lesson 4.2 universal gates - anula vidyalaya

ISHANI NARAHENPITA © 2020 ANULA VIDYALAYA 1

Universal Gates

A universal logic gate is a logic gate that can be used to construct all other logic gates. The NAND gate and NOR

gates can be considered as universal logic gates.

The advantage of universal gates: NAND and NOR gates are economical and easier to fabricate and are the basic

gates used in all IC digital logic families.

NAND Gate

1. How to implement NOT gate using NAND gates?

Simplification of the output of above NAND gate produces A’ at the end.

(A.A)’

A’ + A’

A’

2. How to implement AND gate using NAND gates?

Simplification of the output of above NAND gate produces A.B at the end.

((A.B). (A.B))

(A.B) + (A.B)

(A.B) + (A.B)

A.B

Lesson 4 - Uses logic gates to design basic digital circuits and devices

Lesson 4.2 – Universal Gates

Advanced Level Information & Communication Technology

Ishani Narahenpita (B.Sc. Sp. (Hon) IT, M.Sc. (IT)


3. How to implement OR gate using NAND gates?

Simplification of the output of above NAND gate produces A+B at the end.

NOR Gate

1. How to implement NOT gate using NOR gates?

Simplification of the output of above NOR gate produces A’ at the end.

(A+A)’

A’.A’

A’

2. How to implement OR gate using NOR gates?

Simplification of the output of above NOR gate produces A+B at the end.

((A.A). (B.B))

(A.A) + (B.B)

(A.A) + (B.B)

A + B

((A+B) + (A+B))

(A+B). (A+B)

(A+B). (A+B)

A+B

(A.A)’

(B.B)’

((A.A). (B.B))


3. How to implement AND gate using NOR gates?

Simplification of the output of above NOR gate produces AB at the end.

Represent following Boolean expression using NAND gates

bc + bd

((A+A)+(B+B))

(A+A).(B+B)

(A+A).(B+B)

A.B

𝑏𝑐

𝑏𝑐

𝑏𝑑

𝑏𝑐

𝑏𝑑

𝑏𝑐

𝑏𝑑

𝑏𝑐. 𝑏𝑑

𝑏𝑑

𝑏𝑐. 𝑏𝑑 𝑏𝑐. 𝑏𝑑

𝑏𝑐+𝑏𝑑 De Morgan’s law

𝑏𝑐 + 𝑏𝑑 Double Complement law

Example 1

𝑏𝑐 𝑏𝑐 𝑏𝑐

𝑏𝑐. 𝑏𝑑

AND

AND

OR

EXTRA GATES


METHOD TWO

bc + bd

Step 1- Convert bc + bd to NAND only Boolean expression. For this put two over bars on whole Boolean expression

𝑏𝑐 + 𝑏𝑑

Step 2- Apply De Morgan’s law as below

𝑏𝑐. 𝑏𝑑

Step -Draw the circuit

Represent following Boolean expression using NOR gates

bc + bd

Seven NOR gates are required for above implementation. When trying to implement product terms as it is from NOR gates,

it require more gates. So it is possible to convert bc+bd to a NOR only sum term. For this you can apply De morgan’s law

on bc+bd.


bc + bd

Step 1- Apply distributive law → b(c+d)

Convert b(c+d) to NOR only Boolean expression. For this put two over bars on whole Boolean expression

𝑏(𝑐 + 𝑑)

𝑐

𝑏

𝑑

𝑏𝑐

𝑏𝑑

𝑏𝑐 + 𝑏𝑑 𝑏𝑐 + 𝑏𝑑

Example 2

𝑏𝑐

𝑏𝑑

𝑏𝑐. 𝑏𝑑

𝑏𝑐. 𝑏𝑑



AND

AND

OR



𝑏 + 𝑐 + 𝑑



𝑏𝑐 + 𝑏𝑑

bc bc

bc

d

bd bd

bd

𝑏𝑐. 𝑏𝑑

Example 3

𝑏

𝑐 + 𝑑

𝑏 + 𝑐 + 𝑑

AND

NOT

AND

OR

bc bc

bc

d

bd bd

bd

EXTRA GATES

EXTRA GATES


METHOD 2

𝑏𝑐 + 𝑏𝑑


𝑏𝑐 + 𝑏𝑑


𝑏𝑐. 𝑏𝑑


𝑏𝑐. 𝑏𝑑



𝑑

𝑏𝑐

𝑏𝑑

𝑏𝑐. 𝑏𝑑

𝑑

𝑏𝑐

𝑏𝑑

𝑏𝑐. 𝑏𝑑

𝑏𝑐. 𝑏𝑑





𝑏𝑐 + 𝑏𝑑

Six NOR gates are required for above implementation. When trying to implement product terms as it is from NOR gates, it

require more gates. So it is possible to convert bc+bd to a NOR only sum term. For this you can apply De morgan’s law on

bc+bd as below

Represent following Boolean expression using NOR gates → 𝑏𝑐 + 𝑏𝑑

Step 1- Apply distributive law → b(c+𝑑)

