Binary to Gray Code Conversion Logic

The conversion from a 4-bit binary number ($B_3B_2B_1B_0$) to its equivalent Gray code ($G_3G_2G_1G_0$) follows specific rules:

• The most significant bit (MSB) of the Gray code is the same as the MSB of the binary number:

  $G_3 = B_3$

• For the subsequent bits, the Gray code bit is obtained by XORing the corresponding binary bit with the next higher binary bit:

  $G_2 = B_3 \oplus B_2$

  $G_1 = B_2 \oplus B_1$

  $G_0 = B_1 \oplus B_0$

Here, the symbol $\oplus$ represents the XOR (Exclusive OR) operation.

XOR Gate Requirement Calculation

To implement this conversion using digital logic gates, we need to count the number of XOR gates required:

• For $G_3$: No XOR gate is needed as $G_3$ is directly equal to $B_3$.
• For $G_2$: One XOR gate is required to compute $B_3 \oplus B_2$.
• For $G_1$: One XOR gate is required to compute $B_2 \oplus B_1$.
• For $G_0$: One XOR gate is required to compute $B_1 \oplus B_0$.

Total XOR Gates: Summing the requirements for each bit, we get $0 + 1 + 1 + 1 = 3$ XOR gates.

Conclusion

Therefore, 3 XOR gates are required for the design of a 4-bit Binary to Gray code converter.