步进电机完全旋转360度怎么停止

你是说你用马达驱动IC来驱动马达？哪一个？



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Do you mean you are using some motor driver IC to drive the motor?
Which one?
 

L98N电机驱动器



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    L298N motor driver 

这只是一个拨号H桥驱动芯片，所以它需要PIC来产生步进波形。你的意思是你已经有PIC产生波形使马达旋转了吗？如果你只是施加一个直流电压，它是旋转的，那么它不是步进电机。



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    That is just a dial H bridge driver chip, so it requires the PIC to generate the stepping waveform.
Do you mean you already have the PIC generating the waveform to make the motor rotate?
If you're just applying a DC voltage, and it is rotating, then it is not stepper motor.
 

我写了一段代码，它旋转。但我需要启动电机在一点，停止电机在同一点，当我给一个脉冲。



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    I wrote a piece of code its rotating .. but i need to start motor at one point and stop motor at the same point when i give one pulse..

您想显示您编写的代码吗？如果你给马达“一个脉冲”，那么它只能移动一步。正如我所说的，如果你只是用一个方法来激励它，它就继续运动，那么它就不是步进电机。



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Would you like to show the code you have written?
 

If you give the motor "one pulse", then it should move only one step.
As I said, if you are just energising it one way, and it keeps moving, then it is not a stepper motor.

（1）{A1＝1；B2＝1；A2＝0；B1＝1；A2＝1；B2＝0；A1＝0；B1＝0；A2＝Y；B2＝Y；α1DelayyMS（α）；A1＝Y；B1＝Y；A2＝Y；B2＝Y；γ-DelayyMS（ii）；} i要给出一个脉冲，马达必须从同一位置开始并以相同P结束。部分360度一轮完全旋转



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    part of code
while (1)
{
A1 = 1; B1 = 1; A2 = 0; B2 = 0;
__delay_ms(5);
A1 = 0; B1 = 1; A2 = 1; B2 = 0;
__delay_ms(5);

A1 = 0; B1 = 0; A2 = 1; B2 = 1;
__delay_ms(5);
A1 = 1; B1 = 0; A2 = 0; B2 = 1;
__delay_ms(5);
}
i wanted to give one pulse , motor has to start from one position and end it with same position with 360 deg one complete rotation

这是什么照片？你有LATX寄存器，你在使用吗？



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    Which PIC is this? Do you have LATx registers and are you using them?

步进电机通常被设计成一个完整的循环，例如四个简单的步骤，是实际旋转的60度。这是因为他们设计的磁铁排列在花键上。在HTTPS://Youtu.BE/MHDZ3C6KLRG和HTTPS://Youtu.BE/T-3VNLADBC上可以找到解释步进电机如何工作的视频。



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    Stepper motors are often designed so that a  complete cycle, say four simple steps, is 60 degrees 
of actual rotation. This is because they design the magnets to line up on splines.
Videos on   explaining how stepper motors work can be found at   https://youtu.be/MHdz3c6KLrg and  https://youtu.be/t-3VnLadIbc .

这是一个开始。你测试过一个完整的旋转需要多少个周期吗？A1、B1、A2和B2是如何定义的？正如ChanZIa所提到的，你使用的PIC也是相关的。



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That is a start. Have you tested how many cycles like this you need for a complete rotation?
How EXACTLY are A1, B1, A2, and B2 defined?
As Cinzia mentioned, which PIC you are using is also relevant.
 

你需要看一下电机的数据表。你需要找到每一步的度数，固件中的步长为180度。



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    You need to look at the data sheet for the motor. You need to find the degrees per step, and the in the firmware count the number of steps for 180 degrees

感谢您的答复，电机规格-步进电机步进角- 1.8度每步＝200旋转码-空（）{TrISC0＝0；TrISC1＝0；TrISC2＝0；TrISC3＝0；而（1）{A1= ON；B2= OFF；A2＝OFF；B2＝OFF；α-DelayyMS（1）；A1= ON；B1= ON；A2= OFF；B2= OFF；γ-延迟。Ay= MS；（1）；A2= On；B2= OFF；A1= OFF；B2= ON；B2= OFF；A1＝DelayyMs（1）；A1= OFF；B2= ON；B2= ON；B2＝On；A1= OFF；B1= OFF；A2= OFF；B2= ON；α-DelayyMs（1）；A1= ON；B1= OFF；A2= OFF；B2= ON；B2= ON；αMS（1）；A1= OFF；B1= ON；A2= OFF；B2= OFF；LaysMMS（1）；} AM使用PIC16F87A控制器和XC8编译器。我与马达连接了一个驱动器和一个磁传感器连接到马达。一旦我得到一个触发器，我的马达就应该在固定位置开始旋转一个完整的周期，并以相同的位置结束它的启动。



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    Thanks for you reply,
Motor specification - Bipoloar stepper motor 
                                  Stepping angle - 1.8 Deg per step = 200 rotations
Code - 
void main()
{
   TRISC0 =0;
   TRISC1 =0;
   TRISC2 =0;
   TRISC3 =0;
while(1)
{
     
    A1 =ON; B1= OFF; A2 =OFF; B2= OFF;
   __delay_ms(1);
   A1 =ON; B1= ON; A2 =OFF; B2= OFF;
   __delay_ms(1);
   A1 =OFF; B1= ON; A2 =OFF; B2= OFF;
   __delay_ms(1);
   A1 =OFF; B1= ON; A2 =ON; B2= OFF;
  __delay_ms(1);

