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.