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Numbers comparison circuit (2 - bit)

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**Author: **
Abhinav Deshpande

**Project access type: **
Public

**Description: **

This is a very simple number comparator circuit.

You input 2 numbers and it gives an output

whether **A > B** (both are 2-bit numbers)

(A should be strictly greater than B)

Note that A and B are both 2-bit **binary **numbers. Binary meaning that they are represented with only **1**s and **0**s.

Remark: Even if you don't know about binary numbers, I have got you covered :)

Simply refer to the table given in the circuit to work out the input yourself.

Change the first input number labelled as A1 A0 by toggling the input. Similarly change the second number. Observe the LED for each combination of 2 numbers.

You will see an LED glowing if A > B. Otherwise, you won't see any output.

eg. If you input A1A0 as 11 and B1B0 as 10, the LED would glow.

Or rather you would ask, how can I design such a circuit myself? Well I will explain. So this is called a combinational circuit. You can design your circuit by first making a truth table with inputs as your two 2-bit numbers (4 inputs in total) and thinking about output for each set of inputs. So, you would have 4^{2} =**16 total inputs **in your truth table.

Next, after you have mapped your input and then your output for all those inputs, you would be solving a logical expression for output **Y **for a given set of 4 inputs **A _{1}A_{0} B_{1}B0**

You can use any method to solve for Y. You can use a K-map based reduction or you can simply solve using basic logic identities.

Once you get an **expression (expression **means** **something like this say (for eg. this is not what you may get) Y= A1.B1'+B1'B0'+...**)**

, you would then realize the circuit using logic gates. Just as I have done.

Increase the input to 3-bits , 4-bits and so on...

**Created: **
Sep 22, 2023

**Updated: **
Sep 22, 2023

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