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Fundamentals of digital electronics. Prepared by - Anuradha Tandon Assistant Professor, Instrumentation & Control Engineering Branch, IT, NU. Why go digital?. Analogue signal processing is achieved by using analogue components such as: Resistors.

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Fundamentals of digital electronics

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Fundamentals of digital electronics

Fundamentals of digital electronics

Prepared by - AnuradhaTandon

Assistant Professor,

Instrumentation & Control Engineering Branch,


Why go digital

Why go digital?

  • Analogue signal processing is achieved by using analogue components such as:

    • Resistors.

    • Capacitors.

    • Inductors.

  • The inherent tolerances associated with these components, temperature, voltage changes and mechanical vibrations can dramatically affect the effectiveness of the analogue circuitry.

Why use binary logic only

Why Use Binary Logic Only?

  • Use of transistor as a switch

  • Controlling the transistor operation in either ON and OFF state

  • If use more than two logic levels, transistor needs to be operated in the active region where operating the transistor is difficult

Boolean functions terminology

Boolean Functions: Terminology

  • F(a,b,c) = a’bc + abc’ + ab + c

    • Variable

    – Represents a value (0 or 1), Three variables: a, b, and c

    • Literal

    – Appearance of a variable, in true or complemented form

    – Nine literals: a’, b, c, a, b, c’, a, b, and c

    • Product term

    – Product of literals, Four product terms: a’bc, abc’, ab, c

    • Sum‐of‐products (SOP)

    – Above equation is in sum‐of‐products form.

    – “F = (a+b)c + d” is not.

Boolean logic f unction

Boolean Logic Function

  • Can be represented in two forms:

    • Sum-of-Products (SOP)

      F(A, B, C) = A’BC + BC’ + AB

    • Product-of-Sums (POS)

      F(A, B, C) = (A + B’ + C’).(B’ + C).(A’ + B’)

Boolean logic function cont

Boolean Logic Function cont.….

  • The Boolean function expressed in SOP form can implemented using two levels of basic logic gates:

    • 1st level of AND gates to represent the AND terms and,

    • The 2nd level of OR gates to OR the AND terms

Boolean logic function cont1

Boolean Logic Function cont.….

  • For example the function

    F(X,Y,Z) = XZ+Y’Z+X’YZ

    can be represented using 2-input AND and OR gates as shown in the Fig. 1:

Fig. - 1

Boolean logic function cont2

Boolean Logic Function cont.….

  • The Boolean function expressed in POS form can implemented using two levels of basic logic gates:

    • 1st level of OR gates to represent the OR terms and,

    • The 2nd level of AND gates to AND the OR terms

Boolean logic function cont3

Boolean Logic Function cont.….

  • For example the function


    can be represented using 2-input AND and OR gates as shown in the Fig. 2:

Fig. - 2

Boolean logic function cont4

Boolean Logic Function cont.….

  • SOP or POS form of expression of Boolean logic function is called the standard form

  • The other way to represent the Boolean logic function is the canonical form

Canonical form

Canonical Form

  • The Boolean function is represented as either

    • Sum-of-Minterms (SOM) or

    • Product-of-Maxterms (POM)

Canonical forms

Canonical Forms

  • It is useful to specify Boolean functions in

    a form that:

    – Allows comparison for equality.

    – Has a correspondence to the truth tables

  • Canonical Forms in common usage:

    – Sum of Minterms (SOM)

    – Product of Maxterms (POM)



  • product term is a term where literals are ANDed.

    Example: x’y’, xz, xyz, …

  • Minterm: A product term in which all variables appear exactly once, in normal or complemented form

    Example: F(x,y,z) has 8 minterms

    x’y’z’, x’y’z, x’yz’, ...

Minterms cont

Minterms cont.……

  • Function with n variables has 2n minterms

  • A minterm equals 1 at exactly one input combination and is equal to 0 otherwise

    Example: x’y’z’ = 1 only when x=0, y=0, z=0

  • A minterm is denoted as mi where i corresponds the input combination at which this minterm is equal to 1

2 variable minterms

2 variable minterms

  • Two variables (X and Y) produce 2x2=4


    XY (both normal)

    XY’ (X normal, Y complemented)

    X’Y (X complemented, Y normal)

    X’Y’ (both complemented)



  • Maxterms are OR terms with every variable in true or complemented form.

    X+Y (both normal)

    X+Y’ (x normal, y complemented)

    X’+Y (x complemented, y normal)

    X’+Y’ (both complemented)

2 variable minterms and maxterms

2 Variable Minterms and Maxterms

  • The index above is important for describing which variables in the terms are true and which are complemented.

