DLD - DIGITAL LOGIC DESIGN
Dr. Krishnanaik Vankdoth
B.E(ECE), M.Tech (ECE), Ph.D (ECE)
Professor in ECE Dept
Vaagdevi college of Engineering
Warangal – 506001
krishnanaik.ece@gmail.com
1
Books
1. “ Digital Design” By M. Morris Mano and Michael D.Ciletti
2. Logical Design and Application By
Dr. Krishnanaik
Vankdoth LAP LAMBERT Academic Publishing
Dnfscland/Germany – 2014
3. Complementary Material “ Logic and Computer Design
Fundamentals ” By M. Morris Mano & Charles R Kime.
2
DLD - DIGITAL LOGIC DESIGN
Digital Systems and Binary Numbers
Dr. V. Krishnanaik Ph.D
3
• Digital
• Concerned with the interconnection among digital components and modules
» Best Digital System example is General Purpose
Computer
• Logic Design
• Deals with the basic concepts and tools used to design digital hardware
consisting of logic circuits
» Circuits to perform arithmetic operations (+, -, x, ÷)
4
•
Digital Signal : Decimal values are difficult to represent in
electrical systems. It is easier to use two voltage values than
ten.
•
Digital Signals have two basic states:
1 (logic “high”, or H, or “on”)
0 (logic “low”, or L, or “off”)
•
Digital values are in a binary format. Binary means 2 states.
•
A good example of binary is a light (only on or off)
on
off
Power switches have labels “1” for on and “0” for off.
5
• Bits and Pieces of DLD History
• George Boole
• Mathematical Analysis of Logic (1847)
• An Investigation of Laws of Thoughts; Mathematical Theories of Logic and
Probabilities (1854)
• Claude Shannon
• Rediscovered the Boole
• “ A Symbolic Analysis of Relay and Switching Circuits “
• Boolean Logic and Boolean Algebra were Applied to Digital Circuitry
---------- Beginning of the Digital Age and/or Computer Age
World War II
Computers as Calculating Machines
Arlington (State Machines) “ Control “
6
Motivation
• Microprocessors/Microelectronics have revolutionized our
world
– Cell phones, internet, rapid advances in medicine, etc.
• The semiconductor industry has grown tremendously
7
Digital Systems and Binary Numbers
 Digital age and information age
 Digital computers
– General purposes
– Many scientific, industrial and commercial applications
• Digital systems
– Telephone switching exchanges
– Digital camera
– Electronic calculators, PDA's
– Digital TV
• Discrete information-processing systems
– Manipulate discrete elements of information
– For example, {1, 2, 3, …} and {A, B, C, …}…
8
Analog and Digital Signal
• Analog system
– The physical quantities or signals may vary continuously
over a specified range.
• Digital system
– TheX(t)
physical quantities or signals
can assume only
X(t)
discrete values.
– Greater accuracy
t
Analog signal
t
Digital signal
9
Analog
Technology:
Uses:
Digital
Converts analog waveforms into set of
Analog technology records waveforms as numbers and records them. The numbers are
they are.
converted into voltage stream for
representation.
Can be used in various computing
platforms and under operating systems
like Linux, Unix, Mac OS and Windows.
Analog signal is a continuous signal
which transmits information as a
Signal:
response to changes in physical
phenomenon.
Uses continuous range of values to
Representation:
represent information.
Computing and electronics
Digital signals are discrete time signals
generated by digital modulation.
Uses discrete or discontinuous values to
represent information.
Memory unit:
applications:
not required
Thermometer
required
PCs, PDAs
Data
transmissions:
not of high quality
high quality
Result:
not very accurate
accurate
Storage capacity: limited
high
Process:
processed using OPAMP which uses
electronic circuits
using microprocessor which uses logic circuits
Respose to
Noise:
More likely to get affected reducing
accuracy
Less affected since noise response are analog in
nature
Waves:
Example:
Denoted by sine waves
human voice in air
Denoted by square waves
electronic devices
10
Binary Digital Signal
• An information variable represented by physical quantity.
• For digital systems, the variable takes on discrete values.
– Two level, or binary values are the most prevalent values.
• Binary values are represented abstractly by:
– Digits 0 and 1
– Words (symbols) False (F) and True (T)
– Words (symbols) Low (L) and High (H)
– And words On and Off
V(t)
• Binary values are represented by values
or ranges of values of physical quantities.
Logic 1
undefine
Logic 0
t
Binary digital signal
11
Decimal Number System
• Base (also called radix) = 10
– 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
• Digit Position
– Integer & fraction
• Digit Weight
– Weight = (Base)
1
0
-1
-2
5
1
2
7
4
100
10
1
0.1 0.01
500
10
2
0.7
Position
• Magnitude
– Sum of “Digit x Weight”
• Formal Notation
2
0.04
d2*B2+d1*B1+d0*B0+d-1*B-1+d-2*B-2
(512.74)10
12
• Base = 8
Octal Number System
– 8 digits { 0, 1, 2, 3, 4, 5, 6, 7 }
• Weights
– Weight = (Base)
Position
• Magnitude
– Sum of “Digit x Weight”
• Formal Notation
64
8
1
1/8 1/64
5
1
2
7
4
2
1
0
-1
-2
5 *82+1 *81+2 *80+7 *8-1+4 *8-2
=(330.9375)10
(512.74)8
13
• Base = 2
Binary Number System
– 2 digits { 0, 1 }, called binary digits or “bits”
• Weights
– Weight = (Base)
Position
• Magnitude
– Sum of “Bit x Weight”
• Formal Notation
• Groups of bits
4 bits = Nibble
4
2
1
1/2
1/4
1
0
1
0
1
2
1
0
-1
-2
1 *22+0 *21+1 *20+0 *2-1+1 *2-2
=(5.25)10
(101.01)2
1011
8 bits = Byte
11000101
14
Hexadecimal Number System
• Base = 16
– 16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }
• Weights
– Weight = (Base)
Position
• Magnitude
– Sum of “Digit x Weight”
• Formal Notation
256
16
1
1/16 1/256
1
E
5
7
A
2
1
0
-1
-2
1 *162+14 *161+5 *160+7 *16-1+10 *16-2
=(485.4765625)10
(1E5.7A)16
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The Power of 2
n
2n
n
2n
0
20=1
8
28=256
1
21=2
9
29=512
2
22=4
10
210=1024
3
23=8
11
211=2048
4
24=16
12
212=4096
5
25=32
20
220=1M
Mega
6
26=64
30
230=1G
Giga
7
27=128
40
240=1T
Tera
Kilo
16
Addition
• Decimal Addition
1
+
1
1
Carry
5
5
5
5
1
0
= Ten ≥ Base
 Subtract a Base
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Binary Addition
• Column Addition
1
1
1
1
1
1
1
1
1
1
0
1
= 61
1
0
1
1
1
= 23
1
0
1
0
0
= 84
+
1
0
≥ (2)10
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Binary Subtraction
• Borrow a “Base” when needed
0
1
2
2
0
0
2
1
0
0
1
1
0
1
= 77
1
0
1
1
1
= 23
1
0
1
1
0
= 54
−
0
1
2
= (10)2
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Binary Multiplication
• Bit by bit
1
0
1
1
1
1
0
1
0
0
0
0
0
0
1
0
1
1
1
0
0
0
0
0
1
0
1
1
1
1
1
1
0
0
x
1
1
0
20
Number Base Conversions
Evaluate
Magnitude
Octal
(Base 8)
Evaluate
Magnitude
Decimal
(Base 10)
Binary
(Base 2)
Hexadecimal
(Base 16)
Evaluate
Magnitude
21
Decimal (Integer) to Binary Conversion
• Divide the number by the ‘Base’ (=2)
• Take the remainder (either 0 or 1) as a coefficient
• Take the quotient and repeat the division
Example: (13)10
13 / 2 =
6 /2=
3 /2=
1 /2=
Quotient
Remainder
Coefficient
6
3
1
0
1
0
1
1
a0 = 1
a1 = 0
a2 = 1
a3 = 1
Answer:
(13)10 = (a3 a2 a1 a0)2 = (1101)2
MSB
LSB
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Decimal (Fraction) to Binary Conversion
• Multiply the number by the ‘Base’ (=2)
• Take the integer (either 0 or 1) as a coefficient
• Take the resultant fraction and repeat the division
Example: (0.625)10
Integer
0.625 * 2 = 1 .
