#include
#include
#include
#include
#define weiadd XBYTE[0xfbff]
#define duanadd XBYTE[0xfdff]
unsigned char data dispdata[4];
unsigned code disptab[]={0x01,0x02,0x04,0x08};
unsigned char code xtab[]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f};
sbit P10=P1^0;
sbit adclk=P1^1;
sbit adin=P1^2;
sbit adout=P1^3;
sbit adcs=P1^4;
unsigned char data i;
void display(void) interrupt 3
{
duanadd=0x00;
weiadd=0x00;
i++;i%=4;
duanadd=xtab[dispdata];
weiadd=disptab;
TH1=0xf8;
TL1=0x30;
TR1=1;
}
unsigned int read_ad(unsigned char anx)
{
idata unsigned int ad_v;
idata unsigned char an,i,j;
anx=anxadvalue)
{
t=advalue[a];
advalue[a]=advalue;
advalue=t;
}
}
for(j=1;j<7;j++)
{
sum+=advalue[j];
}
js=(int)(sum/6);
dispdata[0]=js/1000;
dispdata[1]=(js%1000)/100;
dispdata[2]=(js%100)/10;
dispdata[3]=js%10;
sum=0;
delay(250);
delay(250);
delay(250);
delay(250);
delay(250);
}
}
这是pt100测温的程序，但是不完整，还需要再加公式！
公式是求输出模拟量然后转换为电压值，
输出模拟量=4.096*（Rpt100/(R1+Rpt100)-Rvr2/(R1+Rvr2)）
转换成电压的公式：x=5*i/1024 (注：i为变量，也就是上一个公式输出的模拟量)
哪位亲人能帮我解答，必有高分悬赏！！！我的邮箱是[email protected] 先谢谢啦。。。。。
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