1 / 17
文档名称:

宁——英文翻译.doc

格式:doc   大小:100KB   页数:17页
下载后只包含 1 个 DOC 格式的文档,没有任何的图纸或源代码,查看文件列表

如果您已付费下载过本站文档,您可以点这里二次下载

分享

预览

宁——英文翻译.doc

上传人:lixinwxy99999999 2016/12/29 文件大小:100 KB

下载得到文件列表

宁——英文翻译.doc

相关文档

文档介绍

文档介绍:专业文档珍贵文档英文翻译 Overview The 8051 family of micro controllers is based on an architecture which is highly optimized for embedded control systems. It is used ina wide variety of applications from military equipment to automobiles to the keyboard on your PC. Second only to the Motorola 68HC11 in eight bit processors sales, the 8051 family of microcontrol l ers is available ina wide array of variations from manufacturers such as Intel, Philips, and Siemens. These manufacturers have added numerous features and peripherals to the 8051 such asI2C interfaces, analog to digital converters, watchdog timers, and pulse width modulated outputs. Variations of the 8051 with clock speeds up to 40MHz and voltage requirements down volts are available. This wide range of parts based onone core makes the 8051 family an excellent choice as the base architecture for pany's entire line of products since it can perform many functions and developers will only have to learn this one platform. The basic architecture consists of the following features: 1 an eight bit ALU 232 descrete I/O pins (4 groups of8) which can be individually accessed 3 two 16 bit timer/counters 4 full duplex UART 56 interrupt sources with 2 priority levels 6128 bytes ofon board RAM 7 separate 64K byte address spaces for DATA and CODE memory One 8051 processor cycle consists of twelve oscillator periods. Each of the twelve oscillator periods is used for a special function by the 8051 core such asop code fetches and samples of the interrupt daisy chain for pending interrupts. The time required for any 8051 instruction can puted by dividing the clock frequency by12, inverting that result and multiplying itby the number of processor cycles required by the instruction in question. Therefore, if you have a system which is using an clock, you pute the number of instructions per second by dividing this value by12. This gives an instruction frequency of921583 instructions per second. Inverting this will provide th