In a Full Adder what's wrong with [c.(a+b) + a.b] as the expression of the Carry-Out? I checked it with truth table and gives correct output. But in all books the expression is [c.(a XOR b) + a.b].
Electronic – carry out of full adder
digital-logic
Related Solutions
I thought about seeing what my brain could regurgitate of long ago explnatins, as opposed to commin sense observations, BUT here's a page which does it all very well indeed.
They explain a "nice" 'trick'.
- Two's compelment overflow detection can be achieved by XORing the carry-in and the carry-out bo=its of the leftmost full adder.
Their diagram. See above ref for detailed comment:
It can help to add intermediate states to the truth table; there are a lot of letters from the alphabet you haven't used :-). Name the output of the first XOR gate D and add a D column to the truth table. You may want to have clearly separated "inputs", "intermediate" and "outputs". I separate them with a vertical line, and fill out the combinations for the inputs. You have three of those, that's eight combinations.
Then fill out column D, which takes A and B as inputs. Even if there were 15 more inputs, ignore them; only A and B are relevant. Work gate per gate, divide and conquer. When you've finished column D you can fill out column S in the "outputs" section, because that only depends on D and Cin, and you have those.
Repeat for the outputs of the AND gates: add E and F to the "intermediates" section, and fill them out only looking at the columns which are input for each gate.
Related Topic
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Best Answer
You can write \$(A + B)\$ as \$A\cdot B + A \oplus B\$.
Therefore:
\$C \cdot(A+B) + A\cdot B = C \cdot (A\oplus B) + C \cdot A \cdot B + A\cdot B =\\ = C \cdot (A\oplus B) + (1+C) \cdot A \cdot B = C \cdot (A\oplus B) + A \cdot B \$
The latter formula is the one you intuitively get, when you calculate the carry: you have a carry out if at least two inputs are 1.
The books use the latter formula because you don't need to calculate again \$A \oplus B\$, because you can reuse the sum term \$S=A\oplus B\$ (i.e. you save one logic gate).