0.25 * 2 = 0 .
0.5
*2= 1 .
Answer:
Fraction
Coefficient
25
5
0
a-1 = 1
a-2 = 0
a-3 = 1
(0.625)10 = (0.a-1 a-2 a-3)2 = (0.101)2
MSB
LSB
23
Decimal to Octal Conversion
Example: (175)10
175 / 8 =
21 / 8 =
2 /8=
Quotient
Remainder
Coefficient
21
2
0
7
5
2
a0 = 7
a1 = 5
a2 = 2
Answer:
(175)10 = (a2 a1 a0)8 = (257)8
Example: (0.3125)10
Integer
0.3125 * 8 = 2
0.5
*8= 4
Answer:
Fraction
.
.
5
0
Coefficient
a-1 = 2
a-2 = 4
(0.3125)10 = (0.a-1 a-2 a-3)8 = (0.24)8
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Binary − Octal Conversion
•
• Each group of 3 bits represents
an octal digit
8 = 23
Assume Zeros
Example:
( 1 0 1 1 0 . 0 1 )2
( 2
6
. 2 )8
Octal
Binary
0
000
1
001
2
010
3
011
4
100
5
101
6
110
7
111
Works both ways (Binary to Octal & Octal to Binary)
25
Binary − Hexadecimal Conversion
• 16 = 24
• Each group of 4 bits represents
a hexadecimal digit
Assume Zeros
Example:
( 1 0 1 1 0 . 0 1 )2
(1
6
. 4 )16
Hex
Binary
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Works both ways (Binary to Hex & Hex to Binary)
26
Octal − Hexadecimal Conversion
• Convert to Binary as an intermediate step
Example:
( 2
6
.
2 )8
Assume Zeros
Assume Zeros
( 0 1 0 1 1 0 . 0 1 0 )2
(1
6
.
4 )16
Works both ways (Octal to Hex & Hex to Octal)
27
Decimal, Binary, Octal and Hexadecimal
Decimal
Binary
Octal
Hex
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
00
01
02
03
04
05
06
07
10
11
12
13
14
15
16
17
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
28
1.5 Complements
•
•
There are two types of complements for each base-r system: the radix complement and diminished radix
complement.
Diminished Radix Complement - (r-1)’s Complement
– Given a number N in base r having n digits, the (r–1)’s complement
of N is defined as:
(rn –1) – N
•
Example for 6-digit decimal numbers:
– 9’s complement is (rn – 1)–N = (106–1)–N = 999999–N
– 9’s complement of 546700 is 999999–546700 = 453299
•
Example for 7-digit binary numbers:
– 1’s complement is (rn – 1) – N = (27–1)–N = 1111111–N
– 1’s complement of 1011000 is 1111111–1011000 = 0100111
•
Observation:
– Subtraction from (rn – 1) will never require a borrow
– Diminished radix complement can be computed digit-by-digit
– For binary: 1 – 0 = 1 and 1 – 1 = 0
29
Complements
• 1’s Complement (Diminished Radix Complement)
– All ‘0’s become ‘1’s
– All ‘1’s become ‘0’s
Example (10110000)2
 (01001111)2
If you add a number and its 1’s complement …
10110000
+ 01001111
11111111
30
Complements
• Radix Complement
The r's complement of an n-digit number N in base r is defined as
rn – N for N ≠ 0 and as 0 for N = 0. Comparing with the (r  1) 's
complement, we note that the r's complement is obtained by adding 1 to the
(r  1) 's complement, since rn – N = [(rn  1) – N] + 1.
• Example: Base-10
The 10's complement of 012398 is 987602
The 10's complement of 246700 is 753300
• Example: Base-2
The 2's complement of 1101100 is 0010100
The 2's complement of 0110111 is 1001001
31
Complements
• 2’s Complement (Radix Complement)
– Take 1’s complement then add 1
OR
– Toggle all bits to the left of the first ‘1’ from the right
Example:
Number:
10110000
10110000
1’s Comp.:
01001111
+
1
01010000
01010000
32
Complements
• Subtraction with Complements
– The subtraction of two n-digit unsigned numbers M –
N in base r can be done as follows:
33
Complements
• Example 1.5
– Using 10's complement, subtract 72532 – 3250.
• Example 1.6
– Using 10's complement, subtract 3250 – 72532.
There is no end carry.
Therefore, the answer is – (10's complement of 30718) =  69282.
34
Complements
• Example 1.7
– Given the two binary numbers X = 1010100 and Y =
1000011, perform the subtraction (a) X – Y ; and (b) Y 
X, by using 2's complement.
There is no end carry.
Therefore, the answer is Y –
X =  (2's complement of
1101111) =  0010001.
35
Complements
• Subtraction of unsigned numbers can also be done by means of the (r  1)'s
complement. Remember that the (r  1) 's complement is one less then the r's
complement.
• Example 1.8
– Repeat Example 1.7, but this time using 1's complement.
There is no end carry, Therefore,
the answer is Y – X =  (1's
complement of 1101110) = 
0010001.
36
1.6 Signed Binary Numbers
• To represent negative integers, we need a notation for
negative values.
• It is customary to represent the sign with a bit placed in the
leftmost position of the number since binary digits.
• The convention is to make the sign bit 0 for positive and 1 for
negative.
• Example:
• Table 1.3 lists all possible four-bit signed binary numbers in
the three representations.
37
Signed Binary Numbers
38
Signed Binary Numbers
• Arithmetic addition
– The addition of two numbers in the signed-magnitude system follows the rules of
ordinary arithmetic. If the signs are the same, we add the two magnitudes and give
the sum the common sign. If the signs are different, we subtract the smaller
magnitude from the larger and give the difference the sign if the larger magnitude.
– The addition of two signed binary numbers with negative numbers represented in
signed-2's-complement form is obtained from the addition of the two numbers,
including their sign bits.
– A carry out of the sign-bit position is discarded.
• Example:
39
Signed Binary Numbers
• Arithmetic Subtraction
– In
form:
1. 2’s-complement
Take the 2’s complement
of the subtrahend (including the sign bit) and
2.
add it to the minuend (including sign bit).
A carry out of sign-bit position is discarded.
(  A)  (  B )  (  A)  (  B )
(  A)  (  B )  (  A)  (  B )
( 6)  ( 13)
(11111010  11110011)
• Example:
(11111010 + 00001101)
00000111 (+ 7)
40
1.7 Binary Codes
Digital data is represented, stored and transmitted as groups of binary digits also known as
binary code.
Weighted codes: In weighted codes, each digit is assigned a specific weight according to its position.
Non-weighted codes: In non-weighted codes are not appositionally weighted.
Reflective codes: A code is reflective when the code is self complementing. In other words, when the code for 9 is
the complement the code for 0, 8 for 1, 7 for 2, 6 for 3 and 5 for 4.
Sequential codes: In sequential codes, each succeeding 'code is one binary number greater than its preceding
code.
Alphanumeric codes: Codes used to represent numbers, alphabetic characters, symbols
Error defecting and correcting codes: Codes which allow error defection and correction are called error
detecting and' correcting codes.
41
• BCD Code
– A number with k decimal digits will require
4k bits in BCD.
