If you need to quickly add numbers from 0 to 15, and you know how to rapidly convert to binary and back to decimal, this is for you...But if your human then this is just a fun project!!  I'm going to show you how to make a 4 bit (0-15) adding calculator using 74xx series IC chips.

The chips that I used are the very basic gates like OR, AND, XOR, NOR, NAND, etc.  No actual adding chips are used so its one level up from transistors, and it's totally TLL!**

By the way, this is another addition to the unofficial "Macroelectronics" catagory for all of you who have seen my diy ram instructables!

**If you use the 74hc series or 74hct series, then its technically not TLL, but who cares about the techincal details!!

Step 1: The Materials

This project only requires a few parts:

1 Breadboard, maybe 2 - It should have atleast 50-60 rows
2 74xx AND chips - I used the 74hc08
2 74xx XOR chips - I used the 74hc86
1 74xx OR chip - I used the 74hc32
1-2 DIP Switches - I used 2 8 switch DIP switches (I'm gonna expand to 8 bits)
8 1K ohm -or greater- resistors
Lots of wire - Don't yell at me but I use bare copper! Its a bad habit, (some may argue that it's a dangerous one) I know, but I don't care!

<p><strong style="">Connection<br>Problem: IC 74LS83A 4-Bit Binary Full Adder</strong></p><p>I created a 4-bit Adder using IC 74LS83A</p><p><u>Set-up Description</u></p><p>Four breadboards are connected with the + power rails connected via red<br>cables and the &ndash; power rails connected with black cables. I used a multimeter to verify that all the<br>power rails have power. I used two 4-bit<br>Dip Switches (left for A input and right for B input) which were inserted into<br>the E8-11 and E14-17 breadboard positions, respectively, and terminate on the<br>F8-11 and F14-17 positions, respectively. <br>I used a multimeter to verify continuity when the DIP switches are in<br>the closed positions. The two DIP switches<br>(input A and B) are connected to the IC pins as specified in the Motorola<br>Connection Diagram which is attached; specifically: A1-Pin 10, A2-Pin 8, A3-Pin 3, A4-Pin 1,<br>B1-Pin11, B2-Pin7, B3-Pin4, B4-Pin16.</p><p>There are five LED&rsquo;s: Led1 (left most LED) is the one&rsquo;s place, LED4 is<br>in the 8&rsquo;s place, and LED5 is the C4 Carryover place. The four &sum;<br>output pins are connected using dark blue wires. The C4 (Pin14) is connected to<br>LED5 with a white wire. The anode leads<br>of each LED are connected to the IC pins as follows: LED1-Pin9, LED2-Pin6,<br>LED3-Pin2, LED4-Pin15 and LED5-Pin14. <br>Pin13 (C0) pinout is connected to the negative power rail with a black<br>wire, as there is no carry-in value. Pin5 (VCC) is connected to the + Power Rail<br>with a red wire. Pin12 (GND) is<br>connected to the &ndash; Power Rail using a black wire.</p><p>The 4-bit DIP switches are in series with 100K Ohm resistors which in<br>turn are connected to the + Power Rail. <br>There are 10K Ohm resistors in series with the cathode side of the<br>LED&rsquo;s. </p><p>A Power Supply powers the circuits. <br>The Power Supply delivers a constant voltage of 4V and maximum amperage<br>of .3 mA. The emitter-collector circuits<br>are lighting the LED&rsquo;s and the measured amperage is .15 ma. If there were a circuit through the base<br>pins, it would measure 39 micro-amps (I measured it via a multimeter with a<br>simple circuit not through the IC&rsquo;s). </p><p>I am not currently using an IC carrier to mount the IC onto the<br>breadboard. This was ordered; however it<br>will not be received from China for another 2-4 weeks. However, a multimeter continuity test<br>verifies a good connection for each of the 16 IC pins to the breadboard. I have not used anti-static wrist straps.</p><p><u>Problem</u></p><p>When I turn on the power supply and all DIP switches are in the &ldquo;off&rdquo;<br>position, all LED&rsquo;s light. There is no<br>amperage when measuring current going into any of the input A or B pins of the<br>IC. There are no changes when I slide<br>any of the DIP input switches (inputs A and B) to the &ldquo;on&rdquo; or &ldquo;off&rdquo; positions.</p><p>I have ordered two batches of 74LS83A and the results are the same.</p><p>A picture of the setup is attached. <br>A pdf file of the 74LS83A is attached. <br>Note this is a 4-Bit Binary Full Adder with Fast Carry.</p><p><u>Questions</u></p><ul><br><li>1.Can you see any reason why I am getting incorrect<br>results (i.e. all the LED&rsquo;s are lit all the time)?<li>2.Can you suggest any way I can further test the<br>IC with just a multimeter? I do not have<br>access to an automated IC test system?</ul><p></p>
Can you privide binary subtractor circuit??
<p>Sorry I'm 8 months late but if you are still interested: To substract with an adder, add 4 XOR gates and connect their outputs to the B input of every 4 full-adders. Then the B inputs of each of your 4 adders will become one of the XOR gate input. Then connect the carry in and every other XOR input to a switch. That switch should toggle between add and subtract.</p>
Can you add a full circuit schematic for reference.
<p>Can you just give the 4 bit logic circuit of addition , subtraction and multiplication that if user wants to muliply or add numbers on run time, they can do it just like simple calculator???</p>
