单片机点阵式LED简单图形显示技术,Character displayer
关键字:AT89S51,单片机字符显示电路图
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
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 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)%6; 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)%6; P3=tab[cnta]; P1=graph[count][cnta]; cnta++; if(cnta==8) { cnta=0; } }
|