[SOLVED] [Moved]: Problem Constructing 4-Bit Adder Using IC 74LS83A

Connection Problem: IC 74LS83A 4-Bit Binary Full Adder

Set-up Description

I am a volunteer in schools trying to create an electronics project for Middle School children titled "How A Computer Really, Really, Really Works," and am having difficulty getting the IC 4-bit adder to work. The setup I am using and the problems I am having are described below. I attached a picture of the setup and included a link to the Motorola 74LS83A data sheet. Any explanations, hopefully at a very fundamental level, would be appreciated.

Four breadboards are connected with the + power rails connected via red cables and the – power rails connected with black cables. I used a multimeter to verify that all the power rails have power. I used two 4-bit Dip Switches (left for A input and right for B input) which were inserted into the E8-11 and E14-17 breadboard positions, respectively, and terminate on the F8-11 and F14-17 positions, respectively. I used a multimeter to verify continuity when the DIP switches are in the closed positions. The two DIP switches (input A and B) are connected to the IC pins as specified in the Motorola Connection Diagram which is attached; specifically: A1-Pin 10, A2-Pin 8, A3-Pin 3, A4-Pin 1, B1-Pin11, B2-Pin7, B3-Pin4, B4-Pin16.

There are five LED’s: Led1 (left most LED) is the one’s place, LED4 is in the 8’s place, and LED5 is the C4 Carryover place. The four ∑ output pins are connected using dark blue wires. The C4 (Pin14) is connected to LED5 with a white wire. The anode leads of each LED are connected to the IC pins as follows: LED1-Pin9, LED2-Pin6, LED3-Pin2, LED4-Pin15 and LED5-Pin14. Pin13 (C0) pinout is connected to the negative power rail with a black wire, as there is no carry-in value. Pin5 (VCC) is connected to the + Power Rail with a red wire. Pin12 (GND) is connected to the – Power Rail using a black wire.

The 4-bit DIP switches are in series with 100K Ohm resistors which in turn are connected to the + Power Rail. There are 10K Ohm resistors in series with the cathode side of the LED’s.

A Power Supply powers the circuits. The Power Supply delivers a constant voltage of 4V and maximum amperage of .3 mA. The emitter-collector circuits are lighting the LED’s and the measured amperage is .15 ma. If there were a circuit through the base pins, it would measure 39 micro-amps (I measured it via a multimeter with a simple circuit not through the IC’s).

I am not currently using an IC carrier to mount the IC onto the breadboard. This was ordered; however it will not be received from China for another 2-4 weeks. However, a multimeter continuity test verifies a good connection for each of the 16 IC pins to the breadboard. I have not used anti-static wrist straps.

Problem

When I turn on the power supply and all DIP switches are in the “off” position, all LED’s light. There is no amperage when measuring current going into any of the input A or B pins of the IC. There are no changes when I slide any of the DIP input switches (inputs A and B) to the “on” or “off” positions.

I have ordered two batches of 74LS83A and the results are the same.


A picture of the setup is attached. A pdf file of the 74LS83A is attached. Note this is a 4-Bit Binary Full Adder with Fast Carry.

A link to the Motorola data sheet for the 74LS83A is below:
http://pdf.datasheetcatalog.com/datasheet/motorola/SN74LS83D.pdf


Questions

1. Can you see any reason why I am getting incorrect results (i.e. all the LED’s are lit all the time)?
2. Can you suggest any way I can further test the IC with just a multimeter? I do not have access to an automated IC test system?

Re: Problem Constructing 4-Bit Adder Using IC 74LS83A

How A Computer Really, Really, Really Works

Click to expand...

Your adder circuit is an excellent demonstration tool for this.

A picture of the setup is attached.

Click to expand...

Didn't find this. (Did find the list of datasheets for 74LS83.)

There is the 1-bit adder circuit, of course, if you want a basic circuit which will give you less trouble. Guess you'll need different IC's to demonstrate it, however.

Re: Problem Constructing 4-Bit Adder Using IC 74LS83A

I suspect the resistors are the problem. These are my ideas:

1. drop the resistors in series with the LEDs down to 1K, at the moment the LEDs are only going to pass < 1mA.
2. if the banks of switches are wired to 5V (closed = logic high), you need pull-down resistors of around 4.7K to ground at the IC inputs.

I think what is happening is the inputs to the ICs are permanently high. If they are not actively pulled to logic low state, they internally assume the input is high. The ICs are assuming their inputs are '1' all the time and operating the switch is just providing a few microamps more to reinforce the '1' state.

Try wiring the inputs like this:
each input pin goes through it's switch directly to 5V and each input pin also has a 4.7K resistor to 0V. You might be able to re-use those 10K resistors in place of 4.7K but I wouldn't guarantee the value is low enough. Let me know if that works.

Brian.

Brian.

Re: Problem Constructing 4-Bit Adder Using IC 74LS83A

You say:

The 4-bit DIP switches are in series with 100K Ohm resistors which in turn are connected to the + Power Rail. There are 10K Ohm resistors in series with the cathode side of the LED’s.

Click to expand...

So it seems that with the switch open the corresponding input is left floating. It just so happens that floating TTL inputs will drift positive ('1'). The chips read that as all-ones and add them quite correctly. Since you measure from the positive rail to the already positively drifted input there is no p. d. to create a current.

What you need to do is connect 1k0 resistors from each input to ground, and the switches from Vcc to inputs. Alternatively connect the resistors to Vcc and the switches to GND, considering switch open as '1'.

Also, the LED ballasts are too high, reduce them to generate the best current for your LEDs.

Re: Problem Constructing 4-Bit Adder Using IC 74LS83A

BradtheRad said:

Your adder circuit is an excellent demonstration tool for this.



