In the previous post, I talked about how transistors work and how Boolean algebra can be implemented using them. The details of that are not required to read this article, but it may help.
I also mentioned the concept of composability without giving it a formal definition.
A module is an element of hardware or software with a defined interface and functionality. A module is composed of smaller elements, and these may be other modules or fundamental units. For our purposes, the fundamental units are now logic gates – how logic gates are constructed is an electrical engineer’s problem.
Two modules which have the same interface (set of inputs and outpus) and overall functionality (emit the same output for the same input states) can be interchanged without affecting the rest of the design. This is an important goal for designing and selecting modules for use, as there are always performance characteristic tradeoffs to be made. Some may be faster; some may use fewer transistors; others may have less surface area. The best module for a specific scenario can be chosen without having to rework everything connected to it, if a standard interface is used.
I am mainly going to discuss the CPU here, but the principles apply to any processing device, including peripheral devices (single-purpose computers controlled by a CPU, which provide real functionality such as storage or transmission).
- CPU Components
In order to determine what should go in a CPU, we first need to identify what it is.
The most basic functionality of a CPU is this: it is an element of hardware that is capable of performing mathematical or logical operations on input data, and emitting the result as output.
Let us begin there, at the module called the Arithmetic/Logic Unit (ALU).
The ALU is essentially a multi-function calculator. It has two main input buses and one main output bus. These buses must all have the same width; for the purposes of this article, I will be using 8-bit buses.
The first functions to construct for the ALU are the logic functions. These are simple: pick the operations to implement, and then create them in parallel.
Let us implement the three Boolean fundamental operations:
NOT. While we can ask our electrical engineers to make 8-bit-wide logic gates, this has performance penalties, so we will use eight 1-bit-wide gates in parallel.
A[7:0] & B[7:0] => Y[7:0]
A & B => Y
A & B => Y
A & B => Y
A & B => Y
and then the individual output wires recombine into the output bus.
This implements Boolean-
AND logic, but we have yet to implement any of the others. We do exactly the same pattern for Boolean-
OR, swapping out gates, so now we have sixteen logic gates in a line: eight
AND and eight
OR, with each bit of
B inputs splitting to go to two gates, and each output from the gates converging back to a single bit of
This is the part where you stop and ask how we resolve contention between two different logic gates. Suppose that an
AND gate asserts low voltage while the
OR gate on the same line asserts high voltage. The continuous connection from high voltage coming out of
AND, and thence into
GND, will not only waste power, but also destroy the logic gates.
The solution to this problem is called a multiplexer. A multiplexer is a logic element which is capable of acting like a railway switch: multiple input buses enter the device, which uses
AND gates to only connect one of them to output.
There also exists a demultiplexer, which is just a multiplexer facing backwards. One input bus comes in, and gets routed among different output buses. We’ll get back to that later.
So we put a 2:1 8-bit multiplexer between the logic gates and the actual output bus, and use a 1-bit selector line to switch the multiplexer back and forth. When the selector line is high (
1), one of the operations is chosen to pass to output; when the selector line is low (
0), the other operation gets to write.
Do note, however, that in our current design, the ALU is performing both the
AND and the
OR operation simultaneously. This means that we are powering two calculations, but only using one. This wastes power, and is not ideal.
This is where the demultiplexer comes in: put it in front of the collection of logic operators, and tie it to the same select line that controls the multiplexer on the output, and now the input signals will be routed through only one logic cluster, and the rest lie dormant.
Now that we’ve solved the problem of selecting between multiple operations, let’s add in the
NOT operator. Remember that
NOT is a unary operator (that is, it only takes one argument.
OR are binary operators, meaning that they take two arguments. It has nothing to do with the fact that Boolean logic only has two symbols), so we only connect the
A input to it.
This is three operation clusters, but our multiplexer only supports two!
This is where composition becomes most useful. We have established that a 2:1 multiplexer (mux for short, henceforth) and 1:2 demultiplexer (demux) are a simple enough construct, with a 1-bit selection line and n-bit selection buses (we frankly don’t care what the bit width of a multiplexer is, since it has no bearing on its actual operation, as long as all our data goes through).
So, suppose you want to select between more than two things. What do you do?
You take a 2:1 mux, and have each of its inputs be… another mux! This is a “binary tree” structure, and it’s extremely common in computing.
It looks something like this:
1 2 3 4 5 6 7 8 9 10 11 12 13 ┌─────┐ | ══A══╡ │ S[1:0] │ mux ╞══Y0══╗ │ ══B══╡ │ ║ │ └──┬──┘ ║ ┌──┴──┐ └────┐ ╚═══╡ │ ├──S0────┤ mux ╞══Y══ ┌────┘ ╔═══╡ │ ┌──┴──┐ ║ └─────┘ ══C══╡ │ ║ │ mux ╞══Y1══╝ ══D══╡ │ └─────┘
As you can see, every doubling of input buses only requires one more selector line. The new selector line controls where the end mux looks, while the old selector line gets forwarded to each leaf (left) mux.
