Question

急求简易数字电压表课程设计（原理图加汇编c程序）用adc0809 谢谢 qq312188954

Answer 1

#include <AT89X52.H>
unsigned char code dispbitcode[]={0xfe,0xfd,0xfb,0xf7,
0xef,0xdf,0xbf,0x7f};
unsigned char code dispcode[]={0x3f,0x06,0x5b,0x4f,0x66,
0x6d,0x7d,0x07,0x7f,0x6f,0x00};
unsigned char dispbuf[8]={10,10,10,10,0,0,0,0};
unsigned char dispcount;
unsigned char getdata;
unsigned int temp;
unsigned char i;
sbit ST=P3^0;
sbit OE=P3^1;
sbit EOC=P3^2;
sbit CLK=P3^3;
void main(void)
{
ST=0;
OE=0;
ET0=1;
ET1=1;
EA=1;
TMOD=0x12;
TH0=216;
TL0=216;
TH1=(65536-4000)/256;
TL1=(65536-4000)%256;
TR1=1;
TR0=1;
ST=1;
ST=0;
while(1)
{
if(EOC==1)
{
OE=1;
getdata=P0;
OE=0;
temp=getdata*235;
temp=temp/128;
i=5;
dispbuf[0]=10;
dispbuf[1]=10;
dispbuf[2]=10;
dispbuf[3]=10;
dispbuf[4]=10;
dispbuf[5]=0;
dispbuf[6]=0;
dispbuf[7]=0;
while(temp/10)
{
dispbuf[i]=temp%10;
temp=temp/10;
i++;
}
dispbuf[i]=temp;
ST=1;
ST=0;
}
}
}
void t0(void) interrupt 1 using 0
{
CLK=~CLK;
}

void t1(void) interrupt 3 using 0
{
TH1=(65536-4000)/256;
TL1=(65536-4000)%256;
P1=dispcode[dispbuf[dispcount]];
P2=dispbitcode[dispcount];
if(dispcount==7)
{
P1=P1 | 0x80;
}
dispcount++;
if(dispcount==8)
{
dispcount=0;
}
}