电子信息工程外文翻译外文文献英文文献微处理器
外文资料
所译外文资料：
1. 作者G..Bouwhuis, J.Braat, A.Huijser
2. 书名:Principles of Optical Disk Systems
3. 出版时间：1991年9月
4. 所译章节：Session 2/Chapter9, Session 2/Chapter 11
原文：
Microprocessor
One of the key inventions in the history of electronics, and in fact one of the most important inventions ever period, was the transistor. As time progressed after the inven ti on of LSI in tegrated circuits, the tech no logy improved and chips became smaller, faster and cheaper. The functions performed by a processor were impleme nted using several differe nt logic chips. In tel was the first compa ny to in corporate all of these logic comp onents into a si ngle chip, this was the first microprocessor. A microprocessor is a complete computati on engine that is fabricated on a sin gle chip. A microprocessor executes a collecti on of machi ne in struct ions that tell the processor what to do. Based on the in struct ions, a microprocessor does three basic things: https://www.wendangku.net/doc/0d13968067.html,ing the ALU (Arithmetic/Logic Unit), a microprocessor can perform mathematical operatio ns like additi on, subtract ion, multiplicatio n and divisi on; 2.A microprocessor can move data from one memory location to another; 3.A microprocessor can make decisi ons and jump to a new set of in struct ions based on those decisi ons.
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(write) line that lets a clock pulse sequenee the processor and a reset li ne that resets the program coun ter to zero(or whatever) and restarts executi on. And let ' s assume that both the address and data buses are 8 bits wide here.
Here are the comp onents of this simple microprocessor:
1. Registers A, B 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 coun ter is a latch with the extra ability to in creme nt by 1 whe n told
to do so, and also to reset to zero whe n told to do so.
4. The ALU could be as simple as an 8-bit adder, or it might be able to add, subtract, multiply and divide 8- bit values. Let ' s 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, 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 decisi ons.
6. There are six boxes marked “3tate ” .Theseire tri-state buffers. A tri-state buffer can pass a 1, 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 in structio n register and in structio n decoder are resp on sible for con trolli ng all of the
other comp onen ts.
Although they 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 data bus
4. Tell the program counter register to latch the value currently on the data bus
5. Tell the address register to latch the value currently on the data bus
6. Tell the instruction register to latch the value currently on the data bus
7. Tell the program coun ter to in creme nt
8. Tell the program coun ter to reset to zero
9. Activate any of the six tri-state buffers (six separate lin es)
10. Tell the ALU what operati on to perform
11. Tell the test register to latch the ALU ' s test bits
12. Activate the RD line
13. Activate the WR line
Comi ng into the in structi on decoder are the bits from the best register 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 ROM —gen erally 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 an address (2 n) 256 bytes of the 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 starti ng at address 0 and 128 bytes of RAM starti ng at address 128.
ROM sta nds for read-only memory. A ROM chip is programmed with a perma nent 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 cha nges state, the ROM chip prese nts the selected byte onto the data bus.
RAM sta nds for ran dom-access memory. RAM contains bytes of in formati on, and the microprocessor can read or write to those bytes depe nding on whether the RD or WR li ne is sig naled.
One problem with today ' s^AM chips is that they forget everythi ng once the power goes off. That is why the computer n eeds ROM.
By the way, n early all computers con tai n some amount of ROM (it is possible to create a simple computer that contains no RAM —many microcontrollers do this by placi ng a han dful of RAM bytes on the processor chip itself ——but gen erally 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. This boot sector is ano ther small program, and the BIOS store 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 someth ing else from the hard disk into RAM, which the microprocessor the n executes, and so on. This is how the microprocessor loads and executes en tire operati ng system.
Microprocessor In structio ns Eve n the in credibly simple microprocessor show n here will have a fairly large set of in structio ns that it can perform. The collectio n of in structi ons is impleme nted as bit patter ns, each one of which has a differe nt meaning when loaded into the instruction register. Humans are not particularly good at rememberi ng bit patter ns, so a set of short words are defi ned to represe nt the differe nt bit patter ns. This collect ion of words is called the assembly Ian guages of the processor. An assembler can tran slate the words into their bit patter ns very easily, and then the output of assembler is placed in memory for the microprocessor to execute. If you use C Ian guage program ming, a C compiler will tran slate the C code in to assembly Ian guage.
