并列三个16乘16点阵动态图

第八届“蓝天超市杯”电子产品设计竞赛（电子产品论文）『16*16点阵时钟显示屏』作品：时钟显示屏成员：叶勇健、赵云忠时间：2011/11/19摘要：点阵显示屏，它最大的特点在于功能的多样化，为我们带来了极大的方便。

而LED点阵式显示屏，具有亮度高、功耗小、寿命长等优点。

若以LED点阵来作为显示数字、符号的显示屏，则可实现远距离、大视角的时钟显示。

从而非常适合于在公共场所中显示时间。

本设计的重点在于对点阵式显示屏的显示研究。

从宏观上来上，显示屏可分为两个部份：显示器件和微控制器。

由微控制器控制显示器件的亮灭、颜色变化，从而组合出所需要的图相。

在此次设计中，采用了4块8X8个单色LED作为显示器件，用主芯片89c51、译码器74ls154、ds1302作为微控制器，以动态的方式进行图面显示。

由于充分考虑了软硬件的易扩充性，用本设计的构架思想可很方便地进行点阵扩充和多个汉字或图形的显示。

设计中所涉及的知识点主要有：点阵显示原理，字模提取，显示左移算法的实现，单片机、译码器和ds1304的使用。

以及在软硬件设计调试过程中的各种现象及解决方案。

关键词：点阵显示原理；LED点阵；单片机；74ls154；ds1302；Abstract：Dot matrix display, its biggest characteristic is that the function of diversification, bring us a great convenience. While the LED dot matrix display, has the advantages of high brightness, low power consumption, long life etc.. If the LED dot matrix display, symbols as digital display, it can realize long-distance, large visual angle clock display. Thus very suitable for public display time.The design focuses on the dot matrix screen display research. From a macro point, displays can be divided into two parts: display device and a micro controller. The micro controller control display device, light color change, thus combined the chart. In this design, using 4 8X8 LED as a monochrome display device, with the main chip, 89C51 decoder 74ls154, ds1304 as the microcontroller, in a dynamic way to map display. Because of fully considering the hardware and software expandability, the design framework of thought can be conveniently lattice expansion and a plurality of Chinese characters or graphics display.Involved in the design of the knowledge point are: dot matrix display principle, abstraction, showed a left shift algorithm, single chip microcomputer, the decoder and the use of ds1304. And in the software and hardware design and debugging process of a variety of phenomena and solutions.Keywords: Dot matrix display principle; LED matrix; single chip microcomputer; 74ls154; ds1304目录摘要 (2)Abstract (3)目录 (4)第一部分产品分析 (5)1.1 8051系列单片机介绍 (6)1.2 74LS154介绍 (7)1.3 DS1302芯片 (8)1.4 点阵屏工作方式 (9)第二部分产品设计 (10)2.1 设计目的 (11)2.2 相关软件的使用 (12)第三部分设计草图及电路图 (13)3.1 PROTEUS仿真图 (14)3.2PROTEL原理图 (15)3.3 PROTEL PCB图 (16)3.4C语言程序 (17)第四部分产品的显示效果（具体看展示） (18)第六部分：实验总结与体会 (19)小结 (20)附录程序代码 (21)第一部分：产品分析主要芯片介绍1.18051系列的单片机AT89C单片机为很多嵌入式控制系统提供了一种灵活性高且价廉的方案。

七段数码管码显示:
a f g e d c h b
实验说明：
a a a a a a a a
a
h b
bitsel(0) bitsel(7)
segment[7..0]
bitout[7..0]
16×16点阵LED汉字动态显示
4学时 学时 实验内容： 实验内容： 点阵LED动态显示自己的名字 用16×16点阵 × 点阵 动态显示自己的名字
第1行，第1列 行 列
16×16点阵显示 × 点阵显示
第1行，第16列 行 列
r0-r3
第16行，第1列 行 列
第16行，第16列 行 列
点阵LED显示器的每一行发光管的阳极接在一起作为 显示器的每一行发光管的阳极接在一起作为 点阵 数据线D，每一列发光管的阴极接在一起作为列线B， 数据线 ，每一列发光管的阴极接在一起作为列线 ， 因此，要使某一点亮， 因此，要使某一点亮，就必须让该点所在的行线为高 电平，同时让该点所在的列线为低电平。 电平，同时让该点所在的列线为低电平。 实验板上点阵的各行可分别控制， 实验板上点阵的各行可分别控制， 用16个行控制信号 个行控制信号 来控制16行，使用4个列控制信号来控制 列，实验 来控制 行 使用 个列控制信号来控制16列 个列控制信号来控制 板上已经提供了一个4－ 译码器 因此只使用4 译码器， 板上已经提供了一个 －16译码器，因此只使用 个信号即可控制点阵的16列 个信号即可控制点阵的 列。由于不是一个控制信号 控制一个点阵点， 控制一个点阵点，因此需要将需将欲显示的字一列一 列的显示。 列的显示。 若列控制信号r0-r3等于 “0000”，则经过 －16 例：若列控制信号 等于 ，则经过4－ 译 码器后，点阵的第一列会等于‘0’。此时若同时16 码器后，点阵的第一列会等于‘ 。此时若同时 行中的某些行为‘ ，则对应的点阵会亮。 行中的某些行为‘1’，则对应的点阵会亮。

