About Subtractor In digital electronics, Subtractor is a combinational circuit . It is used in the subtraction of inputted bits. Subtractors...
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 Bâ‚’ which is called Borrow.
For 2 inputs we have 2² = 4 combinations.
Inputs | Outputs | ||
X | Y | D | Bâ‚’ |
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 Báµ¢ are inputs. Báµ¢ is the borrow input from any previous Subtractor. X is the subtrahend and Y, Báµ¢ both are minuend.
From these two inputs, we get two outputs. One is D which is the difference. The last one is Bâ‚’ which is called Borrow.
For 3 inputs we have 2³ = 8 combinations in the table.
Input | Output | |||
X | Y | Báµ¢ | D | Bâ‚’ |
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



No comments