新闻中心

EEPW首页 > 嵌入式系统 > 设计应用 > 单片机“叮咚”门铃设计

单片机“叮咚”门铃设计

作者:时间:2013-04-02来源:网络收藏

6.汇编源程序

T5HZ EQU 30H

T7HZ EQU 31H

T05SA EQU 32H

T05SB EQU 33H

FLAG BIT 00H

STOP BIT 01H

SP1 BIT P3.7

ORG 00H

LJMP START

ORG 0BH

LJMP INT_T0

START: MOV TMOD,#02H

MOV TH0,#06H

MOV TL0,#06H

SETB ET0

SETB EA

NSP: JB SP1,NSP

LCALL DELY10MS

JB SP1,NSP

SETB TR0

MOV T5HZ,#00H

MOV T7HZ,#00H

MOV T05SA,#00H

MOV T05SB,#00H

CLR FLAG

CLR STOP

JNB STOP,$

LJMP NSP

DELY10MS: MOV R6,#20

D1: MOV R7,#248

DJNZ R7,$

DJNZ R6,D1

RET

INT_T0: INC T05SA

MOV A,T05SA

CJNE A,#100,NEXT

MOV T05SA,#00H

INC T05SB

MOV A,T05SB

CJNE A,#20,NEXT

MOV T05SB,#00H

JB FLAG,STP

CPL FLAG

LJMP NEXT

STP: SETB STOP

CLR TR0

LJMP DONE

NEXT: JB FLAG,S5HZ

INC T7HZ

MOV A,T7HZ

CJNE A,#03H,DONE

MOV T7HZ,#00H

CPL P1.0

LJMP DONE

S5HZ: INC T5HZ

MOV A,T5HZ

CJNE A,#04H,DONE

MOV T5HZ,#00H

CPL P1.0

LJMP DONE

DONE: RETI

END

7. C语言源程序

#include AT89X51.H>

unsigned char t5hz;

unsigned char t7hz;

unsigned int tcnt;

bit stop;

bit flag;

void main(void)

{

unsigned char i,j;

TMOD=0x02;

TH0=0x06;

TL0=0x06;

ET0=1;

EA=1;

while(1)

{

if(P3_7==0)

{

for(i=10;i>0;i--)

for(j=248;j>0;j--);

if(P3_7==0)

{

t5hz=0;

t7hz=0;

tcnt=0;

flag=0;

stop=0;

TR0=1;

while(stop==0);

}

}

}

}

void t0(void) interrupt 1 using 0

{

tcnt++;

if(tcnt==2000)

{

tcnt=0;

if(flag==0)

{

flag=~flag;

}

else

{

stop=1;

TR0=0;

}

}

if(flag==0)

{

t7hz++;

if(t7hz==3)

{

t7hz=0;

P1_0=~P1_0;

}

}

else

{

t5hz++;

if(t5hz==4)

{

t5hz=0;

P1_0=~P1_0;

}

}

}


上一页 1 2 下一页

评论


相关推荐

技术专区

关闭