fig. represents :
A
B
Y
0
0
0
0
1
1
1
0
1
1
1
1

Accepted Answer

Here   Y = 1   when,
(1)   A = 0  and   B = 1   $$\left( {\overline A B} ight)$$
(2)   A = 1  and  B = 0   $$\left( {A\,\overline B } ight)$$
(3)   A = 1  and  B = 1  $$\left( {A\,B} ight)$$
$$ 	herefore $$   Y = $${\overline A }$$ B + A $${\overline B }$$ + AB
= $${\overline A }$$ B + A ($${\overline B }$$ + B)
= $${\overline A }$$ B + A    [as  $${\overline B }$$ + B = 1]
= (A + $${\overline A }$$) (A + B)
= A + B
So, this represented OR gate.
Note :
X + $${\overline X }$$ Y  =  (X + $${\overline X }$$)  (X + Y)

Answer

AND - Gate

Answer

OR - Gate

Answer

NAND - Gate

Answer

NOR - Gate

Semiconductors and Logic Gates

The truth table given in fig. represents : A B Y 0 0 0 0 1 1 1 0 1 1 1 1

Choose the correct answer: