2020-02-04 09:10发布
Arachne29 发表于 2012-4-12 15:39 我曾经做过一个,你可以看一下 #include
yesno 发表于 2012-4-18 14:11 大叔我想认识下你,,教教我温度传感器吧 ds18的麻烦说下qq好吗?
最多设置5个标签!
大叔我想认识下你,,教教我温度传感器吧 ds18的麻烦说下qq好吗?
嘿嘿 很荣幸,我的qq648855521
/*DS18B20 正常显示 by whj 11 11 18*/
#include<reg52.h>
//#include<intrins.h>
//#include<absacc.h>
//#include <stdio.h>
#define uint unsigned int
#define uchar unsigned char
sbit DS = P2^0; //DS18B20的数据输入/输出脚DQ,根据情况设定
uchar flag1; //正负标志
uchar code table[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};
uchar code table1[]={0xbf,0x86,0xdb,0xcf,0xe6,0xed,0xfd,0x87,0xff,0xef};
uchar code wei[]={0x00,0x01,0x02,0x04,0x08};
uint temp,t;
void delay(uint count) //delay
{
uint i;
while(count)
{
i=200;
while(i>0)
i--;
count--;
}
}
void dsreset(void) //send reset and initialization command
{
uint i;
DS=0;
i=103;
while(i>0)i--;
DS=1;
i=4;
while(i>0)i--;
}
bit tmpreadbit(void) //read a bit
{
uint i;
bit dat;
DS=0;i++; //i++ for delay
DS=1;i++;i++;
dat=DS;
i=8;while(i>0)i--;
return (dat);
}
uchar tmpread(void) //read a byte date
{
uchar i,j,dat;
dat=0;
for(i=1;i<=8;i++)
{
j=tmpreadbit();
dat=(j<<7)|(dat>>1); //读出的数据最低位在最前面,这样刚好一个字节在DAT里
}
return(dat);
}
void tmpwritebyte(uchar dat) //write a byte to ds18b20
{
uint i;
uchar j;
bit testb;
for(j=1;j<=8;j++)
{
testb=dat&0x01;
dat=dat>>1;
if(testb) //write 1
{
DS=0;
i++;i++;
DS=1;
i=8;while(i>0)i--;
}
else
{
DS=0; //write 0
i=8;while(i>0)i--;
DS=1;
i++;i++;
}
}
}
void tmpchange(void) //DS18B20 begin change
{
dsreset();
delay(1);
tmpwritebyte(0xcc); // address all drivers on bus
tmpwritebyte(0x44); // initiates a single temperature conversion
}
uint tmp() //get the temperature
{
float tt;
uchar a,b;
dsreset();
delay(1);
tmpwritebyte(0xcc);
tmpwritebyte(0xbe);
a=tmpread();
b=tmpread();
temp=b;
temp<<=8; //two byte compose a int variable
temp=temp|a;
tt=temp*0.0625;
temp=tt*10+0.5;
return temp;
}
void display(uint temp)
{
uchar A1,A2,A2t,A3;
A1=temp/100;
A2t=temp%100;
A2=A2t/10;
A3=A2t%10;
t=400;
P1= wei[1];;
if (flag1==1)
{
P0=0x40;
}
else
{
P0=0x00;
}
while(t--);
t=400;
P1= wei[2];
P0=table[A1];
while(t--);
t=400;
P1= wei[3];
P0=table1[A2];
while(t--);
t=400;
P1= wei[4];
P0=table[A3];
while(t--);
P1=wei[0];
}
void main()
{
uchar a;
do
{
tmpchange();
for(a=10;a>0;a--)
{
display(tmp());
}
} while(1);
}
一周热门 更多>