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S May 9, 2016 at 11:25 history suggested Karlo CC BY-SA 3.0
typo(s) corrected
May 9, 2016 at 11:08 review Suggested edits
S May 9, 2016 at 11:25
Sep 15, 2015 at 17:40 answer added ACD timeline score: 0
Sep 15, 2015 at 16:51 comment added endolith For a simpler starting point, think about how to combine 2-input AND gates to make a 3-input AND gate. Hint: (a+b+c) = ((a+b)+c)
Sep 15, 2015 at 16:49 history edited endolith CC BY-SA 3.0
prevent link rot
Sep 15, 2015 at 16:48 answer added Alexxx timeline score: 2
Sep 15, 2015 at 7:11 answer added Peter Smith timeline score: 2
Sep 14, 2015 at 14:44 comment added James I have spent over 5 hours on this one question at this point and it must be something obvious im missing because I can't find any help for it online.
Sep 14, 2015 at 14:36 comment added James @Peter smith, would this be correct? i.imgur.com/gIJ0Myc.png it seems correct to me but it could be completely wrong.
Sep 14, 2015 at 14:17 comment added Peter Smith Make two 2 input gates and try cascading one onto the other - you will need an inversion at the cascade.
Sep 14, 2015 at 13:26 comment added James i have written out the truth table for it, I just don't understand how I am supposed to get the 3rd input from two original ones. I have a decent understanding of normal circuit diagrams and making truth tables from them and even simplifying them but I just don't get how I can go from 2 input to 3 only using NAND.
Sep 14, 2015 at 12:09 comment added Peter Smith A good start would be to write out the truth table. I also suggest looking up DeMorgan's theorem which shows the equivalence between NOR and NAND.
Sep 14, 2015 at 11:50 review First posts
Sep 14, 2015 at 12:30
Sep 14, 2015 at 11:46 history asked James CC BY-SA 3.0