Proteus仿真16*16LED点阵显示汉字显示程序：滚动程序：#include <reg52.h>#define uchar unsigned char#define uint unsigned intuchar count,i,j,t;uchar code table[]={0x09,0x00,0xFD,0x08,0x09,0x08,0x09,0x10,0x09,0x20,0x79,0x40,0x41,0x04,0x47, 0xFE,0x41,0x40,0x79,0x40,0x09,0x20,0x09,0x20,0x09,0x10,0x09,0x4E,0x51,0x84,0x21, 0x00,//张0x08,0x20,0x08,0x28,0x7F,0xFC,0x08,0x20,0x0F,0xE0,0x08,0x20,0x08,0x20,0x0F, 0xE0,0x01,0x04,0xFF,0xFE,0x01,0x00,0x03,0x80,0x05,0x60,0x19,0x1C,0x61,0x08,0x01, 0x00,//某0x08,0x20,0x08,0x28,0x7F,0xFC,0x08,0x20,0x0F,0xE0,0x08,0x20,0x08,0x20,0x0F, 0xE0,0x01,0x04,0xFF,0xFE,0x01,0x00,0x03,0x80,0x05,0x60,0x19,0x1C,0x61,0x08,0x01, 0x00,//某0x01,0x00,0x11,0x00,0x11,0x00,0x11,0x08,0x3F,0xFC,0x21,0x00,0x41,0x00,0x41, 0x00,0x81,0x10,0x3F,0xF8,0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x04,0xFF,0xFE,0x00, 0x00,//生0x00,0x10,0x1F,0xF8,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x1F, 0xF0,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x1F,0xF0,0x10,0x10,0x00, 0x00,//日0x10,0x80,0x10,0x80,0x10,0x80,0x10,0x88,0x5B,0xFC,0x54,0x88,0x50,0x88,0x90, 0x88,0x10,0x88,0x1F,0xFE,0x10,0x80,0x11,0x40,0x11,0x20,0x12,0x10,0x14,0x0E,0x18, 0x04,//快0x00,0x20,0x00,0xF0,0x1F,0x00,0x10,0x00,0x11,0x00,0x11,0x00,0x21,0x04,0x7F, 0xFE,0x01,0x00,0x01,0x00,0x09,0x20,0x09,0x10,0x11,0x08,0x21,0x0C,0x45,0x04,0x02, 0x00,//乐0x00,0x00,0x01,0x80,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x01, 0x80,0x01,0x80,0x01,0x80,0x00,0x00,0x01,0x80,0x03,0xC0,0x01,0x80,0x00,0x00,0x00, 0x00//!};uchar codetable1[]={0xf0,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd ,0xfe,0xff}; //列数组void delay(uint a){uint x,y;for(x=10;x>0;x--)for(y=a;y>0;y--);}void init(){TMOD=0X01;TH0=(65536-5000)/256;TL0=(65536-5000)%256;EA=1;ET0=1;TR0=1;}void main (){init();t=1; //控制滚动时长while(1){if(count==t){count=0;j++;if(j==112)j=0;}for(i=j;i<16+j;i++){P1=table1[i%16]; //列扫描，与横扫描顺序不能换P3=table[2*i]; //显示前8行P2=table[2*i+1]; //显示后8行delay(1); //延迟}}}void timer0()interrupt 1{TH0=(65536-50000)/256;TL0=(65536-50000)%256;count++;}移动程序：#include <reg52.h>#define uchar unsigned char#define uint unsigned intuchar count,i,j,t;uchar code table[]={0x09,0x00,0xFD,0x08,0x09,0x08,0x09,0x10,0x09,0x20,0x79,0x40,0x41,0x04,0x47, 0xFE,0x41,0x40,0x79,0x40,0x09,0x20,0x09,0x20,0x09,0x10,0x09,0x4E,0x51,0x84,0x21, 0x00,//张0x08,0x20,0x08,0x28,0x7F,0xFC,0x08,0x20,0x0F,0xE0,0x08,0x20,0x08,0x20,0x0F, 0xE0,0x01,0x04,0xFF,0xFE,0x01,0x00,0x03,0x80,0x05,0x60,0x19,0x1C,0x61,0x08,0x01, 0x00,//某0x08,0x20,0x08,0x28,0x7F,0xFC,0x08,0x20,0x0F,0xE0,0x08,0x20,0x08,0x20,0x0F, 0xE0,0x01,0x04,0xFF,0xFE,0x01,0x00,0x03,0x80,0x05,0x60,0x19,0x1C,0x61,0x08,0x01, 0x00,//某0x01,0x00,0x11,0x00,0x11,0x00,0x11,0x08,0x3F,0xFC,0x21,0x00,0x41,0x00,0x41, 0x00,0x81,0x10,0x3F,0xF8,0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x04,0xFF,0xFE,0x00, 0x00,//生0x00,0x10,0x1F,0xF8,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x1F, 0xF0,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x1F,0xF0,0x10,0x10,0x00, 0x00,//日0x10,0x80,0x10,0x80,0x10,0x80,0x10,0x88,0x5B,0xFC,0x54,0x88,0x50,0x88,0x90, 0x88,0x10,0x88,0x1F,0xFE,0x10,0x80,0x11,0x40,0x11,0x20,0x12,0x10,0x14,0x0E,0x18, 0x04,//快0x00,0x20,0x00,0xF0,0x1F,0x00,0x10,0x00,0x11,0x00,0x11,0x00,0x21,0x04,0x7F, 0xFE,0x01,0x00,0x01,0x00,0x09,0x20,0x09,0x10,0x11,0x08,0x21,0x0C,0x45,0x04,0x02, 0x00,//乐0x00,0x00,0x01,0x80,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x01, 0x80,0x01,0x80,0x01,0x80,0x00,0x00,0x01,0x80,0x03,0xC0,0x01,0x80,0x00,0x00,0x00, 0x00//!