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1、16×16LED点阵滚动显示屏一、硬件原理图二、汇编程序：;多字移动;一个字占32个字节，1K可以存放1K=1024字节/32=32个字，C51=4K=128个字，C52=8K=256个字。;但程序占用160个字节（相当于5个字的空间），最终可以显示的字为C51=123个字，C52=251个字。;超过后单片机将存储不下。ORG000HJMPMIANORG030HMIAN:MOVP3,#00;关闭端口MOVP1,#0FFHMOVP0,#00MOVP2,#00CALLDELAY;延时1msMOVR5,#2;移动列的数量（列数乘2）MOVA,#151;显示几个字MO
2、VB,#32;一个字=32个码MULABMOV22H,A;存放1～8个字的数量MOV23H,B;存放8的倍数个字的数量;如50个字：50×32=1600=0640H,则DPH处为06H=6，DPL处为40H=64;简单算法：50÷8=6余2，6为DPH处数，余的2×32=64为DPL数MOVDPTR,#TABLE1;赋查表初址MOVR0,DPL;保存DPTR初值,R0、R1记录取码进度MOVR1,DPHMOV20H,DPL;20H、21H保持DPTR初值不变MOV21H,DPHLOOP:MOVR7,#5;每屏停留时间L1:MOVR6,#16;每屏16个码MOV
3、R4,#00;扫描指针清零MOVDPL,R0;取码指针存入DPLMOVDPH,R1L16:MOVA,R4;扫描指针存入AMOVP1,A;扫描输出INCR4;扫描指针加1CLRA;A清零MOVCA,@A+DPTR;查表送AMOVP0,A;送P0口显示INCDPTR;取码指针加1CLRA;A清零MOVCA,@A+DPTR;查表送AMOVP2,A;送P2口显示INCDPTR;取码指针加1CALLDELAY;扫描延时1msMOVP0,#00;清除屏幕MOVP2,#00DJNZR6,L16;1屏是否完成?DJNZR7,L1;每屏停留时间是否到了?;MOVR0,DPL;如
4、字不需移动，用该指令;MOVR1,DPHMOVA,R0;如字需移动，用该指令ADDA,R5;如大于256，C=1MOVR0,AJNCDPINCR1DP:MOVA,20H;DPL初值ADDA,22H;如大于256，C=1。因DPL初值在90H以上，显示4个字后DPH将加1，导致判断失误MOVA,DPHSUBBA,21H;已显示字的数量（乘8）CJNEA,23H,LOOP;DPH中的字是否完成？CLRCMOVA,DPLSUBBA,20H;已显示字的数量CJNEA,22H,LOOP;DPL中的字是否完成？JMPMIANDELAY:MOVR2,#2;延时1msD1:M
5、OVR3,#248DJNZR3,$DJNZR2,D1RETTABLE1:;至拾字时为100个字，共150个字,用于测试。;(a3b1H)(１)(8001)DB00H,00H,00H,00H,00H,00H,00H,00HDB00H,00H,10H,02H,20H,02H,7fH,0feHDB00H,02H,00H,02H,00H,00H,00H,00HDB00H,00H,00H,00H,00H,00H,00H,00H;(a3b2H)(２)(8002)DB00H,00H,00H,00H,00H,00H,18H,1eHDB20H,22H,40H,42H,40H,8
6、2H,40H,82HDB41H,02H,41H,02H,21H,02H,1eH,0eHDB00H,00H,00H,00H,00H,00H,00H,00H;(a3b3H)(３)(8003)DB00H,00H,00H,00H,00H,00H,30H,0cHDB40H,02H,40H,02H,40H,02H,41H,02HDB41H,02H,42H,82H,44H,44H,38H,38HDB00H,00H,00H,00H,00H,00H,00H,00H;(a3b4H)(４)(8004)DB00H,00H,00H,00H,00H,30H,00H,50HDB01H,90
7、H,06H,10H,38H,12H,40H,12HDB7fH,0feH,00H,12H,00H,12H,00H,10HDB00H,00H,00H,00H,00H,00H,00H,00H;(a3b5H)(５)(8005)DB00H,00H,00H,00H,00H,00H,7fH,8cHDB41H,02H,42H,02H,42H,02H,42H,02HDB42H,02H,42H,02H,41H,04H,40H,0f8HDB00H,00H,00H,00H,00H,00H,00H,00H;(a3b6H)(６)(8006)DB00H,00H,00H,00H,00H,00
8、H,1fH,0f8HDB20H,84H
