adding bytes with 4-bit full adders and getting the o/p on 7 segment(s)

[Duellist]

hi guys..........
i am still a student and dats the first time to take digital..thought to introduce my self first...
anyway i was asked to use 4-bit full adder(74283 IC) to add 1byte +1byte and have the output on Seven segment(s)...
anyway i passed the stage of connecting the 4-bit full adder on paralel but i faced another problem which is i would have 8 output and i didnt know wat kind of decoders to use or how to connect my seven segment ...

i was wondering if u guys would help...
thx alot

-hussein

 

[ericgibbs]

hi ,

If you are adding a 4 bit byte to a 4 bit byte, how can you have 8 outputs?

Consider:
byte #1 say its 1111
byte #2 1111
Sum is: carry=1 0000

If you have bytes which give a Sum without the Carry bit being set, then there are only 4 bits on the 'Sum' outputs of the adder.

You would connect the 'S' outputs to a 4bit to 7 segment decoder driver ic.

As a student you should be able to work the solution yourself.

 

[Duellist]

i think i didnt explain my self well.....
check this diagram
**broken link removed**
thats adding 4-bit full adder to be acting as 8-bit full adder

consider
byte#1: 11111111
byte#2: 11111111

i would have 0s on all 8 output and carry is one...

i think that counts as 9 different bits or 1 byte and addition bit i dont know how to deal with it

i hope i made my self clear this time

-hussein

 

[ericgibbs]

hi hussein,

In order to drive two 7 segment displays, you connect the 4 'S' pins of the least sig byte to a 7447 ic [ bcd to 7 segment driver] and another 7447 to the most sig byte. That is you will need two driver ic's.

Each of the 7 seg ouput pins will require a resistor for each LED segment, about 220R.

The 7447 has active low outputs, so would you require a common anode LED.

If you want to display the Carry bit, you will need another 7447 LED driver.

Is that clear?

 

[Duellist]

no really....

coz if we connect the 4 'S' pins of the least sig byte to a 7447 IC and another 7447 to the most sig byte... and the carry to another 7447 Led Driver...

regarding the carry led.. i will be watching a max of number 9 on each seven segment and if the out put is like more than 9 les say 11 i would be reading null on the seven segemnt..

i want when i add on
byte#1: 11111111
byte#2: 11111111

i read 254...so wat do u i have to do to read such thing..

thanks for ur effort Eric
-hussein

 

[ericgibbs]

hi,

Thats one of the problems with the standard BCD to 7 seg decoders, the 10,,15 are blanked!.

If you want to display hex numbers 0 thru F [0000 > 1111] you would need a BIN to 7 seg decoder, that would display 0 to 9 and A to F.
I have not seen that type of ic for many years.

You will require more decoding of the binary output SUM for conversion into BCD before you connect the 7447's.

The simple way would be to have a 220R on each 'S' output pin on the adders and drive a single type LED,
the result could be then read in binary.

I will look thru other options and let you know.

Other members, I am sure will also have some ideas.

If you understand let us know.

EDIT: Have a look at this site.
**broken link removed**

 

[Duellist]

i kinda got the general idea of wat u meant
but in this image
**broken link removed**
how can this be actually connected to have units tens hundreds ..coz i find it rather intreseting

but now i kinda need an illustration of how i am gonna connect the circuit and specially starting from the (2) 4-bit full adder

thanks
-hussein

 

[on1aag]

Hi Hussein,

Take a look at this datasheet, especially figure 6 at page 7.

https://www.alldatasheet.com/datasheet-pdf/pdf/102199/NSC/DM74184N.html

on1aag.

 

[ericgibbs]

many thanks on1aag,

I knew that there is such a device, but just couldn't recall it.

Regards

 

[Duellist]

thanks a lot on1aag....

but just to get it straight for the last time .....
**broken link removed**
i am gonna use 4 of the 74185 IC and each of the 4 output coming out of IC i am gonna connect it to a 7447 and then to the seven segment showing the final output...doesnt it go that way???

thanks alot guys
sorry for the continus bother

-hussein

 

[ericgibbs]

hi hussein

That looks OK.

Dont forget you will require three 7447 decoder drivers, one for each BCD digit.

Also 220R resistors on the outputs of the 7447, driving common anode 7 seg LED's.

Good luck.

 

[Duellist]

thanks alot on1aag for pulling me through

and for ericgibbs...thanks a zillion, really
i just wanted to last say that i am requested to have this project on a PCB..which is gonna be tough.....

and btw i use 330R
so now i am gonna be using lots coz for a start i am using 16 R for each input (at the switches)

-but i didnt know that i had to use resistances on the 7447 going to the seven segment...we didnt do it that way ..i think it wont matter....

anyway if i really might ask can anyone guide me through the data sheet of the seven segments common anode and common cathod

thanks again..
-hussein

 

[Duellist]

k i drew a diagram to represent the circuit for the last time...jsu to make sure again
*u must be mad at me ri8 now :d

**broken link removed**

tell me if its correct
-hussein

 

[arod]

Where on earth did you find these 74185 chips. I just finished a project converting an analog signal to a 3 digit BCD equivalent and we had to replace this very chip with an equivalent circuit. The 74185's are considered "legacy" components.

 

[Duellist]

actually i didnt...thats all planning for the project and drawing the guidlines
but wait are you serious man......are they that rare..

wat can i do now to replace them..i would really appreciate it if u helped

 

[arod]

If you have some available to you or can find some cheap, then go ahead with that circuit. I was not able to find any for a reasonable price when i searched. All the sites required you to buy hundreds of them as well and not just a handful.

I would not suggest you go down the same route as my group did as it was a very crude and inaccurate method, but got the job done. If I had the tools at the time (a microcontroller or EEprom programmer), I would buy 3 EEPROM chips that have 8 addresses that hold 4 bits for each address. It would need to be programmed in such a way that one EEPROM outputs the most siginificant digit, one outputs the middle digit, and the last outputs the least significant. I am not sure if this option is even feasible to you or not though.

If anyone else has a better method of doing this, I would love to see your reply. I know a microcontroller could be programmed to do this whole project for you, heh. It would probably also be the most cost-effective solution as well.

 

[on1aag]

Binary to BCD conversion. ( Part two )

Hi Hussein,

Take a look at page two of this datasheet:

https://www.alldatasheet.com/datasheet-pdf/pdf/15643/PHILIPS/74HCT583.html

Are we off the hook now ?

on1aag.

 

[arod]

Also, you have to use resistances going to your LED Displays. Otherwise the LED displays will draw too much current and damage the displays and your supply voltage will drop due to the high current demand of the displays.

 

[ericgibbs]

hi hussein,

>> I have not seen that type of ic for many years.[/I]

I did say, that they are as rare as hens teeth!.

If you are actually going going to build this circuit, rather than a tutorial assignment,
you may have to to build the 74185 logic using standard gates.

A PIC would be ideal for adding two binary byte and displaying BCD on LED's.

Are you allowed to have a free choice on the method used?

 

[Duellist]

hi again...

actually i dont buy my ICs off the internet..i buy them from a store
and i just called it and asked for the 74185 and he said he usually have them but they r out ri8 now...
and i remember seeing this pic
**broken link removed**
so asked him if he had 74184 and he said they r availble...

so are the 74184 like a replacment for the 74185?
coz if it does that would help

-hussein