IT LOVER

24/03/2009

How Microprocessor Work

Filed under: computer zone — टैग्स: — pankaj kuamr @ 11:44 पूर्वाह्न

 

Introduction to How Microprocessors Work

The computer you are using to read this page uses a microprocessor to do its work. The microprocessor is the heart of any normal computer, whether it is a desktop machine, a server or a laptop. The microprocessor you are using might be a Pentium, a K6, a PowerPC, a Sparc or any of the many other brands and types of microprocessors, but they all do approximately the same thing in approximately the same way.

A microprocessor — also known as a CPU or central processing unit — is a complete computation engine that is fabricated on a single chip. The first microprocessor was the Intel 4004, introduced in 1971. The 4004 was not very powerful — all it could do was add and subtract, and it could only do that 4 bits at a time. But it was amazing that everything was on one chip. Prior to the 4004, engineers built computers either from collections of chips or from discrete components (transistors wired one at a time). The 4004 powered one of the first portable electronic calculators.

If you have ever wondered what the microprocessor in your computer is doing, or if you have ever wondered about the differences between types of microprocessors, then read on. In this article, you will learn how fairly simple digital logic techniques allow a computer to do its job, whether its playing a game or spell checking a document!

 

The first microprocessor to make it into a home computer was the Intel 8080, a complete 8-bit computer on one chip, introduced in 1974. The first microprocessor to make a real splash in the market was the Intel 8088, introduced in 1979 and incorporated into the IBM PC (which first appeared around 1982). If you are familiar with the PC market and its history, you know that the PC market moved from the 8088 to the 80286 to the 80386 to the 80486 to the Pentium to the Pentium II to the Pentium III to the Pentium 4. All of these microprocessors are made by Intel and all of them are improvements on the basic design of the 8088. The Pentium 4 can execute any piece of code that ran on the original 8088, but it does it about 5,000 times faster!

The following table helps you to understand the differences between the different processors that Intel has introduced over the years.

 

Name Date Transistors Microns Clock speed Data width MIPS
8080 1974 6,000 6 2 MHz 8 bits 0.64
8088 1979 29,000 3 5 MHz 16 bits
8-bit bus
0.33
80286 1982 134,000 1.5 6 MHz 16 bits 1
80386 1985 275,000 1.5 16 MHz 32 bits 5
80486 1989 1,200,000 1 25 MHz 32 bits 20
Pentium 1993 3,100,000 0.8 60 MHz 32 bits
64-bit bus
100
Pentium II 1997 7,500,000 0.35 233 MHz 32 bits
64-bit bus
~300
Pentium III 1999 9,500,000 0.25 450 MHz 32 bits
64-bit bus
~510
Pentium 4 2000 42,000,000 0.18 1.5 GHz 32 bits
64-bit bus
~1,700
Pentium 4 “Prescott” 2004 125,000,000 0.09 3.6 GHz 32 bits
64-bit bus
~7,000

Microprocessor Progression: Intel

Microprocessor Logic

To understand how a microprocessor works, it is helpful to look inside and learn about the logic used to create one. In the process you can also learn about assembly language — the native language of a microprocessor — and many of the things that engineers can do to boost the speed of a processor.a=1;f=1;while (a <= 5){ f = f * a; a = a + 1;}

 

At the end of the program’s execution, the variable f contains the factorial of 5.

Assembly Language
A C compiler translates this C code into assembly language. Assuming that RAM starts at address 128 in this processor, and ROM (which contains the assembly language program) starts at address 0, then for our simple microprocessor the assembly language might look like this:

 

 

// Assume a is at address 128// Assume F is at address 129 0 CONB 1 // a=1;1 SAVEB 1282 CONB 1 // f=1;3 SAVEB 129 4 LOADA 128 // if a > 5 the jump to 175 CONB 56 COM7 JG 17 8 LOADA 129 // f=f*a;9 LOADB 12810 MUL11 SAVEC 129 12 LOADA 128 // a=a+1;13 CONB 114 ADD15 SAVEC 128 16 JUMP 4 // loop back to if17 STOP

 

ROM
So now the question is, “How do all of these instructions look in ROM?” Each of these assembly language instructions must be represented by a binary number. For the sake of simplicity, let’s assume each assembly language instruction is given a unique number, like this:

  • LOADA – 1
  • LOADB – 2
  • CONB – 3
  • SAVEB – 4
  • SAVEC mem – 5
  • ADD – 6
  • SUB – 7
  • MUL – 8
  • DIV – 9
  • COM – 10
  • JUMP addr – 11
  • JEQ addr – 12
  • JNEQ addr – 13
  • JG addr – 14
  • JGE addr – 15
  • JL addr – 16
  • JLE addr – 17
  • STOP – 18

The numbers are known as opcodes. In ROM, our little program would look like this:

 

 

// Assume a is at address 128// Assume F is at address 129Addr opcode/value 0 3 // CONB 11 12 4 // SAVEB 1283 128 4 3 // CONB 15 16 4 // SAVEB 1297 129 8 1 // LOADA 1289 12810 3 // CONB 511 5 12 10 // COM13 14 // JG 1714 31 15 1 // LOADA 12916 12917 2 // LOADB 12818 128 19 8 // MUL20 5 // SAVEC 12921 129 22 1 // LOADA 12823 12824 3 // CONB 125 1 26 6 // ADD27 5 // SAVEC 12828 128 29 11 // JUMP 430 831 18 // STOP

 

You can see that seven lines of C code became 18 lines of assembly language, and that became 32 bytes in ROM.

