SAU1A FUNDAMENTALS OF DIGITAL COMPUTERS

SAU1A FUNDAMENTALS OF DIGITAL COMPUTERS Unit : I - V

Unit : I Overview Fundamentals of Computers Characteristics of Computers Computer Language Operating Systems Generation of Computers 2

Definition of a Computer A computer is an electronic device that operates under the control of a set of instructions that is stored in its memory unit. A computer accepts data from an input device and processes it into useful information which it displays on its output device. Actually, a computer is a collection of hardware and software components that help you to accomplish many different tasks. 1. Hardware consists of the computer itself, and any equipment connected to it. 2. Software is the set of instructions that the computer follows in performing a task. 3

Components of a Computer 4 4

Functional Components of a Computer 1. INPUT DEVICES - Devices that allow the user to enter information into the PC (keyboard, mouse, scanner, etc.) 2. OUTPUT DEVICES - Devices that allow the computer to communicate with the user (monitor, printer, etc.) 3. CPU - The CPU consists of electronic circuits that interpret and execute instructions; it communicates with the input, output, and storage devices. 4. Memory - or primary storage, works with the CPU to hold instructions and data in order to be processed. 5

Storage Devices Shape Name Capacity Drive Letter (usually) Characteristics 3 ½ Diskette 1.44 Megabytes A: or B: Recordable, Very slow, Tend to lose information easily, Very cheap, Exchangeable, Easy to carry. Hard Disk Drive Between 1 and 16 Gigabytes C: Recordable, Very fast, Boot disk, Safe, Very Expensive. ZIP or LS-120 Between 100 Megabytes and 1 Gigabyte D:, E:, F: Recordable, Fast, Safe, Very Expensive, Exchangeable, Easy to carry. CD-ROM 650 Megabytes (0.65 Gigabytes) D:, E:, F: Some Recordable (expensive), Capacity for Audio, Very Fast, Very Safe, Cheap, Easy to carry. DVD-ROM Up to 17 Gigabytes (Up to 17000 Megabytes) D:, E:, F: New, Capacity for Audio and Video, Very Fast, Very Safe, Very Expensive, Easy to carry. 7

Characteristics of a Computer Speed Memory Diligence Versatility Lack of Decision making 8

Computer Languages Programming languages can be classified as: 1. Machine language The only language understood by computers. It comprises entirely of numbers. 2. Assembly language Similar to machine languages, but enable the use of names instead of numbers. Assembly language is converted to machine language using an interpreter. 3. High-level language Closer to human languages, easier to read, write and maintain. High-level language is converted to Assembly language using a compiler or assembler. Eg. C, C++, Cobol, Fortran, Java 4. Fourth-generation language 4GL is more closer to human languages than typical high-level languages. Most 4GLs are used to access databases. 10

Computer Language 11

Operating System The operating system is the software that provides an interface between the hardware (the computer itself) and the user. It manages the hardware and enables the user to control the computer. Microsoft developed the first operating system for IBM machines, MS DOS (Microsoft Disk Operating System), the grandfather of the current DOS and Windows systems. The IBM family is based on the Intel 8086 family of chips, which evolved into the high-speed Pentium chips of today. At the same time, the Apple Computer Company started Macintosh series based on the Motorola 68000 family of microprocessor chips. 12

Generations of Computers The history of computer development is often referred to in reference to the different generations of computing devices. Each generation of computer is characterized by a major technological development that fundamentally changed the way computers operate, resulting in increasingly smaller, cheaper, more powerful and more efficient and reliable devices. Based on the above mentioned features, we have five generations of computers. Generations of Computers 14

Number System Binary System Binary Code Logic Gates Unit : II Overview 15

The larger the base, more the numerals required: Number System Base Digits or Symbols included Binary 2 0,1 Decimal 10 0,1,2,3,4,5,6,7,8,9 Octal 8 0,1,2,3,4,5,6,7 Hexadecimal 16 0,1,2,3,4,5,6,7,8,9, A,B,C,D,E,F The range of possible numbers for each system is R=BK, where B = Base, K = number of digits For example, a 16-bit PC will have R = 2 16 = 65,536 different values. 16

