Draw a diagram to show how NAND gates can be combined to obtain an OR gate.
Answer
620.4k+ views
Hint: NAND gate and NOR gate are called universal gates since all simple gates like NOT, AND, and OR gates can be constructed using them. In the NAND gate, the two inputs are multiplied and the inverse is taken of the answer.
If A and B are two inputs to a NAND gate, the output is: $\overline {A.B} $
The formula used for converting NAND gate to OR gate is:
\[\overline {\overline A .\overline B } = A + B\]
Complete step by step solution:
If A and B are two inputs to a NAND gate, the output is: $\overline {A.B} $
To construct an OR gate from NAND gates, we need 3 NAND gates. Let us arrange them as shown in the circuit diagram:
We will analyse the circuit by considering each of the NAND gates.
Step 1: NAND Gate -1
Input for the NAND gate = A and A
NAND operation = \[\overline {A.A} \]
Output = $\overline A $
Step 2: NAND Gate -1
Input for the NAND gate = B and B
NAND operation = \[\overline {B.B} \]
Output = $\overline B $
Step 1: NAND Gate -1
Input for the NAND gate = $\overline A $and $\overline B $
NAND operation =
\[\overline {\overline A .\overline B } = \overline{\overline A} + \overline{\overline B} \]
Since double negative is positive,
\[\overline {\overline A .\overline B } = \overline{\overline A} + \overline{\overline B} = A + B\]
Output = $A + B$
The final output is equivalent to the OR gate:
$A||B = A + B$
$\therefore$Three NAND gates can be used to create an OR gate.
Note:
Sometimes, it becomes difficult to remember the conversion of Universal gates. You can easily remember the whole thing by understanding this mnemonic.
NAND Gate:
To convert to AND gate, first negate together because AND means together and negate the result.
To convert to OR gate, first negate them individually because OR means individually and negate the result.
If A and B are two inputs to a NAND gate, the output is: $\overline {A.B} $
The formula used for converting NAND gate to OR gate is:
\[\overline {\overline A .\overline B } = A + B\]
Complete step by step solution:
If A and B are two inputs to a NAND gate, the output is: $\overline {A.B} $
To construct an OR gate from NAND gates, we need 3 NAND gates. Let us arrange them as shown in the circuit diagram:
We will analyse the circuit by considering each of the NAND gates.
Step 1: NAND Gate -1
Input for the NAND gate = A and A
NAND operation = \[\overline {A.A} \]
Output = $\overline A $
Step 2: NAND Gate -1
Input for the NAND gate = B and B
NAND operation = \[\overline {B.B} \]
Output = $\overline B $
Step 1: NAND Gate -1
Input for the NAND gate = $\overline A $and $\overline B $
NAND operation =
\[\overline {\overline A .\overline B } = \overline{\overline A} + \overline{\overline B} \]
Since double negative is positive,
\[\overline {\overline A .\overline B } = \overline{\overline A} + \overline{\overline B} = A + B\]
Output = $A + B$
The final output is equivalent to the OR gate:
$A||B = A + B$
$\therefore$Three NAND gates can be used to create an OR gate.
Note:
Sometimes, it becomes difficult to remember the conversion of Universal gates. You can easily remember the whole thing by understanding this mnemonic.
NAND Gate:
To convert to AND gate, first negate together because AND means together and negate the result.
To convert to OR gate, first negate them individually because OR means individually and negate the result.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Name the crygenes that control cotton bollworm and class 12 biology CBSE

Ribosomal RNA is actively synthesised in A Nucleoplasm class 12 biology CBSE

How many molecules of ATP and NADPH are required information class 12 biology CBSE

In a transcription unit the promoter is said to be class 12 biology CBSE

The number of chromosomes in male grasshopper is a class 12 biology CBSE

Verhulst Pearl logistic growth described by the equation class 12 biology CBSE