Decoding
The instruction decoder needs to turn each of the opcodes into a set of signals that drive the different components inside the microprocessor. Let’s take the ADD instruction as an example and look at what it needs to do:

  1. During the first clock cycle, we need to actually load the instruction. Therefore the instruction decoder needs to:
    • activate the tri-state buffer for the program counter
    • activate the RD line
    • activate the data-in tri-state buffer
    • latch the instruction into the instruction register
  2. During the second clock cycle, the ADD instruction is decoded. It needs to do very little:
    • set the operation of the ALU to addition
    • latch the output of the ALU into the C register
  3. During the third clock cycle, the program counter is incremented (in theory this could be overlapped into the second clock cycle).

Every instruction can be broken down as a set of sequenced operations like these that manipulate the components of the microprocessor in the proper order. Some instructions, like this ADD instruction, might take two or three clock cycles. Others might take five or six clock cycles.

 

 

A microprocessor executes a collection of machine instructions that tell the processor what to do. Based on the instructions, a microprocessor does three basic things:

  • Using its ALU (Arithmetic/Logic Unit), a microprocessor can perform mathematical operations like addition, subtraction, multiplication and division. Modern microprocessors contain complete floating point processors that can perform extremely sophisticated operations on large floating point numbers.
  • A microprocessor can move data from one memory location to another.
  • A microprocessor can make decisions and jump to a new set of instructions based on those decisions.

There may be very sophisticated things that a microprocessor does, but those are its three basic activities. The following diagram shows an extremely simple microprocessor capable of doing those three things:

microprocessor1

 

 This is about as simple as a microprocessor gets. This microprocessor has:

  • An address bus (that may be 8, 16 or 32 bits wide) that sends an address to memory
  • A data bus (that may be 8, 16 or 32 bits wide) that can send data to memory or receive data from memory
  • An RD (read) and WR (write) line to tell the memory whether it wants to set or get the addressed location
  • A clock line that lets a clock pulse sequence the processor
  • A reset line that resets the program counter to zero (or whatever) and restarts execution

Let’s assume that both the address and data buses are 8 bits wide in this example.

Here are the components of this simple microprocessor:

  • Registers A, B and C are simply latches made out of flip-flops. (See the section on “edge-triggered latches” in How Boolean Logic Works for details.)
  • The address latch is just like registers A, B and C.
  • The program counter is a latch with the extra ability to increment by 1 when told to do so, and also to reset to zero when told to do so.
  • The ALU could be as simple as an 8-bit adder (see the section on adders in How Boolean Logic Works for details), or it might be able to add, subtract, multiply and divide 8-bit values. Let’s assume the latter here.
  • The test register is a special latch that can hold values from comparisons performed in the ALU. An ALU can normally compare two numbers and determine if they are equal, if one is greater than the other, etc. The test register can also normally hold a carry bit from the last stage of the adder. It stores these values in flip-flops and then the instruction decoder can use the values to make decisions.
  • There are six boxes marked “3-State” in the diagram. These are tri-state buffers. A tri-state buffer can pass a 1, a 0 or it can essentially disconnect its output (imagine a switch that totally disconnects the output line from the wire that the output is heading toward). A tri-state buffer allows multiple outputs to connect to a wire, but only one of them to actually drive a 1 or a 0 onto the line.
  • The instruction register and instruction decoder are responsible for controlling all of the other components.

Although they are not shown in this diagram, there would be control lines from the instruction decoder that would:

  • Tell the A register to latch the value currently on the data bus
  • Tell the B register to latch the value currently on the data bus
  • Tell the C register to latch the value currently output by the ALU
  • Tell the program counter register to latch the value currently on the data bus
  • Tell the address register to latch the value currently on the data bus
  • Tell the instruction register to latch the value currently on the data bus
  • Tell the program counter to increment
  • Tell the program counter to reset to zero
  • Activate any of the six tri-state buffers (six separate lines)
  • Tell the ALU what operation to perform
  • Tell the test register to latch the ALU’s test bits
  • Activate the RD line
  • Activate the WR line

Coming into the instruction decoder are the bits from the test register and the clock line, as well as the bits from the instruction register.

