0
\$\begingroup\$

A full adder is a combinational circuit that performs the arithmetic sum of three bits: A, B and a carry in, C, from a previous addition

Q) What is the meaning of this bold text? (i.e from a previous addition)

I know what a carry bit is, but when it comes to Full Adder I get confused by this statement...

\$\endgroup\$
4
  • \$\begingroup\$ You can cascade several 1 bit adders to make a multi-bit adder. Except for serial adders, where a 1 bit adder is coupled with shift registers and additions are calculated over many cycles, the "normal" case is that the carry-in input is connected to another adder. \$\endgroup\$
    – Grabul
    Apr 16, 2017 at 7:32
  • \$\begingroup\$ @TEMLIB: I'm really new to this lesson and not familiar with the words either... can you explain this as simple as possible??? \$\endgroup\$
    – JaanuB
    Apr 16, 2017 at 8:19
  • \$\begingroup\$ This is exactly how it works when we learned to add in grade school. 9+6 = 5 carry the 1, then the next column you add the 1 and the other two operands A+B plus the carry in from the prior column. You already knew how this worked... \$\endgroup\$
    – old_timer
    Apr 16, 2017 at 12:57
  • \$\begingroup\$ @old_timer: Thank you very much... Now I can get it \$\endgroup\$
    – JaanuB
    Apr 16, 2017 at 15:02

1 Answer 1

1
\$\begingroup\$

For example if you want to add 2 numbers 2 bits each. $$AB+CD$$ First you preform B+D+0 (because you have no carry from a previous addition) and then you preform A+C+carry from a previous addition, i.e. from adding B+C. You do this the same way you do a long addition you did in school.

\$\endgroup\$
2
  • \$\begingroup\$ And same goes for the Half adder too right??? \$\endgroup\$
    – JaanuB
    Apr 16, 2017 at 8:18
  • \$\begingroup\$ Half adder has only 2 inputs (only the bits you want to add without the carry). You can use it for the first operation of B+D since the carry in is 0. You can just google the difference between half and full adder. There are many simple sources on this topic. \$\endgroup\$ Apr 16, 2017 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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