## About Subtractor

In digital electronics, Subtractor is a combinational circuit. It is used in the subtraction of inputted bits. Subtractors are two types.

### Subtractor types

- Half Subtractor
- Full Subtractor

### Half Subtractor

A half Subtractor is a logical circuit that is used for subtraction of a number (called subtrahend) from another number (called minuend) and generates a number (called difference) and a borrow.

#### A half Subtractor of two inputs

Let X, Y are two inputs. X is the subtrahend and Y is the minuend. From these two inputs, we get two outputs. One is D which is the difference. Another one is Bₒ which is called Borrow.

For 2 inputs we have 2² = 4 combinations.

Inputs | Outputs | ||

X | Y | D | Bₒ |

0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 |

1 | 0 | 1 | 0 |

1 | 1 | 0 | 0 |

From the table, we see that D is true (D=1) only in two cases, and from that, we can write,

D = X̄Y + XȲ = X⊕Y

D = X̄Y + XȲ= X⊕Y

In the table, we also see that the B₀ is true only in a single case. So,

B₀ = X̄Y

Here is the Half Subtractor circuit diagram basing on those two equations-

### Full Subtractor

A full Subtractor is a logic circuit that is used for performing subtraction of two binary numbers taking into account borrow from the previous bit position.

Let X, Y, and Bᵢ are inputs. Bᵢ is the borrow input from any previous Subtractor. X is the subtrahend and Y, Bᵢ both are minuend.

From these two inputs, we get two outputs. One is D which is the difference. The last one is Bₒ which is called Borrow.

For 3 inputs we have 2³ = 8 combinations in the table.

Input | Output | |||

X | Y | Bᵢ | D | Bₒ |

0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 | 1 |

0 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 1 |

1 | 0 | 0 | 1 | 0 |

1 | 0 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 0 |

1 | 1 | 1 | 1 | 1 |

Here the output D is true in 4 cases and output B₀ is true in 4 cases. From here we get two equations basing on these conditions.

D = X̄ȲBi+ X̄YB̄i+ XȲB̄i+ XYBi

= X̄(ȲBi+ YB̄i)+ X(ȲB̄i+ YBi)

= X⊕Y⊕Bi

B₀ = X̄ȲBi+ X̄YB̄i+ X̄YBi+ XYBi

=X̄(ȲBi+ YB̄i)+ YBi (X̄+ X)

= X̄ (Y⊕Bi) + YBi

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