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).


1 | |
2 | |
3 | |
4 |
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}

\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)).
NAND logic can be implemented | |
OR logic cannot be implemented | |
NOR logic can be implemented | |
AND logic cannot be implemented |
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.
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
4 | |
5 | |
6 | |
7 |
Question 3 Explanation:

Question 4 |
The output of the combinational circuit given below is


A+B+C | |
A(B+C) | |
B(C+A) | |
C(A+B) |
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?


Gate 1 is a universal gate. | |
Gate 2 is a universal gate. | |
Gate 3 is a universal gate. | |
None of the gates shown is a universal gate |
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.
Gate 1 and Gate 2 are basic gates and can not be used as universal gates.
There are 5 questions to complete.