Question

我目前正在开展一个arduino项目，我在一个草图中使用两个ds18b20温度传感器和一个Nema 17步进电机。因此，我为ds18b20温度传感器制作了草图，为Nema 17步进电机制作了另一个，两者都完美无缺！然而，当我试图将它们组合在一个草图中时，nema 17将不再运行！经过一段头脑风暴后，我发现我正在使用Delay（）和Delaymicroseconds（）它们阻止arduino接收任何中断，这就是为什么我继续使用micros（）尝试arduino多任务但是我遇到了同样的问题时间。然后我进入调试模式，以确定哪个命令阻止步进电机运行，我发现Serial.print（）和Serial.println（）正在减慢电机速度。 PS：我正在使用TB6560驱动程序和1/8步进模式。这是草图：

#include <OneWire.h>
#include <DallasTemperature.h>

#define ONE_WIRE_BUS 2

OneWire oneWire(ONE_WIRE_BUS);

DallasTemperature sensors(&oneWire);

float Celsius1 = 0;
float Celsius2 = 0;

const int SensorDuration = 500;
unsigned long TempcurrentMillis = 0;
unsigned long TemppreviousLedMillis = 0;
const int stepPin = 9;
const int dirPin = 8;
int motor = 3;
int msg;


const int offDuration = 150;
const int interval = 50;
unsigned long currentMillis = 0;
unsigned long previousLedMillis = 0;
byte targetLedState = HIGH;

void setup() {
  sensors.begin();
  // Sets the two pins as Outputs
  Serial.begin(9600);
  pinMode(stepPin, OUTPUT);
  pinMode(dirPin, OUTPUT);
  pinMode(motor, OUTPUT);
  digitalWrite(motor, LOW);

  digitalWrite(dirPin, HIGH);
}
void loop() {
  currentMillis = micros();
  TempcurrentMillis = micros();
  GetTemp();
  RunMotor();
}
void RunMotor() {
  if (targetLedState == LOW) {
    if ((unsigned long) currentMillis - previousLedMillis >= interval) {
      targetLedState = HIGH;
      digitalWrite(stepPin, HIGH);
      previousLedMillis += interval;

    }
  }
  else {

    if ((unsigned long) currentMillis - previousLedMillis >= offDuration) {
      targetLedState = LOW;
      digitalWrite(stepPin, LOW);

      previousLedMillis += offDuration;

    }
  }
}
void GetTemp() {
  if ((unsigned long) TempcurrentMillis - TemppreviousLedMillis >= SensorDuration) {
    targetLedState = HIGH;
    sensors.requestTemperatures();

    Celsius1 = sensors.getTempCByIndex(0);
    Celsius2 = sensors.getTempCByIndex(1);
    Serial.print("T");
    Serial.println(Celsius1);
    Serial.print("H");
    Serial.println(Celsius2);
    TemppreviousLedMillis += SensorDuration;

  }
}

步进电机和Serial.print（）不会在同一草图中工作

0 个答案: