Digital Design Fundamentals·Combinational Circuits
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Every column past the first has three bits to add: the two number bits a and b, plus a carry-in cin from the column to its right.
A full adder takes all three and produces the same two outputs as before:
cout): the bit passed to the next column.Three inputs means eight rows. Read cout and sum together as the 2-bit count of how many inputs are 1:
| a | b | cin | cout | 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 last row is , which is : sum 1, carry 1.
Figure: A full adder drawn as two half-adder blocks and an OR gate
This is hierarchy at work. A full adder is two half adders plus one OR gate:
a and b, giving a partial sum and a carry.cin, giving the final sum and a second carry.cout.Only one of the two carries can ever be 1, so OR is enough to merge them.
Here is a full adder. Read the two lamps together as cout sum, the 2-bit count of how many inputs are high.
Try it:
a = 1, b = 1, cin = 0. sum drops to 0, cout lights: two 1s make , which is 2.Figure: Four full-adder blocks in a row labelled FA0 to FA3, each taking a pair of input bits
To add two 4-bit numbers, place one full adder per column and wire each cout into the next column's cin. The first column's cin is 0. This is a ripple-carry adder: the carry ripples left along the chain like falling dominoes.
Each stage cannot finish until its carry-in is ready, and that comes from the stage before it. So the carry has to travel through every full adder in turn before the top bit is correct.
A 4-bit ripple-carry adder is four carry delays long. A 32-bit one is 32. This is a textbook critical path: the longest route, straight down the carry chain. It is why wide adders in real processors use faster carry schemes, though they all compute the same sum this simple chain does.