Programming Problems of Two-Bit 595 Driving 8-Bit Common Cathode Digital Tube

chenbingjy

Programming Problems of Two-Bit 595 Driving 8-Bit Common Cathode Digital Tube [Copy link]

Schematic

Code:

#include <reg52.h>
#include <intrins.h>
#define uchar unsigned char 
 

uchar code Table[] = 
{ // 0	 1	  2	   3	4	 5	  6	   7	8	 9	  A	   b	C    d	  E    F    - 
	0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71,0x86,0xFF,0xbf
};
 //-----------------------------------------------------------------------------

sbit SER = P1^0;	//串行数据输入端
sbit STCP = P1^1;	//时钟脉冲信号——上升沿有效 存储寄存器时钟输入端，并行输出
sbit SHCP = P1^2;	//输入信号————上升沿有效 移位寄存器时钟输入端

sbit key1 = P3^0;	//选择键
sbit key2 = P3^1;	//加键
sbit key3 = P3^2;	//减键
sbit key4 = P3^3;

uchar shan,A,K,C,D,E,F;
uchar knum;		//选择键计数变量

void TimerInit();
void KeyScan();
void Display (uchar shi10,uchar shi,uchar fen10,uchar fen,uchar miao10,uchar miao);	// 数码管显示
void SMG_Int(uchar Dat);		// 数码管单字节串行移位函数

//*****************************************************************************
// 主程序
void main () 
{
	TimerInit();
	while(1)
	{
		Display (A,K,C,D,E,F);
		KeyScan();
	} 
}

void KeyScan()
{
	
}


void Display (uchar shi10,uchar shi,uchar fen10,uchar fen,uchar miao10,uchar miao)
{
	uchar i;

	//显示第1位	小时10位
	i = Table[shi10];

	SMG_Int(i);			
	SMG_Int(0x01);		

	STCP = 0;	 //数据并行输出，（借助上升沿）
	_nop_();
	_nop_();
	STCP = 1;

	//显示第2位	小时个位
	i = Table[shi];

	SMG_Int(i);		
	SMG_Int(0x02);		

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;

	//显示第3位	-
	if(shan<10)	 //"-" 闪烁
	i = Table[16]; 
	else 
	i = Table[15];
	SMG_Int(i);			
	SMG_Int(0x04);	

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;

	//显示第4位	分钟10位
	i = Table[fen10];

	SMG_Int(i);			
	SMG_Int(0x08);		

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;

	//显示第5位	分钟个位
	i = Table[fen];

	SMG_Int(i);			
	SMG_Int(0x10);	

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;

	//显示第6位	-
	if(shan<10)	   //"-" 闪烁
	i = Table[16];
	else 
	i = Table[15];

	SMG_Int(i);			
	SMG_Int(0x20);	

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;

	//显示第7位	秒10位
	i = Table[miao10];

	SMG_Int(i);			
	SMG_Int(0x40);	

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;

	//显示第8位	秒个位
	i = Table[miao];

	SMG_Int(i);		
	SMG_Int(0x80);	

	STCP = 0;
	_nop_();
	_nop_();
	STCP = 1;
}

void SMG_Int(uchar Dat)	 //通过8次循环将8位数据移入74HC595
{
	uchar i;
	for(i=8;i>=1;i--)
	{
		if (Dat & 0x80) 
		{
			SER = 1;
		} 
		
		else
		{ 
			SER = 0;
		}
		Dat <<= 1;
		SHCP = 0;
		_nop_();
		_nop_();
		SHCP = 1;
	}
}

void TimerInit()	 //定时器0初始化
{
	TMOD = 0x01;
	TH0 = (65536-50000)/256;	//50ms初值
	TL0 = (65536-50000)%256;
	EA = 1;
	ET0 = 1;
	TR0 = 1;

}

void Timer0() interrupt 1
{
	static uchar cnt,shi=12,fen=59,miao=30;
	
	TH0 = (65536-50000)/256;
	TL0 = (65536-50000)%256;
	
	shan++;
	if(shan==20)
		shan=0;

	cnt++;
	if(cnt==20)
	{
		cnt = 0;
		miao++;
		if(miao == 60)
		{
			miao = 0;
			fen++;
			if(fen==60)
			{
				fen = 0;
				shi++;
				if(shi==24)
				{
					shi=0;
				}
			}
		}
	}

	A=shi/10;
	K=shi%10;
	C=fen/10;
	D=fen%10;
	E=miao/10;
	F=miao%10;
}

It was modified from common anode, and the segment code table was changed, but the program displayed garbled characters, and some digital tubes did not display anything.

Can any experts please tell me how to solve this problem? Thank you

Jacktang

This program and hardware circuit need to be combined to find

chenbingjy

Jacktang posted on 2024-11-22 07:29 This program and the hardware circuit need to be combined to find

Thanks, is there a problem with the hardware?

chineseboyzxy

It is 8 8-digit display, one 595 is responsible for segment code, and one 595 is responsible for bit code. You write all the combinations on paper, and display them one by one from 0 to 9 in the program, add a long delay in between, and then you can find the problem. However, the bit code 595 is not driven by a transistor, so I don’t know if it can drive the current of 7 segments to light up.

wangerxian

Let the segment selector output fully and see if all the digital tubes can light up.

chenbingjy

Thanks, I'll try it back.

chenbingjy

I have another question. I have two 595s driving digital tubes. Are the segment and bit selections mixed up? Thanks.

chenbingjy

chenbingjy

Another question, the position selection does not need to be changed, right?