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4.5 : Full Adder
So the full adder is made by connecting two half adders. The full adder can take three 1-bit inputs and them to create a 2-bit value. Our three values would be the Carry-In, X and Y. Below is the table of the values. Once again, we are adding the 3 bits. This is not a truth table. One thing to notice is that since we have three inputs, there are much more than 4 cases. We still set up the values in increasing order.
| X | Y | Carry In | Carry Out | Sum |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 |
The table looks very intimidating, however, if you look carefully, it's simple bit addition. The first three columns are beng added together to produce the last two. Now let's build our full adder. We're going to use two half-adders. This is rather easy to remember as two halves form a whole. The first half-adder will add X and Y. Now we have two outputs, the sum and carry. We'll call these sum1 and carry1 since they're the results of the first adder. The second adder will add the sum1 and the carry-in. From this we get sum2 and carry2. Now we have three outputs, carry1, sum2, and carry2. However, we only want 2 outputs, a sum and a carry. We have one sum but two carries. This means we need to operate on the two carries. We need there to be an operator who gives outputs a 1 as long as there's at least one 1. Which boolean operator does this? Try not to peek again!
So now that we've used the OR operator to preserve any 1's left over, we finally have two outputs! We have successfully built a full-adder. Down below includes the schematics for a full-adder in 3 different ways; the piece itself, broken down into half adder, and the half adders broken down.
Practice Problems
What boolean operator should we use to operate on carry1 and carry2?