};uchar codetable1[]={0xf0,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd ,0xfe,0xff}; //列数组void delay(uint a){uint x,y;for(x=10;x>0;x--)for(y=a;y>0;y--);}void init(){TMOD=0X01;TH0=(65536-5000)/256;TL0=(65536-5000)%256;EA=1;ET0=1;TR0=1;}void main (){init();t=2; //控制变化时长while(1){if(count==t){count=0;j++;if(j>112)j=0;}for(i=0;i<16;i++){P1=table1[i%16]; //列扫描，与横扫描顺序不能换P3=table[2*i+2*j]; //显示前8行P2=table[2*i+1+2*j]; //显示后8行delay(1); //延迟}}}void timer0()interrupt 1{TH0=(65536-50000)/256;TL0=(65536-50000)%256;count++;}跳动显示：#include <reg52.h>#define uchar unsigned char#define uint unsigned intuchar count,i,j,t;uchar code table[]={0x09,0x00,0xFD,0x08,0x09,0x08,0x09,0x10,0x09,0x20,0x79,0x40,0x41,0x04,0x47, 0xFE,0x41,0x40,0x79,0x40,0x09,0x20,0x09,0x20,0x09,0x10,0x09,0x4E,0x51,0x84,0x21, 0x00,//张0x08,0x20,0x08,0x28,0x7F,0xFC,0x08,0x20,0x0F,0xE0,0x08,0x20,0x08,0x20,0x0F, 0xE0,0x01,0x04,0xFF,0xFE,0x01,0x00,0x03,0x80,0x05,0x60,0x19,0x1C,0x61,0x08,0x01, 0x00,//某0x08,0x20,0x08,0x28,0x7F,0xFC,0x08,0x20,0x0F,0xE0,0x08,0x20,0x08,0x20,0x0F, 0xE0,0x01,0x04,0xFF,0xFE,0x01,0x00,0x03,0x80,0x05,0x60,0x19,0x1C,0x61,0x08,0x01, 0x00,//某0x01,0x00,0x11,0x00,0x11,0x00,0x11,0x08,0x3F,0xFC,0x21,0x00,0x41,0x00,0x41, 0x00,0x81,0x10,0x3F,0xF8,0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x04,0xFF,0xFE,0x00, 0x00,//生0x00,0x10,0x1F,0xF8,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x1F, 0xF0,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x10,0x1F,0xF0,0x10,0x10,0x00, 0x00,//日0x10,0x80,0x10,0x80,0x10,0x80,0x10,0x88,0x5B,0xFC,0x54,0x88,0x50,0x88,0x90, 0x88,0x10,0x88,0x1F,0xFE,0x10,0x80,0x11,0x40,0x11,0x20,0x12,0x10,0x14,0x0E,0x18, 0x04,//快0x00,0x20,0x00,0xF0,0x1F,0x00,0x10,0x00,0x11,0x00,0x11,0x00,0x21,0x04,0x7F, 0xFE,0x01,0x00,0x01,0x00,0x09,0x20,0x09,0x10,0x11,0x08,0x21,0x0C,0x45,0x04,0x02, 0x00,//乐0x00,0x00,0x01,0x80,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x03,0xC0,0x01, 0x80,0x01,0x80,0x01,0x80,0x00,0x00,0x01,0x80,0x03,0xC0,0x01,0x80,0x00,0x00,0x00, 0x00//!};uchar codetable1[]={0xf0,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd ,0xfe,0xff}; //列数组void delay(uint a){uint x,y;for(x=10;x>0;x--)for(y=a;y>0;y--);}void init(){TMOD=0X01;TH0=(65536-5000)/256;TL0=(65536-5000)%256;EA=1;ET0=1;TR0=1;}void main (){init();t=8; //控制滚动时长while(1){if(count==t){count=0;j+=16;if(j>128)j=0;}for(i=j;i<16+j;i++){P1=table1[i%16]; //列扫描，与横扫描顺序不能换P3=table[2*i]; //显示前8行P2=table[2*i+1]; //显示后8行delay(1); //延迟}}}void timer0()interrupt 1{TH0=(65536-50000)/256;TL0=(65536-50000)%256;count++;}4个4*4点阵构成16*16点阵Proteus中只有5×7和8×8等LED点阵，并没有16×16LED点阵，而在实际应用中，要良好地显示一个汉字，则至少需要16×16点阵。