– Decimal 396 is represented in BCD with
12bits as 0011 1001 0110, with each group
of 4 bits representing one decimal digit.
– A decimal number in BCD is the same as its
equivalent binary number only when the
number is between 0 and 9.
– The binary combinations 1010 through 1111
are not used and have no meaning in BCD.
Example:
Consider decimal 185 and its corresponding value in BCD and binary:
BCD addition
42
Binary Codes
• Other Decimal Codes
43
Binary Codes
• Gray Code
– The advantage is that only bit
in the code group changes in
going from one number to the
next.
• Error detection.
• Representation of analog data.
001
• Low 000
power design.
010
011
101
100
110
1-1 and onto!!
111
44
• American Standard Code for Information Interchange (ASCII) Character Code
45
• ASCII Character Code
46
ASCII Character Codes and Properties
• American Standard Code for Information Interchange (Refer to Table 1.7)
• A popular code used to represent information sent as character-based data.
• It uses 7-bits to represent:
– 94 Graphic printing characters.
– 34 Non-printing characters.
• Some non-printing characters are used for text format (e.g. BS = Backspace, CR =
carriage return).
• Other non-printing characters are used for record marking and flow control (e.g.
STX and ETX start and end text areas).
• ASCII has some interesting properties:
– Digits 0 to 9 span Hexadecimal values 3016 to 3916
– Upper case A-Z span 4116 to 5A16
– Lower case a-z span 6116 to 7A16
• Lower to upper case translation (and vice versa) occurs by flipping bit 6.
47
• Error-Detecting Code
– To detect errors in data communication and processing, an
eighth bit is sometimes added to the ASCII character to
indicate its parity.
– A parity bit is an extra bit included with a message to
make the total number of 1's either even or odd.
• Example:
– Consider the following two characters and their even and
odd parity:
48
• Error-Detecting Code
– Redundancy (e.g. extra information), in the form of extra bits, can be
incorporated into binary code words to detect and correct errors.
– A simple form of redundancy is parity, an extra bit appended onto the code
word to make the number of 1’s odd or even. Parity can detect all single-bit
errors and some multiple-bit errors.
– A code word has even parity if the number of 1’s in the code word is even.
– A code word has odd parity if the number of 1’s in the
code word is odd.
– Example:
10001001 1
(even parity)
Message B: 10001001 0
(odd parity)
Message A:
49
Hamming Codes
•
•
•
•
Invented W.B Hamming and Simple 1parity bit can tell us an error occurred
Multiple parity bits can also tell us where it occurred
O(lg(n)) bits needed to detect and correct one bit errors.
In generally we use 7 bits hamming code
– 4 data bits/message bit (m)
and
3 parity bits (2P>= P+m+1)
Example: Byte 1011 0001
Two data blocks, 1011 and 0001.
Expand the first block to 7 bits: _ _ 1 _ 0 1 1.
Bit 1 is 0, because b3+b5+b7 is even.
Bit 2 is 1, b3+b6+b7 is odd.
bit 4 is 0, because b5+b6+b7 is even.
Our 7 bit block is: 0 1 1 0 0 1 1
Repeat for right block giving 1 1 0 1 0 0 1
Error detectings: 0 1 1 0 1 1 1
Re-Check each parity bit
Bits 1 and 4 are incorrect
1 + 4 = 5, so the error occurred in bit 5
Dr. V. Krishnanaik Ph.D
50
Binary Storage and Registers
•
Registers
– A binary cell is a device that possesses two stable states and is capable of storing one of
the two states.
– A register is a group of binary cells. A register with n cells can store any discrete quantity of
information that contains n bits.
n cells
•
A binary cell
•
A register
•
Register Transfer
2n possible states
– Two stable state
– Store one bit of information
– Examples: flip-flop circuits, ferrite cores, capacitor
– A group of binary cells
– AX in x86 CPU
– A transfer of the information stored in one register to another.
– One of the major operations in digital system.
– An example in next slides.
51
A Digital Computer Example
Memory
CPU
Inputs: Keyboard,
mouse, modem,
microphone
Control
unit
Datapath
Input/Output
Synchronous or
Asynchronous?
Outputs: CRT,
LCD, modem,
speakers
52
Transfer of information
Figure 1.1 Transfer of information among register
53
Transfer of information
• The other major
component of a digital
system
– Circuit elements to
manipulate individual
bits of information
– Load-store machine
LD
LD
ADD
SD
Figure 1.2 Example of binary information processing
R1;
R2;
R3, R2, R1;
R3;
54
Binary Logic
• Definition of Binary Logic
– Binary logic consists of binary variables and a set of logical operations.
– The variables are designated by letters of the alphabet, such as A, B, C, x, y, z, etc, with
each variable having two and only two distinct possible values: 1 and 0,
– Three basic logical operations: AND, OR, and NOT.
55
Binary Logic gates
• Truth Tables, Boolean Expressions, and Logic Gates
AND
OR
NOT
x
y
z
x
y
z
x
z
0
0
0
0
0
0
0
1
0
1
0
0
1
1
1
0
1
0
0
1
0
1
1
1
1
1
1
1
z=x•y=xy
x
y
z
z = x = x’
z=x+y
x
y
z
x
z
Universal Gate
• NAND and NOR Gates are called Universal Gates because AND, OR and
NOT gates can be implemented &created by using these gates.
NAND Gate Implementations
NOR Gate Implementations
Dr. V. Krishnanaik Ph.D
Binary Logic
• Logic gates
– Example of binary signals
3
Logic 1
2
Un-define
1
Logic 0
0
Figure 1.3 Example of binary signals
59
Binary Logic
• Logic gates
– Graphic Symbols and Input-Output Signals for Logic
gates:
Fig. 1.4 Symbols for digital logic circuits
Fig. 1.5 Input-Output signals for gates
60
Binary Logic
• Logic gates
– Graphic Symbols and Input-Output Signals for Logic
gates:
Fig. 1.6 Gates with multiple inputs
61
Boolean Algebra
Boolean Algebra : George Boole(English mathematician), 1854
 Invented by George Boole in 1854
 An algebraic structure defined by a set B = {0, 1}, together with two binary
operators (+ and ·) and a unary operator ( )
“An Investigation of the Laws of Thought, on Which Are Founded the Mathematical Theories of Logic and
Probabilities”
Boolean Algebra
{(1,0), Var, (NOT, AND, OR), Thms}
Mathematical tool to expression and analyze digital (logic) circuits
Claude Shannon, the first to apply Boole’s work, 1938
– “A Symbolic Analysis of Relay and Switching Circuits” at MIT
This chapter covers Boolean algebra, Boolean expression and its
evaluation and simplification, and VHDL program
62
Basic Functions and Basic Functions
Boolean functions : NOT, AND, OR,
exclusive OR(XOR) : odd function
exclusive NOR(XNOR) : even function(equivalence)
Basic functions
Z=X  Y or Z=XY
Z=1 if and only if X=1 and Y=1, otherwise Z=0
• AND
• OR
Z=X + Y
Z=1 if X=1 or if Y=1, or both X=1and Y=1. Z=0 if and only if X=0 and Y=0
• NOT
Z=X or
Z=1 if X=0, Z=0 if X=1
63
64
Boolean Operations and Expressions
• Boolean Addition
–Logical OR operation
Ex 4-1) Determine the values of A, B, C, and D that make the sum term A+B’+C+D’
Sol) all literals must be ‘0’ for the sum term to be ‘0’
A+B’+C+D’=0+1’+0+1’=0 A=0, B=1, C=0, and D=1
• Boolean Multiplication
–Logical AND operation
Ex 4-2) Determine the values of A, B, C, and D for AB’CD’=1
Sol) all literals must be ‘1’ for the product term to be ‘1’
AB’CD’=10’10’=1 A=1, B=0, C=1, and D=0
65
Basic Identities of Boolean Algebra
The relationship between a
single variable X, its complement
X, and the binary constants 0
and 1
66
Laws of Boolean Algebra
• Commutative Law: the order of literals does not matter
A+B=B+A
AB=BA
• Associative Law: the grouping of literals does not matter
67
A + (B + C) = (A + B) + C (=A+B+C)
A(BC) = (AB)C (=ABC)
 Distributive Law : A(B + C) = AB + AC
(A+B)(C+D) = AC + AD + BC + BD
Rules of Boolean Algebra
 A+0=A
In math if you add 0 you have changed nothing in Boolean
Algebra ORing with 0 changes nothing
 A•0=0 In math if 0 is multiplied with
anything
you get 0. If you AND anything with 0 you get 0
 A•1 =A ANDing anything with 1 will yield the anything
 A+A = A ORing with itself will give the same result
 A+A’=1
Either A or A’ must be 1 so A + A’ =1
 A•A = A ANDing with itself will give the same result
 A•A’ =0 In digital Logic 1’ =0 and 0’ =1, so AA’=0 since one of the
inputs must be 0.