Microprocessor Memory

The previous section talked about the address and data buses, as well as the RD and WR lines. These buses and lines connect either to RAM or ROM — generally both. In our sample microprocessor, we have an address bus 8 bits wide and a data bus 8 bits wide. That means that the microprocessor can address (28) 256 bytes of memory, and it can read or write 8 bits of the memory at a time. Let’s assume that this simple microprocessor has 128 bytes of ROM starting at address 0 and 128 bytes of RAM starting at address 128.

 

Microprocessor Performance and Trends

The number of transistors available has a huge effect on the performance of a processor. As seen earlier, a typical instruction in a processor like an 8088 took 15 clock cycles to execute. Because of the design of the multiplier, it took approximately 80 cycles just to do one 16-bit multiplication on the 8088. With more transistors, much more powerful multipliers capable of single-cycle speeds become possible.More transistors also allow for a technology called pipelining. In a pipelined architecture, instruction execution overlaps. So even though it might take five clock cycles to execute each instruction, there can be five instructions in various stages of execution simultaneously. That way it looks like one instruction completes every clock cycle.

Many modern processors have multiple instruction decoders, each with its own pipeline. This allows for multiple instruction streams, which means that more than one instruction can complete during each clock cycle. This technique can be quite complex to implement, so it takes lots of transistors.

Trends
The trend in processor design has primarily been toward full 32-bit ALUs with fast floating point processors built in and pipelined execution with multiple instruction streams. The newest thing in processor design is 64-bit ALUs, and people are expected to have these processors in their home PCs in the next decade. There has also been a tendency toward special instructions (like the MMX instructions) that make certain operations particularly efficient, and the addition of hardware virtual memory support and L1 caching on the processor chip. All of these trends push up the transistor count, leading to the multi-million transistor powerhouses available today. These processors can execute about one billion instructions per second!

 

ROM chip

ROM chip

 

ROM stands for read-only memory. A ROM chip is programmed with a permanent collection of pre-set bytes. The address bus tells the ROM chip which byte to get and place on the data bus. When the RD line changes state, the ROM chip presents the selected byte onto the data bus.


<!– Photo courtesy
–>RAM chip

RAM stands for random-access memory. RAM contains bytes of information, and the microprocessor can read or write to those bytes depending on whether the RD or WR line is signaled. One problem with today’s RAM chips is that they forget everything once the power goes off. That is why the computer needs ROM.

By the way, nearly all computers contain some amount of ROM (it is possible to create a simple computer that contains no RAM — many microcontrollers do this by placing a handful of RAM bytes on the processor chip itself — but generally impossible to create one that contains no ROM). On a PC, the ROM is called the BIOS (Basic Input/Output System). When the microprocessor starts, it begins executing instructions it finds in the BIOS. The BIOS instructions do things like test the hardware in the machine, and then it goes to the hard disk to fetch the boot sector (see How Hard Disks Work for details). This boot sector is another small program, and the BIOS stores it in RAM after reading it off the disk. The microprocessor then begins executing the boot sector’s instructions from RAM. The boot sector program will tell the microprocessor to fetch something else from the hard disk into RAM, which the microprocessor then executes, and so on. This is how the microprocessor loads and executes the entire operating system.

Microprocessor Instructions

Even the incredibly simple microprocessor shown in the previous example will have a fairly large set of instructions that it can perform. The collection of instructions is implemented as bit patterns, each one of which has a different meaning when loaded into the instruction register. Humans are not particularly good at remembering bit patterns, so a set of short words are defined to represent the different bit patterns. This collection of words is called the assembly language of the processor. An assembler can translate the words into their bit patterns very easily, and then the output of the assembler is placed in memory for the microprocessor to execute.

Here’s the set of assembly language instructions that the designer might create for the simple microprocessor in our example:

  • LOADA mem – Load register A from memory address
  • LOADB mem – Load register B from memory address
  • CONB con – Load a constant value into register B
  • SAVEB mem – Save register B to memory address
  • SAVEC mem – Save register C to memory address
  • ADD – Add A and B and store the result in C
  • SUB – Subtract A and B and store the result in C
  • MUL – Multiply A and B and store the result in C
  • DIV – Divide A and B and store the result in C
  • COM – Compare A and B and store the result in test
  • JUMP addr – Jump to an address
  • JEQ addr – Jump, if equal, to address
  • JNEQ addr – Jump, if not equal, to address
  • JG addr – Jump, if greater than, to address
  • JGE addr – Jump, if greater than or equal, to address
  • JL addr – Jump, if less than, to address
  • JLE addr – Jump, if less than or equal, to address
  • STOP – Stop execution

If you have read How C Programming Works, then you know that this simple piece of C code will calculate the factorial of 5 (where the factorial of 5 = 5! = 5 * 4 * 3 * 2 * 1 = 120):

 

 

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