电子信息工程 微处理器 外文翻译 外文文献 英文文献

1、外文资料所译外文资料：1.作者 g.bouwhuis, j.braat, a.huijser2.书名：principles of optical disk systems3.出版时间：1991年9月4.所译章节：session 2/chapter9, session 2/chapter 11原文：microprocessorone of the key inventions in the history of electronics, and in fact one of the most important inventions ever period, was the transistor

2、. as time progressed after the invention of lsi integrated circuits, the technology improved and chips became smaller, faster and cheaper. the functions performed by a processor were implemented using several different logic chips. intel was the first company to incorporate all of these logic compon

3、ents into a single chip, this was the first microprocessor. a microprocessor is a complete computation engine that is fabricated on a single chip. a microprocessor executes a collection of machine instructions that tell the processor what to do. based on the instructions, a microprocessor does three

4、 basic things: 1.using the alu (arithmetic/logic unit), a microprocessor can perform mathematical operations like addition, subtraction, multiplication and division; 2.a microprocessor can move data from one memory location to another; 3.a microprocessor can make decisions and jump to a new set of i

5、nstructions based on those decisions.there may be very sophisticated things that a microprocessor does, but those are its three basic activities. microprocessor has an address bus that sends an address to memory, a data bus that can send data to memory or receive data from memory, an rd(read) and wr

6、(write) line that lets a clock pulse sequence the processor and a reset line that resets the program counter to zero(or whatever) and restarts execution. and lets assume that both the address and data buses are 8 bits wide here.here are the components of this simple microprocessor:1. registers a, b

7、and c are simply latches made out of flip-flops.2. the address latch is just like registers a, b and c.3. 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.4. the alu could be as simple as an 8-bit adder, or it mi

8、ght be able to add, subtract, multiply and divide 8-bit values. lets assume the latter here.5. the test register is a special latch that can hold values from comparisons performed in the alu. an alu can normally compare two numbers send determine if they are equal, if one is greater than the other,

9、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.6. there are six boxes marked “3-state”.these are tri-state buffers. a tri-state buffer can pass a 1,

10、 a 0 or it can essentially disconnect its output. a tri-state buffer allows multiple outputs to connect to a wire, but only one of then to actually drive a 1or a 0 onto the line.7. the instruction register and instruction decoder are responsible for controlling all of the other components.although t

11、hey are not shown in this diagram, there would be control lines from the instruction decoder that would:1. tell the a register to latch the value currently on the data bus 2. tell the b register to latch the value currently on the data bus 3. tell the c register to latch the value currently on the d

12、ata bus4. tell the program counter register to latch the value currently on the data bus5. tell the address register to latch the value currently on the data bus6. tell the instruction register to latch the value currently on the data bus 7. tell the program counter to increment 8. tell the program

13、counter to reset to zero9. activate any of the six tri-state buffers (six separate lines)10. tell the alu what operation to perform11. tell the test register to latch the alus test bits12. activate the rd line13. activate the wr linecoming into the instruction decoder are the bits from the best regi

14、ster and clock line, as well as the bits from the instruction register.ram and rom the address and data buses, as well as the rd and wr lines connect either to ram or romgenerally both. in our sample microprocessor, we have an address bus 8 bits wide and a data bus 8 bits wide. that means that the m

15、icroprocessor an address (2n) 256 bytes of the memory and it can read or write 8 bits of the memory at a time. lets assume that this simple microprocessor has 128 bytes of rom starting at address 0 and 128 bytes of ram starting at address 128.rom stands for read-only memory. a rom chip is programmed

16、 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.ram stands for random-access memory. ram contains bytes of information, and the mic

17、roprocessor can read or write to those bytes depending on whether the rd or wr line is signaled. one problem with todays 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

18、 create a simple computer that contains no rammany microcontrollers do this by placing a handful of ram bytes on the processor chip itselfbut 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 b

19、egins 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. this boot sector is another small program, and the bios store it in ram after reading it off the disk. the microproces