So now the questi on is, “ How do all of these in structio ns look in ROM? ” Eac
these assembly Ian guage in struct ions must be represe nted by a binary nu mber.
These nu mbers all know as recodes. The in structi on decoder n eeds to turn each of recodes into a set of sig nals that drive the differe nt comp onents in side the microprocessor. Let ' take the ADD instruction as an example and look at what it n eeds to do.
During the first clock cycle, we need to actually load the instruction. Therefore the in structi on decoder n eeds to: activate the tri-state buffer for the program coun ter; activate the RD line; activate the data-in tri-state buffer; latch the instruction into the in structi on register.
During the second clock cycle, the ADD instruction 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 coun ter is in creme nted (in theory this could be overlapped into the sec ond clock cycle).
Every in struct ion can be broke n dow n as a set of seque need operati ons like these that mani pulate the comp onents 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 Performanee The number of transistors available has a huge effect on the performa nee of a processor. As see n earlier, a typical in struct ion in a processor like an 8088 took 15 clock cycles to execute. Because of the desig n of the multiplier, it took approximately 80 cycles just to do one 16-bit multiplication on the 8088. With more tran sistors, much more powerful multipliers capable of sin gle-cycle speeds become possible.
More tran sistors also allow for a tech no logy called pipeli ning. In a pipeli ned architecture, i nstructio n executi on overlaps. So eve n though it might take five clock cycles to execute each instruction, there can be five instructions in various stages of executi on simulta neously. That way it looks like one in structi on 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 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 also been a tendency toward special instructions that make certain operati ons particularly efficie nt. There has also bee n the additi on of hardware virtual memory support and L1 cach ing 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 billio n in structi ons per sec ond!
The Operational Amplifier will continue to be a vital component of analog design because it is a fun dame ntal comp onent. Each gen erati on of electr onic equipme nt in tegrates more functions on silic on and takes more of the an alog circuitry in side the IC. As digital applicati ons in crease, an alog applicati ons also in crease because the predo minant supply of data and in terface applicati ons are in the real world, and the real world is an an alog world.
The LM386 is a power amplifier designed for use in low voltage consumer applications. The gain is internally set to 20 to keep external part count low, but the additi on of an external resistor and capacitor betwee n pins 1 and 8 will in crease the gai n to any value from 20 to 2OO.The in puts are ground refere need while the output automatically biases to on e-half the supply voltage. The quiesce nt power drain is only 24 mill watts when operating from a 6 volt supply, making the LM386 ideal for battery operati on.
A Crystal is a basic piezoelectric quartz crystal. On its own, it cannot gen erate electrical clocks. It has to be conn ected to a clock oscillator to 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 kinds of oscillators with active or passive feedback comp onen ts, but the crystal oscillator provides the most accurate and stable output freque ncy. Crystal oscillators are the preferred clock source in most high-speed digital systems requiri ng clocks. A chip is a small