Suppose you want to mux 8 things? You take two of these, slap a 2:1 mux between them, and now you're done. The lower two bits of the now-three-bit select line get forwarded to each 4:1 mux, while the new highest bit switches the new root (right) mux.
Repeat ad nauseum for multiplexing more things. Yay modular composition!
A demux system works the exact same way, only flowing right to left in the above diagram rather than left to right. As long as the select buses for both the demux and mux are properly synchronized, traffic flow through the whole system is seamless, and it appears from the outside as if the logic operators are swapped out on the fly.
Braced ALU Core
1 2 3 4 5 6 7 8 9 ┌─────┐ ┌─────┐ │ 1:2 ╞══A══╡ AND │ │ d ╞══B══╡ ╞══Ya══╗ ┌─────┐ ══A══╡ e │ └─────┘ ╚═══╡ 2:1 │ ══B══╡ m │ ┌─────┐ ╔═══╡ mux ╞══Y══ │ u ╞══A══╡ O ╞══Yb══╝ └──┬──┘ │ x ╞══B══╡ R │ │ └──┬──┘ └─────┘ │ ──S─────┴────────────────────────────┘
The 1:2 demux and 2:1 mux can be replaced with 1:n and n:1 elements, where n is whatever power of 2 necessary to route through all the elements contained between them. The only necessary change is to ensure that the S line is n bits wide.
We can imagine adding in a
NOT block in our system, connecting it only to
A and letting
B be discarded, and our ALU gains new functionality without any change to the interface besides widening the mode selector.
This is where things get weird. Honestly, you can skip this.
The way computers do math is the topic of a good solid month of a senior-level design class. I’m not going to attempt to compress that whole month into this article, because honestly I don’t think it can be done, so I’m just going to cover the very simplest method of performing arithmetic, which is not the method used in any modern computer, and if you really want to know, email me asking for it.
Let’s do some basic addition. In base-10 numbers, the order of digits is 0123456789, and after 9, you wrap to 0 and generate a carry digit which propagates to the next more significant column.
When you carry a digit, you perform normal decimal addition on the next column, but also add the 1 generated by the previous column. Then you repeat the process working your way left, until you reach the end of the number. The sum is guaranteed to be one digit wider, at most, than the arguments going in (called addends for addition).
Binary addition works the exact same way, except it only has two digits, so we have the digits 01 and , which means that the bitwise-addition emits a 0 as its bit output and sends a carry to the next column over.
1 2 3 4 1010 1100 +_____ 10110
The left-most column of the addends generates a carry, which rolls to the next column left. In 4-bit addition, the result is also 4-bits, so the carry output (the fifth bit from the right) is not part of the returned sum, but on a carry line, which goes somewhere else.
So, the most obvious algorithm for addition is that which you learned in grade school: start at the right and work your way left, pushing carries as you go. A carry plus a one plus a one is 1 and a carry, so the carries can never build up.
Let’s take a look!
1 2 3 4 5 6 7 8 9 10 11 12 A B Cin ║ ║ │ ║ ╚═══╤════╤════╤════╤════╤════╤════╤════╕ │ ╚════╤╪═══╤╪═══╤╪═══╤╪═══╤╪═══╤╪═══╤╪═══╕│ │ ┌┴┴─┐┌┴┴─┐┌┴┴─┐┌┴┴─┐┌┴┴─┐┌┴┴─┐┌┴┴─┐┌┴┴─┐│ ┌───┤ 7 ├┤ 6 ├┤ 5 ├┤ 4 ├┤ 3 ├┤ 2 ├┤ 1 ├┤ 0 ├┘ │ └──┬┘└──┬┘└──┬┘└──┬┘└──┬┘└──┬┘└──┬┘└──┬┘ │ ╔════╧════╧════╧════╧════╧════╧════╧════╛ │ ║ │ Y │ Cout
Our interface is this: two 8-bit inputs,
B, and one 1-bit input
Cin, which is our carry-input; one 8-bit output
Y, and one 1-bit output
Cout, which is our carry-output. Internally, the carry-out of each adder is attached directly to the carry-in of its next higher neighbor (the pipe running from
Cin horizontally left to
There are two reasons to have a carry-input: wider addition gets broken down into smaller chunks, so 16-bit addition on this machine is addition of the low half, with the carry-out saved and fed to the carry-in when adding the high halves; and subtraction. I’ll cover subtraction later.
Y is also eight bits; since 8-bit addition can generate a 9-bit sum, the 9th bit goes on carry-out.
Let’s take a minute to think about the algorithm this implements.
