# Binary to Excess-3 code conversion

The excess-3 code plays an important role in representing the decimal numbers. The Excess-3 code can also be represented as the XS-3 code. In Excess-3 code, each digit of the decimal number is represented by adding 3 in each decimal digit. There are the following steps to convert the binary number into Excess-3 code:

- Convert the binary number into decimal.
- Add 3 in each digit of the decimal number.
- Find the binary code of each digit of the newly generated decimal number.

We can also add 0011 in each 4-bit BCD code of the decimal number for getting excess-3 code.

Letâ€™s take an example to understand the process of converting a binary number into Excess-3 code.

**Example 1: Convert (11110) _{2} to Excess-3 using binary**

**1. First, convert the given binary number into a decimal number.**

Binary Number: (11110)_{2}

Finding Decimal Equivalent of the number:

Steps | Binary Number | Decimal Number |
---|---|---|

1) Hb | (11110)_{2} | ((1 Ã— 2^{4}) + (1 Ã— 2^{3}) + (1 Ã— 2^{2}) + (1 Ã— 2^{1}) + (0 Ã— 2^{0}))_{10} |

2) | (11110)_{2} | (16 + 8 + 4 + 2 + 0)_{10} |

3) | (11110)_{2} | (30)_{10} |

Decimal number of the Binary number (11110)_{2} is (30)_{10}

**2. Now, we add 3 in each digit of the decimal number.**

The decimal number is 30. Now, we will add 3 into the decimal number 30.

= 30+33

= 63

**3. Now, we find the binary code of each digit of the decimal number 63.**

We write the binary code of each decimal digit in order to get Excess-3 code as:

Steps | Decimal Number | Conversion |
---|---|---|

Step 1 | 63_{10} | (0110)_{2} (0011)_{2} |

Step 2 | 63_{10} | (01100011)_{Excess-3} |

**Result:**

(11110)_{2} = (01100011)_{Excess-3}

Below is the table that contains the excess-3 code of the decimal and BCD.

Decimal Digit | BCD Code | Excess-3 Code |
---|---|---|

A B C D | B_{3}B_{2}B_{1}B_{0} | |

0 | 0 0 0 0 | 0 0 1 1 |

1 | 0 0 0 1 | 0 1 0 0 |

2 | 0 0 1 0 | 0 1 0 1 |

3 | 0 0 1 1 | 0 1 1 0 |

4 | 0 1 0 0 | 0 1 1 1 |

5 | 0 1 0 1 | 1 0 0 0 |

6 | 0 1 1 0 | 1 0 0 1 |

7 | 0 1 1 1 | 1 0 1 0 |

8 | 1 0 0 0 | 1 0 1 1 |

9 | 1 0 0 1 | 1 1 0 0 |

In the above table, the most significant bit of the decimal number is represented by the bit B_{3}, and the least significant bits are represented by B_{2}, B_{1}, and B_{0}.

## Excess-3 to Binary Conversion

The process of converting Excess-3 code into binary is opposite to the process of converting Binary code into Excess-3. There are the following steps to convert the Excess-3 code into binary:

- In the first step, we will make the group of 4 bits and write the equivalent decimal number from the Excess-3 table.
- At last, we find the binary number of the decimal number using a decimal to binary conversion.

**Example 1: (01100011) _{Excess-3}**

**1) Making groups of four bits and write their equivalent decimal number.**

(01100011)_{Excess-3} = (0110 0011)_{Excess-3}

**From the Excess-3 table:**

(0110)_{Excess-3} = (3)_{10}

(0011)_{Excess-3} = (0)_{10}

So, the decimal number of excess-3 code 01100011 is: **(30) _{10}**

**2) Find the binary number.**

Now, find the binary number of the decimal number (30)_{10} using a decimal to binary conversion as:

Divide the number 30 and its successive quotients with base 2.

Operation | Quotient | Remainder |
---|---|---|

30/2 | 15 | 0 (LSB) |

15/2 | 7 | 1 |

7/2 | 3 | 1 |

3/2 | 1 | 1 |

1/2 | 0 | 1(MSB) |

**(30) _{10}=(11110)_{2}**

So, the binary number of excess-3 code 01100011 is: **(11110) _{2}**