Fundamentals of digital electronics

1. Fundamentals of digital electronics Prepared by - AnuradhaTandon Assistant Professor, Instrumentation & Control Engineering Branch, IT, NU

2. 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.

3. 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

4. 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.

5. 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’)

6. 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

7. 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

8. 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

9. Boolean Logic Function cont.…. • For example the function F(X,Y,Z)=(X+Z)(Y’+Z)(X’+Y+Z) can be represented using 2-input AND and OR gates as shown in the Fig. 2: Fig. - 2

10. 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

11. Canonical Form • The Boolean function is represented as either • Sum-of-Minterms (SOM) or • Product-of-Maxterms (POM)

12. 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)

13. Minterms • 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’, ...

14. 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

15. 2 variable minterms • Two variables (X and Y) produce 2x2=4 combinations XY (both normal) XY’ (X normal, Y complemented) X’Y (X complemented, Y normal) X’Y’ (both complemented)

16. Maxterms • 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)

17. 2 Variable Minterms and Maxterms • The index above is important for describing which variables in the terms are true and which are complemented.

18. 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

19. 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)

20. 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, or • ΠMi – for all maxterms that produce 0 in the table • Mintermsand Maxterms are complement of each other.

21. Example: minterm & maxterm

22. Example cont.….

23. SOP and POS Conversion POS SOP F =(A’+B)(A’+C)(C+D) =( A’+BC)(C+D) = A’C+A’D+BCC+BCD = A C+A D+BC+BCD = A’C+A’D+BC SOP POS F = AB + CD = (AB+C)(AB+D) = (A+C)(B+C)(AB+D) = ( A+C)(B+C)(A+D)(B+D)

24. 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.

25. 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

26. 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.

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

28. Truth Table Adjacencies

29. 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

30. Filling of K - map

32. K - map • In a K‐map, physical adjacency does imply gray code adjacency F =A’B’ + A’B = A’ F = A’B + AB = B