1
\$\begingroup\$

The truth table of a full adder is as below.
Cout is given as Cout = AB + Cin(A XOR B).
not sure why is that because when I do my k-map on the Cout, I get Cout = AB + Cin(A+B).
does anyone know why is that? enter image description here

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Hey @kissinger david, can you please accept my answer? \$\endgroup\$
    – kene02
    Dec 18 '21 at 7:59
1
\$\begingroup\$

If you drew up a truth table for the expression you found, you'd realise that it would exactly the same as the one for the expression they've given you (so you're not wrong). Since you would already have circuitry to find A XOR B (to find S), however, the output of the same circuitry should be re-used to find Cout as shown in the diagram below in order to save the extra space and transistors (or whatever hardware is being used) that would be required by an additional OR gate.

enter image description here

(Source: Wikimedia Commons)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.