 A = (A’)’ If you not something twice you are back to the beginning
68
 A + A’B = A + B
If A is 1 the output is 1
If A is 0 the output is B
 A + AB = A
 (A + B)(A + C) = A + BC
• DeMorgan’s Theorem
– F(A,A, , + , 1,0)
= F(A, A, + , ,0,1)
– (A • B)’ = A’ + B’ and (A + B)’ = A’ • B’
– DeMorgan’s theorem will help to simplify digital circuits using NORs
and NANDs his theorem states
69
Boolean Analysis of Logic Circuits
•
Constructing a Truth Table for a Logic Circuit
– Convert the expression into the min-terms containing all the input literals
– Get the numbers from the min-terms
– Putting ‘1’s in the rows corresponding to the min-terms and ‘0’s in the remains
Ex) A(B+CD)=AB(C+C’) (D+D’) +A(B+B’)CD =ABC(D+D’) +ABC’(D+D’) +ABCD+AB’CD
=ABCD+ABCD’+ABC’D+ABC’D’ +ABCD+AB’CD =ABCD+ABCD’+ABC’D+ABC’D’
+AB’CD =m11+m12+m13+m14+m15=(11,12,13,14,15)
A(B+CD) = m11+m12+m13+m14+m15 =(11,12,13,14,15)
70
Standard Forms of Boolean Expressions
The Sum-of-Products(SOP) Form Ex) AB+ABC, ABC+CDE+B’CD’
The Product-of-Sums(POS) Form Ex) (A+B)(A+B+C), (A+B+C)(C+D+E)(B’+C+D’)
Principle of Duality : SOP  POS
Domain of a Boolean Expression : The set of variables contained in the expression
Ex) A’B+AB’C : the domain is {A, B, C}
Standard SOP Form (Canonical SOP Form)
– For all the missing variables, apply (x+x’)=1 to the AND terms of the expression
– List all the min-terms in forms of the complete set of variables in ascending order
Ex : Convert the following expression into standard SOP form: AB’C+A’B’+ABC’D
Sol) domain={A,B,C,D}, AB’C(D’+D)+A’B’(C’+C)(D’+D)+ABC’D
=AB’CD’+AB’CD+A’B’C’D’+A’B’C’D+A’B’CD’+A’B’CD+ABC’D
=1010+1011+0000+0001+0010+0011+1101 =0+1+2+3+10+11+13 =
(0,1,2,3,10,11,13)
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Standard POS Form (Canonical POS Form)
– For all the missing variables, apply (x’x)=0 to the OR terms of the
expression
– List all the max-terms in forms of the complete set of variables in
ascending order
Ex : Convert the following expression into standard POS form:
(A+B’+C)(B’+C+D’)(A+B’+C’+D)
Sol) domain={A,B,C,D},
(A+B’+C)(B’+C+D’)(A+B’+C’+D)
=(A+B’+C+D’D)(A’A+B’+C+D’)(A+B’+C’+D)
=(A+B’+C+D’)(A+B’+C+D)(A’+B’+C+D’)(A+B’+C+D’)(A+B’+C’
+D)=(0100) )(0101)(0110)(1101)= (4,5,6,13)
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Converting Standard SOP to Standard POS
Step 1. Evaluate each product term in the SOP expression. Determine the binary
numbers that represent the product terms
Step 2. Determine all of the binary numbers not included in the evaluation in Step 1
Step 3. Write in equivalent sum term for each binary number Step 2 and expression in
POS form
Ex : Convert the following SOP to POS
Sol) SOP= A’B’C’+A’BC’+A’BC+AB’C+ABC=0+2+3+5+7 =(0,2,3,5,7)
POS=(1)(4)(6) = (1, 4, 6) (=(A+B+C’)(A’+B+C)(A’+B’+C))

SOP and POS Observations
– Canonical Forms (Sum-of-minterms, Product-of-Maxterms), or other standard
forms (SOP, POS) differ in complexity
– Boolean algebra can be used to manipulate equations into simpler forms
– Simpler equations lead to simpler implementations
73
Summary of Minterms and Maxterms
•
There are 2n minterms and maxterms for Boolean functions with n variables.
•
Minterms and maxterms are indexed from 0 to 2n – 1
•
Any Boolean function can be expressed as a logical sum of minterms and as a logical
product of maxterms
•
The complement of a function contains those minterms not included in the original
function
• The complement of a sum-of-minterms is a product-of-maxterms with the same indices
Dual of a Boolean Expression
• To changing 0 to 1 and + operator to – vise versa for a given boolean function
 Example: F = (A + C) · B + 0
dual F = (A · C + B) · 1 = A · C + B
 Example: G = X · Y + (W + Z)
dual G =
 Unless it happens to be self-dual, the dual of an expression does not equal the
expression itself
 Are any of these functions self-dual? (A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC
74
Karnaugh Map
• Simplification methods
– Boolean algebra(algebraic method)
– Karnaugh map(map method))
XY+XY=X(Y+Y)=X
– Quine-McCluskey(tabular method)
–
Three- and Four-input Kanaugh maps
Gray code
75
Dr. V. Krishnanaik Ph.D
Karnaugh Map (K- Map) Steps
Sketch a Karnaugh map grid for the given problem.in power of 2N Squares
Fill in the 1’s and 0’s from the truth table of sop or pos Boolean function
Circle groups of 1’s.
1.
2.
3.


4.
Circle the largest groups of 2, 4, 8, etc. first.
Minimize the number of circles but make sure that every 1 is in a circle.
Write an equation using these circles.
Example) F(X,Y,Z)=m(2,3,4,5) =XY+XY
76
Example) F(X,Y,Z)=m(0,2,4,6) = XZ+XZ  =Z(X+X)=Z
Four-Variable K-Map : 16 minterms : m0 ~ m15
Rectangle group
– 2-squares(minterms) : 3-literals product term
– 4-squares : 2-literals product term
– 8-squares : 1-literals product term
– 16-squares : logic 1
77
F(W, X,Y,Z)=m(0,2,7,8,9,10,11) = WX’ + X’Z’ + W’XYZ
Ex 4-28) Minimize the following expression
AB’C+A’BC+A’B’C+A’B’C’+AB’C’
Sol) B’+A’C
78
Ex Minimize the following expression
B’C’D’+A’BC’D’+ABC’D’+A’B’CD+AB’CD+A’B’CD’+A’BCD’ +ABCD’+AB’CD’
Sol) D’+B’C
Don’t Care Conditions
• it really does not matter since they will never occur(its output is either ‘0’
or ‘1’)
• The don’t care terms can be
used to advantage on the
Karnaugh map
79
Ex K- Map for POS (B+C+D)(A+B+C’+D)(A’+B+C+D’)(A+B’+C+D)(A’+B’+C+D)
Sol) (B+C+D)=(A’A+B+C+D)=(A’+B+C+D)(A+B+C+D)
(1+0+0+0)(0+0+0+0)(0+0+1+0)
(1+0+0+1)(0+1+0+0)(1+1+0+0)
F=(C+D)(A’+B+C)(A+B+D)
 Converting Between POS and
SOP Using the K-map
Ex 4-33) (A’+B’+C+D)(A+B’+C+D)
(A+B+C+D’)(A+B+C’+D’) (A’+B+C+D’)
(A+B+C’+D)
80
81
Quine-McCluskey - Tabular Method
•
Step 1 − Arrange the given min terms in an ascending order and make the groups based on the
number of ones present in their binary representations. - ‘n+1’ groups
•
Step 2 − Compare the min terms present in successive groups. If there is a change in only one-bit
position, then take the pair of those two min terms. Place this symbol ‘_’ in the differed bit position
and keep the remaining bits as it is.
