# Logic Gates

 Question 1
For the circuit shown below, the propagation delay of each NAND gate is $1 \mathrm{~ns}$. The critical path delay, in ns, is ____ (rounded off to the nearest integer). A 1 B 2 C 3 D 4
GATE EC 2023   Digital Circuits
Question 1 Explanation:
The given circuit can be drawn as; $\therefore \quad$ The critical path delay $=1 \mathrm{~ns}+1 \mathrm{~ns}=2 \mathrm{~ns}$
 Question 2
Consider a Boolean gate (D) where the output $Y$ is related to the inputs $A$ and $B$ as, $Y=A+\bar{B}$, where + denotes logical OR operation. The Boolean inputs '0' and '1' are also available separately. Using instances of only D gates and inputs '0' and '1', __________ (select the correct option(s)).
 A NAND logic can be implemented B OR logic cannot be implemented C NOR logic can be implemented D AND logic cannot be implemented
GATE EC 2022   Digital Circuits
Question 2 Explanation:
\begin{aligned} y &=A+\bar{B} =\overline{\bar{A}\cdot B}\\ f(A,B)&=A+\bar{B} \\ f(0,B) &=0+\bar{B}=\bar{B}\Rightarrow NOT\; Gate \\ f(A,\bar{B}) &=A+\bar{\bar{B}} =A+B\Rightarrow OR \;Gate\\ f(0,f(A,\bar{B}))&=0+(\overline{A+B})\Rightarrow NOR \; Gate \\ f(\bar{A},B) &= \bar{A}+\bar{B}=\overline{A\cdot B}\Rightarrow NAND\; Gate\\ f(0,f(\bar{A},B))&=0+\overline{\overline{A\cdot B}} =A\cdot B\Rightarrow AND \; Gate \end{aligned}
Since, we can implement NOT Gate we can also implement OR, AND, NOR and NAND Gate.

 Question 3
The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is
 A 4 B 5 C 6 D 7
GATE EC 2016-SET-3   Digital Circuits
Question 3 Explanation: Question 4
The output of the combinational circuit given below is A A+B+C B A(B+C) C B(C+A) D C(A+B)
GATE EC 2016-SET-1   Digital Circuits
Question 4 Explanation:
\begin{aligned} y &=A B C \oplus A B \oplus B C \\ &=[\overline{A B C} \cdot A B+A B C \cdot \overline{A B}] \oplus B C \\ &=[(\bar{A}+\bar{B}+\bar{C}) \cdot A B+A B C \cdot(\bar{A}+\bar{B})] \oplus B C \\ &=(A B \bar{C}) \oplus(B C) \\ &=\overline{A B \bar{C}} \cdot B C+A B \bar{C} \cdot \overline{B C} \\ &=(\bar{A}+\bar{B}+C) \cdot B C+A B \bar{C} \cdot(\vec{B}+\bar{C}) \\ &=\bar{A} B C+B C+A B \bar{C} \\ &=B C(\bar{A}+1)+A B \bar{C}=B C+A B \bar{C} \\ &=B(C+A \bar{C})=B(C+A) \end{aligned}
 Question 5
A universal logic gate can implement any Boolean function by connecting sufficient number of them appropriately. Three gates are shown. Which one of the following statements is TRUE? A Gate 1 is a universal gate. B Gate 2 is a universal gate. C Gate 3 is a universal gate. D None of the gates shown is a universal gate
GATE EC 2015-SET-3   Digital Circuits
Question 5 Explanation:
Universal gate is a gate by which every other gate can be realized.
Gate 1 and Gate 2 are basic gates and can not be used as universal gates.

There are 5 questions to complete.