Realization of a logic function in SOP form using NOR gate

Let's start this tutorial with some basics of SOP form and NOR gates before getting into the details of how to realize a logic function or Boolean expression in SOP form using NOR gates only.

SOP Form

SOP form stands for Sum of Products form. SOP form is one in which a Boolean expression is expressed as a sum of product terms.

For example,

$$\mathrm{\mathit{f}\lgroup A,B,C\rgroup=AB+ABC+B\overline{C}}$$

This is a Boolean function expressed in SOP (Sum of Products) form.

NOR Gate

NOR Gate is a universal logic gate, i.e., NOR gate can be used for realization of any other type of logic gate or logic function.

NOR means NOT + OR. That means, the OR output is NOTed or inverted. Therefore, the NOR gate is a combination of OR gate and a NOT gate, i.e.,

$$\mathrm{NOR Gate = OR Gate + NOT Gate}$$

A NOR gate is a type of logic gate whose output is HIGH (Logic 1), only when all its inputs are LOW (Logic 0), and it gives an output LOW (Logic 0), even if any of its inputs become HIGH (Logic 1). The logic symbol of a two input NOR gate is shown in Figure 1.

Here, A and B are the input variables to the NOR gate and Y is the output variable of the NOR gate, then the output of the NOR gate is given by,

$$\mathrm{Y=\overline{A+B}=\lgroup A+B\rgroup'}$$

It is read as "Y is equal to A plus B whole bar".

The operation of the NOR gate can be understood from the truth table. The following is the truth table of the NOR gate −

Inputs		Output
A	B	Y = (A+B)'
0	0	1
0	1	0
1	0	0
1	1	0

De Morgan’s Theorem

The implementation of a Boolean function in SOP form using NOR gates requires the knowledge of De Morgan’s theorem of Boolean algebra.

There are two laws of De Morgan’s theorem, which are as follows −

Law 1 − According to first law of the De Morgan’s theorem, the complement of a logical OR operation of variables is equivalent to the logical AND operation of variables in complemented form, i.e.

$$\mathrm{\overline{A+B+C}=\overline{A}. \overline{B}.\overline{C}}$$

Law 2 − This law of De Morgan’s theorem states that the complement of a logical AND operation of variables is equivalent to the logical OR operation of variables in their complemented form, i.e.,

$$\mathrm{\overline{A.B.C}=\overline{A}+ \overline{B}+\overline{C}}$$

So, we have discussed all the basic concepts required for the realization of a logical expression in SOP form using only NOR gates. Now, let us discuss the realization of a logic function in SOP form using NOR gates.

Implementation of Logic Function in SOP Form using NOR Gates

A Boolean function or logic function in SOP (Sum of Products) form can be implemented using only NOR gates. To realize an SOP expression using NOR gates, we first need to transform the given logic function into a form that can be realized using NOR gates. For that the following steps are to be followed −

Step 1 − Double complement the given Boolean or logical function.

For example, consider the following Boolean function,

$$\mathrm{Y=AB+BC}$$

On taking double complement, we get,

$$\mathrm{\overline{\overline{Y}}= \overline{\overline{AB+BC}}}$$

Step 2 − Convert the logical OR operations into the logical AND operation by applying De Morgan’s theorem first law, i.e.,

$$\mathrm{\overline{\overline{Y}}= \overline{\overline{AB}.\overline{BC}}}$$

Step 3 − Convert the product terms into sum terms, use De Morgan’s second theorem, i.e.,

$$\mathrm{\overline{\overline{Y}}= \overline{\lgroup\overline{A}+\overline{B}\rgroup. \lgroup\overline{B}+\overline{C}\rgroup}}$$

Step 4 − Convert the remaining AND operation into OR operation, use De Morgan’s second theorem, i.e.,

$$\mathrm{\overline{\overline{Y}}= \overline{\lgroup\overline{A}+\overline{B}\rgroup}+ \lgroup\overline{\overline{B}+\overline{C}\rgroup}}$$

This is the form of the Boolean which can be implemented using NOR gates.