Expressing functions using minterms

Expressing Functions using Minterms

  • Boolean function can be expressed algebraically from a give truth table

  • Forming sum of ALL the minterms that produce 1 in the function

Expressing functions with maxterms

Expressing Functions with Maxterms

  • Boolean function : Expressed algebraically from a give truth table

  • By forming logical product (AND) of ALL the maxtermsthat produce 0 in the function

  • Example:

  • Consider the function defined by the truth table

  • F(X,Y,Z) = Π M(1,3,4,6)

  • Applying DeMorgan

    • F’ = m + m + m + m = Σm(1 3 4 6)

    • F = F’’ = [m1 + m3 + m4 + m6]’

    • = m1’.m3’.m4’.m6’

    • = M1.M3.M4.M6

    • = Π M(1,3,4,6)

Sum of minterms v s product of maxterms

Sum of Mintermsv/s Product of Maxterms

  • A function can be expressed algebraically as:

    • The sum of minterms

    • The product of maxterms

  • • Given the truth table, writing F as

    • Σmi – for all minterms that produce 1 in the table,


    • ΠMi – for all maxterms that produce 0 in the table

  • Mintermsand Maxterms are complement of each other.

Example minterm maxterm

Example: minterm & maxterm

Example cont

Example cont.….

Sop and pos conversion

SOP and POS Conversion


F =(A’+B)(A’+C)(C+D)

=( A’+BC)(C+D)



= A’C+A’D+BC


F = AB + CD

= (AB+C)(AB+D)

= (A+C)(B+C)(AB+D)

= ( A+C)(B+C)(A+D)(B+D)

Simplification of boolean functions

Simplification of Boolean Functions

  • An implementation of a Boolean Function requires the use of logic gates.

  • A smaller number of gates, with each gate (other then Inverter) having less number of inputs, may reduce the cost of the implementation.

  • There are 2 methods for simplification of Boolean functions.

Simplification of boolean functions cont

Simplification of Boolean Functions cont.….

  • Algebraic method by using Identities & Theorem

  • Graphical method by using KarnaughMap method

    –The K‐map method is easy and straightforward

    –A graphical method of simplifying logic equations or truth tables

    -Also called a K map

Karnaugh map

Karnaugh Map

  • A K‐map for a function of n variables consists of 2n cells, and,

  • in every row and column, two adjacent cells should differ in the value of only one of the logic variables

  • Theoretically can be used for any number of input variables, but practically limited to 5 or 6 variables.

Gray code

Gray Code

  • Gray code is a binary value encoding in which adjacent values only differ by one bit

Truth table adjacencies

Truth Table Adjacencies

K map method

K – map Method

  • The truth table values are placed in the K map

  • Adjacent K map square differ in only one variable both horizontally and vertically.

  • The pattern from top to bottom and left to right must be in the form

  • A SOP expression can be obtained by Oring all squares that contain a 1.

    A’B’, A’B, AB, AB’

    00, 01, 11, 01

Filling of k map

Filling of K - map

K map

K - map

  • In a K‐map, physical adjacency does imply gray code adjacency

    F =A’B’ + A’B = A’ F = A’B + AB = B

Combinational circuits

Combinational Circuits

  • Combinational circuit

    – Output depends on present input

    – Examples: F (A,B,C), FA, HA, Multiplier, Decoder, Multiplexor, Adder, Priority Encoder

    Y = F (a,b)

    Propagation delay

    Y(t+tpd)=F(a(t), b(t))



  • Reception counter : When you reach a Academic Institute

    – Receptionist Ask: Which Dept. to Go ?

    – Receptionist Redirect you to some building according to your Answer.

  • Decoder : knows what to do with this: Decode

    • N input: 2N output

    • Memory Addressing

    – Address to a particular location



  • 2‐input decoder: four possible input binary numbers

  • So has four outputs, one for each possible input binary number



Implementation of boolean function using decoder

Implementation of Boolean Function Using Decoder

  • Using a n‐to‐2n decoder and OR gates any functions of n variables can be implemented.

    • Example:

    S(x,y,z)= Σ(1,2,4,7) , C(x,y,z)=Σ(3,5,6,7)

    • Functions S and C can be implemented using a 3‐to‐8 decoder and two 4‐input OR gates

Implementation of s and c

Implementation of S and C



  • Mux: Another popular combinational building block

    – Routes one of its N data inputs to its one output, based on binary value of select inputs

  • 4 input mux needs 2 select inputs to indicate which input to route through

  • • 8 input mux  3 select inputs

  • • N inputs  log2(N) selects

    – Like a rail yard switch

Mux internal design

MUX Internal Design

Mux internal design cont

MUX Internal Design cont.….

Implementation of logic function using mux

Implementation of Logic Function Using MUX

Sequential circuits

Sequential Circuits

  • Output depends not just on present inputs

  • But also on past sequence of inputs (State)

    • Stores bits, also known as having “state”

    • Simple example: a circuit that counts up in binary

Sequential circuits1

Sequential Circuits

Example needing bit storage

Example Needing Bit Storage

  • Flight attendant call button

    – Press call: light turns on

    • Stays on after button released

    – Press cancel: light turns off

    – Logic gate circuit to implement this?

First attempt at bit storage

First Attempt at Bit Storage

  • We need some sort of feedback

    – Does the right S Q circuit on do what we want?

    • No: Once Q becomes 1 (when S=1), Q stays forever – no value of S can bring back to 0

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