LOOP:MOV R7,#00
;初始化列显示数据
MOV R6,20H
MOV R1,#16H
;是否显示完成一个汉字
MAIN:MOV A,R6
;主程序入口
MOV DPTR,#TAB
MOVC A,@A+DPTR
MOV P0,A
INC R6
MOV A,R6
MOVC A,@A+DPTR
MOV P2,AINC R616*16 Nhomakorabea阵显示汉字
1
判断点阵是共 8*8点阵连接 阴还是共阳级 为16*16点阵
端口连接方法 和显示原理
测试哪个方向 是行，哪个方 向是列
是否可完成一 个汉字显示及 字模软件应用
动态显示汉字 原理
2
8*8点阵连接为16*16点阵
3
4
端口连接方法和显示原理
因为单片机的I/O口是有限的，所以LED汉字的显示 使用到了动态扫描，一般行数据用到了P0,P2口，列 数据用到了P1口并且用了一个4-16线译码器。因为 我们每次只需要点亮一列并延时一段时间，就可以使 我们认为汉字是一直在显示的。
DJNZ 22H,LOOP2
AJMP START
DELAY:MOV R4,#01
;延时程序
D1:MOV R3,#20
DJNZ R3,$
DJNZ R4,D1
RET
TAB: XXXXXXXXXXXXXXXXXXXXXXXXXXX
13
14
制作:W.K.
15
DELAY:MOV R1,#2 D1: MOV R2,#10
DJNZ R2,$ DJNZ R1,D1 RET TAB: DB 0x10,0x80,0x10,0x80,0x50,0x80,0x50, 0xFC,0x7D,0x54,0x52,0x54,0x90,0x54, 0x10,0x94 DB 0x1C,0x94,0xF1,0x24,0x52,0x24,0x10, 0x44,0x10,0x44,0x10,0x84,0x11,0x28, 0x10,0x10 END