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Doubt about BCD Sum circuit using full adders

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data
    I understand this BCD sum circuit. The only thing that I'm not understanding is why last full adder carry out also triggers an 6-sum in the other part of the circuit. I mean, if the nibble is not an valid BCD number, we sum six to the number. Not valid BCD numbers are 10,11,...,15. Using K-maps you simplify this list, use an OR gate and sum six to the nibble.

    So, my question is: why the overflow is also a trigger to sum six to the nibble? And in the same manner, why invalid nibbles also represent carry out?

    In my view, carry-out shouldnt go into that OR gate. It should that be an indicator of overflow, and OR gate would be used only by the invalid nibbles.

    2. Relevant equations
    1323_block_diagram_of_bcd_adder.png
    3. The attempt at a solution

    PS: sorry my english. This is not my native language.
     
  2. jcsd
  3. Apr 22, 2014 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Try adding 15 + 1.

    Code (Text):
      1 1 1 1
    + 0 0 0 1
    ---------
     
    Of course, we know that in straight binary, this should sum to
    Code (Text):
    1 0 0 0 0
    But in binary coded decimal (BCD), the "Output carry" is the 10's numeral, and the rest of the bits represent the one's numeral. In other words, it should be an output carry '1' and the rest should be a '6' to indicate 16. So the BCD result should be
    Code (Text):
    1   0 1 1 0
    Does that make sense?

    [Edit: Of course, it appears that this particular circuit cannot output BCD numbers greater than 19]
     
    Last edited: Apr 22, 2014
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