Conversion from Decimal to other system Step 1 : Repeatedly, divide the integer part by the base to which it has to be converted, and collect the remainders in the reverse direction. Step 2 : Successively, multiply the fractional part by the base to which it has to be converted, and collect the carries from top to bottom. If the fractional part is not zero after 4 or 5 steps, the process can be stopped. Examples: 1. Convert (19.625) 10 to its binary equivalent. 19 / 2 = 9; remainder = 1 9 / 2 = 4; remainder = 1 4 / 2 = 2; remainder = 0 2 / 2 = 1; remainder = 0 1 / 2 = 0; remainder = 1 Thus, (19.625) 10 = (10011.101) 2 0.625 x 2 = 1.250; carry = 1 0.250 x 2 = 0.500; carry = 0 0.500 x 2 = 1.000; carry = 1 18

Conversion from Binary to Octal / Hexadecimal Step 1 : Arrange the given number into groups of 3 bits (Octal) or 4 bits (Hexadecimal), starting from the Least Significant Bit (LSB). If the final group has lesser number of bits, include 0s. Step 2 : Replace each group with its octal or hexadecimal equivalent, as required. Example: Convert (11110.11) 2 to octal and hexadecimal equivalents. (11110.11) 2 = (011 110.110) 2 = ( 3 6. 6) 8 = (36.6) 8 (11110.11) 2 = (0001 1110.1100) 2 = ( 1 E. C) 16 = (1E.C) 16 19

Conversion from Octal / Hexadecimal to Binary These conversions are the simplest of all. Replace each octal or hexadecimal digit with its binary equivalent. Examples: 1. Convert (3D.E8) 16 to its binary equivalent. (3D.E8) 16 = (0011 1101. 1110 1000) 2 2. Convert (45.5) 8 to its binary equivalent. (45.5) 8 = (100 101. 101) 2 20

Binary Arithmetic Rules for Binary Addition 1. 0 + 0 = 0 2. 0 + 1 = 1 ; 1 + 0 = 1 3. 1 + 1 = 0 ; Carry = 1 4. 1 + 1 + 1 = 1 ; Carry = 1 Example: 1. Add 3 and 4. 3 -> 011 Augend 4 -> 100 + Addend -------- --------------- 7 -> 111 Result 2. Add 13 and 7. 1 1 1 1 1 13 -> 1101 Augend 7 -> 111 + Addend --------- ----------------- 20 -> 10100 Result 21

Complements There are 2 types of complements used in Binary System, namely 1 s Complement and 2 s Complement. Complements are also used to represent negative numbers. 1 s Complement: 1 s Complement of a bit can be obtained as 1 minus that bit. i.e. 1 s complement of 1 is 1 1 = 0 and 1 s complement of 0 is 1 0 = 1. In general, invert the given bit to get its 1 s complement. 2 s Complement: 2 s Complement of a bit is obtained by adding 1 to the 1 s complement of that bit. i.e. 2 s Complement of 1010 -> 1 s complement of 1010 + 1 -> 0101 + 1 = 0110 22

Signed Binary Numbers To represent the sign of the given number, we will use a separate bit (sign bit). It becomes the MSB of the binary number. The bit 0 represents (+) or positive numbers and the bit 1 represents (-) or negative numbers. Negative numbers are represented in either of the following ways: 1. Sign-Magnitude The MSB represents the sign and the remaining bits represent the magnitude of the number. Example: +5 = 0 000 0101 +127 = 0 111 1111-5 = 1 000 0101-127 = 1 111 1111 Note: Positive numbers are represented only using the signmagnitude form 23

