Question

高分求单片机多点温度检测系统的程序设计

我做的是基于DS1820的温度检测系统，单片机选用的是AT89C51,9个LED共阳极显示(共阴极也可以~),还有MAX232的串口通讯电路和一些其他的辅助电路~需要的是主程序和一些子程序~急需!!用汇编也可以用C语言也可以~!谢谢!!

Answer 1

只有一个例子,鉴于参考而已,希望对你帮助
//安装目录下的EXE文件打开后可在电脑上显示当前温度值
#include <reg52.h>
#include <INTRINS.H>
unsigned char code displaybit[]={0xf7,0xfb,0xfd,0xfe};
unsigned char code displaycode[]={ 0xC0,/*0*/
0xF9,/*1*/
0xA4,/*2*/
0xB0,/*3*/
0x99,/*4*/
0x92,/*5*/
0x82,/*6*/
0xF8,/*7*/
0x80,/*8*/
0x90,/*9*/};
unsigned char code dotcode[32]={0,3,6,9,12,16,19,22,
25,28,31,34,38,41,44,48,
50,53,56,59,63,66,69,72,
75,78,81,84,88,91,94,97};
unsigned char displaycount;
unsigned char displaybuf[8]={16,16,16,16,16,16,16,16};
unsigned char timecount;
unsigned char readdata[8];

sbit DQ=P1^0;
bit sflag;

bit resetpulse(void)
{
unsigned char i;

DQ=0;
for(i=255;i>0;i--);
DQ=1;
for(i=60;i>0;i--);
return(DQ);
for(i=200;i>0;i--);
}

void writecommandtods18b20(unsigned char command)
{
unsigned char i;
unsigned char j;

for(i=0;i<8;i++)
{
if((command & 0x01)==0)
{
DQ=0;
for(j=35;j>0;j--);
DQ=1;
}
else
{
DQ=0;
for(j=2;j>0;j--);
DQ=1;
for(j=33;j>0;j--);
}
command=_cror_(command,1);
}
}

unsigned char readdatafromds18b20(void)
{
unsigned char i;
unsigned char j;
unsigned char temp;

temp=0;
for(i=0;i<8;i++)
{
temp=_cror_(temp,1);
DQ=0;
_nop_();
_nop_();
DQ=1;
for(j=10;j>0;j--);
if(DQ==1)
{
temp=temp | 0x80;
}
else
{
temp=temp | 0x00;
}
for(j=200;j>0;j--);
}
return(temp);
}

void main(void)
{
TMOD=0x01;
TH0=(65536-4000)/256;
TL0=(65536-4000)%256;
ET0=1;
EA=1;

while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0x44);
TR0=1;
while(1)
{
;
}
}

void t0(void) interrupt 1 using 0
{
unsigned char x;
unsigned int result;

TH0=(65536-4000)/256;
TL0=(65536-4000)%256;
if(displaycount==2)
{
P0=displaycode[displaybuf[displaycount]] | 0x80;
}
else
{
P0=displaycode[displaybuf[displaycount]];
}
P2=displaybit[displaycount];
displaycount++;
if(displaycount==8)
{
displaycount=0;
}

timecount++;
if(timecount==150)
{
timecount=0;
while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0xbe);
readdata[0]=readdatafromds18b20();
readdata[1]=readdatafromds18b20();
for(x=0;x<8;x++)
{
displaybuf[x]=16;
}
sflag=0;
if((readdata[1] & 0xf8)!=0x00)
{
sflag=1;
readdata[1]=~readdata[1];
readdata[0]=~readdata[0];
result=readdata[0]+1;
readdata[0]=result;
if(result>255)
{
readdata[1]++;
}
}
readdata[1]=readdata[1]<<4;
readdata[1]=readdata[1] & 0x70;
x=readdata[0];
x=x>>4;
x=x & 0x0f;
readdata[1]=readdata[1] | x;
x=2;
result=readdata[1];
while(result/10)
{
displaybuf[x]=result%10;
result=result/10;
x++;
}
displaybuf[x]=result;
if(sflag==1)
{
displaybuf[x+1]=17;
}
x=readdata[0] & 0x0f;
x=x<<1;
displaybuf[0]=(dotcode[x])%10;
displaybuf[1]=(dotcode[x])/10;
while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0x44);
}
}
这是汇编
unsigned char code displaybit[]={0xfe,0xfd,0xfb,0xf7,
0xef,0xdf,0xbf,0x7f};
unsigned char code displaycode[]={0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71,0x00,0x40};
unsigned char code dotcode[32]={0,3,6,9,12,16,19,22,
25,28,31,34,38,41,44,48,
50,53,56,59,63,66,69,72,
75,78,81,84,88,91,94,97};
unsigned char displaycount;
unsigned char displaybuf[8]={16,16,16,16,16,16,16,16};
unsigned char timecount;
unsigned char readdata[8];

