Homemade MCU Part 3: Making and driving digital tube circuit

The usage of digital tubes is no different from that of light-emitting diodes. It is just that seven or eight light-emitting diodes are combined on a module to form an 8-shaped figure and a decimal point to display numbers. In order to reduce the number of pins, the same poles of each light-emitting tube are connected together as a common point, thus giving rise to the concept of common anode and common cathode digital tubes. Common anode tubes are to connect the positive poles of each light-emitting tube together, while common cathode tubes are just the opposite. See the figure below:

Generally speaking, the absorption current of most logic ICs is Stronger than the output current. Therefore, everyone likes to use common cathode digital tubes, because there are more optional ICs. Unfortunately, my set of digital tubes is common anode, so I plan to use a three-stage tube to drive the common end. My minimal system board: I use the most commonly used S9012. First I have to plan The best circuit method is to use the most commonly used dynamic scanning display. First build the simplest circuit and debug the parameters of the components to be used. src="data:text/html;charset=utf8,%3Cimg%20id=%22img%22%20src=%22http://hiphotos.baidu.com/txz01/pic/item/61ee2c3e20ee4c3471cf6c8e.jpg?_=3264179% 22%20style=%22border:none;max-width:1333px%22%3E%3Cscript%3Ewindow.onload%20= %20function%20()%20%7Bvar%20img%20=%20document.getElementById('img');%20window.parent.postMessage(%7BiframeId:'iframe_0.8095873957499862',width:img.width,height:img .height%7D,%20'http://www.cnblogs.com');%7D%3C/script%3E" frameborder="0">Don't connect R2 and 74HC244 in the above picture yet. One segment of the digital tube is directly grounded. When adjusting R1, it is measured that when the base current of S9012 is 0.21mA, the collector, that is, the digital tube, has 40mA, indicating that the amplification factor is sufficient. At this time, connect R2 and 74HC244, adjust R2 to control the current of the digital tube at 15mA, so that when the 8 segments are lit together, the transistor needs 120mA of current. The base needs 0.63mA. In order to reduce the load of the transistor, Make the transistor oversaturated, adjust R1 so that the base current is 2mA, and the voltage between the collector and the drain is about 0.1V. Good! At this time, R1 is 2K. R2 is 240 ohms. OK. The next step is to determine the circuit. There are three groups of interfaces between the circuit interface and AT89S51: segment code, bit code and power supply. In order to make AT89S51 independent, these three levels of interfaces are all made of pins, and are freely connected to the P1-P3 ports of AT89S51 with flat cables. The power supply is connected with a short-circuit cap. The completed board is shown in the figure below reverse side: Description: Then write the program. Let's write a query method first!com');%7D%3C/script%3E" frameborder="0"> reverse side: Description: Then write the program. First write a query method Yeah!com');%7D%3C/script%3E" frameborder="0"> reverse side: Description: Then write the program. First write a query method Yeah!
//The six-digit code tube is flashing at an interval of 0.3 seconds. This is a query method, which takes up more CPU resources./************************************************************************Define
pins
: P2_0------upper horizontal a P3_0-------ones
                  P2_1-----upper right vertical b P3_1-------ten
                  P2_2-----lower right vertical c P3_2-------hundreds
                  P2_3-----lower horizontal d P3_3-------thousands
                  P2_4-----lower left vertical e P3_4-------ten thousand
                  P2_5-----upper left vertical f P3_5-------hundred thousand
                  P2_6-----middle horizontal g
                  P2_7-----decimal point H
******************************************************************/
# include
typedef unsigned char uchar;
uchar code bit_num[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf};//bit code value table: 0,1,2,3,4,5
uchar code meg_val[]={0x03,0x9f,0x25,0x0d,0x99,0x49};//segment code value table: 0,1,2,3,4,5
uchar code hello[]={0x03,0xe3,0xe3,0x61,0x91,0xff}; //HELLO
uchar code beybey[]={0x89,0x61,0xc1,0x89,0x61,0xc1};//beybey
uchar code ab6789[]={0xc1,0x11,0x09,0x01,0x1f,0x41};//ab6789
void delay(int n);
void main(void)
{
       uchar i,m;
       P2=0xff; //Turn off the segment code first
