JEE Main, NEET: Logic Gate Made Easy

Edited By Safeer PP | Updated on May 19, 2022 08:58 AM IST

The topic Logic Gate constitutes less than 10 per cent of the Electronic Devices unit. But from this small portion,we selected one question for each year from 2017-21 for NEET. For JEE Main as well the questions from this topic are regularly repeated. This tells us that Logic Gates is an important topic for both the entrance exams. So let’s take a look at the concepts of logic gates that need to be studied for NEET or JEE Main.

JEE Main, NEET: Logic Gate Made Easy

Logic Gate

Logic gates are fundamental blocks of a digital system. There are three basic gates - AND, OR and NOT gates. Also two universal gates NAND and NOR are discussed in the unit Electronic Devices. The questions in the NEET are based on these five logic gates. The input and output of the logic gates can occur in 1 and 0 or High and Low. The truth table shows the output of the logic circuit for various combinations of the inputs. Each basic gate is discussed with its respective truth tables.

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AND Gate

Output Y=A.B

AND gate can have two or more inputs with a single output. The output is High(1) only when all the inputs are High(1).

Equivalent Circuit Of AND Gate

The bulb Y glows only when both switches are closed. That is both A and B are 1. In all the other cases the bulb does not glow as the circuit is incomplete.

Truth Table Of AND Gate

Inputs		Output
A	B	Y
0	0	0
0	1	0
1	0	0
1	1	1

Truth Table For Bulb And Switch Arrangement

Switches		Output
A	B	Bulb
Open	Open	OFF
Open	Closed	OFF
Closed	Open	OFF
Closed	Closed	ON

OR Gate

OR Gate OR Gate

Output Y=A+B

OR gate can have two or more inputs with an only single output. The output is Low (0) only when all the inputs are Law(0).

The Equivalent Circuit Of OR Gate

Equivalent Circuit Of OR Gate

When any of the switches A or B or both are closed the bulb will glow. The bulb will be off when both the switches are open.

Truth Table Of OR GATE

Inputs		Output
A	B	Y
0	0	0
0	1	1
1	0	1
1	1	1

The truth table for the circuit arrangement can be easily made by replacing 0 with open switch and 1 with closed switch here. Try yourself and find out when the bulb will be ON(Glow).

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NOT Gate

1652679789161 NOT Gate

Output

Also Y can be represented as Y=A’

NOT gate has only one input and one output. NOT gate is known as an inverter. The output of the NOT gate is always the complement of the input.

Equivalent Circuit Of NOT GATE

Equivalent Circuit Of NOT Gate

When the switch is open the bulb will glow. When the switch is closed the bulb is not a part of the circuit and it will remain off.

Truth Table Of NOT Gate

Input	Output
A	Y
0	1
1	0

Universal gates is to be discussed next. But before going to it let's have a look at some important relations.

De Morgan's Law -

A and B are input.

1) $\overline{A+B}=\overline{A}\cdot\overline{B}$

2) $\overline{A\cdot B}=\overline{A}+\overline{B}$

3) $\overline{\overline{A}+\overline{B}}=A\cdot B$

4) $\overline{\overline{A}\cdot\overline{B}}=A+B$

Some Important Relations

$A\cdot A=A$

$A\cdot 1=1$

$\overline{\overline{A}}=A$

Universal Gates

Using universal gates we can implement any other basic Gates or Boolean function. NAND and NOR gates are universal gates. Let's have a look at the truth table for these two universal gates.

NAND GATE

NAND gate is the combination of AND gate and NOT gate.

NAND Gate

Output

Truth Table For NAND Gate

Inputs		Output
A	B	Y
0	0	1
0	1	1
1	0	1
1	1	0

NAND gate is nothing but the inverted OR. Also from Demorgan's theorem, we know that

$\overline{A\cdot B }= \bar{A}+ \bar{B}$

So the NAND gate can also be represented as -

1652679778396

Y= $\overline{A\cdot B }= \bar{A}+ \bar{B}$

Here we see that the NAND gate is equivalent to bubbled OR. Also a bubbled NAND gate is equivalent to an OR gate.

NOR Gate

NOR gate is the combination of OR gate and NOT gate.

1652679776358

1652679776825 NOR Gate

Truth Table For NOR Gate

Inputs		Output
A	B	Y
0	0	1
0	1	0
1	0	0
1	1	0

NOR gate is nothing but an inverted AND. Also from Demorgan's theorem, we know that

$\overline{A+B }= \bar{A}\cdot \bar{B}$

So the NOR gate can be represented as-

1652679781140

Y= $\overline{A+B }= \bar{A}\cdot \bar{B}$

We can tell that the NOR gate is equivalent to bubbled AND. That is A’.B’=(A+B)’

. Also bubbled NOR is equivalent to AND gate. That is

$\overline{\overline{A}+\overline{B}}=A\cdot B$

1652679780175

Here the output Y=A.B

Questions based on this are asked in JEE Main 2009 and NEET 2020

JEE Main 2009 Question: The logic circuit shown below has the input waveforms 'A' and 'B' as shown. Find the correct output waveform.