•
Step 3 − Repeat step2 with newly formed terms till we get all prime implicants.
•
Step 4 − Formulate the prime implicant table. It consists of set of rows and columns. Place ‘1’ in
the cells corresponding to the min terms that are covered in each prime implicant.
•
Step 5 − Find the essential prime implicates by observing each column. Those essential prime
implicants will be part of the simplified Boolean function.
•
Step 6 − Reduce the prime implicant table by removing the row of each essential prime implicant
and the columns corresponding to the min terms that are covered in that essential prime implicant.
Repeat step 5 for Reduced prime implicant table. Stop this process when all min terms of given
Boolean function are over.
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83
84
Digital Circuits
• Digital circuits are two types
1. Combinational circuit consists of logic gates whose outputs at any
time are determined directly from the present combination of
inputs without regard to previous inputs.
2. Sequential Circuit employ memory elements in addition to logic
gates. Their outputs are a function of the inputs and the state of
the memory elements.
85
Dr. V. Krishnanaik Ph.D
Combinational Logic
Logic circuits for digital systems may be combinational or sequential.
A combinational circuit consists of input variables, logic gates, and output variables.
Hence, a combinational circuit can be described by:
1.
2.
A truth table that lists the output values for each combination of the input variables, or
m Boolean functions, one for each output variable.
Combinational vs. Sequential Circuits
Combinational circuits are memory-less. Thus, the output value depends ONLY on the current input values.
Sequential circuits consist of combinational logic as well as memory elements (used to store certain circuit states). Outputs depend on BOTH
current input values and previous input values (kept in the storage elements).
Design Procedure
Given a problem statement:
● Determine the number of inputs and outputs
Dept. of Computer Science And Applications, SJCET, Palai
Page 46
7
●Derive the truth table
●Simplify the Boolean expression for each output
●Produce the required circuit
Half Adder :
A combinational circuit that performs the addition of two bits is called a half adder, We need two input and outputs
The truth table for the half adder is listed below:
S(x,y) = x’y + xy’
Dept. of Computer Science And Applications, SJCET, Palai
x y
C S
0 0
0 0
0 1
0 1
1 0
0 1
1 1
1 0
C(x,y) = xy
Page 47
Full Adder :
One that performs the addition of three bits (two significant bits and a previous carry) is a full adder.
x y z
C S
0 0 0
0 0
0 0 1
0 1 0
0 1
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
S = x’y’z + x’yz’ + xy’z’ + xyz
Alternative formulae using algebraic manipulation:
C = X.Y + X.Z + Y.Z
= X.Y + (X + Y).Z
= X.Y + ((X XOR Y) + X.Y).Z
= X.Y + (X XOR Y).Z + X.Y.Z
= X.Y + (X XOR Y).Z
C = xy + xz + yz
S = X'.Y'.Z + X'.Y.Z' + X.Y'.Z' + X.Y.Z
= X‘. (Y'.Z + Y.Z') + X. (Y'.Z' + Y.Z)
= X'. (Y XOR Z) + X. (Y XOR Z)'
= X XOR (Y XOR Z) or (X XOR Y) XOR Z
Full adder using two half adder
Parallel Binary Adders
•Parallel Adder is a digital circuit that produces the arithmetic sum of 2 binary numbers.
•Constructed with full adders connected in cascade, with output carry from each full adder connected to the input carry of next full adder in the
chain.
•The carries are connected in a chain through the full adders.
•Two methods to handle carries in parallel adder
1. Ripple carry: carry out of each FA is connected to the carry input of next FA
2. Carry Look ahead: it anticipates the output carry of each stage, and based on the input bits of each stage, produces the output carry
by either carry generation or carry propagation.
Carry generation: output carry is produced internally by the FA.carry is generated only when both input bits are 1s.the generated
carry C is expressed as the AND function of two input bits A and B so C=AB.
Carry propagation: occurs when the i/p carry is rippled to become the o/p carry.an i/p carrymay be propagated by the full
adder when either or both of the i/p bits are 1s.the propagated carry Cp is expressed as the OR function of the i/p bits
ie Cp=A+B
2- Bit Parallel Adder
•
•
LSB of two binary numbers are represented by A1 and B1.The next higher bit are A2 and B2. The resulting 1 2 and CO, in which the CO
becomes MSB.
The carry output CO of each adder is connected as the carry input of the next higher order.
A2A1
+ B 2 B1
2
C0 ∑ ∑
1
Fig : bit adder using two full adder
Four Bit Parallel Adders
•
An n-bit adder requires n full adders with each output connected to the input carry of the next higher-order full adder.
•
The carry output of each adder is connected to the carry input of next adder called as internal carries.
Fig : bit adder
The logic symbol for a 4-bit parallel adder is shown. This 4-bit adder includes a Carry In (labeled C0) and a Carry Out (labeled C4).
Fig : Truth table for 4 bit parallel adder
74283 Four-bit binary adder
7483 is an older chip that is functionally identical to the 74283, but the pins are laid out differently
Cascading Parallel Adders
When we connect the outputs from one circuit to the inputs of another identical circuit to expand the number of bits being operated on, we say that
the circuits are cascaded together.
For example, you can cascade two 4-bit parallel adders to add two 8-bit numbers. To do this, connect the lower-order adder’s Carry Out to the higherorder adder’s Carry In.
MODULE 3
MCA-102 DIGITAL SYSTEMS & LOGIC DESIGN
Fig : cascading of parallel adders
Binary Subtraction
•
The subtraction A-B can be performed by taking the 2's complement of B and adding to A.
• The 2's complement of B can be obtained by complementing B and adding one to the result.
A-B = A + 2C(B) = A + 1C(B) + 1
= A + B’ + 1
Fig : binary substractor circuit
Adder/Substractor

Design requires:
(i) XOR gates:
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MCA-102 DIGITAL SYSTEMS & LOGIC DESIGN
ADMN 2014-‘17
(ii)S connected to carry-in.
Fig : adder/substractor circuit
•
When S=0, the circuit performs A + B. The carry in is 0, and the XOR gates simply pass B untouched.
•
When S=1, the carry into the least significant bit (LSB) is 1, and B is complemented (1’s complement) prior to the addition; hence, the circuit adds to A the 1’s
complement of B plus 1 (from the carry into the LSB).
Magnitude Comparator :
A magnitude comparator is a combinational circuit that compares two numbers, A and B, and then determines their relative magnitudes.
A>B
A=B
A<B
Problem: Design a magnitude comparator that compares 2 4-bit numbers A and B and determines whether:
A > B, or A = B, or A < B
Inputs
First n-bit number A
and
Second n-bit number B
Outputs:
3 output signals (GT, EQ, LT),
where: GT = 1 IFF A > B
EQ = 1 IFF A = B
Exactly One of these 3 outputs equals 1, while the other 2 outputs are 0`s.
