1．實驗任務
利用8X8點陣顯示數字0到9的數字。
2．電路原理圖

                                                                                            圖4.25.1
3．硬件系統連線
(1)．把“單片機系統”區域中的P1端口用8芯排芯連接到“點陣模塊”區域中的“DR1－DR8”端口上；
(2)．把“單片機系統”區域中的P3端口用8芯排芯連接到“點陣模塊”區域中的“DC1－DC8”端口上；
4．程序設計內容
(1)．數字0－9點陣顯示代碼的形成
　　如下圖所示，假設顯示數字“0”
 

00 00  3E  41  41  41  3E 00
　　因此，形成的列代碼為　00H，00H，3EH，41H，41H，3EH，00H，00H；只要把這些代碼分別送到相應的列線上面，即可實現“0”的數字顯示。
　　送顯示代碼過程如下所示
　　送第一列線代碼到P3端口，同時置第一行線為“0”，其它行線為“1”，延時2ms左右，送第二列線代碼到P3端口，同時置第二行線為“0”，其它行線為“1”，延時2ms左右，如此下去，直到送完最后一列代碼，又從頭開始送。
數字“1”代碼建立如下圖所示

其顯示代碼為　00H，00H，00H，00H，21H，7FH，01H，00H
數字“2”代碼建立如下圖所示
 

00H，00H，27H，45H，45H，45H，39H，00H
數字“3”代碼建立如下圖所示

00H，00H，22H，49H，49H，49H，36H，00H
數字“4”代碼建立如下圖所示
 

00H，00H，0CH，14H，24H，7FH，04H，00H
數字“5”代碼建立如下圖所示
 

00H，00H，72H，51H，51H，51H，4EH，00H
數字“6”代碼建立如下圖所示

00H，00H，3EH，49H，49H，49H，26H，00H
數字“7”代碼建立如下圖所示
 

00H，00H，40H，40H，40H，4FH，70H，00H
數字“8”代碼建立如下圖所示
 

00H，00H，36H，49H，49H，49H，36H，00H
數字“9”代碼建立如下圖所示

00H，00H，32H，49H，49H，49H，3EH，00H
5．匯編源程序
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語言源程序
#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)%6;
  TR0=1;
  ET0=1;
  EA=1;
  while(1)
    {;
    }
}
void t0(void) interrupt 1 using 0
{
  TH0=(65536-3000)/256;
  TL0=(65536-3000)%6;
  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;
        }
    }
}