求大佬帮忙改一下这个单片机程序，通过P3.3口的中断输入控制触发的角度

ID:614734 · 发表于 2021-4-30 13:46

求大佬帮忙改一下这个程序，通过P3.3口的中断输入控制触发的角度，然后通过内部定时器控制触发信号做相应的移相然后输出波形

单片机源程序如下:
#include "reg52.h"
#include"ADC0808.h"
#include "lcd.h"

#define uchar unsigned char
#define uint unsigned int

sbit VT1=P2^0;
sbit VT2=P2^1;
sbit VT3=P2^2;
sbit VT4=P2^3;
sbit VT5=P2^4;
sbit VT6=P2^5;
uchar count;
uchar AD_num,angle_num;//采样AD的值和导通的角度
uchar AD_signal,achieve_signal;
uint delay_num=0;//延时的时间

void Delay10ms()             //@121MHz
{
      unsigned char i, j, k;

      _nop_();
      _nop_();
      i = 5;
      j = 154;
      k = 1;
      do
      {
            do
            {
                     while (--k);
            } while (--j);
      } while (--i);
}

void SCRTrg(uchar Num)    //  脉冲宽度30°
{


            switch(Num)          // 触发相位
      {
            case 1:
                     VT1 = 1;       // 开通
                  VT6 = 0;
                     break;
            case 2:
                     VT1 = 1;          // 关断
                  VT6 = 0;
                     break;
            case 3:
                     VT2 = 0;
                  VT1 = 1;
                     break;
            case 4:
                  VT1 = 1;
                  VT2 = 0;
                     break;
            case 5:
                  VT3 = 0;
         VT2 = 0;
                     break;
      case 6:
                     VT3 = 0;
                  VT2 = 0;
                     break;
            case 7:
                     VT3 = 0;
                  VT4 = 0;
                     break;
            case 8:
                     VT3 = 0;
                  VT4 = 0;
                     break;
            case 9:
                     VT5 = 1;
                  VT4 = 1;
                     break;
            case 10:
                  VT5 = 0;
                     VT4 = 0;
                     break;
            case 11:
                     VT5 = 0;
                  VT6 = 0;
                     break;
      case 12:
                     VT5 = 0;
                  VT6 = 0;
                     break;
            default:
                     break;
      }
}
//定时器和外部中断初始化
void Timer_Init(void)
{
      TMOD=0x11;
      TH1=0xF9;
      TL1= 0x7F;
      ET0=1;
      ET1=1;
      IT0 = 1;       // 下降沿触发
      PX0=1;//外部中断0定义为高优先级
      EX0=1;
}
void main(void)
{
Timer_Init();
      VT1=0;
      /*VT2=0;
      VT3=0;
      VT4=0;
      VT5=0;
      VT6=0;*/
LcdInit();
      AD_num=ADCInit();
      angle_num=(uint)((255-AD_num)/255.0*180);
      delay_num=(uint)(angle_num*10000/180.0);
      LcdPrintf(0,"angle:");
      LcdDisplayI(6,angle_num);
      EA=1;
      while(1)
      {
            if(AD_signal==1)
            {
                     AD_signal=0;
                  if(AD_num!=ADCInit())
                     {
                              TR0=0;
                              TR1=0;
                              AD_num=ADCInit();
                              angle_num=(uint)((255-AD_num)/255.0*180);
                  delay_num=(uint)(angle_num*10000/180.0);
                              LcdDisplayI(6,angle_num);
                              achieve_signal=1;
                              EX0=1;

                     }
            }
      }
}
void exint0() interrupt 0
{
      Delay10ms();
      count=1;
if(angle_num<60)
count=1;
else if(angle_num<120)
            count=11;
      else
            count=9;
      TH0=(65536-delay_num)/256;
      TL0=(65536-delay_num)%256;
      TR0=1;
      EX0=0;//关外部中断
}
void Time0Int()  interrupt 1
{
      TH0=(65536-delay_num)/256;
      TL0=(65536-delay_num)%256;
      TH1=0xF9;
      TL1= 0x82;
      TR0=0;
      TR1=1;
}
void Time1Int()  interrupt 3
{
      TH1=0xF9;
      TL1= 0x82;
      SCRTrg(count++);
      if(count==6)
            AD_signal=1;
      if(count>12)
      {
            count=1;
      }
}

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