Solution:
Inputs: 8-bits (A ⇒ 4-bits, B ⇒ 4-bits).A and B are two 4-bit numbers. Let A = A3A2A1A0, and Let B = B3B2B1B0 .
Design of the EQ
Define Xi = Ai xnor Bi = Ai Bi + Ai’ Bi’
Xi = 1 IFF Ai = Bi ∀ i =0, 1, 2 and 3
Xi = 0 IFF Ai ≠ Bi
Therefore the condition for A = B or EQ=1 IFF
A3= B3 → (X3 = 1), and A2= B2 → (X2 = 1), and A1= B1 → (X1 = 1), and A0= B0 → (X0 = 1).
Thus, EQ=1 IFF X3 X2 X1 X0 = 1. In other words, EQ = X3 X2 X1
Designing GT and LT:
GT = 1 if A > B:
 If A3 > B3 3 = 1 and B3 = 0 If A3 = B3 and A2 > B2

If A3 = B3 and A2 = B2 and A1 > A1

If A3 = B3 and A2 = B2 and A1 = B1 and A0 > B0

Therefore,
GT = A3B3‘+ X3 A2 B2‘+ X3 X2 A1 B1‘+ X3 X2 X1A0 B0‘
Similarly, LT = A3’B3 + X3 A2‘B2 + X3 X2 A1’B1 + X3 X2 X1A0’ B0
X0
LT = 1 IFF A < B
Fig :4 bit magnitude comparator
Decoder : A decoder is a logic circuit that accepts a set of inputs that represents a binary number and activates only the output that corresponds to the input number.
• A decoder has
------
N inputs and
2N outputs
• Exactly one output will be active for each combination of the inputs.
• Each of these input combinations only one of the M outputs will be active HIGH (1), all the other outputs are LOW (0).
• An AND gate can be used as the basic decoding element because it produces a HIGH output only when all inputs are HIGH.
•
If an active-LOW output (74138, one of the output will low and the rest will be high) is required for each decoded number, the entire decoder can be implemented with
NAND gates Inverters
•
If an active-HIGH output (74139, one of the output will high and the rest will be low) is required for each decoded number, the entire decoder can be implemented with
AND gates Inverters
Decoder example :
2-to-4-Line Decoder
fig : 2-4 decoder with active high output
Fig : logical symbol and truth table of 2-4 decoder
3 line to 8 line decoder (active-HIGH)
•
It can be called a 3-line-to- 8-line decoder, because it has three input lines and eight output lines.
•
It could also be called a binary-octal decoder or converters because it takes a three bit binary input code and activates the one of the eight outputs corresponding to that
code. It is also referred to as a 1-of-8 decoder, because only 1 of the 8 outputs is activated at one time.
Application example
A simplified computer I/O port system with a port address decoder with only four address
lines shown.
Fig :peripheral decoding in computer
•Computer must communicate with a variety of external devices called peripherals by sending and/or receiving data through what is known as input/output (I/O) ports
•Each I/O port has a number, called an address, which uniquely identifies it. When the computer wants to communicate with a particular device, it issues the appropriate address
code for the I/O port to which that particular device is connected. The binary port address is decoded and appropriate decoder output is activated to enable the I/O port
•Binary data are transferred within the computer on a data bus, which is a set of parallel lines
BCD -to- Decimal decoders
•
The BCD- to-decimal decoder converts each BCD code into one of Ten Positional decimal digit indications. It is frequently referred as a 4-line -to- 10 line decoder
•
The method of implementation is that only ten decoding gates are required because the BCD code represents only the ten decimal digits 0 through 9.
•
Each of these decoding functions is implemented with NAND gates to provide active -LOW outputs. If an active HIGH output is required, AND gates are used for
decoding
MODULE 3
MCA-102 DIGITAL SYSTEMS & LOGIC DESIGN
ADMN2014-‘17
Fig : a. logic circuit for BCD to Decimal decoder b. logical symbol of BCD to Decimal decoder IC c. truth table
BCD to 7-Segment Display Decoder
1. Used to convert a BCD or a binary code into a 7 segment code used to operate a 7 segment LED display.
2. It generally has 4 input lines and 7 output lines. Here we design a simple display decoder circuit using logic gates.
Fig : logical symbol of BCD-7 segment decoder
Dept. of Computer Science And Applications, SJCET, Palai
Page59
Encoder
Encoders typically have 2N inputs and N outputs.
These are called 2N–to–N encoders.
Encoders can also be devised to encode various symbols and alphabetic characters.
The process of converting from familiar symbols or numbers to a coded format is called encoding.
Fig : Logical diagram of Encoder
8 to-3 encoder Implementation
 Octal-to-Binary
 An octal to binary encoder has 23 = 8 input lines D0 to D7 and 3 output lines Y0 to Y2. Below is the
truth table for an octal to binary encoder.
Fig : Truth table for 8-3 encoder
From the truth table, the outputs can be expressed by following Boolean Function. Y0 = D1 + D3 + D5 + D7
Y1 = D2 + D3 + D6 + D7
Y2 = D4 + D5 + D6 + D7
Note: Above boolean functions are formed by ORing all the input lines for which output is 1. For instance Y0 is 1 for D1, D3, D5, D7 input lines. The encoder can therefore be
implemented with OR gates whose inputs are determined directly from truth table as shown in the image below:
Fig :logic circuit for 8-3 encoder
Decimal – BCD encoder
•Encoder will produce a BCD output corresponding to the highest-order decimal digit input that is active and will ignore any other lower order active
inputs.
Fig :truth table for 10-4 encoder
-- From the truth table, the outputs can be expressed by following Boolean Function.
Note: Below Boolean functions are formed by OR ing all the input lines for which output is 1. For instance A 0 is 1 for 1,3,5,7 or
9 input lines.
A0 = 1+3+5+7+9
A1 = 2+3+6+7
A2 = 4+5+6+7
A3 = 8+9
The decimal to bcd encoder can therefore be implemented with OR gates whose inputs are determined directly from truth table as shown in the
image below.
Fig : Logic circuit and symbol for 10-4 encoder
Multiplexer
A multiplexer or mux is a device that selects one of several analog or digital input signals and forwards the selected input into a single line. A multiplexer of 2n inputs has n
select lines, which are used to select which input line to send to the output.”
The select lines determine which input is connected to the output.
MUX Types
2-to-1 (1 select line)
4-to-1 (2 select lines)
8-to-1 (3 select lines)
16-to-1 (4 select lines)
Fig : multiplexer block diagram
4to-1 Multiplexer (MUX)
Fig : logic circuit, symbol and table for 4-1 multiplexer
The Boolean expression for this 4-to-1 Multiplexer above with inputs D0 to D3 and data select lines A, B is given as:
Y = A’B’D0 + A’B D1 + AB’D2 + ABD3
Demultiplexer
•
A DEMUX is a digital switch with a single input (source) and a multiple outputs (destinations).
•
The select lines determine which output the input is connected to.
•
DEMUX Types
1-to-2 (1 select line)
1-to-4 (2 select lines)
1-to-8 (3 select lines)
1-to-16 (4 select lines)
Fig : Demultiplexer circuit symbol
1 to-4 De-Multiplexer (DEMUX)
Fig: logic circuit for 1-4 demultiplexer, symbol and table
Parity Bit Generator

The most common error detection code used is the parity bit.

A parity bit is an extra bit included with a binary message to make the total number of 1's either odd or even.

