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3个普通IO识别22个按键试验(转)

识别IO 22 普通 按键 试验
2023-09-27 14:28:14 时间

源:http://www.amobbs.com/forum.php?mod=viewthread&tid=2243715

 

  吸取各位前辈的经验,将之前二极管用量多的问题优化一下,目前不用二极管能接6键,2只二极管能接12键,6只二极管能接18键,9只二极管能接21键,第22键要单独占用3只二极管最不化算。

实验用89S51作试验,电路接线就是P1.2, P1.3, P1.4接键盘,P1.0接显示器。









 

/*==================================================================*

 *                   3个IO接识别22键测试程序                         *

 *       ------------------------------------------------            *

 *       MCU:     AT89C2051                                          *

 *       OSC:     12M cysytel                                        *

 *       程序设计:Cowboy                                             *

 *       程序版本:V1.0                                               *

 *==================================================================*/



 #include <reg52.h>



 //================== IO口线连接 ==================

sbit Bus          = P1^0;

 sbit IO_a         = P1^4;

 sbit IO_b         = P1^3;

 sbit IO_c         = P1^2;

   

 //================== 变量声明 ====================

unsigned char Disp_buf[3];

 unsigned char Dig;

 unsigned char Key_count;

 unsigned char bdata Key_state;    

 sbit KB0 = Key_state^0;

 sbit KB1 = Key_state^1;

 sbit KB2 = Key_state^2;

 sbit KB3 = Key_state^3;

 sbit KB4 = Key_state^4;

 sbit KB5 = Key_state^5;



 //================== 表格数据 ====================

code unsigned char LED_font[24]=

 {

         0x84,0x9f,0xa2,0x8a,0x99,0xc8,0xc0,0x9e,0x80, //012345678

         0x88,0x90,0xc1,0xe4,0x83,0xe0,0xf0,0xff,0xfb, //9abcdef -

};



code unsigned char Key_tab[64]=     //键码映射表

{//  0  1  2  3  4  5  6  7  8  9   

         22, 0, 2, 0, 0, 0, 0, 0, 4, 0, //0

          0, 0, 0, 0, 0,18, 0, 0, 0, 0, //1X

          0, 0, 0, 0, 0, 0, 3,14, 0, 0, //2X

         20,10, 6, 0, 0, 0, 0, 0, 1,19, //3X

          0, 5, 0, 0, 0,15, 0,11, 0, 0, //4X

          0,17, 0, 0,13, 8, 0,21, 0, 9, //5X

         16,12, 7, 0                    //6X

 };



 //=============== 检测按键 =================

void Key_scan()

 {   

     unsigned char i;

     Key_count --;                        //扫描次序

    Key_count &= 3;

     switch (Key_count)                //按次序处理

    {

         case 2:                                //第一轮扫描

        KB0 = IO_b; 

         KB1 = IO_c; 

         IO_a = 1;

         IO_b = 0;

         break;

     

         case 1:                                //每二轮扫描

        KB2 = IO_c;

         KB3 = IO_a;

         IO_b = 1;

         IO_c = 0;

         break;

     

         case 0:                                //每三轮扫描

        KB4 = IO_a;

         KB5 = IO_b;

         IO_c = 1;

         break;

     

         default:                        //每四轮扫描

        if (!IO_a) KB0 = 0;

         if (!IO_b) KB2 = 0;

         if (!IO_c) KB4 = 0;

         IO_a = 0;



         //======更新显示缓冲区=======

         i = Key_tab[Key_state];

         if (i == 0)

         {

             Disp_buf[2] = 0x11;                //显示三横

            Disp_buf[1] = 0x11;

             Disp_buf[0] = 0x11;

         }

         else

         {

             Disp_buf[2] = 0x0c;     //字符"C"

             Disp_buf[1] = i / 10;   //键码十位

            Disp_buf[0] = B;于      //键码个位

        }

         Key_state = 0;

     }

 }    



     

 /*===================================================================

                     ONE WIRE 显示总线驱动程序       

 ===================================================================*/



 //=============== 发送一位 =================

void Send_bit(bit Dat)    

 {    

     unsigned char i = 3;

     if (!Dat) Bus = 0;

     else

     {

         Bus = 0;

         Bus = 1;

     }

     while(--i);                 //延时8us    

     Bus = 1;

 }    



 //=============== 总线驱动 =================

void Bus_drive()

 {

     unsigned char i = 0;

     unsigned char Sdat;

     Send_bit(1);                        //Bit6消隐

    do Bus = 1; while(--i);             //延时768us

     do Bus = 0; while(--i);             //延时768us

     Bus = 1;

     Sdat = LED_font[Disp_buf[Dig++]];   //获取显示数据

    Send_bit(Sdat & 0x01);              //发送位0        

     Send_bit(Sdat & 0x02);              //发送位1        

     Send_bit(Sdat & 0x04);              //发送位2        

     Send_bit(Sdat & 0x08);              //发送位3        

     Send_bit(Sdat & 0x10);              //发送位4        

     Send_bit(Sdat & 0x20);              //发送位5        

     Send_bit(Dig  & 0x01);              //发送位选1        

     Send_bit(Dig  & 0x02);              //发送位选2

     while(--i);                         //延时512us

     Send_bit(Sdat & 0x40);              //发送位6

     for (i = 7;i> 0;i--) Send_bit(1);  //位6移至Dout

     if (Dig == 3) Dig = 0;

 }     

         

 /*===================================================================

                     延时 5ms 程序       

 ===================================================================*/

void Delay_5ms()        

 {    

     while(!TF1);    

     TF1 = 0;    

     TH1 = (- 5000) / 256;

     TL1 = (- 5000) % 256;

 }    



 /*===================================================================

                         主程序       

 ===================================================================*/

void main()

 {

     TMOD = 0x10;            //定时器1,16位模式

    TCON = 0xc0;            //TR1=1;TF1=1;

     while(1)                //主循环

    {

         Bus_drive();        //显示总线驱动 

        Key_scan();         //检测按键

        Delay_5ms();        //延时5MS    

     }

 }

 

【29楼】我把22个按键的组态描述一下,看图就不会觉得费劲了

三个IO简称为A,B,C

按键1:A直接接地 

按键2:A、B通过两二极管同时接地

按键3:B直接接地

按键4:B、C通过两二极管同时接地

按键5:C直接接地

按键6:C、A通过两二极管同时接地

按键7:B通过二极管被A拉低

按键8:A通过二极管被B拉低

按键9:C通过二极管被B拉低

按键10:B通过二极管被C拉低

按键11:A通过二极管被C拉低

按键12:C通过二极管被A拉低

按键13:A、B直接短路

按键14:B、C直接短路

按键15:C、A直接短路

按键16:B、C通过两二极管同时被A拉低

按键17:C、A通过两二极管同时被B拉低

按键18:A、B通过两二极管同时被C拉低
按键19:A通过二极管被B或C拉低

按键20:B通过二极管被C或A拉低

按键21:C通过二极管被A或B拉低

按键22:A、B、C通过三个二极管(或电阻)同时接地

 

 

两个二极管的2个IO的情况可参照以前的三菱键盘的帖子:   

http://www.amobbs.com/bbs/bbs_content.jsp?bbs_sn=1280358