Digital Electronics – Unit I.pdf
- 1. EC1371 – DIGITAL ELECTRONICS Dr. K. Kannan, M.E., M.E., Ph.D., Professor & Head, Department of Mechatronics Engineering
- 2. OBJECTIVES • To provide the Digital fundamentals, Boolean algebra and its applications in digital systems • To familiarize with the design of various combinational digital circuits using logic gates • To introduce the analysis and design procedures for synchronous and asynchronous sequential circuits • To explain the various semiconductor memories and related technology • To introduce the electronic circuits involved in the making of logic gates
- 3. UNIT I DIGITAL FUNDAMENTALS (9) Review of Number systems - Logic gates - Boolean algebra - Boolean postulates and laws - De-Morgan’s Theorem, Principle of Duality - Simplification using Boolean algebra - Canonical forms, Sum of product and Product of sum - Minimization using Karnaugh map - NAND and NOR Implementation. CO1:Outline the Boolean functions and various minimization techniques.
- 4. Number Systems • Decimal (Base 10) – 10 digits (0,1,2,3,4,5,6,7,8,9) • Binary (Base 2) – 2 digits (0,1) • Digits are often called bits (binary digits) • Hexadecimal (Base 16) – 16 digits (0-9,A,B,C,D,E,F) • Often referred to as Hex
- 5. Number Systems Decimal Binary Hexadecimal 0 0 0 1 1 1 2 10 2 3 11 3 4 100 4 5 101 5 6 110 6 7 111 7 8 1000 8 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 1110 E 15 1111 F 16 10000 10 17 10001 11 18 10010 12 19 10011 13 20 10100 14
- 6. Positional Notation • Each digit is weighted by the base(r) to the positional power • N = dn-1dn-2 …d0.d1d2…dm = (dn-1x r n-1 ) + (dn-2x r n-1 ) + … + (d0 1x r0 ) + (d1x r1) + (d2 x r2) + … (dm x r m ) • Example : 872.6410 (8 x 102) + (7 x 101) + (2 x 100) + (6 x 10-1) + (4 x 10-2) • Example: 1011.12 = ? • Example :12A16 = ?
- 7. Positional Notation • 872.6410 = 8x102 + 7x101 + 2x100 + 6x10-1 + 4 x10-2 = 800 + 70 + 2 + .6 + .04
- 8. Positional Notation • 1011.12 = 1x23 + 0x22 + 1x21 + 1x20 + 1x2-1 = 8 + 0 + 2 + 1 + .5 = 11.510 Note that positional notation can be used to convert from binary to its equivalent decimal value
- 9. Positional Notation • 12A16 = 1x162 + 2x161 + Ax160 = 256 + 32 + 10 = 29810
- 10. Powers of Bases 2-3= .125 2-2= .25 2-1= .5 160 = 1 20 = 1 161 = 16 = 24 21 = 2 162 = 256 = 28 22 = 4 163 = 4096 = 212 23 = 8 24 = 16 25 = 32 26 = 64 27 = 128 28 = 256 210 = 1024 = 1Kb 29 = 512 220 = 1,048,576 = 1Mb 210 = 1024 230 = 1,073,741,824 = 1Gb 211 = 2048 212 = 4096
- 11. Determining What Base is being Used • Subscripts 87410 10112 AB916 • Prefix Symbols (None)874 %1011 $AB9 • Postfix Symbols AB9H • If I am only working with one base there is no need to add a symbol.
