Implement the circuit in the following.
Question
Implement the circuit in the following. And We want to add the following numbers
A = 10110 and B = 10001
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- What is a circuit that can be used?
- Implement this circuit
Summary
In this, we are going to add two numbers and will draw the circuit. So we will see how to implement the circuit. Numbers are in the form of Binary numbers.
Explanation
If we want to add Binary data we can add it with an adder. And there are two types of adders. And they are Half Adder and Full Adder.
Half Adder will take 2 inputs and then it will perform addition. And After that, it will give us two outputs i.e. sum and carry.
In full Adder. There are 3 inputs and it will perform addition. And then it will give us two outputs. So They are Sum and carry.
The truth table for Full Adder is:
| A | B | Cin | SUM | Cout |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
From this truth table. There is an expression of output and i.e.
Sum = A XOR B XOR Cin
Cout = A.B + Cin(A XOR B)
To add 5-bit binary data, we require 5 full adders.
And to The implementation of the required circuit is:
Since the value of A and B are given
We have values for A0=0, A1=1, A2=1, A3=0 and A4=1.
And B0=1, B1=0, B2=0, B3=0, B4=1.
We get our output as S4, S3, S2, S1, S
Ripple carry adder is not useful here. Because ripple-carry addition does not support the addition of more than 8-bits. Because it takes more time duration to calculate. And also it will delay the output. Because there are many numbers to carry and pass.
Also read, For these situations, just design the h files you need to implement the classes you’d need.