Cininputs arrive at the module and fill the input transistors.
All eight 1-bit adder elements compute their sum and write it to
Pause, ask yourself if we’re done at this point, and why you picked your answer.
The zeroth adder computes its carry output, and pipes it to the first adder.
The first adder, whose inputs just changed, recomputes its sum, writing to
Yand its carry-out.
The second adder recomputes and passes the carry.
The third adder recomputes and passes the carry.
The fourth adder recomputes and passes the carry.
The fifth adder recomputes and passes the carry.
The sixth adder recomputes and passes the carry.
The seventh adder recomputes and sets carry-out.
Coutare now stable and ready to be read.
Think about it: we have to pass the intermediate carry computation between each adder, so the length of time required to wait before the outputs are guaranteed stable is directly proportional to the number of bits being added.
Remember earlier when I said that binary trees come up a lot in computing? It turns out that there is an addition algorithm that uses a tree arrangement to make it so that carry computations can be executed proportionally to the log2 of the width of the adder, rather than the full width. This is faster, but uses far more elements and power.
Callback to modular, composable design: the tree adder, called Look-Ahead Carry, has the exact same interface as the linear adder, called Ripple-Carry Adder, and thus the two adder types can be swapped out without affecting the design of their neighbors. Anything that takes two n-bit and one 1-bit input, and emits one n-bit and one 1-bit output, can be plugged in. As long as the replacement module is mathematically correct, the formal behavior of the system is unchanged. (Obviously, the layout and power consumption are altered, but that’s an electrical engineer’s problem, not ours.)
First, let’s talk about how numbers are represented in binary. 8-bit unsigned numbers (positive only) range from 0 (
0b00000000) to 255 (
0b11111111). Adding 1 to 255 causes an overflow, where the lowest 8 bits roll over to 0 and the carry output goes to where carry-outs go, which is not with the rest of the sum. This is a limiting factor of machine arithmetic, and a common pitfall for programmers.
Notice, though, that these numbers are positive only. Signed integers have two representations. The first is to use the highest bit as a sign bit,
0 for positive and
1 for negative, and the lower seven bits look normal. This is easy for people, but not great for machines.
Recall that the 8-bit integer space is cyclical. We can use this with a notation called 2’s-complement, where the numbers run from -128 (
0b10000000) to 127 (
0b01111111). -1 is
0b11111111, and 0 is
0b00000000, as you’d expect from adding 1 to eight bits of ones. Unfortunately, this means that 127 + 1 is -128, not 128, so signed-arithmetic overflow is still a problem to avoid.
Notice, though, that addition that passes the overflow point winds up being a subtraction of 256. Thanks to properties of arithmetic, we can thusly say that subtraction is actually addition plus 256. The way 2’s-complement notation maps to bits, means that in bit representation,
A - B is actually
A + ¬B + 1.
Let me prove I’m not making this up:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 let minus_one = 0b1111_1111; let plus_one = 0b0000_0001; ~minus_one = ~0b1111_1111 = 0b0000_0000; plus_one = 0b0000_0000 + 0b1 = 0b0000_0001 = ~minus_one + 0b1; let plus_127 = 0b0111_1111; let minus_128 = 0b1000_0000; let minus_127 = 0b1000_0001; ~plus_127 = 0b1000_0000; minus_127 = 0b1000_0000 + 0b1 = 0b1000_0001 = ~plus_one + 0b1 Inductively: ~(number-in-2s-complement) = -number - 1 ~(number-in-2s-complement) + 1 = -number
This works for all
n-bit numbers, which can represent integers from to .
So to implement subtraction, we use the exact same hardware, but stick a
NOT gate in front of
B and force the carry input high. Since we can use the same hardware as from addition, we don’t need to make a new block like we did with the Boolean primitives; we just stick a multiplexer on
Cin rigged to do nothing (addition), or invert
B and assert
Cin (subtraction). Tada, basic arithmetic.
Multiplication and division, and fractional numbers, are an entirely other problem. I will not be discussing them.
Combining Mathematics and Logic
Scroll way back up to the top, and imagine a multiplexer setup rigged to route two data streams through one of our available operations:
NOT, addition, subtraction. This is the operating core of a CPU. A single execution of the machine requires the following pieces of data:
S[2:0]– The three select bits choose which operation is turned on.
A[7:0]– Eight bits of the first argument to our operation.
B[7:0]– Eight bits of the second argument to our operation.
Cin– One bit of carry input (only used for arithmetic).
A twenty bit bus can be split up into
Cin buses, routed through our ALU, and result in 9 bits of output (
Cout). This 20-bit number is a binary instruction, which is exactly what all CPUs execute.
The specification of which bits in the number are which inputs, and which numbers are legal (not all combinations of
Cin result in sensible output), comprise (part of) an instruction set.