Convert b(c+𝑑)to NOR only Boolean expression. For this put two over bars on whole Boolean expression

𝑏(𝑐 + 𝑑)


𝑏 + 𝑐 + 𝑑


𝑐

𝑏

𝑑

𝑐

𝑑

𝑏𝑐

𝑏𝑑 + 𝑏𝑐

𝑏𝑑

𝑏

𝑏𝑑

𝑏𝑐 𝑏𝑑 + 𝑏𝑐 𝑏𝑑 + 𝑏𝑐

𝑏𝑑 + 𝑏𝑐

Example 4

AND

NOT

AND

OR

EXTRA GATES

𝑑

𝑏

𝑐 + 𝑑

𝑏 + 𝑐 + 𝑑



b (c + d)

Five NAND gates are required for above implementation. When trying to implement SUM terms as it is from NAND gates,

it require more gates. So it is possible to convert b(c+d) to a NAND only PRODUCT terms. For this you can apply De

morgan’s law on b(c+d) as below


b (c + d)

Step 1- Apply distributive law → bc + b𝑑


𝑏𝑐 + 𝑏𝑑


𝑏𝑐. 𝑏𝑑


𝑐

𝑑

𝑐. 𝑑

(𝑐. 𝑑)𝑏

𝑏

𝑑 (𝑐. 𝑑)𝑏

(𝑐. 𝑑)𝑏

(𝑐 + 𝑑) . 𝑏 De Morgan’s law

(𝒄 + 𝒅) . 𝒃 Double Complement law

Example 5

OR

AND

(𝑐. 𝑑)𝑏



𝑑

𝑏𝑐

𝑏𝑑

𝑏𝑐. 𝑏𝑑

𝑏𝑐. 𝑏𝑑





b (c + d)

METHOD 2


b (c + d)

Step 1- Apply distributive law → b(c+𝑑)

Convert b(c+𝑑)to NOR only Boolean expression. For this put two over bars on whole Boolean expression

𝑏(𝑐 + 𝑑)


𝑏 + 𝑐 + 𝑑


𝑐 + 𝑑 + 𝑏



Example 6

𝑐 + 𝑑 + 𝑏



𝑐 + 𝑑 𝑐 + 𝑑

𝑐 + 𝑑

𝑏

OR

AND

EXTRA GATES


A⊕B = 𝐴. 𝐵 + 𝐴. 𝐵

Option 1

Option 2

𝐴

𝐵

𝐴. 𝐵

𝐴. 𝐵

(𝐴. 𝐵)(𝐴. 𝐵)

(𝐴. 𝐵) (𝐴. 𝐵)= 𝐴. 𝐵 + 𝐴. 𝐵

(𝐴. 𝐵) + (𝐴. 𝐵) De Morgan’s law

𝐴. 𝐵 + 𝐴. 𝐵 Double Complement law

∴ 𝐴. 𝐵 + 𝐴. 𝐵 = A⊕B

𝐴. 𝐵

𝐴(𝐴. 𝐵)

𝐵(𝐴. 𝐵)

𝐴(𝐴. 𝐵). 𝐴(𝐴. 𝐵)

𝐴(𝐴. 𝐵). 𝐴(𝐴. 𝐵) = 𝐴. 𝐵 + 𝐴. 𝐵

𝐴(𝐴. 𝐵) + 𝐵(𝐴. 𝐵) De Morgan’s law

𝐴(𝐴. 𝐵) + 𝐵(𝐴. 𝐵) Double Complement law

𝐴(𝐴 + 𝐵) + 𝐵(𝐴 + 𝐵) De Morgan’s law

𝐴𝐴 + 𝐴𝐵 + 𝐵𝐴 + B𝐵 Distributive law

𝐴𝐵 + 𝐵𝐴 Inverse Law

∴ 𝐴. 𝐵 + 𝐴. 𝐵 = A⊕B

How to represent XOR gate using NAND gates and NOR gates?

XOR gate using NAND gates


Option 1

Option 2

𝐴

𝐵

𝐴 + 𝐵

𝐴 + 𝐵

𝐴 + 𝐵 + 𝐴 + 𝐵

𝐴 + 𝐵 + 𝐴 + 𝐵 = 𝐴. 𝐵 + 𝐴. 𝐵

(𝐴 + 𝐵)(𝐴 + 𝐵) De Morgan’s law

(𝐴 + 𝐵)(𝐴 + 𝐵) Double Complement law

𝐴𝐴 + 𝐵𝐵 + 𝐵𝐴 + 𝐴𝐵 Distributive Law

𝐵𝐴 + 𝐴𝐵 Inverse Law

∴ 𝐴. 𝐵 + 𝐴. 𝐵 = A⊕B

𝐴 + 𝐵

𝐴 + 𝐴 + 𝐵

𝐵 + 𝐴 + 𝐵

𝐴 + 𝐴 + 𝐵 + 𝐵 + 𝐴 + 𝐵

(𝐴 + 𝐴 + 𝐵 + 𝐵 + 𝐴 + 𝐵)