1 s Co ple e t Su tra tio Contd. Example 1: Subtract 4 from 13. 1. 13 -> 1101 ; 4 -> 0100 2. 1 s complement of 4 -> 1011 1101 + 1011 1 1000 3. Adding carry, result is 1001 i.e. +9 Example 2: Subtract 13 from 4. 1. 4 -> 0100 ; 13 -> 1101 2. 1 s complement of 13 -> 0010 0100 + 0010 0110 3. No carry -> result is negative. Take 1 s complement of 0110 -> 1001 Thus, result is -9 24

2 s Complement Subtraction Contd. Example 1: Subtract 4 from 13. 1. 13 -> 1101 ; 4 -> 0100 2. 2 s complement of 4 -> 011 + 1 = 1100 1101 + 1100 1 1001 3. Ignore carry, result is 1001 i.e. +9 Example 2: Subtract 13 from 4. 1. 4 -> 0100 ; 13 -> 1101 2. 2 s complement of 13 -> 0011 0100 + 0011 0111 3. No carry -> result is negative. Take 2 s complement of 0111 -> 1000 + 1 i.e. 1001. Thus, result is -9 25

Binary Codes Codes are alternate representations for the binary numbers. Different codes are used for binary numbers. Commonly used are 1. BCD (Binary Coded Decimal) Code a) 8421 b) 2421 c) 4221 2. Excess-3 Code 3. Gray Code 26

Logic Gates and Truth Tables Any logic circuit can be implemented using the following basic gates: 1. NOT Gate 2. OR Gate 3. AND Gate Each of the Gates can be represented schematically, or as an expression, or as a truth table. Other gates can be implemented as a combination of the basic gates. For the given logic circuit, we can get the Boolean expression and then form the truth table. Also for the given truth table, we can get the Boolean expression and then construct the logic circuit. 30

NOT Gate This function operates on a single Boolean value. Its output is the complement of its input. i.e. Output Y = X An input of 1 produces an output of 0 and an input of 0 produces an output of 1. X Y = X 0 1 1 0 x x' Output is 1 if at least one input is 1. More than two values can be or-ed together. For example x + y + z = 1 if at least one of the three values is 1. x y out = x+y 0 0 0 0 1 1 1 0 1 x y out 1 1 1 31

AND Gate Output is one if every input has value of 1. More than two values can be and-ed together. For example xyz = 1 only if x=1, y=1 and z=1. x y out = xy 0 0 0 0 1 0 1 0 0 1 1 1 x y out 33

NAND Gate When the output of AND gate is given as input to the NOT gate (or a bubble can be added to the AND gate), the result circuit is the NAND gate. The output of the NAND gate is the complement of the output of the AND gate. x y out = (xy) 0 0 1 0 1 1 1 0 1 1 1 0 When the output of OR gate is given as input to the NOT gate (or a bubble can be added to the AND gate), the result circuit is the NOR gate. The output of the NOR gate is the complement of the output of the OR gate. x y x y out x y out = (x + y) 0 0 1 0 1 0 1 0 0 1 1 0 out 34

EX-OR (Exclusive-OR) Gate The output of the XOR gate is equal to 1, if an odd number of input values are 1 and 0 if an even number of input values are 1. x y out = 0 0 0 0 1 1 1 0 1 1 1 0 x y The output of the XNOR gate is the complement of the output of the XOR gate. x y out x y out = 0 0 0 0 1 1 1 0 1 1 1 0 x out y 36

X Manipulating Boolean Functions Given Logic Circuit: y z xy'+yz 38

We can express it as: f(x,y,z) = xy + yz The truth table is: x y z xy' yz xy + yz 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 1 1 1 0 0 1 0 1 1 0 1 1 0 1 1 1 0 0 0 0 1 1 1 0 1 1 39

Unit : III Overview Boolean Algebra Truth Tables Universal Gates Simplification of Boolean functions Map Methods Mc-Clausky Tabulation Methods 40

