Hopelessly stuck on full-adder

[nblackburn]

I have probably spent 10 hours on the full-adder now, and I think it is obvious if I have not gotten it so far, then I am not going to get it at all.

I have tried visualising it using digital and cannot figure it out. I have tried using 2 half-adder gates, one with A and B and the other with B and C. I have tried A and C and A and B. I have tried inverting B and then feeding it into the half-adder with A and B. There has to be something I am missing because I know it should not take this long.

It seems logical to me to use B in both since in the truth table B is in in the middle and therefore I visualise it as being in both half-adders. What I cannot understand though is how you can half all three inputs and still have sum be 1, since if A B and C were all 1 then both half-adders would output Carry 1 and Sum 0.

I would really appreciate some guidance on this, I have found a solution on YouTube but I do not want to watch it as I do not think I would understand it as well as if I just found the solution on my own.

[dolomiti7]

With a half-adder you can add 2 inputs and you'll get 2 outputs as a result, the sum and the carry. Now if you want to add a third input, you would naturally add it to the result (the sum) of the first addition:

sum=(a+b)+c

The only challenge that you have to solve is what to do with the carry from the first and from the second half-adder. You will need to combine them into the final carry of (a+b+c). If you look to the truth table and add columns for the carry result of the 2 half-adders, it should be straightforward to figure that out.

Hope that helps

[nblackburn]

Thank you for your reply. Your explanation made so much sense, and with it I figured it out in 5 minutes. I think I was thinking about it in terms of just trying to put gates together and get a sum and carry using a not gate for the sum and was not thinking about it in logical steps.