A technology based on 51 microcontroller and using 74HC595 to drive digital tubes

Based on the 51 microcontroller, the eight digital tube patterns are displayed as follows:

xxx11xxx→xx2222xx→x333333x→44444444→x555555x→xx6666xx→x777777x→88888888

Each state lasts one second, and the display cycles repeatedly, where x indicates that the corresponding digital tube is off.

Supplement to the question: It is a digital tube with a common anode, driven by a 595 chip. Using C language.

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Best answer: The program design is completed, and the screenshots of PROTEUS simulation are as follows.

The procedure is as follows:

//==============================================

#include

#define uint unsigned int

#define uchar unsigned char

//HC595 chip

sbit H_ds = P2^0;

sbit H_shcp = P2^1;

sbit H_stcp = P2^2;

char code DISP_8［] = {

0x00, 0x06, 0x5b, 0x4f, 0x66, 0x6d, 0x7d, 0x07, 0x7f};

char code TAB_64［] = {

0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 2, 2, 2, 0, 0,

0, 3, 3, 3, 3, 3, 3, 0, 4, 4, 4, 4, 4, 4, 4, 4,

0, 5, 5, 5, 5, 5, 5, 0, 0, 0, 6, 6, 6, 6, 0, 0,

0, 7, 7, 7, 7, 7, 7, 0, 8, 8, 8, 8, 8, 8, 8, 8};

char i = 1, j = 0;

char DISP_B［] = {0, 0, 0, 0, 0, 0, 0, 0};

//------------------------------------------------ ------------------

void sendbyte_H (uchar aa) //Use HC595 output

{

char z;

aa = ~aa;

for (z = 0; z < 8; z++) { //Loop 8 times to move in data

H_shcp = 0;

H_ds = aa & 128; //The data bits are sent to the HC595 data line

H_shcp = 1; //Input data on rising edge

aa "《= 1;

}

}

//------------------------------------------------ ------------------

void display()

{

char z;

for (z = 0; z < 8; z++) DISP_B［z］ = TAB_64［j * 8 + z］;

j++; j %= 8;

for (z = 0; z < 8; z++) sendbyte_H (DISP_8 [DISP_B [z]]);

H_stcp = 0; H_stcp = 1; //The rising edge causes data to be output in parallel

}

//------------------------------------------------ ------------------

void main()

{

TMOD = 0x01;

TH0 = (65536 - 46080) / 256; //50ms@11.0592MHz

TL0 = (65536 - 46080) % 256;

TR0 = 1;

ET0 = 1;

EA = 1;

while(1);

}

//------------------------------------------------ ------------------

void T0_time()interrupt 1

{

TL0 = (65536 - 46080) % 256; //Reset initial value

TH0 = (65536 - 46080) / 256; //50ms@11.0592MHz

i--;

if (i == 0) {

i = 40;

display(); //pattern display

}

}

//------------------------------------------------ ------------------