Conversion to NOR, Gate Logics !

Hi all..

Finally.. my plus two exams are over. Today it was computer science and there was a question to draw logical circuit for A.B + A.C + A’.B’ using only NOR Gates.

It troubled me.. and I went on try to simplify the expression. It failed. Even though i knew equivalent conversions of NOR to others(AND,OR,NOT) , my logic didn’t allow me to write a large logic circuit. So I missed the Question.

Now, I went through it and made some repeated thoughts and tried applying De Morgans Laws..to solve the problem.

These are the info given in the text book:

A OR B = (A NOR B) NOR (A NOR B)
A AND B = (A NOR A) NOR (B NOR B) NOT A = A NOR A

I found out some interesting and easy method for solving this…with out any confusion.

The method is:

First write the correct combatorial circuit for Boolean expression.

Then replace AND, OR, NOT Gates using following diagram.

Then.. if you see two NOT gates in series…it can be cancelled. The remaining NOT gates are to be replaced with NOT equivalent with NOR shown in the diagram.

Thus you can simply make only NOR circuits :D

Now try the same tactics with NAND Gate too.

It seems De Morgan’s Laws are great. Awesome ;)