请问读写AT24C256的时序是扫描样吗？

1582 AT24c256 单片机

问答对人有帮助，内容完整，我也想知道答案 0 最近在研究IIC通讯，做IIC读写AT24C256的程序，但是单片机用1M晶振可以正常运行，换成12M就出错，这应该就是时序的问题了，哪位大侠能帮我看看12M频率下应该把时序调整成扫描样呢？？[C] 纯文本查看复制代码 #ifndef AT24CXX_H_#define AT24CXX_H_#include #include #define Wait_1us _nop_();_nop_();_nop_();_nop_();_nop_();_nop_();_nop_();_nop_();_nop_();_nop_()#define Wait_2us {Wait_1us;Wait_1us;}#define Sla_Wr 0xa0 /0000 0 0 0 0/#define Sla_Rd 0xa1 /0000 0 0 0 1/#define AT24C01 0x01#define AT24C01A 0x02#define AT24C02 0x03#define AT24C04 0x04#define AT24C08 0x05#define AT24C16 0x06#define AT24C32 0x07#define AT24C64 0x08#define AT24C128 0x09#define AT24C256 0x0a#define AT24C512 0x0b#define Mem_Size AT24C04 // 01/01A/02/04/08/16/32/64/128/256/512#define Chip_1 0x00 //本器件的器件地址，可以第一8个不同地址的芯片使之能用同一段程序进行读写void Start_I2c(void);void End_I2c(void);unsigned char Scan_I2c();void Send_I2c(unsigned char cData);void Send_Ack(void);void Send_Not_Ack(void);void Write_Mem(unsigned char cMem_Size,cChips_Address,cData_Size,unsigned int cStart_Address);void Read_Mem(unsigned char cMem_Size,cChips_Address,cData_Size,unsigned int cStart_Address);#endif[C] 纯文本查看复制代码 #ifndef AT24CXXX_C_#define AT24CXXX_C_#include "at24cxx.h"*it bSda =P0^0;it bScl =P0^1; bit bScan_Err;unsigned char idata cShare_Buf[8]={'P',1,2,3,4,5,6,7}; //I2c read and write buffervoid Start_I2c(void){ // unsigned char /idata/ i; bSda = 1;// Wait_1us; bScl = 1;// for(i=0;i<32;i++); //wait 7us Wait_1us; bSda = 0;// for(i=0;i<32;i++); //wait 7us Wait_1us; bScl = 0; Wait_1us;}void End_I2c(void){// unsigned char /idata/ i; bSda=0; Wait_1us; bScl=1;// for(i=0;i<16;i++); //wait 7us Wait_1us; bSda = 1; Wait_1us;}unsigned char Scan_I2c(){ unsigned char /idata/ i,j;//IIC data bSda = 1; j = 0; for(i=0;i<8;i++) { bScl = 1; j <<= 1; Wait_1us; j \|= bSda; //Scan 0 Scan 1 bScl=0; Wait_1us; } return j;}void Send_I2c(unsigned char cData){ unsigned char /idata/ i; for(i=0;i<8;i++) { bSda=(bit)(cData&0x80); Wait_1us; bScl=1; cData<<=1; Wait_1us; bScl=0; Wait_1us; } bSda=1; Wait_1us; bScl=1; //检测ACK for(i=0;(bSda==1)&&(bScan_Err==0);i++) { if(i==90) bScan_Err=1; //如果101us没有响应就错误 } bScl=0; Wait_1us;}void Send_Ack(void){//unsigned char /idata/ i; bSda=0; Wait_1us; bScl=1;// for(i=0;i<2;i++); //wait 8us Wait_1us; bScl=0; Wait_1us;}void Send_Not_Ack(void){// unsigned char /idata/ i; bSda = 1; Wait_1us; bScl = 