Project Overview
The purpose of the circuit I designed in this project was to determine the outcome of a vote between four representatives of a company. In the case of a tie, the side that the President of the company votes for always wins, and some constraints on this project were that you must use two-input gates, and that the final design must use as few gates as possible.
Problem Conception
![Picture](/uploads/8/9/6/8/89689385/20181010-134243_orig.jpg)
This table shows that the circuit has four inputs and one output, which are the President's vote, the Vice President's vote, the Secretary's vote, and the Treasurer's vote, with the decision as the output. A zero is placed in the table if someone does not vote or the outcome is to not pass the decision, and a one represents a vote or a passed decision. The purpose of this table is to easily visualize what set of input will produce an output of 1, and this information can then be turned into an expression that conveys the information in a more compact way. It should also be noted that the number of variables determines the number of rows through the function (2^x) = R, where X is the number of variables and R is the number of rows.
The un-simplified expression for this truth table is (NotP,V,S,T+P,NotV,NotS,T+P,NotV,S,NotT+P,NotV,S,T+P,V,NotS,NotT+P,V,NotS,T+P,V,S,NotT+P,V,S,T) = D
This expression and it's minterms were derived from the truth table by combining all the states of the variables whenever there was an output of 1. This is in SOP form because that is the standard as well as easiest way to read a logic expression.
This expression and it's minterms were derived from the truth table by combining all the states of the variables whenever there was an output of 1. This is in SOP form because that is the standard as well as easiest way to read a logic expression.
The Un-Simplified Circuit
There are thirty-five gates in this un-simplified circuit, which means you would need one inverter chip, (74LS04), six 'and' chips (74LS08), and two 'or' gates(74LS32). This simulated circuit is in bus form
Boolean Algebra Simplification
The original un-simplified expression is at the top of the work page and the simplified expression is at the bottom. Inbetween is the work I did to get from one to the other.
The Simplified Circuit
There are five AND gates and three OR gates, meaning the circuit will need two 74LS08 AND chips and one 74LS32 OR gate. I determined this by knowing that the chips contain four gates each, so I could divide the number of a specific gate type by four and round any decimals up to find the chip numbers. Lastly, the 330ohm resistor before the LED decreases the voltage of the current passing through the light in order to stop the LED from being damaged.
This simplified circuit is important because it is much more efficient and cost effective than the original circuit. This is because it contains much less gates, 8 versus 35 and thus requires less chips, 3 versus 9, as well.
This simplified circuit is important because it is much more efficient and cost effective than the original circuit. This is because it contains much less gates, 8 versus 35 and thus requires less chips, 3 versus 9, as well.
Majority Vote Circuit Build Materials
This table shows what components are needed, and how many of them are needed, for the simplified circuit. Its purpose is to easily show what parts are necessary to create the circuit without the need for anyone trying to replicate it to do any math or Boolean algebra.
Bread-boarding
Picture one shows my breadboard with the basic necessary connections for the circuit to work at all, like the wires grounding and powering the attached switch board and the wires grounding and powering the gate chips. The positive and negative sides of the bread boards are also linked so both buses can be utilized.
Picture two shows the wiring for some of the gates being completed.
Picture two shows the completed circuit with gates all completed, as well as the LED and resistor placed on the board.
Picture two shows the wiring for some of the gates being completed.
Picture two shows the completed circuit with gates all completed, as well as the LED and resistor placed on the board.
In my first bread boarding experience I would simply wire gates wrong, putting the input where they shouldn't be, or not powering the chips properly. I improved since then because this time I almost made the circuit work on the first try except that I had all of the input wires connected to the LEDs instead of the switches, so I only had to move them over and it functioned perfectly. Through this project I learned the skill of wiring with logic gates and chips, the troubleshooting I did was checking the input wire placement before anything else.
Conclusion
IIn the end I succeeded in creating a circuit which accomplished the task of creating the voting system for the company. The most important take away I found from this project is that it is extremely important to simplify a circuit expression as much as possible or else you may be wasting time and money producing over-designed circuits. This also shows the usefulness of Boolean algebra since that was what allowed for the extreme simplification of the original circuit, saving costs and the ease of making the circuit. The basic process of going from a problem statement to a finished circuit design is, first, you derive a truth table from the information the procedure gives you, and then you turn that into an un-simplified circuit expression. From there, you may draw out the un-simplified circuit to visualize it and compare it to the future simplified circuit. Next, use Boolean algebra to simplify the circuit expression as much as possible, making sure to have its final form in Some of Products form. Lastly, take this simplified circuit expression, visualize it with software or other means in order to test it and compare it to the un-simplified circuit before finally determining how many of what type of gate chips are needed for the final circuit. Now you can recreate the circuit in real life.