- 12. Conversion from Base R to Decimal Use Positional Notation %11011011 = ?10 $3A94 = ?10
- 13. Conversion from Binary to Decimal • Use Positional Notation • %11011011 = ?10 %11011011 = 1x27 + 1x26 + 1x24 + 1x23 + 1x21 + 1x20 = 128 + 64 + 16 + 8 + 2 + 1 = 21910
- 14. Conversion from Hexadecimal to Decimal • $3A94 = ?10 • $3A94 = 3x163 + Ax162 + 9x161 + 4x160 =12288 + 2560 + 144 + 4 =15996
- 15. Conversion from Decimal to Base R • Use Successive Division – Repeatedly divide by the desired base until 0 is reached – Save the remainders as the final answer – The first remainder is the LSB (least significant bit); the last remainder is the MSB (Most significant bit) • 43710 = ?2 = 1101101012 • 43710 = ?16 = 1B516
- 16. Conversion from Decimal to Binary • Use Successive Division – Repeatedly divide by the desired base until 0 is reached – Save the remainders as the final answer – The first remainder is the LSB (least significant bit); the last remainder is the MSB (Most significant bit) • 43710 = ?2 437 / 2 = 218 remainder 1 218 / 2 = 109 remainder 0 109 / 2 = 54 remainder 1 54 / 2 = 27 remainder 0 27 / 2 = 13 remainder 1 13 / 2 = 6 remainder 1 6 / 2 = 3 remainder 0 3 / 2 = 1 remainder 1 1 / 2 = 0 remainder 1 = 1101101012
- 17. Conversion from Decimal to Hexadecimal • 43710 = ?16 437 / 16 = 27 remainder 5 27 / 16 = 1 remainder 11 (11=B) 1 / 16 = 0 remainder 1 • 42710 = 1B516
- 18. Conversion from Binary to Hex • Starting at the LSB working left, group the bits by 4s. Padding of 0s can occur on the most significant group. • Convert each group of 4 into the equivalent HEX value. • %1101110101100 = $? = $1BAC
- 19. Conversion from Hex to Binary • Convert each HEX digit to the equivalent 4- bit binary grouping. • $A73 = %? = %101001110011
- 20. Conversion of Fractions • Conversion from decimal to binary requires multiplying by the desired base (2) • 0.62510 = ?2 • = 0.101 2
- 21. Addition/Subtraction of Binary Numbers 101011 + 1 101100 101011 + 001011 110110 • The carry out has a weight equal to the base (in this case 16). The digit that left is the excess (Base – sum).
- 22. Addition/Subtraction of Hex Numbers • $3A +$28 $62 • The carry out has a weight equal to the base (in this case 16). The digit that left is the excess (Base – sum).
- 23. Signed Number Representation • Three ways to represent signed numbers – Sign-Magnitude – 1s Complement – 2s Complement
- 24. Sign-Magnitude • For an N-bit word, the most significant bit is the sign bit; the remaining bits represent the magnitude • 0110 = +6 • 1110 = -6 • Addition/subtraction of numbers can result in overflow (errors) – (Due to fixed number of bits); two values for zero • Range for n bits: -(2n-1–1) through (2n-1–1)
- 25. 1s Complement • Negative numbers = N’ = (2n-1–1) –P (where P is the magnitude of the number) – For a 5-bit system, -7 = 11111 -00111 11000 • Range for n bits:-(2n-1–1) through (2n-1–1)
- 26. 2s Complement • Negative Numbers = N* = 2n – P (where P is the magnitude of the number) – For a 5-bit system, -7 = 100000 -00111 11001 • Another way to form 2s complement representation is to complement P and add 1 • Range for n bits: -(2n-1) through (2n-1–1)
- 27. Numbers Represented with 4-bit Fixed Digit Representation Decimal Sign Magnitude 1s Complement 2s Complement +8 +7 0111 0111 0111 +6 0110 0110 0110 +5 0101 0101 0101 +4 0100 0100 0100 +3 0011 0011 0011 +2 0010 0010 0010 +1 0001 0001 0001 +0 0000 0000 0000 -0 1000 1111 0000 -1 1001 1110 1111 -2 1010 1101 1110 -3 1011 1100 1101 -4 1100 1011 1100 -5 1101 1010 1011 -6 1110 1001 1010 -7 1111 1000 1001 -8 1000
- 28. Summary of Signed Number Representations • Sign Magnitude – has two values for 0 • - errors in addition of negative and positive numbers • 1s complement – two values for 0 • - additional hardware needed to compensate for this • 2s Complement – representation of choice
- 29. Unsigned/Signed Overflow • You can detect unsigned overflow if there is a carryout of the MSB. • You can detect signed overflow if the sum of two positive numbers is a negative number or if the sum of two negative numbers is a positive number. An overflow never occurs in an addition of a negative and a positive number.
- 30. Codes • Decimal Codes – BCD (Binary Coded Decimal) • Weighted Codes (8421, 2421, etc…) • ASCII Codes – ASCII (American Standard Code for Information Interchange) – Unicode Standard • Unit Distance Codes – Gray • Error Detection Codes – Parity Bit • Error Correction Codes – Hamming Code
- 31. BCD Codes (Decimal Codes) • Coded Representations for the the 10 decimal digits • Requires 4 bits (23 < 10 <24) • Only use 10 combinations of 16 possible combinations (30 billion different coding schemes available)
- 32. BCD Codes (Decimal Codes) • Weighted Code – 8421 code • Most common • Default • The corresponding decimal digit is determined by adding the weights associated with the 1s in the code group. • **** The BCD representation is NOT the binary equivalent of the decimal number ***** – 62310 = 0110 0010 0011 – 2421, 5421,7536, etc… codes • The weights associated with the bits in each code group are given by the name of the code
- 33. BCD Codes (Decimal Codes) • Nonweighted Codes – 2-out-of-5 • Actually weighted 74210 except for the digit 0 • Used by the post office for scanning bar codes for zip codes • Has error detection properties
- 34. BCD Codes (Decimal Codes) U.S. Postal Service bar code corresponding to the ZIP code 14263-1045.
