Wondering how many nand gates are required for xor? To create an XOR (exclusive OR) gate using NAND gates, you need a combination of four NAND gates.

## Here Are All 4 NAND Gates Required For XOR

Gate | Inputs | Output |
---|---|---|

NAND Gate 1 (N1) | A, A | Q1 (A NAND A) |

NAND Gate 2 (N2) | A, Q1 | Q2 (A NAND (A NAND A)) |

NAND Gate 3 (N3) | B, B | Q3 (B NAND B) |

NAND Gate 4 (N4) | B, Q3 | XOR (B NAND (B NAND B)) |

As you can see from the table above, you need them to create an XOR (exclusive OR) gate using NAND gates. Typically, you can use a combination of NAND gates. Specifically, you can use four NAND gates to implement the XOR gate. Here’s a detailed explanation of how to do this and an HTML table to represent the truth table for the XOR gate:

**Creating an XOR Gate with NAND Gates:**

An XOR gate produces a true (1) output when the number of true (1) inputs is odd. To implement this behavior using NAND gates, you can follow these steps:

**Create a 2-input NAND gate (N1):**This gate will take two inputs, A and B, and produce the NAND of A and B.**Create two 2-input NAND gates (N2 and N3):**These gates will take the inputs A and N1 (output of the first NAND gate) as well as B and N1 as their inputs. This will result in two separate outputs, one for A NAND N1 and the other for B NAND N1.**Create a 2-input NAND gate (N4):**This gate takes the outputs of N2 and N3 as its inputs.

The output of N4 is the XOR of the inputs A and B.

**Truth Table for XOR Gate:**

Now, I wouldn’t be doing you any justice here if I did not create a truth table for the XOR gate… So, let’s create one, right?

## XOR Gate Truth Table

A | B | Output (XOR) |
---|---|---|

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 0 |

So, in summary, you can implement an XOR gate using four NAND gates, and the truth table helps visualize its behavior for different input combinations.

## NAND Gate in the XOR Gate Implementation Explained Using Functions And Logic Expressions.

**1. NAND Gate 1 (N1):**

- Inputs: A, A
- Output: Q1
**Function:**Q1 = A NAND A**Logic Expression:**Q1 = (A AND A)’ = (A)’ = NOT A**Explanation:**This gate takes two identical inputs, A and A, and produces the NAND of these inputs. In this case, it’s equivalent to NOT A, meaning it inverts the input A.

**2. NAND Gate 2 (N2):**

- Inputs: A, Q1
- Output: Q2
**Function:**Q2 = A NAND (A NAND A)**Logic Expression:**Q2 = A NAND (NOT A) = (A AND (A)’ = (A AND NOT A) = 0**Explanation:**This gate takes input A and the output Q1 (which is NOT A) and computes the NAND of these inputs. When A is 1 (true) and Q1 is 0 (false), the output is 0.

**3. NAND Gate 3 (N3):**

- Inputs: B, B
- Output: Q3
**Function:**Q3 = B NAND B**Logic Expression:**Q3 = (B AND B)’ = (B)’ = NOT B**Explanation:**Similar to N1, this gate takes two identical inputs, B and B, and produces the NAND of these inputs, which is equivalent to NOT B.

**4. NAND Gate 4 (N4):**

- Inputs: B, Q3
- Output: XOR
**Function:**XOR = B NAND (B NAND B)**Logic Expression:**XOR = B NAND (NOT B) = (B AND (B)’ = (B AND NOT B) = 0**Explanation:**This gate takes input B and the output Q3 (which is NOT B) and computes the NAND of these inputs. When B is 1 (true) and Q3 is 0 (false), the output is 0.

**Explanation of the XOR Gate:**

The XOR output is derived from NAND Gate 4 (N4). It takes input B and the result of NAND Gate 3 (N3), which is NOT B. When both inputs A and B are the same (either both 0 or both 1), N4 produces an output of 0 (false), representing the XOR operation’s behavior where the output is 0 for the same inputs. When A and B are different (one 0 and one 1), N4 outputs 1 (true), representing the XOR operation’s result of 1 for different inputs.