Didn't find this. (Did find the list of datasheets for 74LS83.)

There is the 1-bit adder circuit, of course, if you want a basic circuit which will give you less trouble. Guess you'll need different IC's to demonstrate it, however.

Click to expand...

I uploaded the file for a .jpg picture of the setup. I hope you can see it this time. Unfortunately, I am new at edaboard.com and am not sure how everything works.

Prior to using the IC for the 4-bit adder, the kids will use resistors, transistors and LED's on breadboards to create AND, OR and NAND gates. Then, they assemble thes logic gates to make Half-adders and Full-adders. After that is done, they go to the 4-bit Adder IC.

Regards,

Reid

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I will try to add the image of the setup again.

Prior to using the IC, the kids will use transistors, resistors & LED's to create AND, OR and NAND logic gates. then, they will assemble these logic gates into Half-adders and Full-adders. Only after creating a 1-bit Full-adder, will they start using the IC.

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Sorry, I am having problems using this system. I will reply, again, for the third time. Please forgive me it you have received all 3 replies.

I will attach the jpg file of the setup.

The kids first use transistors, resistors and LED's to create AND, OR and NAND logic gates. Then, they assemble the logic gates into half-adders and full-adders to create a 1-bit full adder. Only, then, do they start the IC. The final stage (if I live that long, I'm an accountant!), they will link two 4-bit Adder IC's to make an 8-bit adder.

We can see the picture now.

It is wired as you described but unfortunately that is wrong! Wire as I suggested earlier and it should work fine.

Connect the switches directly (no resistor) from the input pins to +5V and from the input pins wire a 4.7K resistor to ground. That will ensure the voltages at the logic inputs to the IC meet the required levels for it to recognize.

Also reduce the value of the resistors in series with the LEDs to 1K so they can pass more current. There is plenty of spare current capacity at the output of the IC so there is no need to be so restictive and it will make them much easier to see. Even with 1K resistors they will only pass a meagre 3mA at most.

Brian.

Connect the switches directly (no resistor) from the input pins to +5V and from the input pins wire a 4.7K resistor to ground. That will ensure the voltages at the logic inputs to the IC meet the required levels for it to recognize.

Click to expand...

74LS input current is 0.4 mA, 4.7K won't achieve defined low level, instead a pull-down series resistor must be 1K or below, as mentioned by ydtech.

Normally, you would connect a switch without any series resistor to ground, instead use a pull-up resistor.

My apologies - change my suggestion of 4.7K to 1K as FvM and ydtech suggested. Alternatively you could use 74C or 74HC families instead.

I tried to keep it simple by using positive logic as it is a training aid but as FvM rightly points out, in a producton design it would be normal to to do it the other way around. You can use pull-up resistors if you prefer and wire the switches to ground but turning the switch 'on' will make the logic level = 0 which may at first be counter intuitive.

Brian.

I made the changes as suggested and am making progress. See the picture of the setup which is attached. I changed the resistors in series with the LED’s to 1KΩ and changed the resistors associated with the inputs to 4.7KΩ. The 4.7KΩ resistors are connected to ground. I place small squares of paper between each of the 4.7KΩ resistors to prevent shorting between the resistors. The only thing that has me concerned is, when the Dip switches are open, I get continuity between all the input circuits. I think this is because all the resistors have a common ground.
The results are not as expected. The results as shown in the LED’s are not correct. However, the LED output appears to increment or decrement when Input B is changed. Changes to Input A appear not to affect the LED outputs. Any idea what is happening? Any other ideas on how I can get accurate addition results? I show the results in the table below.

The symptoms suggest it still thinks some of the inputs are high when they should be low. Don't forget that in a real computational circuit the inputs would actively be driven by other devices, here you are trying to simulate them with resistors and switches.

Try changing the 4.7K resistors to 1K and also check you have the C0 pin at 0V or it will try to add an extra 'carry' bit to the output.

You should never test for continuity between inputs - at least not with a normal continuity tester as they pass enough current to damage the IC. Instead, measure the voltage at the input pins, it should be 5V if the switch is closed and almost 0V when the switch is open. Be careful though, if an input is not connected, it may appear to have low voltage on it when measured but still be 'seen' from insde the IC as being high, you actually have to draw current from the pin to be sure it really is logic low.

Brian.

Try changing the 4.7K resistors to 1K

Click to expand...

Or flip the connections, resistors to +5V, switches to ground, the standard TTL way.

Per the datasheet, guaranteed LOW input is 0.7V or less. In the test circuit they use 0.4V with current draw of -0.8mA (meaning drawing current *out* of the pin). Supposing that current, the voltage drop at 4k7 is 3.76V, but probably pretty undefined in practice. Anyway, too high. For a 0.4V drop at recommended current the pull-down is at most 500 ohms, make it 470 to be sure. I recall seeing 1k suggested, but I'd rather you look it up.

TI has documents on how to use each family, look it up. Remember that at board level everything is analog, digital is just a convenient representation.

The setup appears to now work!!! I replace the 4.7K ohm resistors with 1K ohm resistors, and verified the voltage on the C0 pin was 0V. Thank you all for your help!

The setup using an IC appears to operate differently from the logic gates and Half- and Full-Adders I constructed using 2N3904, NPN transistors. The overall polarity seems to be different and they drift to a "high" output. In simple terms that I can understand and explain to the kids, what are the main differences between the two systems?

Regards,

Reid

The technologies are totally different but the polarities should be the same. Anyway, both TI and OnSemi have lots of documentation in the TTL section, including schematics of inputs and outputs. TTL uses a special kind of transistor with multiple emitters which can't be replicated with discretes. What you're doing with discretes is likely to be RTL, and a floating (unconnected) base will be read as 0.