51proteus simulation: adc0804 realizes analog-to-digital conversion

The signals collected by the system are almost all analog signals. It is necessary to learn the analog-to-digital AD conversion of the microcontroller.

This simulation uses adc0804 to convert analog voltage into digital signal, and then output it to the digital tube.

Since the maximum voltage does not exceed 5v, the digital tube only displays the units digit and the digits after the decimal point. I only display 3 digits after the decimal point without rounding.

Simulation circuit diagram:

c program:

#include
#include

#define uchar unsigned char

#define uint unsigned int

sbit duan=P3^0; //u1
sbit wei=P3^1;

sbit ADC_CS=P3^2;
sbit ADC_RD=P3^3;
sbit ADC_WR=P3^4;

uchar code table[]={
0x3f,0x06,0x5b,0x4f,
 0x66,0x6d,0x7d,0x07,
 0x7f,0x6f,0x77,0x7c,
 0x39,0x5e,0x79,0x71 };

void delay(uint xms)
{
  uint i,j;
  for(i=xms;i>0;i--)
  for(j=80;j>0;j--);
 }
 
void show6led(float num)
{
unsigned char i;

  P2=0xff;
 P3=0xff;
 i=num;
 duan=1;
 P2=table[i]+0x80; //Display units and decimal point
 duan=0;
 P2=0xff; //Turn off led first 
 wei=1;
 P2=0xfb; //Segment selection 3
 wei=0;
 delay(5);
 
  P2=0xff;
 P3=0xff;
 num=num-i;
 num=num*10;
 i=num;
 duan=1;
 P2=table[i];
 duan=0;
  P2=0xff;
  wei=1;
 P2=0xf7; //Segment selection 4, tenths
 wei=0;
 delay(5);
   P2=0xff;
 P3=0xff;
 num=num-i;
 num=num*10;
 i=num;
 duan=1;
 P2=table[i];
 duan=0;
 P2=0xff;
 wei=1;
 P2=0xef; //Segment selection 5, percentile
 wei=0;
 delay(5);

  P2=0xff;
 P3=0xff;
 num=num-i;
 num=num*10;
 i=num;
 duan=1;
 P2=table[i];
 duan=0;
 P2=0xff;
 wei=1;
 P2=0xdf; //Segment selection 6, thousandths
 wei=0;
 delay(5);
  
  
 }

void main()
{

uchar val;

 ADC_CS=0;

 
  ADC_WR=1;
  _nop_();
  ADC_WR=0;
  _nop_();
  ADC_WR=1;
  delay(50);
  P1=0xff;
  ADC_RD=1;
 _nop_();
  ADC_RD=0;
  _nop_();
   val=P1;
   ADC_RD =1;
   
  while(1)
  {
   show6led(val*0.02);//adc0832 is 8 bits, 5v/256=0.02v
  
   }

 }