51 MCU Experiment 5: Dynamic Digital Tube

The circuit diagram of the digital tube module of the development board is as follows:

Due to the characteristics of the 74ls138 decoder, that is, only one digital tube can have a cathode level of 0, it is obviously impossible to make multiple digital tubes work at the same time. If you want to get the effect of multiple digital tubes working at the same time, you can only use the dynamic scanning display method. The so-called dynamic scanning display is to use the afterglow of the light-emitting tube and the visual persistence of the human eye to make people feel as if multiple digital tubes are working at the same time.

In the program, the digdisplay function uses for loops and switch statements to continuously scan the 8 digital tubes, complete the bit selection, and assign values ​​to the P0 port to realize the segment selection function. Thus, the 8 digital tubes output 1 to 8 in sequence, and use the delay function to control the output interval time to realize the dynamic output function.

detail:

(1) Set a certain delay after each digital tube displays.

(2) Set P0 to 0 (P0 = 0x00) to achieve the purpose of "erasing the image".

#include

#define uc unsigned 

sbit lsc=P2^2;

sbit lsb=P2^3;

sbit lsa=P2^4;

void delay(uc n)

{

    uc i,j;

    for(i=1;i<=n;i++)

      for(j=110;j>=1;j--);

}

uc code x[16]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,

0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};

void digdisplay()

{

uc i;

for(i=1;i<=8;i++)

{

switch(i)

{

case (1): lsa=0,lsb=0;lsc=0;break;

case (2): lsa=0,lsb=0;lsc=1;break;

case (3): lsa=0,lsb=1;lsc=0;break;

case (4): lsa=0,lsb=1;lsc=1;break;

case (5): lsa=1,lsb=0;lsc=0;break;

case (6): lsa=1,lsb=0;lsc=1;break;

case (7): lsa=1,lsb=1;lsc=0;break;

case (8): lsa=1,lsb=1;lsc=1;break;

}

P0=x[i];

        delay(1);

        P0=0x00;

}

}

void main()

{

    while(1)

    {

digdisplay();

    }

}

Advanced: Use digital tubes to output different digital combinations at intervals.

#include

#define uc unsigned char

sbit lsa=P2^2;

sbit lsb=P2^3;

sbit lsc=P2^4;

uc code x[16]={0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,

0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71};

void delay1(uc n)

{

uc i,j;

for(i=1;i<=n;i++)

  for(j=1;j<=110;j++);

}

void digdisplay(int n)

{

uc i;

if(n==1)

{

for(i=0;i<3;i++)

{

switch(i)

{

case(0): lsc=0;lsb=0;lsa=0; P0=x[0];break;

case(1): lsc=0;lsb=0;lsa=1; P0=x[2];break;

case(2): lsc=0;lsb=1;lsa=0; P0=x[5];break;

}

delay1(1);

}

}

    else

{

for(i=0;i<4;i++)

{

switch(i)

{

case(0): lsc=0;lsb=0;lsa=0; P0=x[4];break;

case(1): lsc=0;lsb=0;lsa=1; P0=x[1];break;

case(2): lsc=0;lsb=1;lsa=0; P0=x[3];break;

case(3): lsc=0;lsb=1;lsa=1; P0=x[1];break;

}

delay1(1);

}

}

}

void delay2(uc n)

{

uc i,j;

for(i=1;i<=10;i++)

  for(j=1;j<=110;j++)

  digdisplay(n);

}

void main()

{

while(1)

{

delay2(1);

delay2(0);

}

}