# Digital Fundamentals

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1 Digital Fundamentals Tenth Edition Floyd Chapter 3 Modified by Yuttapong Jiraraksopakun Floyd, Digital Fundamentals, th 28 Pearson Education ENE, KMUTT ed 29

2 The Inverter Summary The inverter performs the oolean NOT operation. When the input is LOW, the output is HIGH; when the input is HIGH, the output is LOW. Input Output LOW () HIGH () HIGH () LOW() The NOT operation (complement) is shown with an overbar. Thus, the oolean expression for an inverter is =.

3 The Inverter Summary Example waveforms: group of inverters can be used to form the s complement of a binary number: inary number s complement

4 The ND Gate Summary The ND gate produces a HIGH output when all inputs are HIGH; otherwise, the output is LOW. For a 2-input gate, the truth table is Inputs Output The ND operation is usually shown with a dot between the variables but it may be implied (no dot). Thus, the ND operation is written as =. or =. &

5 Summary The ND Gate & Example waveforms: The ND operation is used in computer programming as a selective mask. If you want to retain certain bits of a binary number but reset the other bits to, you could set a mask with s in the position of the retained bits. If the binary number is NDed with the mask, what is the result?

6 The OR Gate Summary The OR gate produces a HIGH output if any input is HIGH; if all inputs are LOW, the output is LOW. For a 2-input gate, the truth table is Inputs Output The OR operation is shown with a plus sign (+) between the variables. Thus, the OR operation is written as = +.

7 Summary The OR Gate Example waveforms: The OR operation can be used in computer programming to set certain bits of a binary number to. SCII letters have a in the bit 5 position for lower case letters and a in this position for capitals. (it positions are numbered from right to left starting with.) What will be the result if you OR an SCII letter with the 8-bit mask? The resulting letter will be lower case.

8 The NND Gate Summary The NND gate produces a LOW output when all inputs are HIGH; otherwise, the output is HIGH. For a 2-input gate, the truth table is Inputs Output The NND operation is shown with a dot between the variables and an overbar covering them. Thus, the NND operation is written as =. (lternatively, =.) &

9 Summary The NND Gate & Example waveforms: The NND gate is particularly useful because it is a universal gate all other basic gates can be constructed from NND gates. How would you connect a 2-input NND gate to form a basic inverter?

10 Negative-OR Equivalent Summary

11 The NOR Gate Summary The NOR gate produces a LOW output if any input is HIGH; if all inputs are HIGH, the output is LOW. For a 2-input gate, the truth table is Inputs Output The NOR operation is shown with a plus sign (+) between the variables and an overbar covering them. Thus, the NOR operation is written as = +.

12 Summary The NOR Gate Example waveforms: The NOR operation will produce a LOW if any input is HIGH. When is the LED is ON for the circuit shown? +5. V 33 Ω The LED will be on when any of the four inputs are HIGH. C D

13 Negative-ND Equivalent Summary

14 The OR Gate Summary The OR gate produces a HIGH output only when both inputs are at opposite logic levels. The truth table is = Inputs Output The OR operation is written as = +. lternatively, it can be written with a circled plus sign between the variables as = +.

15 Summary The OR Gate = Example waveforms: Notice that the OR gate will produce a HIGH only when exactly one input is HIGH. If the and waveforms are both inverted for the above waveforms, how is the output affected? There is no change in the output.

16 The NOR Gate Summary The NOR gate produces a HIGH output only when both inputs are at the same logic level. The truth table is = Inputs Output The NOR operation shown as = +. lternatively, the NOR operation can be shown with a circled dot between the variables. Thus, it can be shown as =..

17 Summary The NOR Gate = Example waveforms: Notice that the NOR gate will produce a HIGH when both inputs are the same. This makes it useful for comparison functions. If the waveform is inverted but remains the same, how is the output affected? The output will be inverted.

