stm8 TIM1输入捕获测频

zfcarlos

测量频率在 1k-2k之间，要求精度0.01%，也就是2kHZ, 0.2HZ。

在做的过程中，发现最大跳动在0.8HZ，不知道代码哪有问题，之后就用avr做了一个，能满足要求，

但我还是想用STM8，希望高人指点，谢谢！

代码如下：

#include "TIM1_Capture.h"

#include "includes.h"

#include "delay.h"

//u16 tab[200];

INT_CHARS ICValue1,ICValue2;

u8  cnt;

u8  TIM1_OVF_Num;

bool flag;

void TIM1_Init(void)

{

    GPIOC->DDR &= ~(1<<1); // 配置PC1为输入

    GPIOC->CR1 |= 1<<1;

    GPIOC->CR2 &= ~(1<<1);    

 

    // 开启TIM1时钟

    CLK->CKENR1 |= 0X80;

   

    TIM1->CNTRH = 0X00;

    TIM1->CNTRL = 0X00;

    TIM1->ARRH = 0Xff;

    TIM1->ARRL = 0Xff;

    TIM1->CR1 = 0X80; // 自动重装  

   

    TIM1->CCMR1 = 0X00;

    TIM1->CCER1 = 0X00;

    TIM1->CCMR1 |= 0X01;

    TIM1->CCER1 &= 0XFD; // 上升沿捕获

    TIM1->IER |= 0X03; // 允许TIM1捕获中断，更新中断

   

    TIM1->SR1 = 0x00;

    TIM1->SR2 = 0x00;

   

    TIM1->CR1 |= 0X01; // 使能TIM1

   

    TIM1->CCER1 = 0X01; // 使能捕获

}

u16 ComputeFreq(void)

{

    u16 temp;

    u32 sum;

    flag = FALSE;

    cnt = 0;

    TIM1_Init();

    while(!flag); // 等待100次捕获结束

   

    sum  = TIM1_OVF_Num;

    sum *= 65536;

    sum += ICValue2.x;

    sum -= ICValue1.x;

    sum /= 10;

    temp = 1600000000/sum;

     

    return temp;

}

#pragma vector=0x0E

__interrupt void TIM1_CAP_COM_IRQHandler(void)

{

    disableInterrupts();

   

    if(cnt==5)

    {

        ICValue1.FJ[0] = TIM1->CCR1H;

        ICValue1.FJ[1] = TIM1->CCR1L;

        flag = FALSE;

        TIM1_OVF_Num = 0;

    }

    else if(cnt==105)  // 100次捕获结束

    {

        ICValue2.FJ[0] = TIM1->CCR1H;

        ICValue2.FJ[1] = TIM1->CCR1L;

        flag = TRUE;

        TIM1->CR1 &= 0XFE; // 失能TIM1

        TIM1->CCER1 = 0X00;

    }

   

    TIM1->SR1 &= 0xFD;

//    TIM1->SR2 = 0x00;

   

    cnt++;

    enableInterrupts();

}

#pragma vector=0x0D

__interrupt void TIM1_OVF_IRQHandler(void)

{

    disableInterrupts();

   

    TIM1_OVF_Num++;

    TIM1->SR1 &= 0xFE;

//    TIM1->SR2 = 0x00;

    enableInterrupts();

}

正点原子

還沒玩STM8....