As it stands, our ALU is barely useful. We have to manually provide the input instruction, and the output result gets dumped in our lap for us to use or throw away.
The solution to this is another digital element, called a register. A register is a component which has two inputs and one output. The registers in modern CPUs are positive-edge triggered, which means that when the control signal switches from low to high, the register takes an instantaneous sample of its data input and then fixes the output line to be that value, until the control line ticks from low to high again.
We can collect a group of registers into a register file, which is a large bank of registers (each of which is as wide as our ALU operators), and connect that register file to the input and output buses of the ALU. Thus, data comes out of the register file, flows through the ALU, and is stored back in the register file.
By signaling the register file, we can choose which registers in the file emit to the
B buses, and which register receives from the
Y bus, of the ALU. Now we only need to be able to put data in and out of the register file to see the ALU’s progress, and the ALU can be set up so that data is stored and, potentially, reused later.
Furthermore, now that our data is provided by the register file, we can take the 16 bits of raw (also called immediate) data out of our instruction numbers and replace it with the register file addresses we wish an execution to use.
Suppose that we have 32 registers. , so we need five bits to specify each register number. We need three registers: the sources for
B, and the destination for
Y, so our instructions come out to 19 bits instead of 20, and look like something like this:
Our CPU now looks like this:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 SCABY Register File ╔════Data access │││││┌──┬──┬──┬──┬──┬──┬──┬──┐║ ││││└┤ │ │ │ │ │ │ │ ╞╝ ││││ ├──┼──┼──┼──┼──┼──┼──┼──┤ │││└─┤ │ │ │ │ │ │ │ ╞══╗ │││ ├──┼──┼──┼──┼──┼──┼──┼──┤ ║ ││└──┤ │ │ │ │ │ │ │ ╞═╗║ ││ ├──┼──┼──┼──┼──┼──┼──┼──┤ ║║ ││ │ │ │ │ │ │ │ │ ╞╗║║ ││ └──┴──┴──┴──┴──┴──┴──┴──┘║║║ ││ ╔═══B══════════════════════╝║║ ││ ║╔══A═══════════════════════╝║ ││ ║║┌───┐ ┌───┐ ┌───┐ ║ ││ ║║│ d │ │ A │ │ m │ ║ ││ ║╚╡ m ╞══╡ L ╞═══════│ u ╞═Y═╣ ││ ╚═╡ u ╞══╡ U ├─Cout─┐│ x │ ║ ││ │ x │ │ │ ││ │ ║ ││ └─┬─┘ └┬─┬┘ │└─┬─┘ ║ │└─────┼─────┼─┴───────┘ │ ╚══to memory controller └──────┴─────┴────────────┘
S bus connects to the demux and mux in the ALU, as well as going into the ALU block (to toggle addition/subtraction), and controls which route through the ALU is active. The
C line connects to the ALU; the
Cout line is shown here as looping back into the ALU; the truth is less clean. Don’t worry about it. The
Y selectors go into the register file, which then sets up internal state to emit the desired registers onto the
B data buses, and receive data from the
Y bus into the specified register.
That’s pretty much it for a (simple) CPU.
A CPU consists of the worker logic in the ALU (arithmetic/logic unit) and data storage in the register file. A multiplexer pair bookends the ALU, so that power waste is minimized by not using the undesired operator units, and the ALU can be set to perform partial addition or subtraction as well as Boolean logic.
There is more to the matter than this, of course: I have neither discussed nor shown the clock line which ticks all the registers, and the data flow from instruction bus and register file, through ALU, back into the register file, is continuous. This is not actually true, but this is a useful approximation for now.
SCABY instruction bus is fetched from instruction memory, which is controlled by a dedicated auto-adder called the Program Counter. This is essentially a memory bank and self-feeding adder, which loads an instruction from memory and then automatically increments so that next clock pulse, the next instruction will be fetched.
Data can be fed into the register file from an external bus, so the CPU has some initial state from which to begin operating.
In order to make the machine Turing complete, the ALU must also be able to compare two numbers, and write to the Program Counter. This permits condition inspection and jumps in program code, which are the two remaining criteria for universal computation.
Additionally, there are other special registers that are kind-of-sort-of part of the register file. One, called the Status Register, is the actual target of
Cout, as well as other status values from the CPU’s operation.
Lastly, there is a memory controller which grants us access to more than 32 memory locations. Some computers (Harvard architecture, from earlier posts) have separate memory banks for instruction codes and data storage, while others (von Neumann architecture, from earlier posts) have only a single memory bank where both code and data are stored.
Von Neumann machines permit a computer to modify its own instructions, which makes it an incredibly powerful system, and unique among all other computer architectures which preceded it.
Post-Script: I will add external references to this in the near future. If you have any questions or confusion about this post, please let me know. I’m sure there are parts I could explain better.