The minimum no. of 2-input NAND gate required to make one XOR gate is

To determine the minimum number of 2-input NAND gates required to make one XOR gate, we can follow these steps:

Step 1: Understand the truth table of an XOR gate. An XOR gate has the following truth table:

Step 2: Express the XOR gate using NAND gates. We can express the XOR gate using NAND gates by combining multiple NAND gates together. One possible way to do this is by using the following logic:

XOR = (A NAND (A NAND B)) NAND (B NAND (A NAND B))

Step 3: Simplify the expression. By applying De Morgan's theorem and simplifying the expression, we can reduce the number of NAND gates required. The simplified expression for XOR gate using NAND gates is:

XOR = (A NAND B) NAND (A NAND (A NAND B))

Step 4: Count the number of NAND gates required. From the simplified expression, we can see that we need a total of 3 NAND gates to construct one XOR gate.

Therefore, the minimum number of 2-input NAND gates required to make one XOR gate is 3.

To determine the minimum number of 2-input NAND gates required to make one XOR gate, we can follow these steps:

Step 1: Understand the truth table of an XOR gate. An XOR gate has the following truth table:

Step 2: Express the XOR gate using NAND gates. We can express the XOR gate using NAND gates by combining multiple NAND gates together. One possible way to do this is by using the following logic:

XOR = (A NAND (A NAND B)) NAND (B NAND (A NAND B))

Step 3: Simplify the expression. By applying De Morgan's theorem and simplifying the expression, we can reduce the number of NAND gates required. The simplified expression for XOR gate using NAND gates is:

XOR = (A NAND B) NAND (A NAND (A NAND B))

Step 4: Count the number of NAND gates required. By analyzing the simplified expression, we can see that it requires a total of 3 NAND gates to implement one XOR gate.

Therefore, the minimum number of 2-input NAND gates required to make one XOR gate is 3.