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I have boolean expression, which was simplified using Karnaugh's map (The first line). And then I used de Morgan's Law to make the expression suitable for using only NAND gates (The second line). But when I create a logic gate circuit it does not work properly and no matter how much I look at this circuit, I can't see where I made a mistake. And sorry for posting expression in a picture, I have no knowledge of how to transfer this expression from paper to computer. Simplified expression and de Morgan's version

Logic circuit using NAND gates

aretas_pau
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1 Answers1

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I checked your circuit and have not been able to spot an error. What is not working?

An alternative solution is:

NAND4(
  NAND3(!X0, !X1, X3), 
  NAND4(X0, X1, X4, X5), 
  NAND4(!X0, X1, !X3, !X5), 
  NAND5(X0, !X1, !X2, X3, X4))

The solution generated by Logic Friday 1 is:

[enter image description here][1]

Update:

I entered the following expression to Logic Friday 1:

INORDER = x5 x4 x3 x2 x1 x0;
F = !(!(!x0 & !(!(!x1 x3) & !(x1 !x3 !x5))) & !(x0 & !(!(x1 x4 x5) & !(!x1 !x2 !x3 x4))));

The resulting 18 minterms are:
enter image description here

Taking X5 as most-significant and X0 as least-significant bit, this can be interpreted as minterm list: 2, 6, 8, 12, 17, 18, 22, 24, 28, 40, 44, 49, 51, 55, 56, 59, 60, 63.

You can quickly convince yourself (minterm 63) that all six inputs set to 1 lead to output 1. Minterm 2: All inputs other than X1 0 leads to output 1 as well. Something might be different with your bit ordering.

Axel Kemper
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  • It should output 1 if all six inputs are numbers f = (4, 5, 6, 7, 12, 13, 14, 15, 16, 18, 24, 26, 34, 35, 49, 53, 57, 61) in binary, but when I run a test, the outputs are wrong and not even close to those numbers. – aretas_pau Feb 27 '17 at 17:01
  • And the expression before de Morgan's simplification outputs correct numbers. – aretas_pau Feb 27 '17 at 17:02
  • @aretas_pau, I agree with Axel that you have some mistake somewhere. All Xi = 1 (63) obviously satisfies this expression as it makes X0*X1*X4*X5 = True which is enough for the whole. But the fact that you have exactly the same number of satisfying solutions suggests that you made somewhere a typo. – SergGr Feb 28 '17 at 22:53