20、sor then begins executing the boot sectors 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 entire operating system.mic

21、roprocessor instructions even the incredibly simple microprocessor shown here 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 a

22、re 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 languages of the processor. an assembler can translate the words into their bit patterns very easily, and then the output

23、 of assembler is placed in memory for the microprocessor to execute. if you use c language programming, a c compiler will translate the c code into assembly language.so now the question is, “how do all of these instructions look in rom?” each of these assembly language instructions must be represent

24、ed by a binary number. these numbers all know as recodes. the instruction decoder needs to turn each of recodes into a set of signals that drive the different components inside the microprocessor. lets take the add instruction as an example and look at what it needs to do.during the first clock cycl

25、e, 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.during the second clock cycle, the add inst

26、ruction is decoded. it needs to do very little: set the operation of alu to addition; latch the output of the alu into the c register.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

27、 as a set of sequenced operations like these that manipulate the components of 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.microprocessor performance the number of transistors avail

28、able 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 transisto

29、rs, 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

30、 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 own pipeline. this allows for multiple instruction streams, which means that more than one instruction

31、can complete during each clock cycle. this technique can be quite complex to implement, so it can be lots of transistors.the trend in processor design has been toward full 32-bit alu with fast floating point processors built in and pipelined execution with multiple instruction streams. there has als

32、o been a tendency toward special instructions that make certain operations particularly efficient. there has also been 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 power

33、houses available today. these processors can execute about one billion instructions per second!the operational amplifier will continue to be a vital component of analog design because it is a fundamental component. each generation of electronic equipment integrates more functions on silicon and take

34、s more of the analog circuitry inside the ic. as digital applications increase, analog applications also increase because the predominant supply of data and interface applications are in the real world, and the real world is an analog world.the lm386 is a power amplifier designed for use in low volt

35、age consumer applications. the gain is internally set to 20 to keep external part count low, but the addition of an external resistor and capacitor between pins 1 and 8 will increase the gain to any value from 20 to 200.the inputs are ground referenced while the output automatically biases to one-ha

36、lf the supply voltage. the quiescent power drain is only 24 mill watts when operating from a 6 volt supply, making the lm386 ideal for battery operation.a crystal is a basic piezoelectric quartz crystal. on its own, it cannot generate electrical clocks. it has to be connected to a clock oscillator t

37、o get a clock waveform. there are two kinds of crystals: series resonant, which can be modeled as a high q series lc circuit, and parallel resonant, which can be modeled as a high q parallel lc circuit. a crystal oscillator is an oscillator with the crystal as the feedback element. there are other k

38、inds of oscillators with active or passive feedback components, but the crystal oscillator provides the most accurate and stable output frequency. crystal oscillators are the preferred clock source in most high-speed digital systems requiring clocks. a chip is a small piece of conducting material on

39、 which an integrated circuit is embedded. a microprocessor is a silicon chip that contains a cpu. in operation, a computer is both hardware and software. one is useless without another. the hardware design specifies the commands it can follow, and the instructions tell it what to do. with the infilt

40、ration in the social field of the computer in recent years, the application of the one-chip computer is moving towards deepening constantly, drive tradition is it measure crescent benefit to upgrade day to control at the same time. in measuring in real time and automatically controlled one-chip comp

41、uter application system, the one-chip computer often uses as a key part, only one-chip computer respect knowledge is not enough, should also follow the structure of the concrete hardware , and direct against and use the software of targets characteristic to combine concretely, in order to do perfect

42、ly.译文：微处理器晶体管是电子学发展史上的关键发明之一，它实际上也是人类历史上最重要的发明之一。集成技术随着时间的推移而提高，芯片也更小，更快，更便宜。处理器完成的功能最早是由几个不同的逻辑芯片实现的，英特尔公司率先将所有这些部件集成到单个芯片中，这就是最早的微处理器，它是在单芯片上制造的完整的运算引擎。微处理器执行一组机器指令，这些指令告诉微处理器去做什么，根据这些指令，微处理器能够完成如下三项基本任务。1.微处理器使用其alu（算术/逻辑单元）可以完成加、减、乘、除等数学运算。2.微处理器可将数据从存储器的一个位置搬移到另一个位置。3.微处理器可做出判断，并根据这些判断跳转到一组新的指令