Boolean Algebra Using Boolean laws helps to reduce the given function to a lesser number of literals. Also a lesser number of gates. Result More efficient and cheaper digital circuits. The simplification can be achieved using the various axioms and laws of Boolean Algebra 41

Laws similar to Ordinary Algebra Commutative Law A + B = B + A A. B = B. A Associative Law A + (B + C) = (A + B) + C A (BC) = (AB) C Distributive Law A (B + C) = AB + AC Boolean Axioms 42

Rules specific to Boolean Algebra 1. a) A + 0 = A b) A. 1 = A When A = 0 When A = 0 A + 0 = 0 + 0 = 0 = A A. 1 = 0. 1 = 0 = A When A = 1 When A = 1 A + 0 = 1 + 0 = 1 = A A. 1 = 1. 1 = 1 = A 2. a) A + 1 = 1 b) A. 0 = 0 When A = 0 When A = 0 When A = 1 A + 1 = 0 + 1 = 1 A + 1 = 1 + 1 = 1 When A = 1 A. 0 = 0. 0 = 0 A. 0 = 1. 0 = 0 43

Rules specific to Boolean Algebra 3. a) A + A = A b) A. A = A When A = 0 A + A = 0 + 0 = 0 = A When A = 1 A + A = 1 + 1 = 1 = A When A = 0 A. A = 0. 0 = 0 = A When A = 1 A. A = 1. 1 = 1 = A 4. a) A + A = 1 b) A. A = 0 When A = 0 A + A = 0 + 1 = 1 When A = 1 A + A = 1 + 0 = 1 When A = 0 A. A = 0. 1 = 0 When A = 1 A. A = 1. 0 = 0 44

Rules specific to Boolean Algebra 5. a) A = A When A = 0 When A = 1 A = 1 ; A = 0 = A A = 0 ; A = 1 = A This double complement rule is also known as Involution 45

Rules specific to Boolean Algebra 6. a) A + AB = A b) A (A + B) = A L.H.S = A + AB = A (1 + B) Distributive law = A. 1 Since, 1 + B = 1 = A Since, A. 1 = A = R.H.S L.H.S = A (A + B) = AA + AB Distributive law = A + AB Since, AA = A = A Since, A + AB = A = R.H.S 46

Rules specific to Boolean Algebra 7. a) A + A B = A + B b) A (A + B) = AB L.H.S = A + A B = A.1 + A B Since, A.1=A = A.(1 + B) + A B Since, 1+B=1 = A + AB + A B Distributive law = A + B (A + A ) Distributive law = A + B Since, A + A = 1 = R.H.S L.H.S = A (A + B) = A.A + AB Distributive law = 0 + AB Since, A. A = 0 = AB Since, 0 + AB = AB = R.H.S 47

Rules specific to Boolean Algebra 8. a) A + BC = (A + B) (A + C) R.H.S = (A + B) (A + C) = AA + BA + AC + BC = A + AB + AC + BC Since, A.A=A ; AB = BA = A (1 + B + C) + BC Distributive Law = A + BC Since, 1 + A + = 1 & A.1 = 1 = L.H.S b) A ( B + C) = AB + AC This is the Distributive Law. 48

Rules specific to Boolean Algebra 9. a) AB + A C + BC = AB + A C Hint: We need to remove BC from L.H.S LHS = AB + A C + BC = AB + A C + BC. 1 Since, 1.BC = BC = AB + A C + BC (A + A ) Since, A + A = 1 = AB + A C + ABC + A BC Distributive law = AB (1 + C) + A C (1 + B) = AB + A C = RHS 49

