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    JEE Main, NEET: Logic Gate Made Easy

    By Safeer PP
    19 May'22  6 min read
    JEE Main, NEET: Logic Gate Made Easy
    Synopsis

    Logic Gates is an important topic from the unit Electronic Devices for NEET and JEE Main. Read on to know in detail about the concepts of logic gates and the questions that need to be studied for NEET and JEE Main.

    JEE Main, NEET: Logic Gate Made Easy
    Synopsis

    Logic Gates is an important topic from the unit Electronic Devices for NEET and JEE Main. Read on to know in detail about the concepts of logic gates and the questions that need to be studied for NEET and JEE Main.

    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.

    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

    AND GateAND 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

    Equivalent Circuit Of AND GateEquivalent 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 GateOR 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 GateEquivalent 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).

    Also Read| NEET Physics: Five-year Analysis, Most Asked Chapters And Concepts

    NOT Gate

    1652679789161NOT Gate

    Output 1652679778599

    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 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+A=A

    A\cdot A=A

    A+1=1

    A\cdot 1=1

    1652679775056

    A+ 0=A

    \overline{\overline{A}}=A

    1652679773913

    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 GateNAND Gate

    Output 1652679775291

    Truth Table For NAND Gate

    Inputs

    Output 1652679775863

    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

    1652679779319

    1652679776825NOR Gate

    Truth Table For NOR Gate

    Inputs

    Output 1652679779041

    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

    1652679783226

    ((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 1652679790900. 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 NANDXOR 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 ORXOR 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 ANDXNOR 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.

    Practice Questions

    NEET 2017

    The given electrical network is equivalent to

    1652679786874

    NEET 2018

    In the combination of the following gates the

    output Y can be written in terms of inputs A and B as

    1652679787450

    NEET 2019

    1652679784689

    The Boolean operation represented by the circuit diagram drawn is:

    NEET 2020

    For the logic circuit shown, the truth table is:

    1652679779862

    NEET 2021

    For the given circuit the input digital signals are applied at the terminals A, B and C, What would be the output at terminal y?

    1652679787197

    1652679786265

    • Technology
    • Electricity
    • Electrical engineering
    • Computer science
    • Computer engineering
    • Electronics
    • Electronic engineering
    • Logic
    • Mathematical logic
    • Theoretical computer science

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