Learn How many nand gates are required for xor In 2023?


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:

  1. Create a 2-input NAND gate (N1): This gate will take two inputs, A and B, and produce the NAND of A and B.
  2. 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.
  3. 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.

Eddie Mcfarren

Eddie Is no stranger to technical writing after spending years in Networking, IT Infrastructure management, and online content marketing. He is an avid researcher, Software and apps dev tester who spends hours solving problems behind the scenes. Get in touch with him via social media and you can email him via contact@gawkygeek.com

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