- 35. Unit Distance Codes • Important when converting analog to digital • Only one bit changes between successive integers • Gray Code is most popular example
- 36. Unit Distance Codes Angular position encoders. (a) Conventional binary encoder. (b) Gray code encoder.
- 37. Unit Distance Codes Angular position encoders with misaligned photosensing devices. (a) Conventional binary encoder. (b) Gray code encoder.
- 38. Alphanumeric Codes • Used to encode Alphabetic and numeric information • ASCII – 7-bit Code • Unicode – 16-bit Code
- 39. How Much Memory? • Memory is purchased in bits – – How many bits do I need if I want to distinguish between 8 colors? – How many bits do I need if I want to represent 16 million different colors?
- 40. How Much Memory? • How many bits do I need if I want to distinguish between 8 colors? 2x-1 < 8 <= 2x x = 3 (3 bits are needed) • How many bits do I need if I want to represent 16 million different colors? 2x-1 < 16 million <= 2x 16M = 1Mx16 = 220x24 = 224 x = 24 (24 bits are needed)
- 41. What do you need to know? • Powers of 2, 16 • Conversions of Hex, Binary to Decimal Conversions of Decimal to Hex, Binary Conversions of Hex to Binary, Binary to Hex • Signed Number Representations Addition of Signed Numbers, Overflows • Codes (BCD, Gray) • Number of bits needed?
- 42. 42 Computers and Electricity Gate A device that performs a basic operation on electrical signals Circuits Gates combined to perform more complicated tasks
- 43. 43 Computers and Electricity How do we describe the behavior of gates and circuits? Boolean expressions Uses Boolean algebra, a mathematical notation for expressing two- valued logic Logic diagrams A graphical representation of a circuit; each gate has its own symbol Truth tables A table showing all possible input value and the associated output values
- 44. 44 Gates Six types of gates – NOT – AND – OR – XOR – NAND – NOR Typically, logic diagrams are black and white with gates distinguished only by their shape
- 45. 45 NOT Gate A NOT gate accepts one input signal (0 or 1) and returns the opposite signal as output
- 46. 46 AND Gate An AND gate accepts two input signals If both are 1, the output is 1; otherwise, the output is 0
- 47. 47 OR Gate An OR gate accepts two input signals If both are 0, the output is 0; otherwise, the output is 1
- 48. 48 XOR Gate An XOR gate accepts two input signals If both are the same, the output is 0; otherwise, the output is 1
- 49. 49 XOR Gate Note the difference between the XOR gate and the OR gate; they differ only in one input situation When both input signals are 1, the OR gate produces a 1 and the XOR produces a 0 XOR is called the exclusive OR
- 50. NAND Gate The NAND gate accepts two input signals If both are 1, the output is 0; otherwise, the output is 1
- 51. 51 NOR Gate Figure 4.6 Various representations of a NOR gate The NOR gate accepts two input signals If both are 0, the output is 1; otherwise, the output is 0
- 52. 52 Review of Gate Processing A NOT gate inverts its single input An AND gate produces 1 if both input values are 1 An OR gate produces 0 if both input values are 0 An XOR gate produces 0 if input values are the same A NAND gate produces 0 if both inputs are 1 A NOR gate produces a 1 if both inputs are 0
- 53. 53 Gates with More Inputs Gates can be designed to accept three or more input values A three-input AND gate, for example, produces an output of 1 only if all input values are 1
- 54. 54 Constructing Gates Transistor A device that acts either as a wire that conducts electricity or as a resistor that blocks the flow of electricity, depending on the voltage level of an input signal A transistor has no moving parts, yet acts like a switch It is made of a semiconductor material, which is neither a particularly good conductor of electricity nor a particularly good insulator
- 55. 55 Constructing Gates A transistor has three terminals – A source – A base – An emitter, typically connected to a ground wire If the electrical signal is grounded, it is allowed to flow through an alternative route to the ground (literally) where it can do no harm
- 56. 56 Constructing Gates The easiest gates to create are the NOT, NAND, and NOR gates
- 57. Boolean Law and Postulates Boolean algebra is the category of algebra in which the variable's values are the truth values, true and false, ordinarily denoted as 1 and 0 respectively. It is also used in set theory and statistics. Postulates are statements that are assumed to be true without proof. Postulates serve two purposes - to explain undefined terms, and to serve as a starting point for proving other statements.