A parity bit added to n-bit code produces (n+1)-bit code with an odd (or even) count of 1s

Odd Parity bit: count of 1s in (n+1)-bit code is odd
o

So use an even function to generate the odd parity bit
Even Parity bit: count of 1s in (n+1)-bit code is even
o
So use an odd function to generate the even parity bit
To check for odd parity
o Use an even function to check the (n+1)-bit code
To check for even parity
o Use an odd function to check the (n+1)-bit code
Even Parity Generators and Checkers for 3-bit codes
An even parity bit could be added to n-bit code to produce an n + 1 bit code:
• Use an odd function to produce codes with even parity
• Use odd function circuit to check code words with even parity
Example: n = 3. Generate even parity code words of length 4 with an odd function circuit (parity generator):
The design procedure is made simple by writing the truth table for the circuit.
Fig : truth table fon parity bit generator
Fig : K-map for the truth table in fig 3.35 From this the minimal output equation is
This function can be implemented using exclusive-or gates, shown in fig 3.37.Similarly the checker circuit can be designed using XOR gates, where
C
Fig : even parity generator and checker
Operation: (X,Y,Z) = (0,0,1) gives (X,Y,Z,P) = (0,0,1,1) and E = 0.If Y changes from 0 to 1 between generator and checker, then E = 1 indicates an error.
Odd Parity Generators and Checkers
Similarly, an odd parity bit could be added to n-bit code to produce an n + 1 bit code
•
•
Use an even function to produce codes with odd parity
Use even function circuit to check code words with odd parity
Message
X
Odd Parity
Y
Z
Checker Bit
Generator P
C
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
1
0
1
1
1
0
0
Fig : Truth table for Odd parity generator
MODULE 3
MCA-102 DIGITAL SYSTEMS & LOGIC DESIGN
ADMN 2014-‘17
0
1
1
00
01
1
1
11
10
1
Fig : k-map for fig 3.39 From this the minimal output equation is
P=X’Y’Z’+X’YZ’+XYZ’+XY’Z
This function can be implemented using XOR and XNOR gates, shown in fig . Similarly the checker circuit can be designed using XOR and XNOR gates, where
X Y
Z
PLA [Programmable Logic Array]
Combination of a programmable AND array followed by a programmable OR array.
This layout allows for a large number of logic functions to be synthesized in the sum of products (and sometimes product of sums) canonical forms.
Fig : block diagram of PLA
Example :Design a PLA to realize the following logic functions f0 (A'.B'+ A. C')
f1 (A. C'+ B)
f2 (A'.B'+ B. C’)
Fig : PLA table
Fig : realisation of PLA
f3 (B+ AC)

Example: Design a PLA to realise the following three logic functions f1 (A, B, C, D, E) = A'.B'.D' + B'.C.D' + A'.B.C.D.E'
f2 (A, B, C, D, E) = A'.B.E + B'.C.D'.E
f3 (A, B, C, D, E) = A'.B'.D' + B'.C'.D'.E + A'.B.C.D
Fig :realisation of PLA
SEQUENTIAL LOGIC CIRCUITS
Sequential circuits are constructed using combinational logic and a number of memory elements with some
or all of the memory outputs fed back into the combinational logic forming a feedback path or loop.
Sequential circuit = Combinational logic + Memory Elements
Fig 3.45 sequential circuit
There are two types of sequential circuits:
 synchronous: outputs change only at specific time
 asynchronous: outputs change at any time
A state variable in a sequential circuit represents the single-bit variable Q stored in a memory element in circuit.
– Each memory element may be in state 0 or state 1 depending on the current value stored in the memory element.
•The State of A sequential Circuit:
–The collection of all state variables (memory element stored values) that at any time contain all the information about the past necessary to account
for the circuit’s future behavior.
–A sequential circuit that contains n memory elements could be in one of a maximum of 2n states at any given time depending on the stored values in
the memory elements.
–Sequential Circuit State transition: A change in the stored values in memory elements thus changing the sequential circuit from one state to another.
Clock Signals & Synchronous Sequential Circuits
•A clock signal is a periodic square wave that indefinitely switches values from 0 to 1 and 1 to 0 at fixed intervals.
Clock cycle time or clock period: The time interval between two consecutive rising or falling edges of the clock.
Fig : clock signal
Synchronous Sequential Circuits:
Sequential circuits that have a clock signal as one of its inputs:
– All state transitions in such circuits occur only when the clock value is either 0 or 1 or happen at the rising or falling edges of the clock depending on the type of memory
elements used in the circuit.
Sequential Circuit Memory Elements: Latches, Flip-Flops
• Latches and flip-flops are the basic single-bit memory elements used to build sequential circuit with one or two inputs/outputs, designed using
individual logic gates and feedback loops.
•Latches: – The output of a latch depends on its current inputs and on its previous inputs and its change of state can happen at
any time when its inputs change.
• Flip-Flop: – The output of a flip-flop also depends on current and previous input but the change in output (change of state or
state transition) occurs at specific times determined by a clock input.
S-R Latch :
An S-R (set-reset) latch can be built using two NOR gates forming a feedback loop.
•The output of the S-R latch depends on current as well as previous inputs or state, and its state (value stored) can change as soon as its inputs change.
When Q is HIGH, the latch is in SET state.
When Q is LOW, the latch is in RESET state.
S
R
Fig : SR latch circuit diagram, truth table and symbol
Q
Q'
SR Flip flop


Since the S-R latch is responsive to its inputs at all times an enable line C is used to disable or enable state transitions.
Behaves similar to a regular S-R latch when enable C=1
S
Q
EN
Q'
R
Fig : gated SR flip flop
S = 0, R = 0; this is the normal resting state of the circuit and it has no effect of the output states. Q and Q’
will remain in whatever state they were in prior to the occurrence of this input condition. It works in HOLD (no change) mode operation. • S = 0, R
= 1; this will reset Q to 0, it works in RESET mode operation.
S = 1, R = 0; this will set Q to 1, it works in SET mode operation.
S = 1, R = 1; this condition tries to set and reset the NOR gate latch at the same time, and it produces Q = ¯
= 0. This is an unexpected condition
and are not used.
Since the two outputs should be inverse of each other. If the inputs are returned to 1 simultaneously, the
output states are unpredictable. This input condition should not be used and when circuits are constructed, the design should make this condition
SET=RESET = 1 never arises.
Clocked SR Flip Flop
Additional clock input is added to change the SR flip-flop from an element used in asynchronous sequential circuits to one,
which can be used in synchronous circuits.
Fig: truth table for clocked SR flip flop
Clocked SR Flip Flop Circuit with PRESET and CLEAR
Some flip-flops have asynchronous preset Pr and clear Cl signals.
Output changes once these signals change, however the input signals must wait for a change in clock to change the output
Fig : SR flip flop with preset and clear
JK Flip Flop
: Another types of Flip flop is JK flip flop.
It differs from the RS flip flops when J=K=1 condition is not indeterminate but it is defined to give a very useful changeover (toggle) action.
Toggle means that Q and Q ¯
will switch to their opposite states.
The JK Flip flop has clock input Cp and two control inputs J and K.
Operation of Jk Flip Flop is completely described by truth table
Fig : JK flip flop
JK Flip Flop with preset and clear : This flip flop can also have other inputs called Preset (or SET) and clear that
can be used for setting the flip Flop to 1 or resetting it to 0 by applying the appropriate signal to the Preset and Clear inputs.
These inputs can change the state of the flip flop regardless of synchronous inputs or the clock.
Fig :JK flip flop with PRESET and CLEAR
T Flip Flop :
The T flip flop has only the Toggle and Hold Operation. If Toggle mode operation. The output
will toggle from 1 to 0 or vice versa.
Fig: T flip flop
D Flip Flop
Also Known as Data Flip flop
Can be constructed from RS Flip Flop or JK Flip flop by addition of an inverter.
Inverter is connected so that the R input is always the inverse of S (or J input is always complementary of K).
The D flip flop will act as a storage element for a single binary digit (Bit).