piece of con duct ing material on which an in tegrated 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 desig n specifies the comma nds it can follow, and the in structi ons tell it what to do. With the infiltration in the social field of the computer in recent years, the applicati on of the on e-chip computer is moving towards deepe ning con sta ntly, drive traditi on is it measure cresce nt ben efit to upgrade day to con trol at the same time. In measuri ng in real time and automatically con trolled on e-chip computer applicati on system, the one-chip computer often uses as a key part, only one-chip computer respect kno wledge is not eno ugh, should also follow the structure of the con crete hardware , and direct against and use the software of target's characteristic to comb ine con cretely, in order to do perfectly. 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。
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8位。构成这个简易处理器的组件如下：
1.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。
2.地址锁存器：和寄存器A，B，C一样。
3.程序计数器：一种具备加一”功能和置零”功能的锁存器。
4.算术逻辑单元：可以简单到只是一个8位加法器，也可以是能够完成8位加、减、乘、除的单元（此处我们假定为后者）。
5 .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。
6. 3 - State是三态缓冲器。它可以传送逻辑1,逻辑0,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。
7.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：
1.通知寄存器A锁定当下出现在数据总线上的数值
2.通知寄存器B锁定当下出现在数据总线上的数值。
3.通知寄存器C锁定当下出现在数据总线上的数值
4.通知程序计数器锁定当下出现在数据总线上的数值
5 .通知地址寄存器锁定当下出现在数据总线上的数值
6 ?通知指令寄存器锁定当下出现在数据总线上的数值
7?通知程序计数器增加
8?通知程序计数器复位置零
9?激活任何一个三态缓冲器
10 .通知ALU需要完成的操作
11.通知测试寄存器锁定ALU的测试位
12.激活RD信号线
13.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8位地址总线和一套8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8 位数据。假定该微处理器有128字节（地址从0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的计算机一一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。这就是微处理器装载和执行整个操作系统的过程。
微处理器指令甚至这里给出的简单得难以置信的微处理器也拥有一套相当大的指令集。指令的集合是以比特组合的方式实现的；每一条指令在装载到指令寄存器的时候，都有不同的涵义。人类不善于记忆比特组合，因此定义了一组短字来代表不同的比特组合。这些短字的集合就称为处理器汇编语言。汇编器可以很容易地将这些短字翻译成与其对应的比特组合，汇编器的输出被放置到存储器中以便微处理器执行。假如使用C语
言进行编程，那么编译器会将C代码翻译为汇编语言。
微处理器性能可用晶体管数量对于微处理器性能有很大的影响。正如先前看到的那样，像8088这样的处理器执行一条典型指令需要15个时钟周期。由于要设计乘法器，在8088上完成一次16位乘法需要约80个时钟周期。晶体管越多，具备单周期乘法能力的乘法器就会越多。
更多的晶体管允许使用流水线技术。在流水线结构中，指令的执行是重叠的。这样的话，尽管执行每条指令可能需要5个周期，却可以在不同阶段同时执行5条指令。这样看上去好像每个周期都能完成一条指令。
许多现代处理器有多个指令译码器，而每个指令译码器都有各自的流水线。这样，就可以实现多指令流——即在一个周期内可以完成多条指令。该技术实现起来相当复杂，所以使用大量的晶体管。
处理器设计的趋势已经是全32位ALU、内置快速浮点处理器和多指令流水线。还有一个趋势是采用能使特定操作高效执行的特殊指令。此外，还有一种趋势是在处理器芯片附加上硬件虚拟存储器和
L1高速缓存。所有这些趋势都需要增加晶体管，这导致了今天集成度高达几百万晶体管芯片的出现。这些处理器在1秒内可以执行约10亿
条指令。
运放是一种基础的部件，它将作为模拟设计达到的关键部件，每一代电子设备在硅片上集成了更多的功能，将更多的模拟电路置于集成电路内部，随着数字应用的增加，模拟应用也会增加，因为大量的数据应用和接口应用都在现实世界中，而现实世界是一个模拟的世界。
晶体是一种基本的压电石英晶体，它本身是不能产生时钟信号的，必须和时钟振荡器连接在一起才能得到时钟波形。晶体有两种：串联谐振晶体（可视做高品质因数的串联LC电路）和并联谐振晶体（可视做高品质因数的并联LC电路）o晶体振荡器是一种用
晶体做反馈元件的振荡器。而其他类型的振荡器采用有源，无源元件作为反馈元件，但晶体振荡器的输出频率最为精确和稳定。晶体振荡器是多数高速数字系统时钟源的首先。
LM386是美国国家半导体公司生产的音频功率放大器，主要应用于低电压消费类产品。为使外围元件最少，电压增益内置为20。但在1脚和8脚之间增加一只外接电阻和电容,便可将电压增益调为任意值，直至200 o输入端以地为参考，同时输出端被自动偏置到电源电压的一半,在6V电源电压下,它的静态功耗仅为24mW,使得LM386特别适用于电池供电的场合。
芯片是嵌入了集成电路的一小片半导体材料，微处理器是包含CPU的一片硅片。
一部计算机在运转上既有硬件又有软件，没有对方，哪个也没有用。硬件设计指定了计算机能够遵循的命令，而指令告诉计算机该做什么。近年来随着计算机在社会领域的渗透，单片机的应用正在不断地走向深入，同时带动传统控制检测，日新月益更新。在实时检测和自动控制的单片机应用系统中，单片机
往往是作为一个核心部件来使用，仅单片机方面知识是不够的，还应根据具体硬件结构，以及针对具体
应用对象特点的软件结合，以作完善。
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