- 58. 58 Interpretation of P1 and P2 in Logic P1: Commutative • a is true OR b is true = b is true OR a is true • a is true AND b is true = b is true AND a is true P2: Distributive • a is true OR (b is true AND c is true) = (a is true OR b is true) AND (a is true OR c is true) • a is true AND (b is true OR c is true) = (a is true AND b is true) OR (a is true AND c is true)
- 59. 59 Interpretation of P3 and P4 in Logic P3: Identity 0: one false statement, 1: one true statement • a is true OR one false statement = a is true • a is true AND one true statement = a is true P4: Complement Negate the statement • a is true OR a is false = one true statement • a is true AND a is false = one false statement
- 61. 61 Distributive Laws • a·(b+c) = (a·b)+(a·c) • a+(b·c) = (a+b)·(a+c) ID a b c b+c a· (b+c) a· b a· c (a· b)+(a· c) 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 2 0 1 0 1 0 0 0 0 3 0 1 1 1 0 0 0 0 4 1 0 0 0 0 0 0 0 5 1 0 1 1 1 0 1 1 6 1 1 0 1 1 1 0 1 7 1 1 1 1 1 1 1 1
- 62. 62 Distributive Laws a· (b+c) (a·b)+(a·c) a+(b·c) (a+b)·(a+c)
- 63. 63 Identity a+0 = a, 0 input to OR is passive a·1 = a, 1 input to AND is passive A 1 A A 0 A
- 64. 64 Complement a+a' = 1 a·a' = 0
- 65. 65 Theorems and Proofs Theorem 1: Principle of Duality • Every algebraic identity that can be proven by Boolean algebra laws, remains valid if we swap all ‘+’ and ‘·’, 0 and 1. Proof: • Visible by inspection – all laws remain valid if we interchange all ‘+’ and ‘·’, 0 and 1
- 67. 67 Theorem 2 & 3 Uniqueness of Complement: For every a in B, its complement a' is unique. Boundedness: For all elements a in B, a+1=1; a*0=0. Proof: a+1 = 1 *(a+1) = (a + a')*(a+1) = a + a'*1 = a + a' = 1 A 1 1 A 0 0
- 68. 68 Theorem 4 & 5 • The complement of element 1 is 0 and vice versa, i.e. 0' = 1, 1' = 0. • For every a in B, a + a = a and a * a = a. Proof: a + a = (a + a) * 1 = (a + a) * (a + a') = a + (a*a') = a + 0 = a A A A A A A
- 69. Operator Precedence The operator precedence for evaluating Boolean expressions is (1) parentheses, (2) NOT, (3) AND, and (4) OR.
- 70. Boolean Functions Boolean algebra is an algebra that deals with binary variables and logic operations. A Boolean function described by an algebraic expression consists of binary variables, the constants 0 and 1, and the logic operation symbols. For a given value of the binary variables, the function can be equal to either 1 or 0. F1 = x + y’z
- 71. Boolean Functions F1 = x + y’z
- 73. Simplification of Boolean Functions
- 74. Complement of a Function The complement of a function F is F’ and is obtained from an interchange of 0’s for 1’s and 1’s for 0’s in the value of F. The complement of a function may be derived algebraically through DeMorgan’s theorems
- 75. Complement of a Function
- 76. Minterms / Standard Product A binary variable may appear either in its normal form (x) or in its complement form (x). Now consider two binary variables x and y combined with an AND operation. Since each variable may appear in either form, there are four possible combinations: x’y’, x’y, xy’, and xy. Each of these four AND terms is called a Minterm, or a standard product.
- 77. A binary variable may appear either in its normal form (x) or in its complement form (x). Now consider two binary variables x and y combined with an OR operation. Since each variable may appear in either form, there are four possible combinations: x’y’, x’y, xy’, and xy. Each of these four OR terms is called a Maxterm, or a standard sum. Maxterms / Standard Sum
- 79. Functions of Three Variables
- 80. Functions of Three Variables
- 81. Canonical Form • Boolean functions expressed as a sum of minterms or product of maxterms are said to be in canonical form • The minterms whose sum defines the Boolean function are those which give the 1’s of the function in truth table. • The maxterms whose product defines the Boolean function are those which give the 0’s of the function in truth table.
