A brief introduction to designing timers using microcontrollers

　　1. Experimental tasks

　　Use the timer/counter T0 of the AT89S51 microcontroller to generate a one-second timing time as the second counting time. When one second is generated, the second count is increased by 1. When the second count reaches 60, it automatically starts from 0. Now I will start to introduce how to make this microcontroller timer. The hardware circuit is shown in the figure below

　　2. Circuit diagram

Figure 4.15.1

　　3. Hardware connections on the system board

　　(1. Connect the P0.0/AD0-P0.7/AD7 ports in the "MCU System" area to any a-h ports in the "Four-way Static Digital Display Module" area using an 8-core cable; requirements: P0.0/AD0 corresponds to a, P0.1/AD1 corresponds to b, ..., P0.7/AD7 corresponds to h.

　　(2. Use an 8-core cable to connect the P2.0/A8-P2.7/A15 ports in the "MCU System" area to any a-h ports in the "Four-way Static Digital Display Module" area; requirements: P2.0/A8 corresponds to a, P2.1/A9 corresponds to b, ..., P2.7/A15 corresponds to h.

　　4. Programming content

　　The internal 16-bit timer/counter of the AT89S51 microcontroller is a programmable timer/counter. It can work in 13-bit timing mode, 16-bit timing mode and 8-bit timing mode. This can be done by setting the special function register TMOD. The timer/counter works by setting the TCON special function register through software.

　　Now we choose the 16-bit timing mode. For T0, the maximum timing is only 65536us, that is, 65.536ms, which cannot reach the 1-second timing we need. Therefore, we must use software to handle this problem. Suppose we take the maximum timing of T0 as 50ms, that is, to achieve 1-second timing, 20 times of 50ms timing are required. For these 20 times, we can use software methods to count.

　　Therefore, we set TMOD = 00000001B, that is, TMOD = 01H

　　Next, we need to load the preset initial values ​​into TH0 and TL0 of T0 timer/counter. The following formula can be used to calculate

　　TH0 = (216 - 50000) / 256

　　TL0 = (216 - 50000) MOD 256

　　When T0 is working, how do we know that the 50ms timing time has arrived? This time we detect the TF0 flag bit in the TCON special function register. If TF0=1, it means that the timing time has arrived.

　　5. Flowchart

Figure 4.15.2

　　6. Assemble the source program (query method)

　　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 language source program (query method)

　　#include 《AT89X51.H》

　　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;

　　}

　　}

　　}

　　1. Assemble source program (interrupt method)

　　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：RT

　　TABLE: DB 3FH，06H，5BH，4FH，66H，6DH，7DH，07H，7FH，6FH

　　END

　　2. C language source program (interrupt method)

　　#include 《AT89X51.H》

　　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;

　　}