目录

单片机实例26点阵式LED简单图形显示技术硬件电路图汇编程序C语言程序

目录

单片机实例26——点阵式LED简单图形显示技术(硬件电路图+汇编程序+C语言程序)

26. 点阵式LED简单图形显示技术

1. 实验任务

在8X8点阵式LED显示“★”、“●”和心形图,通过按键来选择要显示的图形。

2. 电路原理图

https://i-blog.csdnimg.cn/blog_migrate/34434d70a468b222bbbe068e74e83565.png

图4.26.1

3. 硬件系统连线

(1). 把“单片机系统”区域中的P1端口用8芯排芯连接到“点阵模块”区域中的“DR1-DR8”端口上;

(2). 把“单片机系统”区域中的P3端口用8芯排芯连接到“点阵模块”区域中的“DC1-DC8”端口上;

(3). 把“单片机系统”区域中的P2.0/A8端子用导线连接到“独立式键盘”区域中的SP1端子上;

4. 程序设计内容

(1). “★”在8X8LED点阵上显示图如下图所示

1 2 3 4 5 6 7 8

12H,14H,3CH,48H,3CH,14H,12H,00H

(2). “●”在8X8LED点阵上显示图如下图所示

1 2 3 4 5 6 7 8

00H,00H,38H,44H,44H,44H,38H,00H

(3). 心形图在8X8LED点阵上显示图如下图所示

1 2 3 4 5 6 7 8

30H,48H,44H,22H,44H,48H,30H,00H

5. 汇编源程序

CNTA EQU 30H

COUNT EQU 31H

ORG 00H

LJMP START

ORG 0BH

LJMP T0X

ORG 30H

START: MOV CNTA,#00H

MOV COUNT,#00H

MOV TMOD,#01H

MOV TH0,#(65536-4000) / 256

MOV TL0,#(65536-4000) MOD 256

SETB TR0

SETB ET0

SETB EA

WT: JB P2.0,WT

MOV R6,#5

MOV R7,#248

D1: DJNZ R7,$

DJNZ R6,D1

JB P2.0,WT

INC COUNT

MOV A,COUNT

CJNE A,#03H,NEXT

MOV COUNT,#00H

NEXT: JNB P2.0,$

SJMP WT

T0X: NOP

MOV TH0,#(65536-4000) / 256

MOV TL0,#(65536-4000) MOD 256

MOV DPTR,#TAB

MOV A,CNTA

MOVC A,@A+DPTR

MOV P3,A

MOV DPTR,#GRAPH

MOV A,COUNT

MOV B,#8

MUL AB

ADD A,CNTA

MOVC A,@A+DPTR

MOV P1,A

INC CNTA

MOV A,CNTA

CJNE A,#8,NEX

MOV CNTA,#00H

NEX: RETI

TAB: DB 0FEH,0FDH,0FBH,0F7H,0EFH,0DFH,0BFH,07FH

GRAPH: DB 12H,14H,3CH,48H,3CH,14H,12H,00H

DB 00H,00H,38H,44H,44H,44H,38H,00H

DB 30H,48H,44H,22H,44H,48H,30H,00H

END

6. C语言源程序

#include <AT89X52.H>

unsigned char code tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
unsigned char code graph[3][8]={{0x12,0x14,0x3c,0x48,0x3c,0x14,0x12,0x00},
{0x00,0x00,0x38,0x44,0x44,0x44,0x38,0x00},
{0x30,0x48,0x44,0x22,0x44,0x48,0x30,0x00}
};
unsigned char count;
unsigned char cnta;

void main(void)
{
	unsigned char i,j;

	TMOD=0x01;
	TH0=(65536-4000)/256;
	TL0=(65536-4000)%256;
	TR0=1;
	ET0=1;
	EA=1;

	while(1)
	{
		if(P2_0==0)
		{
			for(i=5;i>0;i--)
				for(j=248;j>0;j--);
			if(P2_0==0)
			{
				count++;
				if(count==3)
				{
					count=0;
				}
				while(P2_0==0);
			}
		}
	}
}

void t0(void) interrupt 1 using 0
{
	TH0=(65536-4000)/256;
	TL0=(65536-4000)%256;
	P3=tab[cnta];
	P1=graph[count][cnta];
	cnta++;
	if(cnta==8)
	{
		cnta=0;
	}
}