The Binary Subtractor is another type of combinational arithmetic circuit that is the opposite of the Binary Adder we looked at in a previous tutorial.
As their name implies, a Binary Subtractor is a decision making circuit that subtracts two binary numbers from each other, for example, X – Y to find the resulting difference between the two numbers.
Unlike the Binary Adder which produces a SUM and a CARRY bit when two binary numbers are added together, the binary subtractor produces a DIFFERENCE, D by using a BORROW bit, B from the previous column. Then obviously, the operation of subtraction is the opposite to that of addition.
We learnt from our maths lessons at school that the minus sign, “–” is used for a subtraction calculation, and when one number is subtracted from another, a borrow is required if the subtrahend is greater than the minuend. Consider the simple subtraction of the two denary (base 10) numbers below.
123  X  (Minuend) 
– 78  Y  (Subtrahend) 
45  DIFFERENCE 
We can not directly subtract 8 from 3 in the first column as 8 is greater than 3, so we have to borrow a 10, the base number, from the next column and add it to the minuend to produce 13 minus 8. This “borrowed” 10 is then return back to the subtrahend of the next column once the difference is found. Simple school math’s, borrow a 10 if needed, find the difference and return the borrow.
The subtraction of one binary number from another is exactly the same idea as that for subtracting two decimal numbers but as the binary number system is a Base2 numbering system which uses “0” and “1” as its two independent digits, large binary numbers which are to be subtracted from each other are therefore represented in terms of “0’s” and “1’s”.
Binary Subtraction can take many forms but the rules for subtraction are the same whichever process you use. As binary notation only has two digits, subtracting a “0” from a “0” or a “1” leaves the result unchanged as 00 = 0 and 10 = 1. Subtracting a “1” from a “1” results in a “0”, but subtracting a “1” from a “0” requires a borrow. In other words 0 – 1 requires a borrow.
0  1  1  (borrow)1→ 0 
– 0  – 0  – 1  – 1 
0  1  0  1 
For the simple 1bit subtraction problem above, if the borrow bit is ignored the result of their binary subtraction resembles that of an ExclusiveOR Gate. To prevent any confusion in this tutorial between a binary subtractor input labelled, B and the resulting borrow bit output from the binary subtractor also being labelled, B, we will label the two input bits as X for the minuend and Y for the subtrahend. Then the resulting truth table is the difference between the two input bits of a single binary subtractor is given as:
Symbol  Truth Table  
2input ExOR Gate

Y  X  Q 
0  0  0  
0  1  1  
1  0  1  
1  1  0 
As with the Binary Adder, the difference between the two digits is only a “1” when these two inputs are not equal as given by the ExOR expression. However, we need an additional output to produce the borrow bit when input A = 0 and B = 1. Unfortunately there are no standard logic gates that will produce an output for this particular combination of X and Y inputs.
But we know that an AND Gate produces an output “1” when both of its inputs X and Y are “1” (HIGH) so if we use an inverter or NOT Gate to complement the input X before it is fed to the AND gate, we can produce the required borrow output when X = 0 and Y = 1 as shown below.
Then by combining the ExclusiveOR gate with the NOTAND combination results in a simple digital binary subtractor circuit known commonly as the Half Subtractor as shown.
A half subtractor is a logical circuit that performs a subtraction operation on two binary digits. The half subtractor produces a sum and a borrow bit for the next stage.
Symbol  Truth Table  
Y  X  DIFFERENCE  BORROW  
0  0  0  0  
0  1  1  0  
1  0  1  1  
1  1  0  0 
From the truth table of the half subtractor we can see that the DIFFERENCE (D) output is the result of the ExclusiveOR gate and the Borrowout (Bout) is the result of the NOTAND combination. Then the Boolean expression for a half subtractor is as follows.
For the DIFFERENCE bit:
D = X XOR Y = X Y
For the BORROW bit
B = notX AND Y = X.Y
If we compare the Boolean expressions of the half subtractor with a half adder, we can see that the two expressions for the SUM (adder) and DIFFERENCE (subtractor) are exactly the same and so they should be because of the ExclusiveOR gate function. The two Boolean expressions for the binary subtractor BORROW is also very similar to that for the adders CARRY. Then all that is needed to convert a half adder to a half subtractor is the inversion of the minuend input X.
One major disadvantage of the Half Subtractor circuit when used as a binary subtractor, is that there is no provision for a “Borrowin” from the previous circuit when subtracting multiple data bits from each other. Then we need to produce what is called a “full binary subtractor” circuit to take into account this borrowin input from a previous circuit.
The main difference between the Full Subtractor and the previous Half Subtractor circuit is that a full subtractor has three inputs. The two single bit data inputs X (minuend) and Y (subtrahend) the same as before plus an additional Borrowin (Bin) input to receive the borrow generated by the subtraction process from a previous stage as shown below.