43、。一个微处理器可以做非常复杂的工作，但上述三项是最基本的。微处理器有一套地址总线（向存储器发送地址），一套数据总线（向存储器发送数据或者接收存储器数据），一条读信号线rd和一条写信号线wr（用于通知存储器是从寻址地址读取数据还是写入数据），一条时钟信号线（为处理器安排时序的时钟脉冲）和一条复位信号线（将程序计数器置零和重新开始执行）。这里假定数据总线和地址总线的宽度都是位。构成这个简易处理器的组件如下:寄存器a,寄存器b和寄存器c：它们是由触发器构成的简易锁存器。地址锁存器：和寄存器a，一样。程序计数器：一种具备“加一”功能和“置零”功能的锁存器。算术逻辑单元：可以简单到只是一个位加法器，也可

44、以是能够完成位加、减、乘、除的单元（此处我们假定为后者）。测试寄存器：一种保存alu比较结果的专用锁存器。通常，alu能够将两个数进行比较，并判断出二者是否相等或者一个比另一个更大。测试寄存器也可以保存加法运算最后一步的进位。这些数值保存在触发器当中，指令译码器利用这些数值做出判决。“state”是三态缓冲器。它可以传送逻辑，逻辑，或者和输出断开。三态缓冲器允许在一条信号线上连接多个输出信号，但只有一个信号输出。指令寄存器和指令译码器负责控制所有其他组件。从指令译码器引出完成如下功能的控制信号线：通知寄存器锁定当下出现在数据总线上的数值通知寄存器锁定当下出现在数据总线上的数值。通知寄存器锁定当

45、下出现在数据总线上的数值通知程序计数器锁定当下出现在数据总线上的数值通知地址寄存器锁定当下出现在数据总线上的数值通知指令寄存器锁定当下出现在数据总线上的数值通知程序计数器增加通知程序计数器复位置零激活任何一个三态缓冲器通知alu需要完成的操作通知测试寄存器锁定alu的测试位激活rd信号线激活wr信号线指令译码器的数据位不仅来自指令寄存器，而且来自测试寄存器和时钟信号线。只读存储器和随机存取存储器数据总线、地址总线、读写信号线都连接到rom上或者连接到ram上（通常两者都有）。在这个微处理器例子中，有一套位地址总线和一套位数据总线。这意味着微处理器可寻址256字节的存储器，一次可以读写位数据。假

46、定该微处理器有128字节（地址从开始）的ram和128字节（地址从128开始）的ram。rom是只读存储器。rom芯片是用一组永久的预设字节进行编程得到的。地址总线告知rom芯片要将哪个字节取出并置于数据总线上。当rd信号线改变状态时，rom芯片将选中的字节输出到数据总线上。rom是随机存取存储器。rom中包含着以字节为单位的信息，微处理器能够依据rd/wr信号哪个有效来决定字节的读写。当前ram芯片的一个问题是：掉电后，所有保存在ram上的内容全部丢失。这就是计算机需要rom的原因。顺便提一下，几乎所有计算机都有一定数量的rom（可以建造一种简单的不含ram的计算机许多微控制器在片内集成了一定数量的ram但是一般不可能建造出一种不含rom的计算机）。在pc机中，rom被称作bios基本输入输出系统）。当计算机启动时，它就执行在bios中找到的指令。这些 bios指令完成对机内硬件的测试，然后从硬盘中读取引导扇区。引导扇区也是一个小程序，bios将其从硬盘中读出来之后，这个小程序就存储在ram中。然后，微处理器开始从ram执行引导扇区的指令。这个程序将告知微处理器从硬盘其他位置读取信息到ram中，然后微处理器执行相应的指令等。这就是微处理