Rules specific to Boolean Algebra b) (A+B) (A +C) (B+C) = (A+B) (A +C) LHS = (A + B) (A + C) (B + C) = (AA + A B + AC + BC) (B + C) = (A B + AC + BC) (B + C) Since, AA =0 = A BB + ABC + BBC + A BC + ACC + BCC = A B + A BC + ABC + BC + AC + BC Since, BB=B = A B (1 + C) + BC (A + 1) + AC + BC = A B + BC + AC + BC Since, 1+C=1 = A B + AC + BC RHS = (A + B) (A + C) = AA + A B + AC + BC = A B + AC + BC = LHS 50

Demorgan s Laws The AND and OR functions can be shown to be related to each other through the following equations. Theorem 1: The complement of a product is equal to the sum of individual complements. In other words, (AB) = A + B NAND = Bubbled OR or Theorem 2: The complement of a sum is equal to the product of individual complements. In other words, (A + B) = A. B or NOR = Bubbled AND 51

For Example: 1. Simplify: Find the complement of AB + C D and simplify. Solution: (AB + C D ) = (AB ). (C D ) = (A + B). (C + D ) = (A + B). (C + D) 2. Show that (A B + AB ) = A B + AB Solution: LHS = (A B + AB ) = (A B). (AB ) = (A + B ). (A + B ) = (A + B ). (A + B) = AA + A B + AB + BB = A B + AB Since, AA = 0 and BB = 0 52

Universal Gates Universal gates are the ones which can be used for implementing any gate like AND, OR and NOT, or any combination of these basic gates. NAND and NOR gates are universal gates. But there are some rules that need to be followed when implementing NAND or NOR based gates. NAND as Universal Gate Any logic function can be implemented using NAND gates. To achieve this, first the logic function has to be written in Sum of Product (SOP) form. Once logic function is converted to SOP, the logic circuit with AND gates in first level and OR gates in second level can be converted into a NAND- NAND gate circuit. 53

Simplification of Boolean functions Simplification of Boolean functions is mainly used to reduce the gate count of a design. Less number of gates means less power consumption, sometimes the circuit works faster. When number of gates is reduced, cost also comes down. There are many ways to simplify a logic design. 1. Algebraic Simplification - Simplify symbolically using theorems. 2. Karnaugh Maps Simplifies using Diagrammatic technique. 3. Quine-McCluskey Simplifies using Tabulation technique. 54

Karnaugh Maps (K-Maps) Karnaugh maps provide a systematic method to obtain simplified sum-ofproducts (SOPs) Boolean expressions. This is a compact way of representing a truth table and is a technique that is used to simplify logic expressions. It is ideally suited for four or less variables, becoming cumbersome for five or more variables. A K-map of n variables will have 2 n squares. Each square represents either a minterm or maxterm. For a Boolean expression, product terms are denoted by 1's, while sum terms are denoted by 0's - but 0's are often left blank. 55

2 Variable K-Map A 2-variable K-Map will have 4 cells. The function can have 4 possible combinations. a) 00, 01, 10, 11 or b) A B, A B, AB, AB or c) m0, m1, m2, m3 The K-Map can be drawn in either of the following ways: 56

For example : F = X'Y+XY+XY' 1. Draw the k-map for function F with marking 1 for X'Y, XY and XY position. 2. Now combine two 1's as shown in figure to form the two single terms. F = X + Y 57

3 Variable K-Map A 3-variable K-Map will have 8 cells. The function can have 8 possible combinations. a) 000, 001, 010, 011, 100, 101, 110, 111 or b) A B C, A B C, A BC, A BC, AB C, AB C, ABC, ABC or c) m0, m1, m2, m3, m4, m5, m6, m7 The K-Map can be drawn as below: Using gray code arrangement ensures that minterms of adjacent cells differ by only ONE literal. SAE2B Digital Electronics & Microprocessor 58

For example : F(X,Y,Z) = Σ(1,3,4,5,6,7) 1. Draw the k-map for function F with marking 1 for m1, m3, m4, m5, m6 and m7. 2. Now combine 1 s as shown in figure to form the reduced terms. F = X + Z 59