- 82. Express the Boolean function F = A + B’C as a sum of minterms. Canonical Form
- 83. Canonical Form Express the Boolean function F = A + B’C as a sum of minterms.
- 86. Conversion between Canonical Forms
- 87. Conversion between Canonical Forms
- 88. Standard Forms The two canonical forms of Boolean algebra are basic forms that one obtains from reading a given function from the truth table. These forms must contain all the variables, either complemented or uncomplemented. Another way to express Boolean functions is in standard form. In this configuration, the terms that form the function may contain one, two, or any number of literals. There are two types of standard forms: the sum of products and products of sums.
- 89. Sum of Products The sum of products is a Boolean expression containing AND terms, called product terms/minterms, with one or more literals each. The sum denotes the ORing of these terms. F1 = y’ + xy + x’yz’ The expression has three product terms, with one, two, and three literals. Their sum is an OR.
- 90. Product of Sums A product of sums is a Boolean expression containing OR terms, called sum terms. Each term may have any number of literals. The product denotes the ANDing of these terms. F2 = x(y’ + z)(x’ + y + z’) This expression has three sum terms, with one, two, and three literals. The product is an AND operation
- 92. Logic Gates
- 93. Logic Gates
- 94. Positive and Negative Logic Choosing the high‐level H to represent logic 1 defines a positive logic system. Choosing the low‐level L to represent logic 1 defines a negative logic system.
- 95. Gate Level Minimization Gate-level minimization is the design task of finding an optimal gate-level implementation of the Boolean functions describing a digital circuit. The complexity of the digital logic gates that implement a Boolean function is directly related to the complexity of the algebraic expression from which the function is implemented. However, the procedure of minimization using Boolean Law is awkward because it lacks specific rules to predict each succeeding step in the manipulative process. The Karnaugh map or K-map method provides a simple, straightforward procedure for minimizing Boolean functions. This method may be regarded as a pictorial form of a truth table.
- 96. K - Map A K-map is a diagram made up of squares, with each square representing one minterm of the function that is to be minimized. Since, any Boolean function can be expressed as a sum of minterms, it follows that a Boolean function is recognized graphically in the map from the area enclosed by those squares whose minterms are included in the function. In fact, the map presents a visual diagram of all possible ways a function may be expressed in standard form. By recognizing various patterns, we can derive alternative algebraic expressions for the same function, from which the simplest can be selected.
- 100. Four-Variable K-Map
- 101. Four-Variable K-Map
- 102. Four-Variable K-Map
- 103. Prime Implicants A prime implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map. If a minterm in a square is covered by only one prime implicant, that prime implicant is said to be essential. The prime implicants of a function can be obtained from the map by combining all possible maximum numbers of squares.
- 104. Prime Implicants
- 105. Four-Variable K-Map
- 106. Don’t- Care Conditions The logical sum of the minterms associated with a Boolean function specifies the conditions under which the function is equal to 1. The function is equal to 0 for the rest of the minterms. Functions that have unspecified outputs for some input combinations are called incompletely specified functions It is customary to call the unspecified minterms of a function don’t-care conditions . These don’t-care conditions can be used on a map to provide further simplification of the Boolean expression.
- 108. NAND and NOR Implementation
- 109. NAND and NOR Implementation
- 110. 1. Simplify the function and express it in sum-of-products form. 2. Draw a NAND gate for each product term of the expression that has at least two literals. The inputs to each NAND gate are the literals of the term. This procedure produces a group of first-level gates. 3. Draw a single gate using the AND-invert or the invert-OR graphic symbol in the second level, with inputs coming from outputs of first-level gates. 4. A term with a single literal requires an inverter in the first level. However, if the single literal is complemented, it can be connected directly to an input of the second level NAND gate. Two-Level NAND Implementation
- 113. Multilevel NAND Circuits 1. Convert all AND gates to NAND gates with AND-invert graphic symbols. 2. Convert all OR gates to NAND gates with invert-OR graphic symbols. 3. Check all the bubbles in the diagram. For every bubble that is not compensated by another small circle along the same line, insert an inverter (a one-input NAND gate) or complement the input literal.
- 117. Thank You