In my last post, I promised that I’d explain how we can build interesting mathematical operations using logic gates. In this post, I’m going to try do that by walking through the design of a circuit that adds to multi-bit integers together.

As I said last time, most of the time, when we’re trying to figure out how to do something with gates, it’s useful to with boolean algebra to figure out what we want to build. If we have two bits, what does it mean, in terms of boolean logic to add them?

Each input can be either 0 or 1. If they’re both one, then the sum is 0. If either, but not both, is one, then the sum is 1. If both are one, then the sum is 2. So there’s three possible outputs: 0, 1, and 2.

This brings us to the first new thing: we’re building something that operates on single bits as inputs, but it needs to have more than one bit of output! A single boolean output can only have two possible values, but we need to have three.

The way that we usually describe it is that a single bit adder takes two inputs, X and Y, and produces two outputs, SUM and CARRY. Sum is called the “low order” bit, and carry is the “high order” bit – meaning that if you interpret the output as a binary number, you put the higher order bits to the left of the low order bits. (Don’t worry, this will get clearer in a moment!)

Let’s look at the truth table for a single bit adder, with a couple of extra columns to help understand how we intepret the outputs:

X | Y | SUM | CARRY | Binary | Decimal |
---|---|---|---|---|---|

0 | 0 | 0 | 0 | 00 | 0 |

0 | 1 | 1 | 0 | 01 | 1 |

1 | 0 | 1 | 0 | 01 | 1 |

1 | 1 | 0 | 1 | 10 | 2 |

If we look at the SUM bit, it’s an XOR – that is, it outputs 1 if exactly one, but not both, of its inputs is 1; otherwise, it outputs 0. And if we look at the carry bit, it’s an AND. Our definition of one-bit addition is thus:

We can easily build that with gates:

This little thing is called a half-adder. That may seem confusing at first, because it is adding two one-bit values. But we don’t really care about adding single bits. We want to add numbers, which consist of multiple bits, and for adding pairs of bits from multibit numbers, a half-adder only does half the work.

That sounds confusing, so let’s break it down a bit with an example.

- Imagine that we’ve got two two bit numbers, 1 and 3 that we want to add together.
- In binary 1 is 01, and 3 is 11.
- If we used the one-bit half-adders for the 0 bit (that is, the lowest order bit – in computer science, we always start counting with 0), we’d get 1+1=0, with a carry of 1; and for the 1 bit, we’d get 1+0=1 with a carry of 0. So our sum would be 10, which is 2.
- That’s wrong, because we didn’t do anything with the carry output from bit 0. We need to include that as an input to the sum of bit 1.

We could try starting with the truth table. That’s always a worthwile thing to do. But it gets really complicated really quickly.

X | X0 | X1 | Y | Y0 | Y1 | OUT | OUT0 | CARRY0 | OUT1 | CARRY1 |
---|---|---|---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |

0 | 0 | 0 | 2 | 0 | 1 | 2 | 0 | 0 | 1 | 0 |

0 | 0 | 0 | 3 | 1 | 1 | 3 | 1 | 0 | 1 | 0 |

1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |

1 | 1 | 0 | 1 | 1 | 0 | 2 | 0 | 1 | 1 | 0 |

1 | 1 | 0 | 2 | 0 | 1 | 3 | 1 | 0 | 1 | 0 |

1 | 1 | 0 | 3 | 1 | 1 | 4 | 0 | 1 | 0 | 1 |

2 | 0 | 1 | 0 | 0 | 0 | 2 | 0 | 0 | 1 | 0 |

2 | 0 | 1 | 1 | 1 | 0 | 3 | 1 | 0 | 1 | 0 |

2 | 0 | 1 | 2 | 0 | 1 | 4 | 0 | 0 | 0 | 1 |

2 | 0 | 1 | 3 | 1 | 1 | 5 | 1 | 0 | 1 | 1 |

3 | 1 | 1 | 0 | 0 | 0 | 3 | 1 | 0 | 1 | 0 |

3 | 1 | 1 | 1 | 1 | 0 | 4 | 0 | 1 | 1 | 1 |

3 | 1 | 1 | 2 | 0 | 1 | 5 | 1 | 0 | 1 | 1 |

3 | 1 | 1 | 3 | 1 | 1 | 6 | 0 | 1 | 1 | 1 |

This is a nice illustration of why designing CPUs is so hard, and why even massively analyzed and tested CPUs still have bugs! We’re looking at one of the simplest operations to implement; and we’re only looking at it for 2 bits of input. But already, it’s hard to decide what to include in the table, and to read the table to understand what’s going on. We’re not really going to be able to do much reasoning here directly in boolean logic using the table. But it’s still good practice, both because it helps us make sure we understand what outputs we want, and because it gives us a model to test against once we’ve build the network of gates.

And there’s still some insight to be gained here: Look at the row for 1 + 3. In two bit binary, that’s 01 + 11. The sum for bit 0 is 0 – there’s no extra input to worry about,

but it does generate a carry out. The sum of the input bits for bit

one is 1+1=10 – so 0 with a carry bit. But we have the carry from bit

0 – that needs to get *added* to the sum for bit1. If we do that – if we do another add step to add the carry bit from bit 0 to the sum from bit 1, then we’ll get the right result!

The resulting gate network for two-bit addition looks like:

The adder for bit 1, which is called a *full adder*, adds the input bits X1 and Y1, and then adds the *sum* of those (produced by that first adder) to the carry bit from bit0. With this gate network, the output from the second adder for bit 1 is the correct value for bit 1 of the sum, but we’ve got two different carry outputs – the carry from the first adder for bit 1, and the carry from the second adder. We need to combine those somehow – and the way to do it is an OR gate.

Why an OR gate? The second adder will only produce a carry if the first adder produced a 1 as its output. But there’s no way that adding two bits can produce both a 1 as its sum output *and* a 1 as its carry output. So the carry bit from the second adder will only ever be 1 if the output of the first adder is 0; and the carry output from the first adder will only ever be 1 if the sum output from the first carry is 0. Only one of the two carries will ever be true, but if either of them is true, we should produce a 1 as the carry output. Thus, the or-gate.

Our full adder, therefore, takes 3 inputs: a carry from the next lower bit, and the two bits to sum; and it outputs two bits: a sum and a carry. Inside, it’s just two adders chained together, so that first we add the two sum inputs, and then we add the sum of that to the incoming carry.

For more than two bits, we just keep chaining full adders together. For example,

here’s a four-bit adder.

This way of implementing sum is called a *ripple carry adder* – because the carry bits ripple up through the gates. It’s not the most efficient way of producing a sum – each higher order bit of the inputs can’t be added together until the next lower bit is done, so the carry *ripples* through the network as each bit finishes, and the total time required is proportional to the number of bits to be summed. More bits means that the ripple-carry adder gets slower. But this works, and it’s pretty easy to understand.

There are faster ways to build multibit adders, by making the gate network more complicated in order to remove the ripple delay. You can imagine, for example, that instead of waiting for the carry from bit 0, you could just build the circuit for bit 1 so that it inputs X0 and Y0; and similarly, for bit 2, you could include X0, X1, Y0, and Y1 as additional inputs. You can imagine how this gets complicated quickly, and there are some timing issues that come up as the network gets more complicated, which I’m really not competent to explain.

Hopefully this post successfully explained a bit of how interesting operations like arithmetic can be implemented in hardware, using addition as an example. There are similar gate networks for subtraction, multiplication, etc.

These kinds of gate networks for specific operations are parts of real CPUs. They’re called *functional units*. In the simplest design, a CPU has one functional unit for each basic arithmetic operation. In practice, it’s a lot more complicated than that, because there are common parts shared by many arithmetic operations, and you can get rid of duplication by creating functional units that do several different things. We might look at how that works in a future post, if people are interested. (Let me know – either in the comments, or email, or mastodon, if you’d like me to brush up on that and write about it.)