1;// for(i=0;i<2;i++); //wait 8us Wait_1us; bScl=0; Wait_1us;} void Write_Mem(unsigned char cMem_Size,cChips_Address,cData_Size,unsigned int cStart_Address){ unsigned char /idata/ i,cAddress[2]; //0-L;1-H bScan_Err=0; cAddress[0]=cStart_Address; cStart_Address>>=8; cAddress[1]=cStart_Address; i=Sla_Wr; i\|=cChips_Address; if(cMem_Size==AT24C256) i&=0xf7; Start_I2c(); Wait_1us; Send_I2c(i); //SLA:1010 xxx0 Wait_1us; if(cMem_Size==AT24C256) //1-16用1个字节地址，32-512用2个字节地址，匹配地址AT24C32、64 Send_I2c(cAddress[1]); //先发送高字节地址 Send_I2c(cAddress[0]); //如果是双字节地址则为低地址否则为高地址 for(i=0;(bScan_Err==0)&&(i!=cData_Size);i++) Send_I2c(cShare_Buf); //写一个字节数据 End_I2c(); }void Read_Mem(unsigned char cMem_Size,cChips_Address,cData_Size,unsigned int cStart_Address){ unsigned char /idata*/ i,cAddress[2]; bScan_Err=0; cAddress[0]=cStart_Address; //将16位地址分解为高低字节 cStart_Address>>=8; cAddress[1]=cStart_Address; i = Sla_Wr; i \|= cChips_Address; if(cMem_Size==AT24C256) i&=0xf7; Start_I2c(); Send_I2c(i);//SLA:1010 xxx0 if(cMem_Size==AT24C256) //1-16用1个字节地址，32-512用2个字节地址，匹配地址AT24C32、64 Send_I2c(cAddress[1]); //先发送高字节地址 Send_I2c(cAddress[0]); //address and write i\|=0x01; //Read Start_I2c(); Send_I2c(i); //address and read for(i=0;(bScan_Err==0)&&(i!=cData_Size);i++) { cShare_Buf=Scan_I2c(); //读入第一个字节 if(i==(cData_Size-1)) Send_Not_Ack(); else Send_Ack(); } End_I2c();}#endif 0
2019-5-10 04:14:59　　评论淘帖0 邀请回答您可以邀请以下用户，快速回答问题 × heks 该类别下有 20 个回答。邀请回答新星之火12138 该类别下有 18 个回答。邀请回答 wang21cj 该类别下有 17 个回答。邀请回答 hgimtk 该类别下有 16 个回答。邀请回答科源机电该类别下有 15 个回答。邀请回答 liutiefu 该类别下有 15 个回答。邀请回答 kghfh 该类别下有 14 个回答。邀请回答 dgfdf 该类别下有 13 个回答。邀请回答 723662364d 该类别下有 13 个回答。邀请回答一巷清苑该类别下有 13 个回答。邀请回答 chm5 该类别下有 13 个回答。邀请回答 hjhdf 该类别下有 12 个回答。邀请回答 efwedfd 该类别下有 12 个回答。邀请回答小华同学该类别下有 12 个回答。邀请回答 jsqueh 该类别下有 12 个回答。邀请回答红旧衫该类别下有 12 个回答。邀请回答早知该类别下有 12 个回答。邀请回答 Ehunt 该类别下有 12 个回答。邀请回答张峰9998 该类别下有 12 个回答。邀请回答 hy381 该类别下有 12 个回答。邀请回答举报杨平相关推荐 • AT24c256 3139 • 对什么对EEPROM进行连续写操作时程序会跑飞 4232 • 铁电存储器FM24C256在电能表中的使用概述 14318 • AT24C256能不能写入16位的数字？ 1745 • 请问CH368芯片是否支持外接AT24C256C型号的EEPROM？ 459 • 怎样使用STM32的IO口模拟IIC并对AT24C128进行读写呢 1369 • 利用软件模拟i2c读写EEPROM，超过256个字节就读写不了了的原因？如何处理？ 256 • 关于PIC24FJ256GB206读写sd卡的程序 3586 • spi读写时序里,图中圆圈处MISO和MOSI为什么不一样代表什么时序 6778 • FM24C64和EEPROM AT24C04 在读写上有什么区别 11921 2个回答

答案对人有帮助，有参考价值 0 就是调延时了.... 12M说明你MCU速度变快了,延时要改长. 说白了,就是要Wait_1us是真正的1us.

2019-5-10 09:39:23 评论举报莫循虎