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Creating a full adder using a 3-to-8 decoder

  1. Oct 24, 2010 #1
    I'm trying to create a full adder using one 3-to-8 decoder and some nand gates. As of now I know I will have X, Y, and C_in as my inputs. I am having trouble with figuring out what the 8 outputs of the decoder should be, so I am unsure about where and how to use the nand gates. Anyone able to give me a nudge in the right direction?
  2. jcsd
  3. Oct 24, 2010 #2
    so you want to create a full adder for the three bits? Or is one of the bits a carry?

    edit: guessing C is carry.

    Draw out the map for X, Y C, and L0 through L7 first

    Then draw out the map for X, Y and C.

    Then you can correspond the values e.g. L1 would correspond to an output of 1 for the sum and 0 for the carry. L7 corresponds to bit x added to bit y added to the carry which the output should be
    sum of one carry one

    So you feed sum into a NAND and get the COMPLEMENT of what you want since you will eventually feed it into another NAND gate
    ditto for c

    should get you started might be different from what I say just off the top of my head.
    Last edited: Oct 24, 2010
  4. Oct 24, 2010 #3
    Right. C_in is the carry in and X and Y are the two bits I want to add together....
  5. Oct 24, 2010 #4
    Scratch that last thing about getting the compliment

    anyway just realize that
    for input L0 you want an output of s=0 c=0
    L1 s=1 c=0
    L2 s=1 c=0
    L3 s=0 c=1
    then minimize the function

    I missed this

    So a 3 - 8 decoder has 3 inputs and 8 outputs. each output corresponds to a combination of the input. so there are 2^3 combinations of x,y,c there will be one and only one output for each combination.

    input of
    000 turns on the L0 line
    001 turns on the L1 line
    010 L2
    011 L3
    100 L4
    Last edited: Oct 24, 2010
  6. Oct 24, 2010 #5
    Could you be a little clearer with the first part about the maps? Should I have a K-map for inputs X Y Cin and output Cout, and then one for X Y Cin and output Sum? What exactly will my 8 outputs from the decoder be?
  7. Oct 24, 2010 #6
    Edited my last post.

    So L3 being on represents X=0 Y=1 C=1
  8. Oct 24, 2010 #7
    If you are selecting decoder outputs with a nand gate, I have to assume the selected output of your decoder is active low.

    Adding X, Y and carry_in is really just adding three bits. X_out of your full adder will be high whenever you have an odd number of input ones. So for this you need a 4 input nand gate fed with the decoder outputs generated from these inputs.
    Last edited: Oct 24, 2010
  9. Oct 24, 2010 #8
    Okay, that makes sense. I guess my big problem is the gates. Do I need to add these 8 outputs a certain way to yield one sum and one carry out?
  10. Oct 24, 2010 #9
    Make an input output chart of
    L0-L7 and Sum Carry

    Remember that one and only one output will be active at a time no matter what combination of X Y C you choose.

    Also the answer is yes you do. You need to turn 1 signal into 2 depending on the input and desired output.
    Last edited: Oct 24, 2010
  11. Oct 24, 2010 #10
    I'm sorry you are helping a lot but I am very slow with this stuff. I am not sure how to make a K-map with s and c_out with the Li's since each output of the decoder L has three components. Eh.
  12. Oct 24, 2010 #11
    no doesn't need to be a kmap no need to reduce it yet

    Code (Text):

    Input        intermediate                                   output
    x | y | c | L0 | L1 | L2 | L3 | L4 | L5 | L6 | L7 | Sum | Carry
    0   0    0    1     0     0    0     0     0     0     0     0        0
    0   0    1    0     1     0    0      0    0     0     0     1        0
    0   1    0    0     0     1    0      0    0     0     0     1        0

    You know the basic design of an adder right?

    edit "quote" this to make it easier to read the spacing is off
  13. Oct 24, 2010 #12
    Yes, I know how to make one using two XCORs, two ANDs, and an OR gate.
  14. Oct 24, 2010 #13
    oh so its just the NANDs throwing you off?

    You can make any function using NAND using boolean algerbra.

    so, if you want to make OR

    Code (Text):

    a ---          a`
            NAND ---- --------------         !(a`b`) apply DeMorgans  a+b
    a-----                                 NAND------------------------------
    b`----        b`                   |

    quote this as well
  15. Oct 24, 2010 #14
    Ok so now I have the graph and the decoder with the outputs that yield a sum of 1 hooked up with a nand gate.

    I understand how to use the gates to make other gates. What is confusing me is just taking the particular outputs of the decoder that have sums or whatever and translating that using nand gates. The whole idea of using a decoder for this confuses me since I know how to do it with just gates... the function and relationship between the decoder and the gates is what is throwing me off.
  16. Oct 24, 2010 #15
    Alright so what you want to do is

    L -> Sum, Carry
    L0 -> 0 0
    but since you can only use NAND

    L0 -> 0 0
    L1 -> 1 0
    L2 -> 1 0

    Okay I don't know if you can actually do this but I don't see why you couldn't (someone confirm?)

    write out the equation for this function in sums of products form

    and partially apply DeMorgan's
    ABC + CDF

    applying DeMorgan's we get
    !(ABC) * !(CDF)

    look familiar?
    Last edited: Oct 24, 2010
  17. Oct 24, 2010 #16
    No you are helping a lot. But I am most clueless on how to use these gates after the decoder to complete the circuit.
  18. Oct 24, 2010 #17
    OK OK this is looking familiar. I will need to formulate an equations that relates L to S and C, and then implement this with the outputs from the decoder? Will I treat L1 as 001, or just as L1? Is this just notation at this point? And also, am I just adding up the sums that go to 1?

    Also, should this k-map have the first inputs X and Y Cin also?
  19. Oct 24, 2010 #18
    Since you are using the decoder outputs you want to relate the Ls to Cout and Sum

    yes sums of products does add to 1
  20. Oct 24, 2010 #19
    Alright I was wrong, you definitely can't partially apply Dmorgans thinking it over.
    Sorry I am really tired and am making a lot of mistakes :(

    but finding POS is a way to go. If you can find that, you can simply make equivalent OR and AND gates that are needed.
    Last edited: Oct 24, 2010
  21. Oct 24, 2010 #20
    Also for so many variables you probably would like to use Quine-McClusky's or a computer.

    http://www-home.fh-konstanz.de/~voland/QMC/index.html [Broken]
    Last edited by a moderator: May 5, 2017
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