What is the mastering procedure for 28BYJ-48 stepper motor?

Although we have completed the program of using interrupts to control motor rotation, in fact this program still has little practical value. We can't power it on and off every time we want it to rotate, right? . Also, it must be able to not only rotate forward but also reverse, that is to say, it must not only be able to rotate around, but it must also be able to rotate back. Okay, let's make an example program. Combined with the key program in Chapter 8, we design a functional program like this: press the numeric keys 1 to 9 to control the motor to rotate 1 to 9 circles; use the up and down keys to change the direction of rotation. , press the up button to rotate forward 1 to 9 times, and the down button to rotate 1 to 9 times in the reverse direction; the left button will rotate forward 90 degrees, and the right button will rotate 90 degrees; the Esc key will terminate the transition. Through this program, we can also further understand how to use buttons to control programs to complete complex functions, and how to coordinate tasks between control and execution modules, and your programming level can also be exercised and improved in such theoretical exercises.

			#include <reg52.h> sbit KEY_IN_1 = P2^4; sbit KEY_IN_2 = P2^5; sbit KEY_IN_3 = P2^6; sbit KEY_IN_4 = P2^7; sbit KEY_OUT_1 = P2^3; sbit KEY_OUT_2 = P2^2; sbit KEY_OUT_3 = P2^1; sbit KEY_OUT_4 = P2^0; unsigned char code KeyCodeMap[4][4] = { //矩阵按键编号到规范键盘键码的映射表 { 0x31, 0x32, 0x33, 0x26 }, //数字键 1、数字键 2、数字键 3、向上键 { 0x34, 0x35, 0x36, 0x25 }, //数字键 4、数字键 5、数字键 6、向左键 { 0x37, 0x38, 0x39, 0x28 }, //数字键 7、数字键 8、数字键 9、向下键 { 0x30, 0x1B, 0x0D, 0x27 } //数字键 0、ESC 键、 回车键、 向右键 }; unsigned char KeySta[4][4] = { //全体矩阵按键的以后形态 {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1} }; signed long beats = 0; //电机迁移转变节奏总数 void KeyDriver(); void main(){ EA = 1; //使能总中缀 TMOD = 0x01; //设置 T0 为形式 1 TH0 = 0xFC; //为 T0 赋初值 0xFC67，准时 1ms TL0 = 0x67; ET0 = 1; //使能 T0 中缀 TR0 = 1; //启动 T0 while (1){ KeyDriver(); //挪用按键驱动函数 } } /* 步进电机启动函数，angle-需转过的角度 */ void StartMotor(signed long angle){ //在盘算前封闭中缀，完成后再翻开，以防止中缀打断盘算进程而形成毛病 EA = 0; beats = (angle * 4076) / 360; //实测为 4076 拍迁移转变一圈 EA = 1; } /* 步进电机中止函数 */ void StopMotor(){ EA = 0; beats = 0; EA = 1; } /* 按键举措函数，依据键码履行响应的操作，keycode-按键键码 */ void KeyAction(unsigned char keycode){ static bit dirMotor = 0; //电机迁移转变偏向 //掌握电机迁移转变 1-9 圈 if ((keycode>=0x30) && (keycode<=0x39)){ if (dirMotor == 0){ StartMotor(360*(keycode-0x30)); }else{ StartMotor(-360*(keycode-0x30)); } }else if (keycode == 0x26){ //向上键，掌握迁移转变偏向为正转 dirMotor = 0; }else if (keycode == 0x28){ //向下键，掌握迁移转变偏向为反转 dirMotor = 1; }else if (keycode == 0x25){ //向左键，固定正转 90 度 StartMotor(90); }else if (keycode == 0x27){ //向右键，固定反转 90 度 StartMotor(-90); }else if (keycode == 0x1B){ //Esc 键，中止迁移转变 StopMotor(); } } /* 按键驱动函数，检测按键举措，调剂响应举措函数，需在主轮回中挪用 */ void KeyDriver(){ unsigned char i, j; static unsigned char backup[4][4] = { //按键值备份，保管前一次的值 {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1} }; for (i=0; i<4; i++){ //轮回检测 4*4 的矩阵按键 for (j=0; j<4; j++){ if (backup[i][j] != KeySta[i][j]){ //检测按键举措 if (backup[i][j] != 0){ //按键按下时履行举措 KeyAction(KeyCodeMap[i][j]); //挪用按键举措函数 } backup[i][j] = KeySta[i][j]; //刷新前一次的备份值 } } } } /* 按键扫描函数，需在准时中缀中挪用，引荐挪用距离 1ms */ void KeyScan(){ unsigned char i; static unsigned char keyout = 0; //矩阵按键扫描输入索引 static unsigned char keybuf[4][4] = { //矩阵按键扫描缓冲区 {0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF} }; //将一行的 4 个按键值移入缓冲区 keybuf[keyout][0] = (keybuf[keyout][0] << 1) | KEY_IN_1; keybuf[keyout][1] = (keybuf[keyout][1] << 1) | KEY_IN_2; keybuf[keyout][2] = (keybuf[keyout][2] << 1) | KEY_IN_3; keybuf[keyout][3] = (keybuf[keyout][3] << 1) | KEY_IN_4; //消抖后更新按键形态 for (i=0; i<4; i++){ //每行 4 个按键，所以轮回 4 次 if ((keybuf[keyout][i] & 0x0F) == 0x00){ //延续 4 次扫描值为 0，即 4*4ms 内多是按下形态时，可以为按键已波动的按下 KeySta[keyout][i] = 0; }else if ((keybuf[keyout][i] & 0x0F) == 0x0F){ //延续 4 次扫描值为 1，即 4*4ms 内多是弹起形态时，可以为按键已波动的弹起 KeySta[keyout][i] = 1; } } //履行下一次的扫描输入 keyout++; //输入索引递增 keyout = keyout & 0x03; //索引值加到 4 即归零 //依据索引，释放以后输入引脚，拉低下次的输入引脚 switch (keyout){ case 0: KEY_OUT_4 = 1; KEY_OUT_1 = 0; break; case 1: KEY_OUT_1 = 1; KEY_OUT_2 = 0; break; case 2: KEY_OUT_2 = 1; KEY_OUT_3 = 0; break; case 3: KEY_OUT_3 = 1; KEY_OUT_4 = 0; break; default: break; } } /* 电机迁移转变掌握函数 */ void TurnMotor(){ unsigned char tmp; //暂时变量 static unsigned char index = 0; //节奏输入索引 unsigned char code BeatCode[8] = { //步进电机节奏对应的 IO 掌握代码 0xE, 0xC, 0xD, 0x9, 0xB, 0x3, 0x7, 0x6 }; if (beats != 0){ //节奏数不为 0 则发生一个驱动节奏 if (beats > 0){ //节奏数大于 0 时正转 index++; //正转时节奏输入索引递增 index = index & 0x07; //用&操作完成到 8 归零 beats--; //正转时节奏计数递加 }else{ //节奏数小于 0 时反转 index--; //反转时节奏输入索引递加 index = index & 0x07; //用&操作异样可以完成到-1 时归 7 beats++; //反转时节奏计数递增 } tmp = P1; //用 tmp 把 P1 口以后值暂存 tmp = tmp & 0xF0; //用&操作清零低 4 位 tmp = tmp | BeatCode[index]; //用|操作把节奏代码写到低 4 位 P1 = tmp; //把低 4 位的节奏代码和高 4 位的原值送回 P1 }else{ //节奏数为 0 则封闭电机一切的相 P1 = P1 | 0x0F; } } /* T0 中缀效劳函数，用于按键扫描与电机迁移转变掌握 */ void InterruptTimer0() interrupt 1{ static bit div = 0; TH0 = 0xFC; //从新加载初值 TL0 = 0x67; KeyScan(); //履行按键扫描 //用一个静态 bit 变量完成二分频，即 2ms 准时，用于掌握电机 div = ~div; if (div == 1){ TurnMotor(); } }

