25． 点阵式LED“0－9”数字显示技术 1． 实验任务 利用8X8点阵显示数字0到9的数字 2． 电路原理图 图4.25.1 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； 4． 程序设计内容 (1)． 数字0－9点阵显示代码的形成 如下图所示，假设显示数字“0” 1 2 3 4 5 6 7 8 ●●● ● ● ● ● ● ● ● ● ● ● ●●● 00 00 3E 41 41 41 3E 00 因此，形成的列代码为 00H，00H，3EH，41H，41H，3EH，00H，00H；只要把这些代码分别送到相应的列线上面，即可实现“0”的数字显示 送显示代码过程如下所示 送第一列线代码到P3端口，同时置第一行线为“0”，其它行线为“1”，延时2ms左右，送第二列线代码到P3端口，同时置第二行线为“0”，其它行线为“1”，延时2ms左右，如此下去，直到送完最后一列代码，又从头开始送。
数字“1”代码建立如下图所示1 2 3 4 5 6 7 8 ● ●● ● ● ● ● ●●● 其显示代码为 00H，00H，00H，00H，21H，7FH，01H，00H 数字“2”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●● ● ● ● ● ●●●● ● ●●●●● 00H，00H，27H，45H，45H，45H，39H，00H 数字“3”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●● ● ● ● ●●● ● ● ● ●●● 00H，00H，22H，49H，49H，49H，36H，00H 数字“4”代码建立如下图所示 1 2 3 4 5 6 7 8 ● ●● ● ● ● ● ●●●●● ● ● 00H，00H，0CH，14H，24H，7FH，04H，00H 数字“5”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●●●● ● ●●●● ● ● ● ● ●●● 00H，00H，72H，51H，51H，51H，4EH，00H 数字“6”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●● ● ● ● ●●●● ● ● ● ● ●●● 00H，00H，3EH，49H，49H，49H，26H，00H 数字“7”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●●●● ● ● ● ● ● ● 00H，00H，40H，40H，40H，4FH，70H，00H 数字“8”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●● ● ● ● ● ●●● ● ● ● ● ●●● 00H，00H，36H，49H，49H，49H，36H，00H 数字“9”代码建立如下图所示 1 2 3 4 5 6 7 8 ●●● ● ● ● ● ●●●● ● ● ● ●●● 00H，00H，32H，49H，49H，49H，3EH，00H 5． 汇编源程序TIM EQU 30HCNTA EQU 31HCNTB EQU 32HORG 00HLJMP STARTORG 0BHLJMP T0XORG 30HSTART: MOV TIM,#00HMOV CNTA,#00HMOV CNTB,#00HMOV TMOD,#01HMOV TH0,#(65536-4000)/256MOV TL0,#(65536-4000) MOD 256SETB TR0SETB ET0SETB EASJMP $T0X:MOV TH0,#(65536-4000)/256MOV TL0,#(65536-4000) MOD 256MOV DPTR,#TABMOV A,CNTAMOVC A,@A+DPTRMOV P3,AMOV DPTR,#DIGITMOV A,CNTBMOV B,#8MUL ABADD A,CNTAMOVC A,@A+DPTRMOV P1,AINC CNTAMOV A,CNTACJNE A,#8,NEXTMOV CNTA,#00HNEXT: INC TIMMOV A,TIMCJNE A,#250,NEXMOV TIM,#00HINC CNTBMOV A,CNTBCJNE A,#10,NEXMOV CNTB,#00HNEX: RETITAB: DB 0FEH,0FDH,0FBH,0F7H,0EFH,0DFH,0BFH,07FHDIGIT: DB 00H,00H,3EH,41H,41H,41H,3EH,00HDB 00H,00H,00H,00H,21H,7FH,01H,00HDB 00H,00H,27H,45H,45H,45H,39H,00HDB 00H,00H,22H,49H,49H,49H,36H,00HDB 00H,00H,0CH,14H,24H,7FH,04H,00HDB 00H,00H,72H,51H,51H,51H,4EH,00HDB 00H,00H,3EH,49H,49H,49H,26H,00HDB 00H,00H,40H,40H,40H,4FH,70H,00HDB 00H,00H,36H,49H,49H,49H,36H,00HDB 00H,00H,32H,49H,49H,49H,3EH,00HEND6． C语言源程序#include unsigned char code tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};unsigned char code digittab[10][8]={ {0x00,0x00,0x3e,0x41,0x41,0x41,0x3e,0x00}, //0{0x00,0x00,0x00,0x00,0x21,0x7f,0x01,0x00}, //1{0x00,0x00,0x27,0x45,0x45,0x45,0x39,0x00}, //2{0x00,0x00,0x22,0x49,0x49,0x49,0x36,0x00}, //3{0x00,0x00,0x0c,0x14,0x24,0x7f,0x04,0x00}, //4{0x00,0x00,0x72,0x51,0x51,0x51,0x4e,0x00}, //5{0x00,0x00,0x3e,0x49,0x49,0x49,0x26,0x00}, //6{0x00,0x00,0x40,0x40,0x40,0x4f,0x70,0x00}, //7{0x00,0x00,0x36,0x49,0x49,0x49,0x36,0x00}, //8{0x00,0x00,0x32,0x49,0x49,0x49,0x3e,0x00} //9};unsigned int timecount;unsigned char cnta;unsigned char cntb;void main(void){TMOD=0x01;TH0=(65536-3000)/256;TL0=(65536-3000)%256;TR0=1;ET0=1;EA=1;while(1){;}}void t0(void) interrupt 1 using 0{TH0=(65536-3000)/256;TL0=(65536。