       P3=0xff; //Turn off the bit code
       delay(20);//Wait for a
       whilewhile(1)
        {
          for (m=30;m>0;m--) //Display 30 times, about 0.3 seconds
{
        for(i=0;i<=5;i++)
               {
                P2=0xff;
                P3=bit_num[i]; //Output bit code to P3 port
                P2=ab6789[i]; //Output segment code to P2 port
                delay(5);
        }
      }
          P2=0xff; //Turn off segment code
          P3=0xff; //Turn off bit code
          delay(1000); //Wait for 0.3 seconds
         }
}void delay(int n) //Subroutine
{
       int j;
       uchar k;
       for(j=0;j        {
          for(k=255;k>0;k--);
        }
}
=========================================
When I write the program into the chip and plug it in to run, it is garbled. Guess what happened?
It turns out that the direction of the P2 port is reversed. Have you noticed that in the AT89S51 pin arrangement, P0--P1 and P3 are all PX_0 on the top. Only the P2 port pin arrangement is P2_0 on the bottom. The direction is reversed. Since it is reversed, I will rewrite the segment code table. Try again, everything is normal.
Here I will talk about the arrangement of the segment codes. Many people asked how the digital tube segment codes are arranged. I also checked on the Internet, and it seems that there is no standard arrangement. It depends on your own connection method. This is also the reason why some digital tube programs downloaded on the Internet cannot be displayed normally on your own board. Generally speaking, the top diagram I have is the most marked. In the above program, it was originally intended that P2_0 corresponded to segment code a (that is, the horizontal line above). Until P2_7 corresponds to segment h (that is, the decimal point). As a result, who knew that the P2 port was just reversed. In this way, it is reversed, and P2_0 corresponds to segment h (the decimal point). For example, the digital tube I originally defined to display "2" segment code is 10100100B. Once it is connected in reverse, it is no longer "2". If you want to display "2" again, you have to reverse the high and low bits of the segment code. Change it to 00100101B and it will be OK. Here is another one that uses interruption to display: //This uses interruption mode and also flashes.
/************************************************************************
Define pins: P2_0------decimal point P3_0------units  
                  P2_1------middle horizontal P3_1------tens
                  P2_2------upper left vertical P3_2------hundreds
                  P2_3------lower left vertical P3_3------thousands
                  P2_4------lower horizontal P3_4------ten thousand
                  P2_5------lower right vertical P3_5------hundred thousand
                  P2_6------upper right vertical
                  P2_7------upper horizontal
******************************************************************/
# include
typedef unsigned char uchar;
uchar code bit_num[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf};//bit code: 0,1,2,3,4,5
uchar code meg_val[]={0x49,0x99,0x0d,0x25,0x9f,0x03};//segment code: 0,1,2,3,4,5
uchar i,aa; //define global variable
bit fg; //define a light on and off flag void timer0(void) interrupt 1 using 1 //interrupt program
{
      if (fg) //light up the 6-bit digital tube when fg is 1
       { P2=0xff;
         if (i>=6)
           {
             i=0;
            }
         else
           {
             P3=bit_num[i]; //output bit code to P3 port
             P2=meg_val[i]; //output segment code to P2 port
             i++;
            }
       }
      else //When fg is 0, turn off the digital tube
        {
           if (aa==0)
         {
           P3=0xff;
           P2=0xff;
                }
         }
       aa++;
       if (aa>=254) //When the value of aa is accumulated to 254, the fg flag flips.
         {
           fg=~fg; 
           aa=0;
}
       TH0=0xf8; //Reinstall the initial value of the timer, 2ms, the value is 65536-2000
       TL0=0x30;
}
void main(void)
{
       P2=0xff; //First turn off the segment code
       P3=0xff; //Turn off the bit code
       TMOD=0x01; //Set T0 to mode 1
       TH0=0xf8; //Load the initial count high bit
       TL0=0x30; //Load the initial count low bit
       EA=1; //Total charge allowed
       ET0=1; //T0 charge allowed
       fg=1; //Set the on/off flag to on
       TR0=1; //Start interrupt
       while(1);
}

OK!