使用WinAVR的微控制器AVR中的基本I/O [英] Basic I/O in Microcontroller AVR using WinAVR

问题描述

我想驱动BLDC电机，我使用ATMEGA32作为控制器的CPU，从BLDC电机读取霍尔效应传感器时遇到问题

I want to drive a BLDC motor, i use ATMEGA32 as CPU of controller , i have a problem in reading hall effect sensor from BLDC motor

这是我的代码:

DDRB=(1<<0)|(1<<1)|(1<<2)|(1<<3)|(1<<4)|(1<<5);
DDRD=(0<<0)|(0<<1)|(0<<2)|(0<<3);
PORTD=(1<<0)|(1<<1)|(1<<2)|(1<<3);

PORTB作为输出，PORTD作为输入 这是主程序:

PORTB as OUTPUT and PORTD as INPUT this is main program :

       if (~(PORTD &(1<<0)) && PORTD &(1<<1) && ~(PORTD &(1<<2))) /// 0   1   0
        {
            PORTB=(1<<0)|(0<<1)|(0<<2)|(1<<3)|(0<<4)|(0<<5);
        }
        else if (~(PORTD &(1<<0)) && PORTD &(1<<1) && PORTD &(1<<2)) /// 0   1   1
        {
            PORTB=(0<<0)|(0<<1)|(0<<2)|(1<<3)|(0<<4)|(1<<5);
        }
        ...........

我的回答是该程序无法从霍尔效应传感器读取逻辑零(0)，因此BLDC无法正常运行，如何在AVR ATMEGA32中读取多个输入? 我使用Codeblock和winavr....

my answer is this program cannot read logic zero (0) from hall effect sensor so BLDC cannot run well, how to read multiple input in AVR ATMEGA32 ? i use Codeblock and winavr....

非常感谢您

推荐答案

我不太了解您要做什么，但是如果您想从atmega 32的Port D中读取内容，则无需处理"PORTD"寄存器，还有另一个称为"PIND"的寄存器，这是您从中读取数字信号的寄存器. 也没有>> DDRD =(0<< 0)|(0<< 1)|(0<< 2)|(0<< 3);如果您想清除一位(将位设置为逻辑零)，则只使用按位而不是移位的位(例如DDRD = ~(1<<bitnumber)

I don't really understand what you want to do, but if you want to read from Port D in the atmega 32, you do not deal with the 'PORTD' register, there is another Register called 'PIND', this is the one you read digital signals from. Also there nothing as ">DDRD=(0<<0)|(0<<1)|(0<<2)|(0<<3);" If you want to clear a bit (set the bit to logic zero), you just use the bitwise not with the shifted one (e.g. DDRD = ~(1<<bitnumber)

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