(𝐴 + 𝐴 + 𝐵 + 𝐵 + 𝐴 + 𝐵)= 𝐴. 𝐵 + 𝐴. 𝐵

((𝐴. (𝐴 + 𝐵) + 𝐵. (𝐴 + 𝐵)) De Morgan’s law

(𝐴(𝐴 + 𝐵) + 𝐵(𝐴 + 𝐵)) Double Complement law

𝐴𝐴 + 𝐴𝐵 + 𝐵𝐴 + 𝐵𝐵 Distributive Law

𝐴𝐵 + 𝐵𝐴 Inverse Law

∴ 𝐴. 𝐵 + 𝐴. 𝐵 = A⊕B

XOR gate using NOR gates


A ⊕ B = A. 𝐵 + 𝐴. 𝐵

Option 1

Option 2

𝐴

𝐴. 𝐵

𝐵

𝐴. 𝐵

(𝐴. 𝐵) (𝐴. 𝐵)

(𝐴. 𝐵) (𝐴. 𝐵)= A. 𝐵 + 𝐴. 𝐵

(𝐴. 𝐵) + (𝐴. 𝐵) De Morgan’s law

𝐴. 𝐵 + 𝐴. 𝐵 Double Complement law

∴ 𝐴. 𝐵 + 𝐴. 𝐵 = A ⊕ B

𝐴. 𝐵

𝐴(𝐴. 𝐵)

𝐵(𝐴. 𝐵)

𝐴(𝐴. 𝐵). 𝐵(𝐴. 𝐵)

𝐴(𝐴. 𝐵). 𝐵(𝐴. 𝐵)

𝐴(𝐴. 𝐵). 𝐵(𝐴. 𝐵)= A. 𝐵 + 𝐴. 𝐵

𝐴 + (𝐴. 𝐵) . 𝐵 + (𝐴. 𝐵) De Morgan’s law

(𝐴 + 𝐴. 𝐵) ( 𝐵 + 𝐴. 𝐵) Double Complement law

𝐴. 𝐵 + 𝐴. 𝐴. 𝐵+ 𝐴. 𝐵. 𝐵 + 𝐴. 𝐵. 𝐴. 𝐵 Distributive law

𝐴. 𝐵 +0.B+ 0.B + 𝐴. 𝐵. 𝐴. 𝐵 Inverse law

𝐴. 𝐵 +0+ 0 + 𝐴. 𝐵. 𝐴. 𝐵 Identity law

𝐴. 𝐵 + 𝐴. 𝐴. 𝐵. 𝐵 Associative law

𝐴. 𝐵 + 𝐴. 𝐵 Idempotent law

∴ 𝐴. 𝐵 + 𝐴. 𝐵 = A ⊕ B

How to represent XNOR gate using NAND gates and NOR gates?

XNOR gate using NAND gates


Option 1

Option 2

𝐴 + 𝐵

𝐴 + 𝐴 + 𝐵

𝐵 + 𝐴 + 𝐵

𝐴 + 𝐴 + 𝐵 + 𝐵 + 𝐴 + 𝐵

𝐴 + 𝐴 + 𝐵 + 𝐵 + 𝐴 + 𝐵

(𝐴 + 𝐴 + 𝐵). (𝐵 + 𝐴 + 𝐵) De Morgan’s law

(𝐴 + 𝐴 + 𝐵). (𝐵 + 𝐴 + 𝐵) Double Complement law

(𝐴 + 𝐴. 𝐵)( 𝐵 + 𝐴. 𝐵) De Morgan’s law

(𝐴 + 𝐵) (𝐵 + 𝐴) Redundancy Law

𝐴𝐵 + 𝐵. 𝐴 + 𝐵. 𝐵 + 𝐴𝐴 Distributive law

𝐴𝐵 + 𝐵. 𝐴 Inverse Law

∴ 𝐴𝐵 + 𝐵. 𝐴= A ⊕ B

𝐴

𝐵

𝐴 + 𝐵

𝐵 + 𝐴

𝐴 + 𝐵 + 𝐵 + 𝐴

𝐴 + 𝐵 + 𝐵 + 𝐴

(𝐴 + 𝐵). (𝐵 + 𝐴) De Morgan’s law

(𝐴 + 𝐵). (𝐵 + 𝐴) Double Complement law

𝐴𝐴 + 𝐴. 𝐵 + 𝐵𝐵 + 𝐵𝐴 Distributive law

𝐴. 𝐵 + 𝐵𝐴 Inverse Law

∴ 𝐴𝐵 + 𝐴. 𝐵= A ⊕ B

XNOR gate using NOR gates

lesson 4.2 universal gates - anula vidyalaya

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