18 Summary Fixed Function Logic Two major fixed function logic families are TTL and CMOS. third technology is icmos, which combines the first two. Packaging for fixed function logic is shown in in ±. in in Pin no. identifiers 4 Lead no. identifier DIP package SOIC package

19 Summary Fixed Function Logic Some common gate configurations are shown. V CC V CC V CC V CC ' GND ' GND ' GND ' GND V CC V CC V CC V CC ' GND ' GND ' GND ' GND V CC V CC V CC V CC ' GND ' GND ' GND ' GND

20 Fixed Function Logic Summary Logic symbols show the gates and associated pin numbers. () (2) (4) (5) (9) () (2) (3) V CC (4) (3) (6) (8) () () & (3) (2) (4) (5) (6) (9) () (8) (2) (3) () (7) GND

21 Fixed Function Logic Summary Data sheets include limits and conditions set by the manufacturer as well as DC and C characteristics. For example, some maximum ratings for a 74HC are: V CC MIMUM RTINGS Symbol Parameter Value Unit DC Supply Voltage (Referenced to GND).5 to + 7. V V V in DC Input Voltage (Referenced to GND).5 to V CC +.5 V V V out DC Output Voltage (Referenced to GND).5 to V CC +.5 V V I in DC Input Current, per pin ± 2 m DC Output Current, per pin ± 25 m I CC DC Supply Current, V CC and GND pins ± 5 m P D Power Dissipation in Still ir, Plastic or Ceramic DIP 75 mw T stg SOIC Package TSSOP Package Storage Temperature to + 5 C T L Lead Temperature, mm from Case for Seconds C Plastic DIP, SOIC, or TSSOP Package 26 Ceramic DIP 3

22 Digital Fundamentals Tenth Edition Floyd Chapter 4 Modified by Yuttapong Jiraraksopakun Floyd, Digital Fundamentals, th 28 Pearson Education ENE, KMUTT ed 29

23 Summary oolean ddition In oolean algebra, a variable is a symbol used to represent an action, a condition, or data. single variable can only have a value of or. The complement represents the inverse of a variable and is indicated with an overbar. Thus, the complement of is. literal is a variable or its complement. ddition is equivalent to the OR operation. The sum term is if one or more if the literals are. The sum term is zero only if each literal is. Determine the values of,, and C that make the sum term of the expression + + C =? Each literal must = ; therefore =, = and C =.

24 oolean Multiplication Summary In oolean algebra, multiplication is equivalent to the ND operation. The product of literals forms a product term. The product term will be only if all of the literals are. What are the values of the, and C if the product term of.. C =? Each literal must = ; therefore =, = and C =.

25 Summary Commutative Laws The commutative laws are applied to addition and multiplication. For addition, the commutative law states In terms of the result, the order in which variables are ORed makes no difference. + = + For multiplication, the commutative law states In terms of the result, the order in which variables are NDed makes no difference. =

26 Summary ssociative Laws The associative laws are also applied to addition and multiplication. For addition, the associative law states When ORing more than two variables, the result is the same regardless of the grouping of the variables. + ( +C) = ( + ) + C For multiplication, the associative law states When NDing more than two variables, the result is the same regardless of the grouping of the variables. (C) = ()C

27 Summary Distributive Law The distributive law is the factoring law. common variable can be factored from an expression just as in ordinary algebra. That is + C = (+ C) The distributive law can be illustrated with equivalent circuits: C + C (+ C) C C + C

28 Rules of oolean lgebra Summary. + = 2. + = 3.. = 4.. = 5. + = 6. + = 7.. = 8.. = = 9. =. + =. + = + 2. ( + )( + C) = + C

29 Rules of oolean lgebra Summary Rules of oolean algebra can be illustrated with Venn diagrams. The variable is shown as an area. The rule + = can be illustrated easily with a diagram. dd an overlapping area to represent the variable. The overlap region between and represents. = The diagram visually shows that + =. Other rules can be illustrated with the diagrams as well.

30 Summary Rules of oolean lgebra Illustrate the rule + = + with a Venn diagram. This time, is represented by the blue area and again by the red circle. The intersection represents. Notice that + = +

31 Rules of oolean lgebra Summary Rule 2, which states that ( + )( + C) = + C, can be proven by applying earlier rules as follows: ( + )( + C) = + C + + C = + C + + C = ( + C + ) + C =. + C = + C This rule is a little more complicated, but it can also be shown with a Venn diagram, as given on the following slide

32 Summary Three areas represent the variables,, and C. The area representing + is shown in yellow. The area representing + C is shown in red. The overlap of red and yellow is shown in orange. The overlapping area between and C represents C. ORing with gives the same area as before. + + C = C ( + )( + C) + C C C

33 DeMorgan s Theorem DeMorgan s st Theorem Summary The complement of a product of variables is equal to the sum of the complemented variables. = + pplying DeMorgan s first theorem to gates: + Inputs Output + NND Negative-OR

34 Summary DeMorgan s Theorem DeMorgan s 2 nd Theorem The complement of a sum of variables is equal to the product of the complemented variables. + =. pplying DeMorgan s second theorem to gates: + Inputs Output + NOR Negative-ND

35 DeMorgan s Theorem Summary pply DeMorgan s theorem to remove the overbar covering both terms from the expression = C + D. To apply DeMorgan s theorem to the expression, you can break the overbar covering both terms and change the sign between the terms. This results in = = C. D. Deleting the double bar gives = C. D.