Step 5 − Finally, determine the number of NOR gates required to implement the expression, and connect them according to logic function to get the logic circuit.

Now, let us discuss some solved examples to understand the concept in depth.

Example 1

Realize the following logic function in SOP form using NOR gate.

$$\mathrm{Y=AB+ABC+BC}$$

Solution

The given logic function is,

$$\mathrm{Y=AB+ABC+BC}$$

Double complementing the function, we get,

$$\mathrm{\overline{\overline{Y}}= Y=\overline{\overline{AB+ABC+BC}}}$$

Applying De Morgan’s theorem $\mathrm{\lgroup\overline{A+B+C}= \overline{A}.\overline{B}.\overline{C}\rgroup}$, we get,

$$\mathrm{Y=\overline{ \overline{AB}.\overline{ABC}. \overline{BC}}}$$

Using De Morgan’s theorem $\mathrm{\lgroup\overline{A.B.C}= \overline{A}+\overline{B}+\overline{C}\rgroup}$, we have,

$$\mathrm{Y=\overline{\lgroup\overline{A}+ \overline{B}\rgroup.\lgroup\overline{A} +\overline{B}+\overline{C}\rgroup. \lgroup\overline{B}+\overline{C}\rgroup}}$$

Again, using De Morgan’s theorem $\mathrm{\lgroup\overline{A.B.C}= \overline{A}+\overline{B}+\overline{C}\rgroup}$, we get,

$$\mathrm{Y=\overline{\lgroup\overline{A}+ \overline{B}\rgroup}+\overline{\lgroup\overline{A} +\overline{B}+\overline{C}\rgroup}+ \overline{\lgroup\overline{B}+\overline{C}\rgroup}}$$

Hence, this is the form of the given logic function which can be implemented using NOR gates. The realization of this logic function Y in SOP form using NOR gates is shown in Figure 2.

Note − A' is same as A̅.

Example 2

Implement the following logic function in SOP form using NOR gates.

$$\mathrm{Y=A\overline{B}+B\overline{C}+ABC}$$

Solution

The given logic function is,

$$\mathrm{Y=A\overline{B}+B\overline{C}+ABC}$$

As the function in the given form cannot be implemented using NOR gate. So first we will convert it into a form that can be realized using NOR gates as follows −

Double complementing on the both sides, we get,

$$\mathrm{\overline{\overline{Y}}=Y=\overline{\overline{A\overline{B}+B\overline{C}+ABC}}}$$

Using De Morgan’s theorem $\mathrm{\lgroup\overline{A+B+C}= \overline{A}.\overline{B}.\overline{C}\rgroup}$, we get,

$$\mathrm{Y=\overline{\overline{A\overline{B}}.\overline{B\overline{C}}.\overline{ABC}}}$$

Again, apply De Morgan’s theorem $\mathrm{\lgroup\overline{A.B.C}= \overline{A}+\overline{B}+\overline{C}\rgroup}$ to convert AND operations into OR operations, i.e.,

$$\mathrm{Y=\overline{\lgroup\overline{A}+B\rgroup.\lgroup\overline{B}+C\rgroup.\lgroup\overline{A}+\overline{B}+\overline{C}\rgroup}}$$

Applying De Morgan’s theorem $\mathrm{\lgroup\overline{A.B.C}= \overline{A}+\overline{B}+\overline{C}\rgroup}$ again to convert the AND operations into OR operations, i.e.

$$\mathrm{Y=\lgroup\overline{\overline{A}+B}\rgroup+\lgroup\overline{\overline{B}+C}\rgroup+\lgroup\overline{\overline{A}+\overline{B}+\overline{C}}\rgroup}$$

This is the form of the given logic function which can be implemented using NOR gates. Now, determine the number of NOR gates and connect them according to the logic expression to get the logic implementation of the function. The realization of the given logic function in SOP form using NOR gates is shown in Figure 3.

Conclusion

This is all about the realization of a logic function in SOP form using NOR gates. From the above discussion, we may conclude that a logic function in SOP form cannot be realized using NOR gates directly, but it first needs to be transformed in a realizable form. Then, NOR gates are connected together to implement the desired logic function.