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Digital Electronics

Introduction to Digitial Electronics
WHAT IS DIGITAL SIGNAL
DIGITAL ELECTRONICS
DIGITAL SIGNAL OVER ANALOG
ADVANTAGES OF DIGITAL SIGNAL
BINARY SYSTEM
BINARY NUMBER
BIT
BYTE
NUMBER SYSTEM
DECIMAL NUMBERS
BINARY NUMBERS
DECIMAL TO BINARY CONVERSION
BINARY TO DECIMAL CONVERSION
HEXADECIMAL TO DECIMAL
OCTAL TO DECIMAL
BINARY ARITHMETIC OPERATIONS
BINARY ADDITION
BINARY SUBTRACTION
BINARY MULTIPLICATION
BINARY DIVISION
BINARY COMPLIMENTS
1’S COMPLIMENT
2’S COMPLIMENT
ADDITION USING 1’S COMPLIMENT
SUBTRACTION USING 1’S COMPLIMENT
9’s compliments
10’s compliment
R’S COMPLIMENTS
(R-1)’S COMPLIMENTS
BOOLEAN ALGEBRA
INTRODUCTION
BOOLEAN ALGEBRA OPERATIONS
BOOLEAN ALGEBRA TERMINOLOGIES
LAWS OF BOOLEAN ALGEBRA
DE MORGAN’S THEOREM
DE MORGAN’S FIRST LAW
DE MORGAN’S FIRST LAW
MINIMISATION TECHNIQUE
TRUTH TABLE
CANONICAL MINTERM
CANONICAL MAXTERM
SOLVED EXAMPLE
KARNAUGH MAP
2 VARIABLE K-MAP
3 VARIABLE K-MAP
4 VARIABLE K-MAP
5 VARIABLE K-MAP
LOGIC GATES
INTRODUCTION
AND GATE
OR GATE
NOT GATE
NAND GATE
NOR GATE
XOR GATE
X-NOR GATE
DIGITAL LOGIC FAMILY
INTRODUCTION
DIODE TRANSISTOR LOGIC
RESISTOR TRANSISTOR LOGIC
3 INPUT RTL NOR GATE
TRANSISTOR TRANSISTOR LOGIC
CHARACTERSTICS OF DIGITAL LOGIC FAMILY
FAN IN FAN OUT
POWER DISSIPATION
FIGURE OF MERIT
EMITTER COUPLED LOGIC
ARITHMATIC OPERATIONS
HALF ADDER
FULL ADDER
HALF SUBTRACTOR
FULL SUBTRACTOR
PARALLEL BINARY SUBTRACTOR
DERIVING FULL SUBTRACTOR
DIFFERENT BINARY CODES
INTRODUCTION
WEIGHTED BINARY SYSTEMS
8421 CODE OR BCD CODE
2421 CODE
5211 CODE
REFLECTIVE CODE
SEQUENTIAL CODE
EXCESS 3 CODE
EXCESS 3 CODE EXAMPLE
GRAY CODE
MORSE CODE CHART
COMBINATIONAL CIRCUITS
MULTIPLEXER
4 TO 1 MULTIPLEXER
APPLICATIONS OF MULTIPLEXER
DEMULTIPLEXER
1 TO 4 DEMUX
APPLICATIONS OF DEMUX
ENCODERS
INTRODUCTION
PRIORITY ENCODER
8 TO 3 PRIORITY ENCODER
4 TO 2 PRIORITY ENCODER
DECODERS
INTRODUCTION
BASIC BINARY DECODER
CASCADING DECODERS
APPLICATION OF DECODER
PARITY GENERATORS
PARITY BIT
PARITY GENERATOR
EVEN PARITY GENERATOR
ODD PARITY GENERATOR
EVEN PARITY CHECKER
ODD PARITY CHECKER
FLIP FLOP
INTRODUCTION
S-R FLIP FLOP
D FLIP FLOP
J-K FLIP FLOP
T FLIP FLOP
SHIFT REGISTERS
INTRODUCTION
SERIAL IN SERIAL OUT SHIFT REGISTER
SERIAL IN PARALLEL OUT
PARALLEL IN PARALLEL OUT
BIDIRECTIONAL SHIFT REGISTER
COUNTERS IN SHIFT REGISTER
COUNTERS
COUNTERS
SYNCHRONOUS COUNTERS
4 BIT SYNCHRONOUS UP COUNTERS
4 BIT SYNCHRONOUS UP COUNTER OPERATION
4 BIT SYNCHRONOUS DOWN COUNTER
APPLICATION OF SYNCHRONOUS COUNTERS
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ADDITION USING 1’S COMPLIMENT

ADDITION USING 1’S COMPLIMENT

There are three different cases possible when we add two binary numbers which are as follows:

Case 1: Addition of the positive number with a negative number when the positive number has a greater magnitude.

Initially, calculate the 1’s complement of the given negative number. Sum up with the given positive number. If we get the end-around carry 1, it gets added to the LSB.

Example: 1101 and -1001

  1. First, find the 1’s complement of the negative number 1001. So, for finding 1’s complement, change all 0 to 1 and all 1 to 0. The 1’s complement of the number 1001 is 0110.

  2. Now, add both the numbers, i.e., 1101 and 0110;

    1101+0110=1 0011

  3. By adding both numbers, we get the end-around carry 1. We add this end around carry to the LSB of 0011.

    0011+1=0100

Case 2: Adding a positive value with a negative value in case the negative number has a higher magnitude.

Initially, calculate the 1’s complement of the negative value. Sum it with a positive number. In this case, we did not get the end-around carry. So, take the 1’s complement of the result to get the final result.

Example: 1101 and -1110

  1. First find the 1’s complement of the negative number 1110. So, for finding 1’s complement, we change all 0 to 1, and all 1 to 0. 1’s complement of the number 1110 is 0001.

  2. Now, add both the numbers, i.e., 1101 and 0001;

    1101+0001= 1110

  3. Now, find the 1’s complement of the result 1110 that is the final result. So, the 1’s complement of the result 1110 is 0001, and we add a negative sign before the number so that we can identify that it is a negative number.

Case 3: Addition of two negative numbers

Example: -1101 and -1110 in five-bit register

  1. Firstly find the 1’s complement of the negative numbers 01101 and 01110. So, for finding 1’s complement, we change all 0 to 1, and all 1 to 0. 1’s complement of the number 01110 is 10001, and 01101 is 10010.

  2. Now, we add both the complement numbers, i.e., 10001 and 10010;

    10001+10010= 1 00011

  3. By adding both numbers, we get the end-around carry 1. We add this end-around carry to the LSB of 00011.

    00011+1=00100

  4. Now, find the 1’s complement of the result 00100 that is the final answer. So, the 1’s complement of the result 00100 is 110111, and add a negative sign before the number so that we can identify that it is a negative number.