This program is a synthesis of the knowledge in Chapter 8 and this chapter - using buttons to control the rotation of stepper motors. There are several points worth noting in the program, which we describe as follows:

• In order for the motor to complete two different operations of forward and reverse rotation, we did not use the forward start function and The reverse startup function is implemented by these two functions, and a method parameter is not added when defining the startup function to indicate its direction. The only difference between our start function void StartMotor (signed long angle) and the start function for one-way forward rotation is that the type of the method parameter angle is changed from unsigned long to signed long. We use the inherent positive and negative characteristics of signed numbers. To distinguish between forward rotation and reversal, a negative number represents the angle of forward rotation, and a positive number represents the angle of reverse rotation. Isn't this solution very simple and clear? And do you have a better understanding of the different usages of signed numbers and unsigned numbers?

• In order to terminate the operation of motor rotation, we have defined a separate StopMotor function to complete it. Although this function is very complex, although it is only called in the Esc key branch. , but we still put it forward as a function. This approach is based on such a programming principle: use separate functions to complete certain operations of the hardware as much as possible. When a piece of hardware includes multiple operations, organize these operation functions together to form a unified interface to the underlying layer. Such hierarchical processing will make the entire program clearly hierarchical, which is conducive to debugging and maintenance of the program and expansion of functions.

• The infix function has to handle two tasks: key scanning and motor driving. In order to prevent the infix function from being too complicated, we have separated two functions: key scanning and motor driving. (This is also in line with the programming principles of 2 above), and the logic of the interrupt function becomes simple and clear. There is another contradiction here, that is, the timing time we choose for button scanning is 1ms, while in the examples before this chapter, the motor rhythm duration may be 2ms; obviously, using a timing of 1ms can determine the distance of 2ms, but using a timing of 2ms can determine the distance of 2ms. But we can't get the accurate 1ms distance; so what we do is, the timer is still on time 1ms, and then mark it with a bit variable, change its value every 1ms, and we only choose to execute the action once when the value is 1, This is the distance of 2ms; if I want 3ms or 4ms, just change the bits to char or int type, then increment them, and determine which value should be reset to zero. That's it. This is to achieve accurate software timing based on the hardware timer.

The above is the detailed content of What is the mastering procedure for 28BYJ-48 stepper motor?. For more information, please follow other related articles on the PHP Chinese website!