Fig : D flip flop
EDGE TRIGGERED FLIP FLOP
Edge triggered flip-flop changes only when the clock C changes
The three basic types are introduced here: S-R, J-K and D.
Edge-triggered S-R flip-flop : The basic operation is illustrated below, along with the truth table for this type of flip-flop. The
operation and truth table for a negative edge-triggered flip-flop are the same as those for a positive except that the falling edge
of the clock pulse is the triggering edge.
As S = 1, R = 0. Flip-flop SETS on the rising clock edge.
Fig : edge triggered SR flip flop
Edge-triggered J-K flip-flop: The J-K flip-flop works very similar to S-R flip-flop. The only difference is that this flip-flop has NO
invalid state. The outputs toggle (change to the opposite state) when both J and K inputs are HIGH.
Edge-triggered D flip-flop : The operations of a D flip-flop is much simpler. It has only one input addition to the clock. It is very
useful when a single data bit (0 or 1) is to be stored.
If there is a HIGH on the D input when a clock pulse
is applied, the flip-flop SETs and stores a 1. If there is a LOW on the D input when a clock pulse is applied, the flip-flop RESETs and stores a 0. The truth table
below summarize the operations of the positive edge-triggered D flip-flop. As before, the negative edge-triggered flip-flop works the same except that the falling
edge of the clock pulse is the triggering edge.
MASTER-SLAVE FLIP FLOP
Is designed using two separate flip flops. Out of these, one acts as the master and the other as a slave. The figure of a master-slave J-K flip flop is shown below.
Fig : master slave flip flop
From the above figure you can see that both the J-K flip flops are presented in a series connection. The output of the master J-K flip flop is fed to the
input of the slave J-K flip flop. The output of the slave J-K flip flop is given as a feedback to the input of the master J-K flip flop. The clock pulse
[Clk] is given to the master J-K flip flop and it is sent through a NOT Gate and thus inverted before passing it to the slave J-K flip flop.
When Clk=1, the master J-K flip flop gets disabled. The Clk input of the master input will be the
opposite of the slave input. So the master flip flop output will be recognized by the slave flip flop only when the Clk value becomes 0. Thus, when
the clock pulse males a transition from 1 to 0, the locked outputs of the master flip flop are fed through to the inputs of the slave flip-flop making this
flip flop edge or pulse-triggered.
Programmable Logic: PALs and GALs
• Basic PAL Operation
– Programmable array of AND gates
– Fixed OR gate
117
-- Implementing a Sum-of-Product Expression
Ex : Show how a PAL is programmed for the following function : X=AB’C+A’BC’+A’B’+AC
Sol)
PAL Block Diagram
118
PAL Output Combinational Logic
X0=X
X1=X’
119
Digital System Application : 7-Segment LED Driver
Seven-Segment LED driver
120
Figure: The minimum logic implementation for segment a of the 7-segment display.
121
Registers
CLK – Clock
CD – Clear/Reset
LD – Load
• Capable of storing a set of bits
• Built using Flip Flops
Universal Shift Register
» Parallel-in Parallel-out
» Serial-in Serial-out
» Serial-in Parallel-out
» Parallel-in Serial-out
122
Dr. V. Krishnanaik Ph.D
4-bit Register
• This is called a parallel-in, parallel-out register
123
4-bit Serial-In, Serial-Out Register
Sift Registers
Multiply 5 by 2, 5 by 4, 5 by 8
Divide 48 by 2, by 4, by 8
Can sift bits either to left or right
Basic shift register is same as serial-in, serial-out register
124
Serial-In, Parallel-Out Shift Register
• There are also
– Parallel-in, serial out shift registers
– Bidirectional shift registers
125
Binary Counters
• Register that goes through a predetermined sequence of states is called
a counter
– e.g., 000, 001, 010, …. 110, 111
– e.g., 001, 010, 100, 001, 010, …
– Useful in counting, timing, generating patterns, etc.
Example – 3-bit Binary Counter
• Suppose we want to count from 0 to 7
126
Example – Counter (Cont.)
• Suppose we want to count from 0 to 7
– 000
– 001
– 010
– 011
– 100
– 101
– 110
– 111
– 000
Alternate
Change if all lower order bits (in previous
round) were 1s
Build counter using JK Flip Flops
127
Excitation Tables
• Truth tables tell us what will be the output for a given combination of inputs
• Sometimes we want to find what inputs to give to achieve a desired output
– Can be achieved via excitation tables
JK FLIP FLOT EX-TABLE
J
Qt+
K
Qt+
Qt
1
J
K
1
0
0
Qt
0
0
0
1
0
0
1
1
0
1
1
0
1
1
Q /t
1
1
Qt
Qt+1
J
K
Qt
Qt+1
J
K
0
0
0
0/1
0
0
0
X
0
1
1
0/1
0
1
1
X
1
0
0/1
1
1
0
X
1
1
1
0/1
0
1
1
x
0
128
Excitation Table FOR SR,D,T Flip Flops
Qt
Qt+1
D
Qt
Qt+1
S
R
0
0
0
0
0
0
X
0
1
1
0
1
1
0
1
0
0
1
0
0
1
1
1
1
1
1
x
0
Qt
Qt+1
T
0
0
0
0
1
1
1
0
1
1
1
0
129
Example – Counter
Qt
Qt+1
Q2
Q1
Qt
Q0
Qt+1
Q2
Q1
Q0
Q Q Q Q Q Q J K J K J K
Q Q Q Q Q Q J2 K J1 K J0 K
2
2
1
0
2
1
0
2
2
1
1
0
0
1
0
2
1
0
2
1
0
0 0 0 0 0 1
0 0 0 0 0 1 0 X 0 X 1 X
0 0 1 0 1 0
0 0 1 0 1 0 0 X 1 X X 1
0 1 0 0 1 1
0 1 0 0 1 1 0 X X 0 1 X
0 1 1 1 0 0
0 1 1 1 0 0 1 X X 1 X 1
1 0 0 1 0 1
1 0 0 1 0 1 X 0 0 X 1 X
1 0 1 1 1 0
1 0 1 1 1 0 X 0 1 X X 1
1 1 0 1 1 1
1 1 0 1 1 1 X 0 X 0 1 X
1 1 1 0 0 0
1 1 1 0 0 0 X 1 X 1 X 1
130
• Now draw K-maps for all 6 inputs
J0 = K0 = 1
J1 = K1 = Q0
J2 = K2 = Q1 * Q0
131
Exercise – Counter
• Build a 0 to 7 counter using T Flip Flops. Show all steps
• Your circuit should be similar to following
132
Example – Counter With External Input
• Suppose we want to build a 0 to 3 counter that increments
only when a push button is pressed
! Pressed
00
00
Pressed
Pressed
! Pressed
01
11
01
11
! Pressed
Pressed
10
Pressed
10
! Pressed
133
Example – Counter With External Input (Cont.)
Qt
Q Q
Qt+1
B
Q1
Q0
Q Q J1
K
1
0
1
J0
K
1
0
0
0
0
0
0
0
X
0
X
0
0
1
0
1
0
X
1
X
0
1
0
0
1
0
X
X
0
0
1
1
1
0
1
X
X
1
1
0
0
1
0
X
0
0
X
1
0
1
1
1
X
0
1
X
1
1
0
1
1
X
0
X
0
1
1
1
0
0
X
1
X
1
0
Another solution
-- Use a normal 0-3 counter &
connect push button to clock input
J1 = K1 = Q0B
J0 = K0 = B
134
State Diagrams
• State transition diagram for a washing machine
135
136
Dr. Krishnanaik Vankdoth
B.E(ECE), M.Tech (ECE), Ph.D (ECE)
Professor
Vaagdevi College of Engineering
Electronics and Communications Engineering
Warangal – 506001
krishnanaik.ece@gmail.com
krishnanaik_ece@yahoo.com
Phone : +919441629162
137