Understanding Full-Adder Circuit

Discussion in 'Homework Help' started by jegues, Nov 22, 2010.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I'm just having some difficulty understanding the full-adder and the how the carry-in/out are obtained and what effect it has on the circuit.

    See figure attached.

    The original question was to design the simpliest circuit that could determine how many bits in a three-bit unsigned number are equal to 1.

    With the circuit posted in mind lets consider the scenario where z2=z1=z0=1.

    With this we should obtain Cout = 1 and S = 1, giving us a binary value of 3, thus indicating 3 ones were counted in the input.

    The part I don't understand is how the full adder obtains these results.

    How does z2 + z1 = s = 1, and why is Cout = 1? Also, what is the significance of the carry in? If I change the values of the carry in how does it change the output?

    If anyone could clarify it would be greatly appreciated.

    Thanks again!
     
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  2. bertus

    Administrator

    Apr 5, 2008
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  3. renotenz

    New Member

    Oct 25, 2010
    11
    0
    Well... binary numbers only consist of 1 and 0.

    The Full Adder is using the method of binary addition

    0 + 0 = 0
    0 + 1 = 1
    1 + 0 = 1
    1 + 1 = 10

    So if your input Z1 = Z2 = 1, the addition result will be 1 + 1 = 10

    Since there is a Z0 Carry In which is also 1, the result will be 10 + 1 = 11.

    Then the 11 divided as 1 for S and 1 for Carry Out.

    Another example is if you give 0 for the carry in, the result will be 10 (1 + 1 + 0 = 10), the S will be 0 and Carry Out will be 1.

    If the result is only 1 or 0, then it goes to S, and the value of Carry Out is 0.

    Cheers :)
     
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