sbit DQ=P3^7;
bit sflag;

bit resetpulse(void)
{
unsigned char i;

DQ=0;
for(i=255;i>0;i--);
DQ=1;
for(i=60;i>0;i--);
return(DQ);
for(i=200;i>0;i--);
}

void writecommandtods18b20(unsigned char command)
{
unsigned char i;
unsigned char j;

for(i=0;i<8;i++)
{
if((command & 0x01)==0)
{
DQ=0;
for(j=35;j>0;j--);
DQ=1;
}
else
{
DQ=0;
for(j=2;j>0;j--);
DQ=1;
for(j=33;j>0;j--);
}
command=_cror_(command,1);
}
}

unsigned char readdatafromds18b20(void)
{
unsigned char i;
unsigned char j;
unsigned char temp;

temp=0;
for(i=0;i<8;i++)
{
temp=_cror_(temp,1);
DQ=0;
_nop_();
_nop_();
DQ=1;
for(j=10;j>0;j--);
if(DQ==1)
{
temp=temp | 0x80;
}
else
{
temp=temp | 0x00;
}
for(j=200;j>0;j--);
}
return(temp);
}

void main(void)
{
TMOD=0x01;
TH0=(65536-4000)/256;
TL0=(65536-4000)%256;
ET0=1;
EA=1;

while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0x44);
TR0=1;
while(1)
{
;
}
}

void t0(void) interrupt 1 using 0
{
unsigned char x;
unsigned int result;

TH0=(65536-4000)/256;
TL0=(65536-4000)%256;
if(displaycount==2)
{
P0=displaycode[displaybuf[displaycount]] | 0x80;
}
else
{
P0=displaycode[displaybuf[displaycount]];
}
P2=displaybit[displaycount];
displaycount++;
if(displaycount==8)
{
displaycount=0;
}

timecount++;
if(timecount==150)
{
timecount=0;
while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0xbe);
readdata[0]=readdatafromds18b20();
readdata[1]=readdatafromds18b20();
for(x=0;x<8;x++)
{
displaybuf[x]=16;
}
sflag=0;
if((readdata[1] & 0xf8)!=0x00)
{
sflag=1;
readdata[1]=~readdata[1];
readdata[0]=~readdata[0];
result=readdata[0]+1;
readdata[0]=result;
if(result>255)
{
readdata[1]++;
}
}
readdata[1]=readdata[1]<<4;
readdata[1]=readdata[1] & 0x70;
x=readdata[0];
x=x>>4;
x=x & 0x0f;
readdata[1]=readdata[1] | x;
x=2;
result=readdata[1];
while(result/10)
{
displaybuf[x]=result%10;
result=result/10;
x++;
}
displaybuf[x]=result;
if(sflag==1)
{
displaybuf[x+1]=17;
}
x=readdata[0] & 0x0f;
x=x<<1;
displaybuf[0]=(dotcode[x])%10;
displaybuf[1]=(dotcode[x])/10;
while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0x44);
}
}

Answer 2

帮你顶一下.做这个只给分不给钱是很难的..

Answer 3

参考《基于PROTBUSR的电路及单片机系统设计与仿真》 周润景 张丽娜编著 北京航空航天大学出版社。
有原码，有图纸.