Then the combinational circuit of a “full subtractor” performs the operation of subtraction on three binary bits producing outputs for the difference D and borrow Bout. Just like the binary adder circuit, the full subtractor can also be thought of as two half subtractors connected together, with the first half subtractor passing its borrow to the second half subtractor as follows.
As the full subtractor circuit above represents two half subtractors cascaded together, the truth table for the full subtractor will have eight different input combinations as there are three input variables, the data bits and the Borrowin, B_{IN} input. Also includes the difference output, D and the Borrowout, B_{OUT} bit.
Symbol  Truth Table  
Bin  Y  X  Diff.  Bout  
0  0  0  0  0  
0  0  1  1  0  
0  1  0  1  1  
0  1  1  0  0  
1  0  0  1  1  
1  0  1  0  0  
1  1  0  0  1  
1  1  1  1  1 
Then the Boolean expression for a full subtractor is as follows.
For the DIFFERENCE (D) bit:
D = (X.Y.B_{IN}) + (X.Y.B_{IN}) + (X.Y.B_{IN}) + (X.Y.B_{IN})
which can be simplified too:
D = (X XOR Y) XOR B_{IN} = (X Y) B_{IN}
For the BORROW OUT (B_{OUT}) bit:
B_{OUT} = (X.Y.B_{IN}) + (X.Y.B_{IN}) + (X.Y.B_{IN}) + (X.Y.B_{IN})
which will also simplify too:
B_{OUT} = X AND Y OR (X XOR Y)B_{IN} = X.Y + (X Y)B_{IN}
As with the binary adder, we can also have n number of 1bit full binary subtractor connected or “cascaded” together to subtract two parallel nbit numbers from each other. For example two 4bit binary numbers. We said before that the only difference between a full adder and a full subtractor was the inversion of one of the inputs.
So by using an nbit adder and n number of inverters (NOT Gates), the process of subtraction becomes an addition as we can use two’s complement notation on all the bits in the subtrahend and setting the carry input of the least significant bit to a logic “1” (HIGH).
Then we can use a 4bit fulladder ICs such as the 74LS283 and CD4008 to perform subtraction simply by using two’s complement on the subtrahend, B inputs as X – Y is the same as saying, X + (Y) which equals X plus the two’s complement of Y.
If we wanted to use the 4bit adder for addition once again, all we would need to do is set the carryin (C_{IN}) input LOW at logic “0”. Because we can use the 4bit adder IC such as the 74LS83 or 74LS283 as a fulladder or a fullsubtractor they are available as a single adder/subtractor circuit with a single control input for selecting between the two operations.
good work.
May I ask about “how does 4 bit binary to bcd converter?”
Great idea and very simple ,i simply understood all ,thanks u very much
Please give the application of full subtract or 3 bit circuit and the limitations and scope of the full subtract or circuit
why is there no separate gate for subtraction???
it would be simpler for the circuit… for ex.. the above 4bit adder is used to add only no.s upto 15, after that 5 bit is needed… hence for 2’s complement we need to add 64 different not gates…… is there any way we take ve numbers as inputs without converting to positive??
Just to let you know, the circuit given for the full subtractor is wrong, In the second half of the circuit, the not gate should connect to the output of the XOR gate, rather than to the BorrowIn. With the circuit as it is, it will never match up to the correct truth table
Hello Shane, Yes you are correct there is an error in the Full Subtractor circuit. I can not copy my own notes. It has been corrected and thank you for letting me know.
Uhm… The full subtractor still isn’t right. First, the table doesn’t show the correct outputs. Just look at the second line: (XY)Bin becomes (10)0 = 1, no borrow, but it says 1 with borrow. Conveniently, if you just change the column headings to X,Y,Bin instead of Bin,Y,X, then the rest of the table doesn’t have to change.
Then, the second borrow (AND gate in the middle) is taking its inputs from the wrong places. The inverting input should be coming from the XOR in the middle (X+Y) and the noninverting input should be tied to Bin. This will then match the first stage halfsubtractor, and the macrodiagram that shows two halves backtoback. I didn’t see the original diagram, but judging from the comments above, and the label on the output, I think you originally had the outputs in the right place, but the inverter on the wrong line. The label on the output of this output should be !(X+Y)&B, not !B&(X+Y).
Wow, now I look like a huge jerk for pointing out all that. Please double check before making changes, just to make sure I’m not steering you wrong.
Hello Eric, you are right. I incorrectly drew the full subtractor circuit in my notes and developed the truth table from that, error and all. Thanks for spotting the error and letting me know. Its been amended. 🙂
sorry no comments
English only
pls, i just want you guys to take note of the half adder tutorial you posted. i think it is supposed to be (difference and borrow at the output) and not sum and borrow. thanks.
Hello Mike, Yes your right. Officially, the addition gives the Sum while Subtraction gives the Difference as I mentioned numerous times in the tutorial. In the UK we just call them sums as a generalised expression whether its addition or subtraction, but your right it should be labelled correctly. I will amend.
please set adder and subtractor using ic 7483 and using 1`s and2`s complement