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THE 4-BIT ADDER SUBTRACTOR
by Isai Damier

Interactive 4 bit adder-subtractor digital logic boolean circuit.

Introduction

To be able to perform arithmetic, you must first be familiar with numbers. Therefore, although we give a few helping examples, this article is not about binary numerals.

The main interactive circuit at the top of this page is an arithmetic circuit capable of performing both addition and subtraction on any two 4-bit binary numbers. The circuit has a Mode switch that allows you to choose between adding (M=0) and subtracting (M=1). To understand why this circuit works, let’s review binary addition and binary subtraction. We use 4-bit numbers in the examples because the main interactive circuit is a 4-bit adder–subtractor.

Binary addition is certainly easier than decimal addition. You just add 0s and 1s. For example to add the numbers five (0101) and six (0110) together, we just add the respective bits:

Decimal numerals Binary numerals
  6
+5
  0110
+0101
  11   1011

For binary subtraction, we use 2’s complement to keep things simple. For instance, to perform the operation six (0110) minus five (0101), we first obtain the 2’s complement of five and then add it to six:

Step one: Getting the 2’s complement of 5

  1. Flip every bit in five to get 1010.
  2. Add one to 1010 to get: 1010 + 1 = 1011.
Step two: Adding the 2’s complement of 5 to 6:
  1. Decimal numerals Binary numerals
       6
    +(-5)
      0110
    +1011
       1   10001.

We show the carry bit in green because normally it does not count towards the result.

Design

Now that we have reviewed our binary addition and subtraction skills, let’s look at what the circuit is doing. If you fix your gaze on the full–adders while you play around with the input switches, you will see that no matter what operation you think you are doing, the full–adder’s job never changes: the full–adder adds up the signals that appear at its inputs. This is surely not a strange behavior. In reality, whether you are doing binary addition or binary subtraction, the last thing you always do is add the numbers (e.g. “Step two” above tells you to add).

So how does the circuit know whether you want to add or subtract? The XOR gates and the Mode switch, of course! The circuit assumes that your subtraction will always be in the form A — B. So because one input of each XOR gate is attached to the Mode switch while the other input of each XOR gate is attached to a B bit, when the Mode switch is equal to 1, the output of the XOR is the complement of B. This is because B xor symbol: a cirle with a plus sign inside 0 = B and B xor symbol: a cirle with a plus sign inside 1 = B’. In addition to going into the XOR gates, the Mode signal also goes into the carry input of the least significant full–adder (C0), which basically adds a 1 to B’ when the Mode switch is ON. Hence, when the Mode switch is equal to 1, the XOR gates and the Mode switch take the 2’s complement of B and pass it to the full–adders.

To recap, here is a brief summary of how the circuit works: When M = 0, the full–adders see A + B because B xor symbol: a cirle with a plus sign inside 0 = B. However, when M = 1, the full–adders see A + B’ + 1 because B xor symbol: a cirle with a plus sign inside 1 = B’ and C0 = M. And, yes, (B’ + 1) is the 2’s complement of B.

Although our presentation focuses on a 4–bit adder–subtractor, the circuit can be extended to any arbitrary size by simply incorporating more XOR gates and full–adders.

Using the Adder–Subtractor

The following four tables show you how to use the adder-subtractor to perform different type of arithmetic operations. At the time of publication no similar breakdown exists in the literature. Hence, we advice that you pay attention to the distinctions.

Operation Note calculation
Unsigned addition Ignore the V-bit
A3A2A1A0
B3B2B1B0

CS3S2S1S0


Operation Note calculation
Signed addition
  • The addends A & B and the sum S are all 2’s complements.
  • Ignore C & V
A3A2A1A0
B3B2B1B0

S3S2S1S0


Operation Note calculation
Subtraction without overflow (V=0)
  • The minuend A, the subtrahend B, and the difference S are all 2’s complements.
  • Ignore C
A3A2A1A0
B3B2B1B0

S3S2S1S0


Operation Note calculation
Subtraction with overflow (V=1) The minuend A, the subtrahend B, and the difference S are all 2’s complements.
A3A2A1A0
B3B2B1B0

CS3S2S1S0


2’s Complement Table


Binary 2’s complement Binary 2’s complement
01111 +15 10000 -16
01110 +14 10001 -15
01101 +13 10010 -14
01100 +12 10011 -13
01011 +11 10100 -12
01010 +10 10101 -11
01001 +9 10110 -10
01000 +8 10111 -9
00111 +7 11000 -8
00110 +6 11001 -7
00101 +5 11010 -6
00100 +4 11011 -5
00011 +3 11100 -4
00010 +2 11101 -3
00001 +1 11110 -2
00000 +0 11111 -1