3 common IOs recognize 22 buttons test

Learning from the experience of predecessors, the problem of too many diodes was optimized. At present, 6 keys can be connected without diodes, 12 keys can be connected with 2 diodes, 18 keys can be connected with 6 diodes, and 21 keys can be connected with 9 diodes. The 22nd key should occupy 3 diodes alone, which is the least cost-effective.

The experiment uses 89S51 for the test. The circuit wiring is P1.2, P1.3, P1.4 connected to the keyboard, and P1.0 connected to the display.





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

* 3 IO connection recognition 22 keys test program *

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

* MCU: AT89C2051 *

* OSC: 12M cysytel *

* Program design: Cowboy *

* Program version: V1.0 *

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



#include



//================== IO port line connection==================

sbit Bus = P1^0;

sbit IO_a = P1^4;

sbit IO_b = P1^3;

sbit IO_c = P1^2;

  

//==================== Variable declaration====================

unsigned char Disp_buf[3];

unsigned char Di g;

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;



//================= Table data====================

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]= //Key code mapping table

{// 0 1 2 3 4 5 6 7 8 9   

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

         , 5, 0, 0, 0, 0, 0, 0, //5X

         16,12

        , 7, 0, 0, 0, //6X 0, 0, 0, 0, 0, 0, 0,

         0, //7X 0, 0, 0, 0, 0, 0, 0, 0,         0

         ,     // 8X 0 ,    ... i;     Key_count --; //Scanning order     Key_count &= 3;     switch (Key_count) //Process in order     {         case 2: //First round of scanning         KB0 = IO_b;          KB1 = IO_c;          IO_a = 1;         IO_b = 0;         break;         case 1: //Every second round of scanning



































    



        KB2 = IO_c;

        KB3 = IO_a;

        IO_b = 1;

        IO_c = 0;

        break;

    

        case 0: //Scan every three rounds

        KB4 = IO_a;

        KB5 = IO_b;

        IO_c = 1;

        break;

    

        default: //Scan every four rounds

        if (!IO_a) KB0 = 0;

        if (!IO_b) KB2 = 0;

        if (!IO_c) KB4 = 0;

        IO_a = 0;



        //======Update display buffer=======

        i = Key_tab[Key_state];

        if (i == 0)

        {

            Disp_buf[2] = 0x11; //Display three horizontal lines

            Disp_buf[1] = 0x11;

            Disp_buf[0] = 0x11;

        }

        else

        {

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

            Disp_buf[1] = i / 10; //ten digits of key code

            Disp_buf[0] = B; //one digit of key code                         }       

        Key_state     =

        0     ;

    }

}     /



    

* = ...     (!Dat) Bus = 0;     else     {         Bus = 0;         Bus = 1;     }     while(--i); //Delay 8us         Bus = 1; }     //=============== Bus driver ================= void Bus_drive() {     unsigned char i = 0;     unsigned char Sdat;     Send_bit(1); //Bit6 blanking     do Bus = 1; while(--i); //Delay 768us     do Bus = 0; while(--i); //Delay 768us     Bus = 1;     Sdat = LED_font[Disp_buf[Dig++]]; //Get display data     Send_bit(Sdat & 0x01); //Send bit 0             Send_bit(Sdat & 0x02); //Send bit 1             Send_bit(Sdat & 0x04); //Send bit 2             Send_bit(Sdat & 0x08); //Send bit 3             Send_bit(Sdat & 0x10); //Send bit 4             Send_bit(Sdat & 0x20); //Send bit 5             Send_bit(Dig & 0x01); //Send bit select 1             Send_bit(Dig & 0x02); //Send bit select 2     while(--i); //Delay 512us     Send_bit(Sdat & 0x40); //Send bit 6     for (i = 7;i> 0;i--) Send_bit(1); //Move bit 6 to Dout     if (Dig == 3) Dig = 0; }     

















































































        

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