I am trying to build a 1-bit full adder that outputs a 2-bit sum.

I know that the standard 1-bit FA outputs a 1-bit sum and a carry bit, but I was wondering how can I modify the FA such that the carry bit output is discarded and the only output is a 2-bit sum.

  • 3
    \$\begingroup\$ Does the carry bit not just become the highest bit of your 2 bit sum? \$\endgroup\$
    – K H
    Jan 27, 2021 at 0:33
  • \$\begingroup\$ @KH Yes, but if the carry bit is the highest bit, it means that if I merge the carry and a 1 sum bit into a 2-bit number, the result is 11, when the result should be 10 (2). \$\endgroup\$
    – Adam Lee
    Jan 27, 2021 at 0:46
  • \$\begingroup\$ For input 0 0, output 0 carry 0 > 00. For input 1 0, output 1 carry 0 > 01. For input 0 1, output 1 carry 0 > 01. For input 1 1, output 0 carry 1 > 10. Where is the missing case? \$\endgroup\$
    – K H
    Jan 27, 2021 at 0:56
  • \$\begingroup\$ @KH Got it, thanks for the quick response :D \$\endgroup\$
    – Adam Lee
    Jan 27, 2021 at 1:21
  • \$\begingroup\$ I'd write you an answer but I don't actually know how a full adder works. I only know how it must work to do what it says it does. And now I've looked up a full adder and it appears you have an additional carry in input, so taking that into account 11 is a possible output as well. \$\endgroup\$
    – K H
    Jan 27, 2021 at 1:25

2 Answers 2


You could feed the carry bit to another full adder that has zeroes on its other two inputs, but that seems kind of pointless because it will just get you a copy of the carry bit.

The carry bit is the second bit.

If you don't need a carry input you can just use a half adder.


So I looked up what constitutes a full adder and found its truth table.

Taken from wikipedia:

enter image description here

So you can see that if the adder has the correct output, your sum and carry bits are simply the 2 bit number you want. So I guess all the full adder really does is add up to 3 binaries at a time instead of 2.


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