How many NOR gates are used to get AND gate form them?
3
Implementing AND Gate Using Only NOR Gates
The question asks how many NOR gates are required to construct an AND gate. NOR gates are known as universal gates because any logic gate (AND, OR, NOT, NAND, XOR, XNOR) can be implemented using only NOR gates.
Understanding the Basic NOR Gate
A NOR gate produces an output that is the inverse of the logical OR of its inputs. For two inputs A and B, the output Y is given by the Boolean expression:
\( Y = \overline{A+B} \)
Implementing a NOT Gate using a NOR Gate
A NOT gate (inverter) takes one input A and produces the inverse \(\overline{A}\). We can implement a NOT gate using a NOR gate by connecting both inputs of the NOR gate together. If the input is A, the NOR gate output will be:
\( \overline{A+A} \)
Using the property of Boolean algebra (\(A+A = A\)), this simplifies to:
\( \overline{A} \)
Thus, a single NOR gate with its inputs tied together acts as a NOT gate.
Implementing an AND Gate using NOR Gates
The output of an AND gate for inputs A and B is \(A \cdot B\). We want to achieve this output using only NOR gates. We can use De Morgan's theorem to help us. De Morgan's theorem states that:
\( \overline{A \cdot B} = \overline{A} + \overline{B} \)
Taking the complement of both sides, we get:
\( A \cdot B = \overline{\overline{A} + \overline{B}} \)
The expression \( \overline{\overline{A} + \overline{B}} \) looks like a NOR operation where the inputs are \(\overline{A}\) and \(\overline{B}\). So, we need to perform the following steps:
- Obtain \(\overline{A}\) from input A.
- Obtain \(\overline{B}\) from input B.
- Perform a NOR operation on \(\overline{A}\) and \(\overline{B}\).
Let's see how many NOR gates are needed for each step:
- To get \(\overline{A}\) from A: We need one NOR gate acting as a NOT gate (inputs A tied together).
- To get \(\overline{B}\) from B: We need one NOR gate acting as a NOT gate (inputs B tied together).
- To get \( \overline{\overline{A} + \overline{B}} \) from \(\overline{A}\) and \(\overline{B}\): We need one NOR gate with inputs \(\overline{A}\) and \(\overline{B}\).
Summing the NOR gates required:
- 1 NOR gate for \(\overline{A}\)
- 1 NOR gate for \(\overline{B}\)
- 1 NOR gate for the final NOR operation
Total number of NOR gates = \(1 + 1 + 1 = 3\).
Therefore, three NOR gates are required to implement an AND gate.
Verification with Truth Tables
Let's verify with a truth table:
| A | B | \(\overline{A}\) (using NOR) | \(\overline{B}\) (using NOR) | \(\overline{\overline{A} + \overline{B}}\) (using NOR) | A \(\cdot\) B (AND Output) |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | \(\overline{1+1} = \overline{1} = 0\) | 0 |
| 0 | 1 | 1 | 0 | \(\overline{1+0} = \overline{1} = 0\) | 0 |
| 1 | 0 | 0 | 1 | \(\overline{0+1} = \overline{1} = 0\) | 0 |
| 1 | 1 | 0 | 0 | \(\overline{0+0} = \overline{0} = 1\) | 1 |
As the truth table shows, the output of the circuit made of three NOR gates matches the output of a standard AND gate.
Conclusion
To implement an AND gate using only NOR gates, we need three NOR gates: two to invert the inputs A and B, and one to perform the final NOR operation on the inverted inputs.
Revision Table: Logic Gate Implementations with NOR
| Desired Gate | Boolean Expression | Implementation with NOR Gates | Number of NOR Gates |
|---|---|---|---|
| NOT | \(\overline{A}\) | Single NOR with inputs tied together. | 1 |
| OR | \(A+B\) | Two NOR gates: First NOR for \(\overline{A+B}\), second NOR as inverter for \(\overline{\overline{A+B}}\). | 2 |
| AND | \(A \cdot B\) | Three NOR gates: Two NORs as inverters for \(\overline{A}\) and \(\overline{B}\), one NOR for \(\overline{\overline{A}+\overline{B}}\). | 3 |
| NAND | \(\overline{A \cdot B}\) | Four NOR gates: Two NORs as inverters for \(\overline{A}\) and \(\overline{B}\), one NOR for \(\overline{\overline{A}+\overline{B}}\) (this is AND), and one more NOR as inverter for \(\overline{A \cdot B}\). OR Simplified: Two NORs for OR \((A+B)\), one NOR as inverter for \(\overline{A+B}\) (NOT OR = NOR), apply De Morgan's? Wait, easier approach: Since NAND is \(\overline{A \cdot B}\), using De Morgan's, \(\overline{A \cdot B} = \overline{A} + \overline{B}\). This is like an OR gate followed by a NOT gate. An OR gate can be implemented with 2 NOR gates. So, OR (2 NORs) followed by a NOT (1 NOR) = 3 NOR gates for \(\overline{A} + \overline{B}\)? This does not directly give \(\overline{A \cdot B}\). Let's re-evaluate NAND. We implemented AND using 3 NORs. NAND is just AND followed by NOT. So, AND (3 NORs) + NOT (1 NOR) = 4 NOR gates for NAND from AND. A different approach for NAND: \(\overline{A \cdot B}\). We know \(A+B\) is implemented with 2 NORs. How about \(\overline{A} + \overline{B}\)? This is \(\overline{A \cdot B}\). \(\overline{A}\) is 1 NOR, \(\overline{B}\) is 1 NOR. Need to OR them using NORs. An OR gate is 2 NORs. So \(\overline{A} + \overline{B}\) needs \(\overline{A}\) (1 NOR), \(\overline{B}\) (1 NOR), and then OR them (2 NORs). Total 1+1+2 = 4. |
4 |
The implementation of a NAND gate using NOR gates can be shown to require 4 NOR gates.
Additional Information: Universal Gates
NOR and NAND gates are called universal gates because any other logic gate can be constructed using only one type of these gates. This is important in digital circuit design as it allows for the simplification of manufacturing processes; complex circuits can be built using only one or two types of basic gates.
Knowing how to convert between different logic gates using only universal gates is a fundamental concept in digital electronics and logic design. It demonstrates the flexibility and power of NOR and NAND gates.
Important Questions from Boolean Algebra
The probability of a toothache, given evidence of a cavity, P(toothache | cavity) is ________.
P(cavity V toothache) is ________.
The probability for Cavity, given that either Toothache or Catch is true, P(Cavity | toothache V catch) is _______.
How many different Boolean functions of degree n are there?
Consider a Boolean function of ‘n’ variables. The order of an algorithm that determines whether the Boolean function produces a output 1 is:
