How to make 4 bit binary adder using IC 7483?

Discussion in 'Homework Help' started by Shreyash, Oct 5, 2016.

  1. Shreyash

    Thread Starter New Member

    Oct 5, 2016
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    its my project.
    we have the following: IC7483, IC 7400(NAND), IC 7432(OR GATE), BREADBOARD, WIRES. two of each and leds.
    how to implement it?
     
  2. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    Have you read the datasheet? Which bits did you not understand?
     
  3. Shreyash

    Thread Starter New Member

    Oct 5, 2016
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  4. Shreyash

    Thread Starter New Member

    Oct 5, 2016
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    why do we need the a1,a2,a3,a4 and b1-4 again below.
    is just the upper half not enough?why does this have only one set of ai's and bi's (i=1to4)
     
    Last edited: Oct 5, 2016
  5. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    A four bit adder adds two four bit numbers to a four bit sum and a carry.
     
  6. ci139

    Member

    Jul 11, 2016
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    we belive you, but what you actually have to do with those chips?
    HD74HC83RPEL
     
  7. ci139

    Member

    Jul 11, 2016
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    you list NAND 7400 but #3 uses 7408 AND - ? so you need to rewire this thing for NAND
    and i don't think #3 works as BCD adder coz the carry is set at outputs (12 to 15) or (6,7,14,15) or (16 to 31)
    wy to se t carry at 6,7 ?
    you need to set carry extra to 16 and beyond at 10 to 15 that'd be
    10 1010
    11 1011 -- AND(S4,S2) // the lower AND in Fig #3 is connected wrong or shown wrong or both
    12 1100 -- OR AND(S4,S3)
    13 1101
    14 1110
    15 1111
    -- suppose we connect the lower and "right"
    and our carry is now correct
    so for nubers >=10 the sum is extra incremented by 6 as 10=10*((carry=1) OR 1)+((10+6) MOD 16)
    10 + i = 10*carry + (((10+i) mod 16 + 6*carry) MOD 16) =10*carry + i
    --- so remake it for NAND instead of AND
    your carry is currently carry = C10 = CO or (S4 and S2) or (S4 and S3)
     
    Last edited: Oct 5, 2016
  8. ci139

    Member

    Jul 11, 2016
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    the carry cant be converted using boolean algebra
    but paying attention to function where NAND=0 if both inputs are 1 otherwise it's 1
    so we need to avoid settin our NAND-s while SUM <=9 ... obvously we need 1 NAND to invert a bit for that
    0 0000 0
    1 0001 0
    2 0010 0
    3 0011 0
    4 0100 0
    5 0101 0
    6 0110 0
    7 0111 0
    8 1000 0
    9 1001 0
    A 1010 1
    B 1011 1
    C 1100 1
    D 1101 1
    E 1110 1
    F 1111 1
    it likely is more likely if part of the sum is formed in one adder and part in an other - but it's not shure it can be done so
    it may also be your teacher was drunk? giving such a task ...
     
  9. ci139

    Member

    Jul 11, 2016
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    i dont think it has solution using the components you listed with partial summing eighter coz if we bit mask one operand to 1001 0110 then
    the condition that has to correct the final modulus would had to change it's input causing indefinit/strobing output if we bit mask the mod 10 e.g. 1010 0101 we'd be adding hopefully 0 or10 to our sum but also 8 or 2 the least likely can't be set to form what is required
    change the school ? "We teach here everything that can't be done!" -- weird motto
     
  10. ci139

    Member

    Jul 11, 2016
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    • Do NOT just provide solutions to student's problems. This is known as "cheating".
    so it turns out you don't have to change the school
    new carry = OR( c16 , NANDasNOT( NAND( S4 , OR( S3 , S2 ) ) ) )
    BCD-SUM-2.png
    it was tricky to see
     
  11. ci139

    Member

    Jul 11, 2016
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    cdxv BCD TEST-sx.png
    confirmed it with (if there's no more flaws ...)
     
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