 Digital Electronics NUMBER SYSTEM BINARY CODES BOOLEAN ALGEBRA K MAPS COMBINATIONAL CKT INTRODUCTION ADDER FULL ADDER(FA) FA using HAs BINARY ADDER SERIAL ADDER PARALLEL ADDER CARRY LOOK AHEAD ADDER (CLA) QUESTION (BCD to Excess-3 using ADDER) SUBTRACTORS FULL SUBTRACTOR FS using HSs SERIAL SUBTRACTOR PARALLEL SUBTRACTOR SUBTRACTION using ADDER 4-bit ADDER & SUB. in a SINGLE CIRCUIT COMPARATORS 2-bit COMPARATOR HIGHER COMPARITOR from LOWER COMPARATORS QUESTION (10-bit using 4-bit Comparator) DECODER FA USING DECODER HIGHER DECODER from LOWER DECODERS DEMULTIPLEXER ENCODER QUESTION (Octal to Binary Encoder) MULTIPLEXER(MUX) HIGHER MUXes from LOWER MUX Implementation of BOOLEAN FUNCTION using MUXes-I Implementation of BOOLEAN FUNCTION using MUXes-II QUESTION (Implement function using MUX) QUESTION (Implement function using MUX) Implementation of GATES using MUXes BINARY to GRAY converter GRAY to BINARY converter PARITY GENERATOR(4-bit message) PARITY GENERATOR(3-bit message) MORE QUESTIONS Q1 (Timing Diagram) Q2 (Timing Diagram) Q3 (Implement equation using Half Adder) Q4 (Error in 2 to 1 MUX) Q5 (Palindrome Circuit) Q6 (Implement function using MUX & ADDER) Q7 (Implement function using ADDER & MUX) Q8 (Implement function using ADDER & MUX) Q9 (4 to 1 MUX using 2 to 1 MUX) Q10 (Implement ALU using MUX & ADDER) SEQUENTIAL CIRCUITS TIMING CIRCUITS

COMPARATORS

Here we’ll be designing circuits to compare different binary numbers.  Suppose we have two numbers A & B at the input and 3 output as A>B, A=B, A<B and only one of the three outputs would be high accordingly if A is greater than or equal to or less than B.

1-bit comparator: Let’s begin with 1 bit comparator and from the name we can easily make out that this circuit would be used to compare 1 bit binary numbers. If we list all the input combinations at the input then we get the following table describing the corresponding outputs.

A             B             f (A>B)  f (A=B) f (A<B)

0              0              0              1              0

1              0              1              0              0

0              1              0              0              1

1              1              0              1              0

And now we find the equations using K-maps each for f (A>B), f (A=B) and f (A<B) as follow: 