Comparator – Designing 1-bit, 2-bit and 4-bit comparators using logic gates

A Comparator is a combinational circuit that gives output in terms of A>B, A<B, and A=B. This is quite expected from the name as the purpose of comparing numbers is to represent their relationship with each other.

How to design a simple 1-bit comparator?

A 1-bit comparator compares two single bits.

 A B A>B A

Let’s apply a shortcut to find the equations for each of the cases. Normally, we can use a K-map. But this shortcut is efficient and handy when you understand it.

For A>B, there is only one case when the output is high when A=1 and B=0. Let’s call this X. We can write the equation as follows

X(which stands for A>B) = AB’

Similarly, denote A<B with Y and A=B with Z.

Since Y is high when A=0 and B=1, we get the following equation

Y = A’B

Since Z is high in two cases, there will be an OR gate. This is because the logic behind an OR gate is that a high output can be achieved in one or more cases. Z is high when A=0 and B=0, it is also high when A=1 and B=1. Therefore,

Z = A’B’ + AB

This is similar to the equation of an EXNOR gate. Hence,

Z = A $\oplus$B

How to design a 2-bit comparator?

Let’s plot the truth table for a 2-bit comparator

 A1 A0 B1 B0 A>B A

The shortcut that we saw above can be used here too. But notice that since we have four variables (A1,A0,B1,B0) and each of the three outputs is high at least 4 times, the equations that we will get will have 4 terms of 4 variables. At least. So, though applying the shortcut is possible, we won’t. K-maps come in handy in situations like these. Because it is possible to achieve the simplest equation using them. And remember, the simpler the equation, lesser the gates required. That is the aim of any designing process. To obtain the simplest hardware implementation.

Using K-maps

we get the following equations

A>B = A1B1′ + B0′(A0B1′ + A0A1)

A<B = B1A1′ + B0B1A0′ + A1’A0’B0

A=B = $\overline { A0\oplus B0 }$ $\overline { A1\oplus B1 }$

How to design a 4–bit comparator?

The truth table for a 4-bit comparator would have 4^4 = 256 rows. So we will do things a bit differently here. We will compare each bit of the two 4-bit numbers and based on that comparison and the weight of their positions we will draft a truth table.

 A3B3 A2B2 A1B1 A0B0 A>B AB3 x x x 1 0 0 A3B2 x x 1 0 0 A3=B3 A2B1 x 1 0 0 A3=B3 A2=B2 A1B0 1 0 0 A3=B3 A2=B2 A1=B1 A0

Now let’s derive the equations for the three outputs.

A=B

Recall the 1-bit comparator circuit we saw above. The equation for the A=B condition was A $\oplus$B. Let’s call this x. Taking a look at the truth table above, A=B is true only when A3=B3 and A2=B2 and A1=B1 and A0=B0. Hence

Z (A=B) = A3B3 . A2B2 . A1.B1 . A0.B0 = x3x2x1x0

A>B

Since there are multiple occasions where this particular condition is high, we will OR (add) each of those individual occasions. We find the first instance of A>B at the top of the table where A3>B3. For this to be possible in a binary system A3 has to be equal to 1 and B3 has to be equal to 0. Since there are only 0s and 1s in a binary system. We can represent this as A3.B3′. Moving on to the next instance of A>B, we can see that it occurs at A3=B3 and A2>B2. From the equation for A=B above, A3=B3 can be represented as x3. And this entire instance can be written as x3A2B2′. Similarly, deriving equations for the remaining instances, we get the following equation

X(A>B) = A3B3′ + x3A2B2′ + x3x2A1B1′ + x3x2x1A0B0′

A<B

Employing the same principles we used above, we get the following equation

Y(A<B) = A3’B3 + X3A2’B2 + X3X2A1’B1 + X3X2X1A0’B0

Designing the circuit based on the above equations, we get the following logic diagram for a 4-bit comparator.