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How many number of 2-input NAND gates are required to realise a half adder circuit?
- a)5
- b)6
- c)4
- d)8
Correct answer is option 'A'. Can you explain this answer?
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How many number of 2-input NAND gates are required to realise a half a...
Understanding Half Adder
A half adder is a digital circuit that computes the sum of two binary digits. It has two main outputs:
- Sum (S): Represents the addition of A and B.
- Carry (C): Represents the carry from the addition.
The truth table for a half adder is as follows:
| A | B | Sum (S) | Carry (C) |
|---|---|---------|-----------|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
Designing Half Adder Using NAND Gates
To realize a half adder using 2-input NAND gates, we need to derive the Sum and Carry expressions based on the truth table.
- Sum (S) can be expressed as: A XOR B
- Carry (C) can be expressed as: A AND B
Using NAND gates, we can derive these functions as follows:
Sum (S) Logic
1. A NAND B gives us the complement of A AND B.
2. To achieve A XOR B using NAND gates, we can manipulate the outputs of NAND gates:
- S = (A NAND (A NAND B)) NAND (B NAND (A NAND B))
Carry (C) Logic
1. To find Carry (C = A AND B):
- C = (A NAND B) NAND (A NAND B)
Counting the NAND Gates
- For Sum (S): 4 NAND gates are needed.
- For Carry (C): 1 NAND gate is needed.
Thus, the total number of 2-input NAND gates required to realize a half adder is:
- Total = 4 (for S) + 1 (for C) = 5 NAND gates.
Conclusion
Therefore, the correct answer is option 'A': 5 NAND gates are required to realize a half adder circuit.
A half adder is a digital circuit that computes the sum of two binary digits. It has two main outputs:
- Sum (S): Represents the addition of A and B.
- Carry (C): Represents the carry from the addition.
The truth table for a half adder is as follows:
| A | B | Sum (S) | Carry (C) |
|---|---|---------|-----------|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
Designing Half Adder Using NAND Gates
To realize a half adder using 2-input NAND gates, we need to derive the Sum and Carry expressions based on the truth table.
- Sum (S) can be expressed as: A XOR B
- Carry (C) can be expressed as: A AND B
Using NAND gates, we can derive these functions as follows:
Sum (S) Logic
1. A NAND B gives us the complement of A AND B.
2. To achieve A XOR B using NAND gates, we can manipulate the outputs of NAND gates:
- S = (A NAND (A NAND B)) NAND (B NAND (A NAND B))
Carry (C) Logic
1. To find Carry (C = A AND B):
- C = (A NAND B) NAND (A NAND B)
Counting the NAND Gates
- For Sum (S): 4 NAND gates are needed.
- For Carry (C): 1 NAND gate is needed.
Thus, the total number of 2-input NAND gates required to realize a half adder is:
- Total = 4 (for S) + 1 (for C) = 5 NAND gates.
Conclusion
Therefore, the correct answer is option 'A': 5 NAND gates are required to realize a half adder circuit.
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Half Adder using NAND Gates:
The half adder can also be designed with the help of NAND gates. NAND gate is considered as a universal gate. A universal gate can be used for designing any digital circuitry. It is always simple and efficient to use the minimum number of gates in the designing process of our circuit. The minimum number of NAND gates required to design a half adder is 5.
- The first NAND gate takes the inputs which are the two 1-bit numbers.
- The resultant NAND-operated inputs will be again given as input to 3- NAND gates along with the original input.
- Out of these 3 NAND gates, 2-NAND gates will generate the output which will be given as input to the NAND gate connected at the end.
- The gate connected at the end will generate the
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How many number of 2-input NAND gates are required to realise a half adder circuit?a)5b)6c)4d)8Correct answer is option 'A'. Can you explain this answer?
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