1652679780438

1652679774552

Solution: The given logical circuit is a bubbled NOR gate which is equivalent to AND gate. So the output will be HIGH when both the inputs are High. From the above-given graph it is clear that both the outputs are High in two cases only. So the output will be

1652679777767

And the NEET 2020 question was to find the truth table of the given logic circuit in the above question. So the truth table will be equivalent to that of an AND gate.

Realization Of Basic Gates Using Universal Gates

Here the term realization means how to implement a basic gate or boolean expression using the combination of only a single type of universal gate. That is either only NAND or NOR to get a desired output. Let us see some common examples.

Realization Of NOT Gate Using NAND Gate

1652679780656

Realizing AND Gate Using NAND Gate

1652679778833

$Y=\overline{\overline{AB}}=A.B\\$

Y = A.B represents the AND gate. This was a question in JEE Main 2014

Realization Of OR Gate Using NAND Gate

1652679783829

From De Morgan's law

$X=\overline{\bar{A}\ .\ \bar{B}}=A+B$

The above circuit is nothing but the bubbled NAND gate, which is equivalent to OR gate

This was a question in JEE Main 2010 and JEE Main, 2020 and 2021

JEE Main 2020 Question: Identify the correct o/p signal Y in the given combination of gates (as shown) for the given I/P A and B

1652679779587

1652679786615

The given circuit is nothing but the bubbled NAND, which is equivalent to an OR gate. So the output will be LOW when both the inputs are LOW. In all other cases, the output will be High.

So the output graph will look like this -

1652679781912

Realization Of NOR Gate Using NAND Gate

1652679785352

The above circuit represents a bubbled NAND given to an inverter. Bubbled NAND is nothing but OR gate. An OR gate given to the inverter is the NOR gate. JEE Main 2021 had a question to identify the truth table of the above circuit

Realization Of NOT Gate Using NOR Gate

1652679784973

Realization Of AND Gate Using NOR Gate

1652679785623

This logic circuit is nothing but the bubbled OR gate. The same is asked in JEE Main 2021. But the first two inverters (NOT gate) are realized using NAND gate.

Please read the question given below.

Question:

Identify the logic operation carried out.

1652679782190

Solution: AND gate(bubbled OR)

Realization Of OR Gate Using NOR Gate

1652679788639

((A+B)’ )’=A+B since $\bar{\bar{A}}=A$ . The sign ‘ represents compliment

Realization Of NAND Gate Using NOR Gate

1652679785833

The above logic circuit is nothing but the output of bubbled NOR (AND) is given to an inverter gate. This was a question in JEE Main 2021.

Some Familiar Logic Operations Using Universal Gates

XOR

XOR output is given by the Boolean expression . This function can be realized using different combinations of gates. The realization of A’B+AB’ using universal gates is discussed here.

XOR Using NAND Gate

XOR Using NAND

1652679786026

So we got the output AB’+A’B

Truth Table For A’B+AB’

Input A	Input B	Output
0	0	0
0	1	1
1	0	1
1	1	0

XOR Using NOR Gate

XOR Using OR

$\\\overline{\overline{\overline{A+\overline{A+B}}+\overline{B+\overline{A+B}}}}\\={\overline{A+\overline{A+B}}+\overline{B+\overline{A+B}}}\\using\ \overline{A+B}=\overline{A}.\overline{B}\\=\overline{A}.(A+B)+\overline{B}(A+B)=\\\overline{A}B+A\overline{B}$

XNOR

XNOR output is $AB+\overline{A}. \overline{B}$

Truth Table For AB+A’B’

Input A	Input B	Output
0	0	1
0	1	0
1	0	0
1	1	1

XNOR Using NAND Gate

XNOR Using AND

The output of the combination will be $AB+\overline{A}. \overline{B}$

XNOR Using NOR Gate

1652679788892

The output of the combination will be $AB+\overline{A}. \overline{B}$ . This was a question in JEE Main 2021. The question was to find the truth table of the above circuit.

Diode Circuits For AND And OR Gate

AND Gate

1652679787689

OR Gate

1652679782974

Study all the basic and universal logic gates discussed, and for combinations of logic gates apply Demorgan's theorem or make a truth table to get the output.

But if you familiarise yourself with the above logic circuits it will be helpful to save time during the examination. You can arrive at the solutions by simply observing the circuits.