1. NAND & NOR as Universal gatesBy
Sushma chopde

2. Objectives:
1. To understand what are universal gates 2. To know the construction of other gates using NAND & NOR gates

3. We can construct any other logic gate or circuit by using
Combination of only NAND or NOR gates ,and therefore they
are called universal gates.
Lets see how to construct other gates using NAND gates first.
NOT gate:
We know that the Boolean equation of NAND gate is y=
And Boolean equation for NOT gate is y=A̅
This can be written as y = =A̅ ( i.e by law (A*A)=A ) A
Y=A̅ A y

4. Boolean equation for AND gate is y=A*B
Starting with NAND gate eq. Y=(A*B)’
Take double complement of it i.e y=(A*B)’’=A*B
And this gives us AND gate A
B (A*B)’ y=(A*B)’’
=A*B A B Y 1

5. OR gate: OR gate is given by y=A+B
Using double complement law y’’=(A+B)’’
Will not change the equation
Thus y= (A̅ *B̅)’ using Demorgan’s theorem and now this can be implemented
using NAND gates only as shown below.
A A̅
y=(A̅.B̅)’=A+B
B B̅

6. Truth table for OR gate . A B Y 1

7. For making NOR gate just connect one more NAND
Desired gate NOR
For making NOR gate just connect one more NAND
gate in the above figure as is shown in the following
equation Y=
Y= A̅ . B̅
Y’’ =( A̅ . B̅)’’

8. Ex-OR gate Eq. of EX-OR is y=A̅ .B + B̅ . A
Take double complement of this equation which gives
Y=(((A.B)’.B)’.((A.B)’.A)’)’

9. Ex-OR gate
A Y B

10. Truth Table of Ex-OR gateY 1

11. Lets try to construct other gates using NOR gates only
NOT gate Equation for NOT gate is y=A̅ Therefore By OR law y= A Y A Y 1

12. OR gate OR gate eq. is Y=A+B Y= = (double complement law) A Y B A B Y1

13. AND gate
Eq.for AND gate is Y=A.B Y̅=A̅ + B̅ =(A̅ +B̅)’ A Y B A B Y 1

14. NAND gate
NAND gate eq. is Y= Applying Demorgan’s law Y=A̅ +B̅ =(A̅ +B̅)’’

15. NAND gate
A Y B A B Y 1

16. Ex-OR gate
Now try to construct Ex-OR gate using NOR gates only using the same strategy which we have used in our discussion. Thank you