51单片机T0定时器应用1

1．实验任务
　　用AT89S51单片机的定时/计数器T0产生一秒的定时时间，作为秒计数时间，当一秒产生时，秒计数加1，秒计数到60时，自动从0开始。硬件电路如下图所示
2．电路原理图

图4.15.1
3．系统板上硬件连线
（1)．把“单片机系统”区域中的P0.0/AD0－P0.7/AD7端口用8芯排线连接到“四路静态数码显示模块”区域中的任一个a－h端口上；要求：P0.0/AD0对应着a，P0.1/AD1对应着b，……，P0.7/AD7对应着h。
（2)．把“单片机系统”区域中的P2.0/A8－P2.7/A15端口用8芯排线连接到“四路静态数码显示模块”区域中的任一个a－h端口上；要求：P2.0/A8对应着a，P2.1/A9对应着b，……，P2.7/A15对应着h。
4．程序设计内容
　　AT89S51单片机的内部16位定时/计数器是一个可编程定时/计数器，它既可以工作在13位定时方式，也可以工作在16位定时方式和8位定时方式。只要通过设置特殊功能寄存器TMOD，即可完成。定时/计数器何时工作也是通过软件来设定TCON特殊功能寄存器来完成的。
　　现在我们选择16位定时工作方式，对于T0来说，最大定时也只有65536us，即65.536ms，无法达到我们所需要的1秒的定时，因此，我们必须通过软件来处理这个问题，假设我们取T0的最大定时为50ms，即要定时1秒需要经过20次的50ms的定时。对于这20次我们就可以采用软件的方法来统计了。
　　因此，我们设定TMOD＝00000001B，即TMOD＝01H
　　下面我们要给T0定时/计数器的TH0，TL0装入预置初值，通过下面的公式可以计算出
TH0＝（216－50000）　/　256
TL0＝（216－50000）　MOD　256
　　当T0在工作的时候，我们如何得知50ms的定时时间已到，这回我们通过检测TCON特殊功能寄存器中的TF0标志位，如果TF0＝1表示定时时间已到。
5．程序

图4.15.2
6．汇编源程序（查询法）
SECOND            EQU 30H
TCOUNT           EQU 31H
                            ORG 00H
START:               MOV SECOND,#00H
                           MOV TCOUNT,#00H
                            MOV TMOD,#01H
                            MOV TH0,#(65536-50000) / 256
                           MOV TL0,#(65536-50000) MOD 256
                            SETB TR0
DISP:                  MOV A,SECOND
                            MOV B,#10
                           DIV AB
                            MOV DPTR,#TABLE
                            MOVC A,@A+DPTR
                           MOV P0,A
                            MOV A,B
                            MOVC A,@A+DPTR
                           MOV P2,A
WAIT:                 JNB TF0,WAIT
                            CLR TF0
                            MOV TH0,#(65536-50000) / 256
                           MOV TL0,#(65536-50000) MOD 256
                            INC TCOUNT
                            MOV A,TCOUNT
                           CJNE A,#20,NEXT
                            MOV TCOUNT,#00H
                            INC SECOND
                           MOV A,SECOND
                            CJNE A,#60,NEX
                           MOV SECOND,#00H
NEX:                   LJMP DISP
NEXT:                LJMP WAIT
TABLE:              DB 3FH,06H,5BH,4FH,66H,6DH,7DH,07H,7FH,6FH
                            END
7．C语言源程序（查询法）
#include
unsigned char code dispcode[]={0x3f,0x06,0x5b,0x4f,
                               0x66,0x6d,0x7d,0x07,
                               0x7f,0x6f,0x77,0x7c,
                               0x39,0x5e,0x79,0x71,0x00};
unsigned char second;
unsigned char tcount;
void main(void)
{
  TMOD=0x01;
  TH0=(65536-50000)/256;
  TL0=(65536-50000)%6;
  TR0=1;
  tcount=0;
  second=0;
  P0=dispcode[second/10];
  P2=dispcode[second];
  while(1)
    {
      if(TF0==1)
        {
          tcount++;
          if(tcount==20)
            {
              tcount=0;
              second++;
              if(second==60)
                {
                  second=0;
                }
              P0=dispcode[second/10];
              P2=dispcode[second];             
            }
          TF0=0;
          TH0=(65536-50000)/256;
          TL0=(65536-50000)%6;
        }
    }
}
8．汇编源程序（中断法）
SECOND                     EQU 30H
TCOUNT                    EQU 31H
                                     ORG 00H
                                     LJMP START
                                     ORG 0BH
                                     LJMP INT0X
START:                        MOV SECOND,#00H
                                     MOV A,SECOND
                                     MOV B,#10
                                     DIV AB
                                     MOV DPTR,#TABLE
                                     MOVC A,@A+DPTR
                                     MOV P0,A
                                     MOV A,B
                                     MOVC A,@A+DPTR
                                     MOV P2,A
                                     MOV TCOUNT,#00H
                                     MOV TMOD,#01H
                                     MOV TH0,#(65536-50000) / 256
                                     MOV TL0,#(65536-50000) MOD 256
                                     SETB TR0
                                     SETB ET0
                                     SETB EA
                                     SJMP $
INT0X:
                                     MOV TH0,#(65536-50000) / 256
                                     MOV TL0,#(65536-50000) MOD 256
                                     INC TCOUNT
                                     MOV A,TCOUNT
                                     CJNE A,#20,NEXT
                                     MOV TCOUNT,#00H
                                     INC SECOND
                                     MOV A,SECOND
                                     CJNE A,#60,NEX
                                     MOV SECOND,#00H
NEX:                                     MOV A,SECOND
                                     MOV B,#10
                                     DIV AB
                                     MOV DPTR,#TABLE
                                     MOVC A,@A+DPTR
                                     MOV P0,A
                                     MOV A,B
                                     MOVC A,@A+DPTR
                                     MOV P2,A
NEXT:                          RETI
                  
TABLE:                        DB 3FH,06H,5BH,4FH,66H,6DH,7DH,07H,7FH,6FH
                                     END
9．C语言源程序（中断法）
#include
unsigned char code dispcode[]={0x3f,0x06,0x5b,0x4f,
                               0x66,0x6d,0x7d,0x07,
                               0x7f,0x6f,0x77,0x7c,
                               0x39,0x5e,0x79,0x71,0x00};
unsigned char second;
unsigned char tcount;
void main(void)
{
  TMOD=0x01;
  TH0=(65536-50000)/256;
  TL0=(65536-50000)%6;
  TR0=1;
  ET0=1;
  EA=1;
  tcount=0;
  second=0;
  P0=dispcode[second/10];
  P2=dispcode[second];
  while(1);
}
void t0(void) interrupt 1 using 0
{
  tcount++;
  if(tcount==20)
    {
      tcount=0;
      second++;
      if(second==60)
        {
          second=0;
        }
      P0=dispcode[second/10];
      P2=dispcode[second];             
    }
  TH0=(65536-50000)/256;
  TL0=(65536-50000)%6;
}