so these gates are each quad versions, which means if I wanted to do this completely with discrete components, I would need to make 4 pf each gate per chip, right?
perfect project, no other word
<p>Ok I got it !!!</p>
<p>did you do this project as it is above given circuit</p><p>then how to give inputs and check for outputs</p>
<p>how to check the output</p>
<p>how to check the output</p>
<p>Hey man i want to build this one but I couldn't understand which wire goes where ... so I am asking you for an on board instructions ( what you wired where and to ic which leg ) if you can give a diagram of that 1(2) breadboard ! please ? If you can :) It'll be a great thing for people like me :) thank you :) (Y)</p>
<p>I used another complicated system of using 555 timers and 4017 ICs to convert binary into decimals...</p>
how could I connect this to an LCD that would tell me the value instead of LEDs so it would actually read me the sum in binary?
I, personally, prefer the 4000 family of ICs for these types of projects over the 7400 family. The 4000 family has a wider operating voltage range of 3(?)V - 15V. And at 9V, the outputs can push enough current to drive LEDs directly (just remember the resistors). And to answer my month old question, yes you can start with a half adder.
By &quot;TLL&quot;, do you mean &quot;TTL&quot; (Transistor-Transistor Logic)? But I do have one question I think I have the answer to: since we do not have any carry bits in the very first part of the adder, could we start off with a half-adder? I don't see why not, as it would always be low anyways, but I just want to be sure.
oh ok. gotya. <br>I have one bit put together right now and i'm pretty sure the schematic matches yours, but I'm having a good deal of trouble and seem to be inducting power into it or something when i bring my hand close or something. I end up with something like: <br> <br>0 + 0 = 1 <br>0 + 1 = 10 <br>1 + 0 = 10 <br>1 + 1 = 11 <br> <br>any ideas?
Do you need that OR gate? ether one or the other AND gates has an output so you could just have both the AND's output go right to the carry out, right? <br>also, if you powered a input with Vcc, would it fry? I'm not quite understanding, but nice job!
Yes the OR gate is necessary and no powering an input with Vcc won't fry it. Let me explain: The chips use <a href="http://en.wikipedia.org/wiki/CMOS" rel="nofollow">CMOS logic</a>. CMOS has to important properties: 1) inputs are voltage dependent not current dependent (no current flows into or out of the input); 2) outputs source current when high and sink current when low. The first property means that even if you connect an input to Vcc, no current will flow. The second property means that if you have two outputs connected together and one is high and the other is low, then a ton of current will flow from one to the other, frying them both. The solution is to put an OR gate there so that both of the outputs are connected to inputs that do not sink or source current.<br> <br> I hope this clears things up!
Where can you actually buy 74hc chips? <br>
The 74hc chips can be found at various electronics suppliers such as Mouser, Digikey, or Jameco.
May I ask you which battery terminals have you connected to the vcc and gnd points on the chips? <br>Also I did not understand the way you connected the dip switches. <br>Sorrry if I sound a little dumb but this is my first electronics project with transistors. <br>
Hey I tried making this project but when I switch on the dips to make inputs the current bypasses the entire circuit. <br>P.S. <br>can you please tell me what is your supply voltage, <br>I used two 9v batteries in series. <br>P.P.S. <br>When I dont have a input all the LEDs glow even if all the dip switch keys are open.
I think your supply may be the issue. According to the datasheets, the chips have a maximum supply of 6v. two 9v batteries gives 18v which is way more than enough to fry the chips. This is probably why the LEDs are glowing, the chips are dead. The best thing you can try is replacing all of the chips and lowering your supply voltage. I have a converted computer PSU that gives stable 5v but a 2 or 4 battery pack would work. Just keep your supply below 6v. <br> <br>Good Luck!
hey i have all the same parts as you except my or gate is a 74LS32 and i cant get it to work when i test it so any trouble shooting or things i need to know when using gates? ps its not a bad gate because i have tried 2 of them now <br> <br>
I think that might be the problem. 74hc is CMOS logic levels whereas 74ls is TTL logic so there are incompatibilities when interfacing between the two. If you want, you can do a little more research on the issue, but what you need to know for this is that 74ls won't work with 74hc.<br><br>I believe that 74hct can work at both CMOS and TTL levels.
You could make this subtract by putting a signal into the carry-in of the first and unused carry-in input and then inverting your b value. e.g.: 1011=0100. That is how I made the calculator I built in minecraft to be able to also subtract. When you do the subtract thing, the carry-out of the last adder will turn on but will not mean anything, you could put a switch there to turn it off while subtracting.
sorry its been awile but a npn transitor would work very well with a 100k resistor on its base and with that combo the data could last maybe ten seconds which is more than enough. i dont quite know quite how to show the the circuit.
