Semiconductor Electronics Materials Devices and Simple Circuits - Result Question 100
102. The circuit
[2008]
is equivalent to
(a) AND gate
(b) NAND gate
(c) NOR gate
(d) OR gate
Show Answer
Answer:
Correct Answer: 102. (c)
Solution:
- (c) Let $A$ and $B$ be inputs and $Y$ the output.
Then $Y_1=(\overline{A+B})=\overline{A} \cdot \overline{B}$
(By De-Morgan’s theorem)
$Y_2=\overline{Y_1}=(\overline{\overline{A} \cdot \overline{B}})=\overline{\overline{A}+B}=A+B$
$\therefore Y=\overline{Y_2}=(\overline{A+B})$
Hence, the given circuit is equivalent to a NOR gate.
NOR and NAND gates is a universal gates as they can be used to perform the basic logic functions. AND, OR and NOT gate.
