原创 15．定时计数器T0作定时应用技术（一）

1． 实验任务<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

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。

2． 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。

3、实现程序

1． 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

2． C语言源程序（查询法）

#include <AT89X51.H>

unsigned char code dispcode[]={={0x3f,0x06,0x05b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f};

unsigned char second;

unsigned char tcount;

void main(void)

{

  TMOD=0x01;

  TH0=(65536-50000)/256;

  TL0=(65536-50000)%256;

  TR0=1;

  tcount=0;

  second=0;

  P0=dispcode[second/10];

  P2=dispcode[second%10];

  while(1)

    {

      if(TF0==1)

        {

          tcount++;

          if(tcount==20)

            {

              tcount=0;

              second++;

              if(second==60)

                {

                  second=0;

                }

              P0=dispcode[second/10];

              P2=dispcode[second%10];             

            }

          TF0=0;

          TH0=(65536-50000)/256;

          TL0=(65536-50000)%256;

        }

    }

}

1．      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

2．      C语言源程序（中断法）

#include <AT89X51.H>

unsigned char code dispcode[]={={0x3f,0x06,0x05b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f};

unsigned char second;

unsigned char tcount;

void main(void)

{

  TMOD=0x01;

  TH0=(65536-50000)/256;

  TL0=(65536-50000)%256;

  TR0=1;

  ET0=1;

  EA=1;

  tcount=0;

  second=0;

  P0=dispcode[second/10];

  P2=dispcode[second%10];

  while(1);

}

void t0(void) interrupt 1 using 0    //定时器T0中断程序

{

  tcount++;

  if(tcount==20)

    {

      tcount=0;

      second++;

      if(second==60)

        {

          second=0;

        }

      P0=dispcode[second/10];

      P2=dispcode[second%10];             

    }

  TH0=(65536-50000)/256;

  TL0=(65536-50000)%256;

}