I dont mean to brag but I happen to have a very wide knowledge of the inner workings of computers, logic gates, and computer arithmatic and data flow and i was wondering I already know how to do it but i dont really have the time. could you make an instructible on making a byte of d-ram memory? if you want to make or have questions please leave a comment to talk to me.
Awesome! I actually started some work on that and another project after I published the results of my SRAM project. I've been working on a dram cell and a 555 timer circuit. I developed a fairly reliable long tail differential comparator for both and I'm in the process of simulating the circuits. I don't know if I can do a full byte, but I would be happy to attempt a few bits.
its actually very simple for just one byte just 16 n channel mosfets, 8 small value capacitors 0.1 or 1 uf, and 8 comparators. since there is only one byte it does not need a transistor per cell because that is just for selection. how long the data will last is based on the size of the capacitors and the amount of current drawn into the comparator. so a regular comparator will work fine but a comparator with low input current draw would work the best.
Actually, the mosfets have reverse current body diodes in them, so any transistor that will have current going both directions will need two mosfets per gate. (32 total mosfets) I would then need 3 bjts per comparator (2 pnp and an npn). It's going to be large and I don't have the room without ripping up some other circuits I've been working on.<br><br>No offense, but why do you want me to make a full byte? Personally, I would think a few bits or a nybble would suffice.
actually it just need to amplify a signal so you could just use an npn transistor with a 1 k resistor on the input no pull up resistor required. but yes a byte is sorta large so how about a nybble? also there is one thing i missed to tell you an OR gate per bit is also required but you can just used two resistors ten k would be ideal.
The problem with a single npn is that if 3 bits are 1 and one is 0, there is going to be leakage to that one. If the voltage is high enough to be amplified enough to be a 1 (&lt;.5 volt is needed) With a comparator, however, the threshold could be 2.5v (5v supply) so anything less is a 0 and anything over is a 1. It's a less temperamental solution, and a bit of a personal preference. Besides, real dram uses a comparator, but using an IC just doesn't sit well with the spirit of it, for me at least.<br><br>A nybble will definitely work, but if I have the room, I will try a byte too :)
3V to 5V should work fine. 2 or 3 AA will do it, or if you can get power from USB that is regulated 5v.
I cant seem to find a XOR GATE with the model number 74HC86 on jemeco thats an OR gate.
Are you sure? 74HC86N - Quad 2 Input Exclusive OR Gate??
oooo whoops sorry bout that!!!!!!!!!
Damn! After seeing your DIY RAM instructables I was going to do just this myself. Oh well, beat to the punch. Maybe I'll post a more complete version, with subtraction and such. Good job.
Thanks man! Yeah, I have the entire schematic for an 8-bit adder/subtractor (number range 127 to -128) I just need another breadboard and a few hours of free time (the hard part) and I can start. But if you beat me to it, good luck to you!!
Could I possibly use your schematics for the 8 bit I would like to build that!!!!<br><br>And also correct me if i'm wrong but this requires NO programmer right????
Yeah, just give me a day or to to draw it up on the computer. I have it all in a sketch book, so it should be only like a day's wait. <br> <br>And no, it requires no programming what so ever, it's just logic chips and wires!
Thats great that theres no programmer!!!!!<br>Im seriously excited to build this!
These are the pics that I took of the sketchbook. I hope you can read them. <br> <br>Basically, the first picture is the 2's complementor. So the Bx input (x being the bit) would go into that to be made negative (or positive if already negative). <br> <br>The second picture is of the add/subtract selector. The Bx input is the normal, &quot;un-negatized&quot; B input, and the /Bx is the &quot;negatized&quot; input. When the select line is pulled high, the calculator adds, and when it's pulled low, the calculator subtracts. The ADDx outputs are connected to the Add B input. You may want to put diodes on the outputs before connecting them together and then you will want to tie the line low through a 10K ohm resistor where both outputs are connected together. <br> <br>The third x inputs come from the selector's ADDx outputs. COi is the Carry In which is connected to the last bit's COo, which is the Carry Out. On the LSB, the COi is tied to ground. <br> <br>The Last picture is of the gate and chip count. Yes, you will need 10 AND Chips! <br> <br>Good luck and Have fun!!
Very nice design thank you so much for taking time to take the photos.
Yeah, no problem. I would like to see it, so when you finish it, send some pix! Or better yet, make an 'ible!
I would be happy to!!!!!!!

About This Instructable


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Bio: I like to make things that move, sense, calculate, compute, blink, and make noise. I like making things that create high voltages, electrical arcs, and ... More »
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