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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 30H CNTA EQU 31H CNTB EQU 32H ORG 00H LJMP START ORG 0BH LJMP T0X ORG 30H START: MOV TIM,#00H MOV CNTA,#00H MOV CNTB,#00H MOV TMOD,#01H MOV TH0,#(65536-4000)/256 MOV TL0,#(65536-4000) MOD 256 SETB TR0 SETB ET0 SETB EA SJMP $ T0X: 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,#DIGIT MOV A,CNTB MOV B,#8 MUL AB ADD A,CNTA MOVC A,@A+DPTR MOV P1,A INC CNTA MOV A,CNTA CJNE A,#8,NEXT MOV CNTA,#00H NEXT: INC TIM MOV A,TIM CJNE A,#250,NEX MOV TIM,#00H INC CNTB MOV A,CNTB CJNE A,#10,NEX MOV CNTB,#00H NEX: RETI TAB: DB 0FEH,0FDH,0FBH,0F7H,0EFH,0DFH,0BFH,07FH DIGIT: DB 00H,00H,3EH,41H,41H,41H,3EH,00H DB 00H,00H,00H,00H,21H,7FH,01H,00H DB 00H,00H,27H,45H,45H,45H,39H,00H DB 00H,00H,22H,49H,49H,49H,36H,00H DB 00H,00H,0CH,14H,24H,7FH,04H,00H DB 00H,00H,72H,51H,51H,51H,4EH,00H DB 00H,00H,3EH,49H,49H,49H,26H,00H DB 00H,00H,40H,40H,40H,4FH,70H,00H DB 00H,00H,36H,49H,49H,49H,36H,00H DB 00H,00H,32H,49H,49H,49H,3EH,00H END 6. 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=(65536-3000)%256; P3=tab[cnta]; P1=digittab[cntb][cnta]; cnta++; if(cnta==8) { cnta=0; } timecount++; if(timecount==333) { timecount=0; cntb++; if(cntb==10) { cntb=0; } } } |
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