4 Variable K-Map A 4-variable K-Map will have 16 cells. The function can have 16 possible combinations m0 to m15. The K-Map can be drawn as below: 60

For example : F(A, B, C, D) = Σ(0, 1, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15) 1. Draw the k-map for function F with marking 1 for the given minterms. 2. Now combine 1 s as shown in figure to form the reduced terms. AB CD 00 01 11 10 00 01 1 1 1 1 1 1 F = B + D 11 1 1 10 1 1 1 1 61

5 Variable K-Map A 5-variable K-Map will have 32 cells m0 to m31. The 5-Variable Map can be drawn as 2 4-variable maps-m0 to m15 and m16 to m31. If the reduced expressions for the 2 individual maps is Y1 and Y2, we can write the final expression as Y = Y1 + Y2. If necessary, apply Boolean laws for further simplification. 62

For Example: F(A,B,C,D,E) = Σ(0,1,3,5,7,,9,11,13,15,16,17,20,21) A = 0 BC DE A = 1 DE 00 01 11 10 BC 00 01 11 10 00 1 1 1 00 1 1 01 1 1 01 1 1 11 1 1 11 10 1 1 10 The reduced expressions for the 2 maps are: Y1 = AB C D + AE and Y2 = A B D Thus, Y = AB C D + AE + A B D = B D (A + AC ) + AE = B D (A + C ) + AE = A B D + B C D + AE 63

Quine McCluskey Method Represent the expression in the canonical SOP form. Consider the function F(A,B,C,D) = Σ(0, 1, 4, 5, 10, 11, 14, 15) Step 1: Represent each minterm in binary as shown: Decimal 0 1 4 5 10 11 14 15 Binary 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 1 0 1 0 1 1 1 1 1 0 1 1 1 1 64

Step 2: Arrange the minterms in different groups, so that the first group has no 1 s, second group contains only one 1, third group contains two 1 s and so on. 0 0 0 0 0 ; First Group 1 0 0 0 1 ; Second Group 4 0 1 0 0 5 0 1 0 1 ; Third Group 10 1 0 1 0 11 1 0 1 1 ; Fourth Group 14 1 1 1 0 15 1 1 1 1 ; Fifth Group 65

Step 3 : a) Each minterm of one group is compared with each minterm in the group immediately below. b) Each time a number is found in one group which is the same as a number in the group below except for one digit, the numbers pair is ticked and a new composite is created. c) This composite number has the same number of digits as the numbers in the pair except the digit different which is replaced by an "x". 66

Column 1 - A B C D Column 2 - A B C D 0 0 0 0 0 1 0 0 0 1 4 0 1 0 0 5 0 1 0 1 10 1 0 1 0 11 1 0 1 1 14 1 1 1 0 15 1 1 1 1 (0,1) 0 0 0 x (0,4) 0 x 0 0 (1,5) 0 x 0 1 (4,5) 0 1 0 x (10,11) 1 0 1 x (10,14) 1 x 1 0 (11,15) 1 x 1 1 (14,15) 1 1 1 x 67

Step 4 : Repeat Step 3 on Column 2. Column 1 - A B C D Column 2 - A B C D Column 3 - A B C D 0 0 0 0 0 1 0 0 0 1 4 0 1 0 0 (0,1) 0 0 0 x (0,4) 0 x 0 0 (1,5) 0 x 0 1 (4,5) 0 1 0 x (0,1,4,5) 0 x 0 x (0,4,1,5) 0 x 0 x 5 0 1 0 1 10 1 0 1 0 11 1 0 1 1 14 1 1 1 0 15 1 1 1 1 (10,11) 1 0 1 x (10,14) 1 x 1 0 (11,15) 1 x 1 1 (14,15) 1 1 1 x (10,11,14,15) 1 x 1 x (10,14,11,15) 1 x 1 x 68

