Question

我正在尝试使用按钮开关来控制步进电机的软件。

我有三个电机功能，它将以三个不同的角度（从0-180度向前，90度反向，90度反向再次回到0）以一个接一个的顺序旋转。

我想使用一个按钮开关按顺序逐步执行这些功能。

有人会对如何做到这一点有任何建议吗？我想象我必须创建一个案例陈述，但一直在圈子里没有进展。

这是我下面的当前代码，此时按下开关将运行第一个“if”功能，顺时针旋转电机180度：

#define F_CPU 1000000

#include <avr/io.h>
#include <avr/interrupt.h>
#include <util/delay.h>

#define A PINB0
#define C PINB1
#define B PINB4
#define D PINB3

int16_t i=0; // Auxiliary variable ()

void HalfStep(uint8_t step);
void Forward_180(void);
void Reverse_90(void);
void Reverse_90_2(void);

/********************************* MAIN ********************************/
int main(void)
{
    DDRB |= (1<<A) | (1<<B) | (1<<C) | (1<<D);  // Set output pins

    DDRB &= ~(1<<PINB5);        // Set input pin for switch 

    while(1)
    {

        if(bit_is_clear(PINB, 5))
        {
            Forward_180();              // Clockwise 180 degrees 
        }

        if(bit_is_clear(PINB, 5))       // Anti-clockwise 90 degrees
        {
            Reverse_90();
        }

        if(bit_is_clear(PINB, 5))       // Anti-clockwise 90 degrees again 
        {
            Reverse_90_2();
        }
    }

    return 0;
}

/***************** HALF STEPPING & DIRECITON FUNCTIONS *****************/
void HalfStep(uint8_t step)
{
    switch (step)
    {
        case 0: // whole number after i%8
        PORTB |= (1<<A) | (1<<D);
        PORTB &= ~((1<<B) | (1<<C));
        break;

        case 1: // 0.125
        PORTB |= (1<<A);
        PORTB &= ~((1<<B)|(1<<C)|(1<<D));
        break;

        case 2: // 0.25
        PORTB |= (1<<A)|(1<<C);
        PORTB &= ~((1<<B)|(1<<D));
        break;

        case 3: // 0.375
        PORTB |= (1<<C);
        PORTB &= ~((1<<A)|(1<<B)|(1<<D));
        break;

        case 4: // 0.5
        PORTB |= (1<<B)|(1<<C);
        PORTB &= ~((1<<A)|(1<<D));
        break;

        case 5: // 0.625
        PORTB |= (1<<C);
        PORTB &= ~((1<<A)|(1<<B)|(1<<D));
        break;

        case 6: // 0.75
        PORTB |= (1<<D)|(1<<C);
        PORTB &= ~((1<<A)|(1<<B));
        break;

        case 7: // 0.875
        PORTB |= (1<<D);
        PORTB &= ~((1<<A)|(1<<B)|(1<<C));
        break;
    }
}

void Forward_180(void)
{
    while (i<=48)
    {
        HalfStep(i%8);
        i++;
        _delay_ms(5);
    };
}

void Reverse_90(void)
{
    while (i>=24)
    {
        HalfStep(i%8);
        i--;
        _delay_ms(10);
    };
}

void Reverse_90_2(void)
{
    while (i>=0)
    {
        HalfStep(i%8);
        i--;
        _delay_ms(10);
    };
}

Answer 1

我建议使用状态机。

这样的事情（我假设您需要编写一些代码才能进行演示）：

/* States */
typedef enum EnumStates
{
    ST_IDLE,
    ST_FORWARD_180,
    ST_WAIT_FOR_STEP_2,
    ST_REVERSE_90,
    ST_WAIT_FOR_STEP_3,
    ST_REVERSE_90_2,
}
EState;

/* to know if the switch was pressed */
char _keypressed = 0;

/* state variable */
EState _state = ST_IDLE;

/* very simple keyboard process */
void SwitchTask(void)
{
    /* The keypressed flag will be valid for one cycle of the while loop */
    _keypressed = 0;
    if (bit_is_clear(PINB, 5))
    {
        _keypressed = 1; 
        /* this is a simple way to implement debouncing */
        /* I assume you do not have timing requirements */
        _delay_ms(100);
    }
}

/* demo task */
/* This task implements a simple state machine */
void DemoTask(void)
{
    switch(_state)
    {
        case ST_FORWARD_180:
            Forward_180();
            _state = ST_WAIT_FOR_STEP_2;
            break;
        case ST_REVERSE_90:
            Reverse_90();
            _state = ST_WAIT_FOR_STEP_3;
            break;
        case ST_REVERSE_90_2:
            Reverse_90_2();
            _state = ST_IDLE;
            break;
        case ST_IDLE:
        case ST_WAIT_FOR_STEP_2:
        case ST_WAIT_FOR_STEP_3:
            if (_keypressed > 0)
            {
                _state++;
            }
            break;
        default:
            break;
    }
}

int main(void)
{
    /* initialization */
    DDRB |= (1<<A) | (1<<B) | (1<<C) | (1<<D);  // Set output pins
    DDRB &= ~(1<<PINB5);        // Set input pin for switch 
    _keypressed = 0;

    while(1)
    {
        SwitchTask();
        DemoTask();
    }   
    return 0;
}

使用按钮开关移动代码

1 个答案: