Arduino 开关以打开继电器计时器 [英] Arduino Switch to Turn a Relay timer

问题描述

简而言之:我想在打开开关后打开继电器 30 秒.

Briefly: I would like to turn on a relay for 30 seconds, after I toggle a switch on.

我正在尝试在家中进行百叶窗自动化.

I'm trying to do a blinds automation at home.

我有一个简单的 ON-OFF-ON 开关，连接到连接到继电器的 Arduino.

I have a simple ON-OFF-ON switch, attached to an Arduino connected to Relays.

如果我从中心向下拨动开关，我想打开继电器#1 最多 30 秒.换句话说，当我切换时继电器打开，当定时器达到 30 秒时继电器关闭.

I want to turn on Relay#1 for a maximum of 30 seconds if I toggle the switch down from center. In other words, relay turns on when I switch, and when timer reaches 30 seconds relay turns off.

同样，如果我从中心向上拨动开关，我想打开 Relay#2 正好 30 秒

similarly I want to turn on Relay#2 for exactly 30 seconds if I toggle the switch up from center

当我切换回中心时，我希望计时器重置.

And when I switch back to center, I would like the timer to reset.

我不知道怎么做.有人可以帮忙吗?

I could not figure out how. Could anyone help?

为此我一直在尝试使用 elapsedMillis 库，这是一个很好的库，可以帮助我避免使用延迟:http://playground.arduino.cc/Code/ElapsedMillis

I have been trying to use elapsedMillis library for this, which is a nice library that helps me avoid using Delays: http://playground.arduino.cc/Code/ElapsedMillis

然而，即使我可以在没有 30 秒限制的情况下工作继电器，我也无法找出结束继电器工作的代码.这是我当前的代码:

However even though I could work the relays without the 30 second limitation, I couldn't figure out the code to end working of the relays. Here is my current code:

      #include <elapsedMillis.h>
    #define RELAY_ON 0
    #define RELAY_OFF 1

    #define RELAY1_TURNS_ON_BLINDS  5
    #define RELAY2_SHUTS_DOWN_BLINDS 6

    #define shutswitch A0
    #define openswitch A1

    bool LocalCommandToOpen;
    bool LocalCommandToShut;

    void setup() ////////SETUP////////
    {
    digitalWrite(RELAY1_TURNS_ON_BLINDS, RELAY_OFF);
     digitalWrite(RELAY2_SHUTS_DOWN_BLINDS, RELAY_OFF);

     pinMode(RELAY1_TURNS_ON_BLINDS, OUTPUT);
     pinMode(RELAY2_SHUTS_DOWN_BLINDS, OUTPUT);

     pinMode(shutswitch, INPUT);
     pinMode(openswitch, INPUT);
        } ////SETUP

  void loop() { ///////LOOP
         if (digitalRead(shutswitch) == 1)
     {
       LocalCommandToOpen = 1;
     }
     else
     {
       LocalCommandToOpen = 0;
     }

      if ( digitalRead(openswitch) == 1)
     {
       LocalCommandToShut = 1;
     }
     else
     {
       LocalCommandToShut = 0;
     }

    unsigned int CloseInterval = 14000;
     elapsedMillis timeElapsedSinceCloseButtonPush = 0;
     unsigned int OpenInterval = 14000;
     elapsedMillis timeElapsedSinceOpenButtonPush = 0;



     //MANUAL SWITCH OPERATION

     if ( LocalCommandToShut == 1 )
     {
       digitalWrite(RELAY1_TURNS_ON_BLINDS, RELAY_OFF);
       digitalWrite(RELAY2_SHUTS_DOWN_BLINDS, RELAY_ON);
     }
     else
     {
       digitalWrite(RELAY2_SHUTS_DOWN_BLINDS, RELAY_OFF);
     }

    //MANUEL DUGME ILE ACMA
     if ( LocalCommandToOpen == 1)
     {
       digitalWrite(RELAY2_SHUTS_DOWN_BLINDS, RELAY_OFF);
       digitalWrite(RELAY1_TURNS_ON_BLINDS, RELAY_ON);
     }
     else
     {
       digitalWrite(RELAY1_TURNS_ON_BLINDS, RELAY_OFF);
     }

     delay(500);

    } /////////////////LOOP////////////////////////////////////

推荐答案

您可能会使用状态机；这让事情更容易理解.

You might use a state machine; this makes things a bit easier to follow.

类似于:

Arduino 开关以打开继电器计时器

关于状态机的精彩讨论在这里:

A nice discussion of state machines is here:

复杂的状态转换:最佳实践

这篇关于Arduino 开关以打开继电器计时器的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！