EKT 121 / 4 ELEKTRONIK DIGIT 1

Pgt 104ELEKTRONIK DIGIT CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC1K-Map (1) Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit. This will replace Boolean reduction when the circuit is large. Write the Boolean equation in a SOP form first and then place each term on a map. 2K-Map (2) A graphical tool for simplifying Boolean expression. Divided into cells, each representing a specific combination of variable values. Number of cells = total possible combinations of the variables in an expressionEg: 3 variables = 23 = 8 cells 4 variables = 24 = 16 cells

3 The map is made up of a table of every possible SOP using the number of variables that are being used. If 2 variables are used then a 2X2 map is used If 3 variables are used then a 4X2 map is used If 4 variables are used then a 4X4 map is used If 5 Variables are used then a 8X4 map is usedK-Map (3)4K-Map SOP Minimization5Step 1: Place a 1 in each cell corresponding to a minterm in the SOP expression.Step 2: Group all 1s that are adjacent cells into groupings of the maximum number of cells. A group can only contain a number of 1s equal to a power of two or groups can overlap.Step 3: For each group, determine the variables that are the same. A complemented and an uncomplemented variable are not the same and cancel each other.Step 4: Write the resulting product terms for each group, using only the variables that are the same within the group.Step 5: Combine the resulting product terms into an SOP expression which will be equivalent to the original SOP expression.Simplifying an SOP Expression

AAB BNotice that the map is going false to true, left to right and top to bottomThe upper right hand cell is A B if X= A B then put an X in that cellAAB B1This show the expression true when A = 0 and B = 00 12 32 Variables K-Map (1)7If X=AB + AB then put an X in both of these cellsAAB B11From Boolean reduction we know that A B + A B = BFrom the Karnaugh map we can circle adjacent cell and find that X = B AAB B112 Variables K-Map (2)83 Variables K-Map (1)Gray Code

00A B01A B11 A B10A B 0 1 C C 0 1

2 3

6 7

4 5

9X = A B C + A B C + A B C + A B CGray Code

00A B01A B11 A B10A B 0 1 C CEach 3 variable term is one cell on a 4 X 2 Karnaugh map11113 Variables K-Map (2)10X = A B C + A B C + A B C + A B CGray Code

00A B01A B11 A B10A B 0 1 C COne simplification could be X = A B + A B11113 Variables K-Map (3)11X = A B C + A B C + A B C + A B CGray Code

00A B01A B11 A B10A B 0 1 C CAnother simplification could be X = B C + B CA Karnaugh Map does wrap around11113 Variables K-Map (4)12X = A B C + A B C + A B C + A B CGray Code

00A B01A B11 A B10A B 0 1 C CThe Best simplification would be X = B11113 Variables K-Map (5)133 Variables K-Map (6) Use a K-Map to simplify the following 3-variable minterm SOP expression: X = A B C + A B C + A B C + A B C + A B C

Answer: X = B + AC

On a 3 Variables K-Map One cell requires 3 Variables Two adjacent cells require 2 variables Four adjacent cells require 1 variable Eight adjacent cells is a 1154 Variables K-MapGray Code

00A B01A B11A B10A B0 0 0 1 1 1 1 0C D C D C D C D 0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10164 Variables K-MapGray Code

00A B01A B11A B10A B0 0 0 1 1 1 1 0C D C D C D C D 0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10BBAACCDDD17Gray Code

00A B01A B11A B10A B0 0 0 1 1 1 1 0C D C D C D C D111111X = ABD + ABC + CDNow try it with Boolean reductions Simplify: X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D 18On a 4 Variables K-Map One Cell requires 4 variables Two adjacent cells require 3 variables Four adjacent cells require 2 variables Eight adjacent cells require 1 variable Sixteen adjacent cells give a 1 or true19 Simplify :

Z = B C D + B C D + C D + B C D + A B CGray Code

00A B01A B11A B10A B00 01 11 10C D C D C D C D111111111111Z = C + A B + B D 20Simplify using K-Map (1)Firstly, change the circuit to an SOP expression

21Y= A + B + B C + ( A + B ) ( C + D)Y = A B + B C + A B ( C + D )

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

Simplify using K-Map (2)SOP expressionHow about standard SOP expression???22Gray Code

00A B01A B11A B10A B00 01 11 10C D C D C D C D11111111Y = 1 11111111Simplify using K-Map (3)23K-Map POS Minimization243 Variables K-Map (1)Gray Code

0 00 11 1 1 0 0 1 ABC 0 1

2 3

6 7

4 5

25

3 Variables K-Map (2)260 00 11 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 104 Variables K-Map (1)27

4 Variables K-Map (2)28

4 Variables K-Map (3)29 Mapping a Standard SOP expression Example: Answer:

Mapping a Standard POS expression Example: Using K-Map, convert the following standard POS expression into a minimum SOP expression

Answer:

Y = AB + AC or standard SOP:

K-Map - Examples

30K-Map with Dont Care Conditions (1)

Input OutputExample : 3 variables with output dont care (X)31

4 variables with output dont care (X)K-Map with Dont Care Conditions (2)32 Dont Care Conditions Example: Determine the minimal SOP using K-Map:

Answer:

K-Map with Dont Care Conditions (3)33 Solution :

ABCD0001111000 01 11 100 1 1 01 X 1 0X X X X0 0 1 00 1 3 2

5 7 6

13 15 14

8 9 11 10

Minimum SOP expression is CDADBC34Exercise Minimize this expression with a K-Map ABCD + ACD + BCD + ABCD

35K-map POS & SOP SimplificationExample: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in (a) S-of-P (b) P-of-S

Using the minterms (1s)F(ABCD)= BD+BC+ACD Using the maxterms (0s) and complimenting F Grouping as if they were minterms, then using De- Morgens theorem to get F. F(ABCD)= BD+CD+AB F(ABCD)= (B+D)(C+D)(A+B)

365 variable K-map (1)5 variables -> 32 minterms, hence 32 squares required

375 variable K-map (2)Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11.The centre line must be considered as the centre of a book, each half of the K-map being a pageThe centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.385 variable K-Map (3)Example: Simplify the Boolean functionF(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31)Soln: F(ABCDE) = BE+ADE+ABE

396 variable K-map 6 variables -> 64 minterms, hence 64 squares required

40End of Chapter 1- tqvm -41