Sequential Logic Flip-Flops Shift Register Counters Unit : IV Overview 69

Sequential Logic The outputs of a sequential logic circuit depend on both the current inputs and on previous inputs and outputs of the circuit. Sequential elements have storage elements that record the state of the circuit. In other words, the state information combined with the inputs is generating the outputs. The state and inputs also combine to generate a new state of the circuit. The same inputs in a sequential circuit may generate different outputs and different new states, depending on the circuit s current state. 70

Flip-Flops A bi-stable device i.e. a circuit with only 2 stable states, namely the 0 state and the 1 state. Ability to retain its state and store a bit of information. It is one-bit memory cell. A flip-flop has 2 outputs and they complement each other. Types of Flip Flop SR Flip Flop, JK Flip Flop, D Flip Flop, T Flip Flop. 71

Shift Register A common form of register used in computers and in many other types of logic circuits is a shift register. It is simply a set of flip flops (usually D latches or RS flip-flops) connected together so that the output of one becomes the input of the next, and so on in series. It is called a shift register because the data is shifted through the register by one bit position on each clock pulse. 72

Serial-in Serial-out Register On the leading edge of the first clock pulse, the signal on the D input is latched in the first flip flop. On the leading edge of the next clock pulse, the contents of the first flip-flop is stored in the second flip-flop, and the signal which is present at the D input is stored is the first flip-flop, etc. Because the data is entered one bit at a time, this called a serial-in shift register. Since there is only one output, and data leaves the shift register one bit at a time, then it is also a serial out shift register. 73

Parallel-in Parallel-out Register Parallel input can be provided through the use of the preset and clear inputs to the flip-flop. The parallel loading of the flip-flop can be synchronous (i.e., occurs with the clock pulse) or asynchronous (independent of the clock pulse) depending on the design of the shift register. Parallel output can be obtained from the outputs of each flip-flop as shown in Figure. 74

Counters Counter is a register which counts the sequence in binary form. The state of counter changes with application of clock pulse. The counter is binary or non-binary. The total number of states in counter is called as modulus. If counter is modulus-n, then it has n different states. State diagram of counter is a pictorial representation of counter states directed by arrows in graph. 111 000 001 110 010 State diagram of mod-8 counter 101 100 011 75

Asynchronous (Ripple) Counters All Flip-Flops are in toggle mode. The clock input is applied. Count enable = 1. Counter counts from 0000 to 1111. 76

Synchronous Counter In synchronous counters, the clock inputs of all the flip-flops are connected together and are triggered by the input pulses. Thus, all the flip-flops change state simultaneously (in parallel). The J and K inputs of FF0 are connected to HIGH. FF1 has its J and K inputs connected to the output of FF0, and the J and K inputs of FF2 are connected to the output of an AND gate that is fed by the outputs of FF0 and FF1. After the 3rd clock pulse, both outputs of FF0 and FF1 are HIGH. The positive edge of the 4th clock pulse will cause FF2 to change its state due to the AND gate. 77

Up / Down Counter Bidirectional counters, also known as Up/Down counters, are capable of counting in either direction through any given count sequence and they can be reversed at any point within their count sequence by using an additional control input 78

Unit : IV Overview Adders,Subtractors Decoders,Encoders Multiplexer,Demultiplexer ROM PLA 79

HALF ADDER From our truth table, we can derive the following statement: Our adding circuit should consist of an XOR gate (for R output, the Sum line) and an AND gate (for Q output, the Carry line). INPUTS OUTPUTS Q R A B A B A B 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 The XOR circuit delivers the Sum Line (R) of adding two single digits. The AND circuit produces the Carry Line (Q). We ve just created a fundamental circuit called a Half Adder, which provides the capability of adding two single bit numbers. 80

