Table of Contents
How many 2 input NAND gates are required to realize a 2 input or gates?
It can take from 2 NAND gates (in case of AND gate) to 5 NAND gates (in case of XOR gate). For your convenience I have provided the implementations. Here circuit 2 is 2-input AND gate, circuit 3 is 2-input OR gate, circuit 4 is 2-input NOR gate, and circuit 5 is 2 input XOR gate.
How many NAND gates are required to realize an XNOR gate?
The number of 2-input NAND gates required to implement a 2-input XOR gate is 4.
| Logic Gates | Min. number of NOR Gate | Min. number of NAND Gate |
|---|---|---|
| NAND | 4 | 1 |
| NOR | 1 | 4 |
| Half-Adder | 5 | 5 |
| Half-Subtractor | 5 | 5 |
How many 2 inputs NAND gates are required to implement a 3 input NAND gate *?
Four. The second solution keeps the propagation delays the same for all three inputs. You can do the same thing with NOR gates, just substitute them for the NANDs. The second solution keeps the propagation delays the same for all three inputs.
How many NOR gates are required for Xnor?
four NOR gates
An XNOR gate is made by connecting four NOR gates as shown below. This construction entails a propagation delay three times that of a single NOR gate.
How many NOR gates are required for AND gate?
, noting from de Morgan’s Law that a NOR gate is an inverted-input AND gate. This construction uses five gates instead of four….XNOR.
| Input A | Input B | Output Q |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
How many and gates are required to realize?
How many AND gates are required to realize Y = CD + EF + G? Explanation: To realize Y = CD + EF + G, two AND gates are required and two OR gates are required.
How many and gates are required to realize the following expression Y AB BC?
2 AND gates
6. How many AND gates are required to realize the following expression Y=AB+BC? Explanation: 2 AND gates are required to realize the expression.