  A1 =OFF; B1= OFF; A2 =ON; B2= OFF;
  __delay_ms(1);
  A1 =OFF; B1= OFF; A2 =ON; B2= ON;
  __delay_ms(1);
  A1 =OFF; B1= OFF; A2 =OFF; B2= ON;
  __delay_ms(1);
  A1 =ON; B1= OFF; A2 =OFF; B2= ON;
  __delay_ms(1);
}
}
 
Am using pic16f877a controller and XC8 compiler .
I have connected motor with driver and one magnetic sensor connected to motor . 
As soon as i get one trigger my motor should start at fixed position rotates complete one cycle and end it with same position where it started.
 

好的，你回答了我的一个问题，但是忽略了“A1，A2，B1，B2定义”的问题。我可以猜测看你的代码，但我不需要。我猜你有磁性传感器连接到另一个PIC管脚。哪一个？这个“扳机”是什么？另一个PIC PIN？你必须使用PIC16F87A吗？这确实是一个古老的芯片，当试图同时改变同一个端口上的多个引脚时，需要非常小心。PIN兼容的PIC16F1XXX设备更便宜，更容易使用，并且没有16F87A的问题。



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ok, you answered one of my questions, but ignored the "how are A1, A2, B1, B2 defined" question.
I can probably guess looking at your code, but I shouldn't have to.
 

I guess you have the magnetic sensor connected to another PIC pin. Which one?
 

What is this "trigger"?
Another PIC pin?
 
Do you have to use a PIC16F877A?
That really is an ancient chip, and requires great care when trying to change multiple pins on the same port all at once.
There are pin compatible PIC16F1xxx devices available that are cheaper, easier to use, and don't have the problems that the 16F877A does.
 
 
 

定义了A1 RC0//输出α定义A2 RC1//输出α定义B1 RC2//输出α定义B2 RC3/输出



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    #define A1 RC0 // OUTPUT
#define A2 RC1 // OUTPUT
#define B1 RC2 // OUTPUT
#define B2 RC3 // OUTPUT
 
 

磁传感器我还没有连接…我现在正在和那部分一起工作



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    magnetic sensor still i have not connected ... Am working with that part right now

触发器可以是来自接近传感器（检测金属）的一个脉冲。



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    trigger may be one pulse from proximity sensor (which detects the metal)
 

我强烈建议您一次一个地写入POTC，而不是一个引脚，以避免在没有LAT寄存器的情况下在PICS上进行读写修改问题。您可以写入整个端口，因为如果写入PIN为输入，则无所谓。01；//A1= ON；B1= OFF；A2= OFF；B2= OFF；A1= ON；B1= ON；A2= OFF；B2= OFF；SimePosic＝0B0101；//A1= ON；B1= ON；A2= OFF；B2= OFF；等等。



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I would strongly advise you to write PORTC all in one instruction, not one pin at a time, to avoid read-modify-write problems on PICs without LAT registers.
You can write the whole port, as it does not matter if you write to pins set as inputs.
 
 
So
A1 =ON; B1= OFF; A2 =OFF; B2= OFF;  
becomes
PORTC = 0b0001;    //A1 =ON; B1= OFF; A2 =OFF; B2= OFF;  
 
A1 =ON; B1= ON; A2 =OFF; B2= OFF; 
becomes
PORTC = 0b0101;    //A1 =ON; B1= ON; A2 =OFF; B2= OFF; 
 
and so on.

当你指的是高和低时，使用和关闭是令人困惑的。L29 8N既能提供电流，又能吸收电流，因此，基本上对两对电机绕组施加两种不同的极性。如果设置B1和B2低，它将基本上缩短绕组，这可以用作制动器。这里有一篇文章介绍了使用ARDUION:http://www.指导表……的L29 8N。电机B真值表3BIN 4IN描述符0N/AN/AMotor B是OF100B电机停止（制动器）101电机B打开，后转110电机B打开，转转11B电机停止（制动器），来自http://www. BANANAROBOICTIC…-H桥式电动驱动器



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    It is confusing to use ON and OFF when you probably mean HIGH and LOW. The L298N can both source and sink current, so you are basically applying two different polarities to each pair of motor windings. If you set B1 and B2 low, it will essentially short the winding, which may be used as a brake.
 
Here is an article on using the L298N with an Arduino:
 
http://www.instructables....-with-L298N-Dual-Moto/
 
Motor A truth tableENAIN1IN2Description0N/AN/AMotor A is off100Motor A is stopped (brakes)101Motor A is on and turning backwards110Motor A is on and turning forwards111Motor A is stopped (brakes) 
Motor B truth tableENBIN3IN4Description0N/AN/AMotor B is off100Motor B is stopped (brakes)101Motor B is on and turning backwards110Motor B is on and turning forwards111Motor B is stopped (brakes) 
That is from https://www.bananarobotic...-H-Bridge-Motor-Driver

不能正确编辑我的文章。这就是我想要展示的：



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    Casn't properly edit my post. Here is what I was trying to show:
 

这似乎是一个直流电动机，而不是步进电机。



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    That appears to be for a DC motor, not a stepper motor.