Circuit Diagram for the Half Adder A R SUM B Q CARRY 81

FULL ADDER The Half Adder operates as the key component in a divide-and-conquer approach to adding n digits. Three single digit numbers could be added two at a time: 1. Add the first two digits together using a half-adder. 2. Add the result of this sum to the third number using another half-adder. To achieve this, 1. We need 3 input lines, one for each number. 2. Two input lines feed into the first half adder; the third input line functions as a carry-in line. 3. The second half-adder has two input lines: the output sum from the first half-adder, plus the third, carry-in input line. 82

Truth Table for the Full Adder INPUTS OUTPUTS A B C IN Q CARRY R SUM 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 83

Subtractor What we have is a circuit that can add numbers. We want to make this circuit also work for subtracting two numbers Let s rethink our problem, with a specific example: 5-2 = 3, or, put another way: 5 + (-2) = 3 Now, we have an addition problem, instead of subtraction. This means our existing circuit will work if we could unlock how to represent negative numbers, like -2. This could be achieved using 2 s Complement. 84

Half Subtractor SAE2B Digital Electronics & Microprocessor 85

Full Subtractor To subtract the higher order columns you also need to subtract the borrow. The circuit can be constructed out of 2 half subtractors SAE2B Digital Electronics & Microprocessor 86

DECODERS A decoder is a combinational circuit that converts binary information from n inputs to exactly one output of a maximum of 2^n outputs. The maximum number of outputs is directly related to the number of inputs. A decoder with n inputs supports 2^n outputs. When n = 2, there are 2^2 = 4 outputs that can be decoded. When n = 3, there are 2^3 = 8 outputs that can be decoded. 87

Example: Design a 3 x 8 decoder using a truth table. Y0 = A B C Y1 = A B C Y2 = A BC Y3 = A BC Y4 = AB C Y5 = AB C Y6 = ABC Y7 = ABC 88

DEMULTIPLEXER A demultiplexer, sometimes abbreviated demux, is a circuit that has one input and more than one output. It is used when a circuit wishes to send a signal to one of many devices. This description sounds similar to the description given for a decoder, but a decoder is used to select among many devices while a demultiplexer is used to send a signal among many devices. 89

The truth table for a 1-to-2 demultiplexer is I A D 0 D 1 0 0 1 1 0 1 0 1 Using our 1-to-2 decoder as part of the circuit, we can express this circuit easily 0 0 1 0 0 0 0 1 90

Design of Circuit using MUX Multiplexers can be used in the design of both complex circuits and also realization of primitive gates such as XOR and AND gates. XOR gate: If we look at an XOR gate s functionality a little differently, we can easily see that when one input is 0 the other is simply propagated through and when one is 1 the other is complemented. Using a multiplexer to realize this, is shown in the following figure. This realization uses only 6 transistors if we use the second method used for a 2-to-1 MUX whereas XOR gates usually consist of 8 transistors. a b y a 0 0 0 1 1 0 1 1 0 1 1 0 b b - 0 1 y 91

Read Only Memory (ROM) A read-only memory (abbreviated ROM) consists of an array of semiconductor devices that are interconnected to store an array of binary data. Once binary data is stored in the ROM, it can be accessed or read out whenever desired, but it cannot be changed under normal operating conditions. Shown next is an example of a truth table for an 8X4 ROM. The 8 indicates that there are 8 words of data stored in the ROM. The 4 indicates that each word of data stored in the ROM consists of 4 bits. Each word of data has a size of 4 bits. SAE2B Digital Electronics & Microprocessor 92

Each combination of the input lines A, B, and C is used to access or output one stored word value of 4 bits using the output lines F0, F1, F2, and F3. Each input combination serves as an address that is used to access one of the words of data stored in the ROM. 93

Programmable Array Logic (PAL) The PAL (programmable array logic) is a special case of the programmable logic array in which the AND array is programmable and the OR array is fixed. The basic structure of the PAL is the same as the PLA. Because only the AND array is programmable, the PAL is less expensive than the more general PLA. In general, the PAL is easier to program. 94

Shown below is a segment of an unprogrammed PAL: 95