36 C D Summary oolean nalysis of Logic Circuits Combinational logic circuits can be analyzed by writing the expression for each gate and combining the expressions according to the rules for oolean algebra. pply oolean algebra to derive the expression for. Write the expression for each gate: ( + ) C ( + ) = C ( + )+ D pplying DeMorgan s theorem and the distribution law: = C ( ) + D = C + D

37 Summary SOP and POS forms oolean expressions can be written in the sum-of-products form (SOP) or in the product-of-sums form (POS). These forms can simplify the implementation of combinational logic, particularly with PLDs. In both forms, an overbar cannot extend over more than one variable. n expression is in SOP form when two or more product terms are summed as in the following examples: C + C + C D C D + E n expression is in POS form when two or more sum terms are multiplied as in the following examples: ( + )( + C) ( + + C)( + D) ( + )C

38 Summary SOP Standard form In SOP standard form, every variable in the domain must appear in each term. This form is useful for constructing truth tables or for implementing logic in PLDs. You can expand a nonstandard term to standard form by multiplying the term by a term consisting of the sum of the missing variable and its complement. Convert = + C to standard form. The first term does not include the variable C. Therefore, multiply it by the (C + C), which = : = (C + C) + C = C + C + C

39 Summary POS Standard form In POS standard form, every variable in the domain must appear in each sum term of the expression. You can expand a nonstandard POS expression to standard form by adding the product of the missing variable and its complement (rule 8: =) and applying rule 2, which states that ( + )( + C) = + C. Convert = ( + )( + + C) to standard form. The first sum term does not include the variable C. Therefore, add C C and expand the result by rule 2. = ( + + C C)( + + C) = ( + + C )( + + C)( + + C)

40 Summary Converting Standard SOP to Standard POS C = ++C C = ++C C = ++C C = ++C SOP active-high logic POS active-low logic

41 Karnaugh maps Summary The Karnaugh map (K-map) is a tool for simplifying combinational logic with 3 or 4 variables. For 3 variables, 8 cells are required (2 3 ). The map shown is for three variables labeled,, and C. Each cell represents one possible product term. Each cell differs from an adjacent cell by only one variable. C C C C C C C C

42 Summary Gray code Karnaugh maps Cells are usually labeled using s and s to represent the variable and its complement. C The numbers are entered in gray code, to force adjacent cells to be different by only one variable. Ones are read as the true variable and zeros are read as the complemented variable.

43 Summary Karnaugh maps lternatively, cells can be labeled with the variable letters. This makes it simple to read, but it takes more time preparing the map. Read the terms for the yellow cells. C C C C C C C C C The cells are C and C. C C C C

44 Summary SOP Standard form in Karnaugh Map

45 Summary Karnaugh maps K-maps can simplify combinational logic by grouping cells and eliminating variables that change. Group the s on the map and read the minimum logic. changes across this boundary C C changes across this boundary. Group the s into two overlapping groups as indicated. 2. Read each group by eliminating any variable that changes across a boundary. 3. The vertical group is read C. 4. The horizontal group is read. = C +

46 Karnaugh maps 4-variable map has an adjacent cell on each of its four boundaries as shown. CD CD CD CD Summary Each cell is different only by one variable from an adjacent cell. Grouping follows the rules given in the text. The following slide shows an example of reading a four variable map using binary numbers for the variables

47 Summary Karnaugh maps Group the s on the map and read the minimum logic. changes changes CD C changes across outer boundary C changes. Group the s into two separate groups as indicated. 2. Read each group by eliminating any variable that changes across a boundary. 3. The upper (yellow) group is read as D. 4. The lower (green) group is read as D. = D +D

48 Summary Converting Standard SOP to Standard POS with Karnaugh SOP = C + + C POS = (+C) (+) (+C) SOP = C+ C+

49 Don t Care Conditions Summary

50 Don t Care Conditions Summary

51 Summary Five-Variable Karnaugh Maps

52 Homework 7 Chapter 3 (2, 2, 8, 45) Chapter 4 (, 3, 2, 36 a, 44)

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