8051 Counter Programming

8051 Counter Programming

When the timer/counter is used as a counter, the TMOD, TH and TL registers are used, functioning the same as for the timer studied in the last section.

 Bit in TMOD Register : As seen earlier, the  bit in the TMOD register decides the timer/counter functioning as a counter or a timer. When  bit in the TMOD register is 0, the timer mode is selected. When timer/counter is used as a timer, the 8051's crystal is used as a source of the frequency. When  bit in the TMOD register is 1, the counter mode is selected. When timer/counter is used as a counter, it gets its pulses from outside the 8051. The pin P 3.4 (pin number 14) and pin 3.5 (pin number 15) of 8051 are used for applying pulses counter 0 and counter 1 respectively. These two pins belong to port 3. The counter counts up for each clock pulse applied at this pin. These pins are called T0 (timer 0 clock input) and Tl(timer 1 clock input).

Counter 0 in Mode 1 : The Fig. 16.5.1 (a) shows the block diagram of counter 0 in mode 1 and the Fig. 16.5.1 (b) shows the block diagram of counter 0 in mode 1 when GATE = 0 and  = 1. Here, counter 0 counts up when the logic signal on pin T0 goes from high level to low level.

To operate counter 0 in mode 1 we have to perform following steps :

1.  bit in TMOD register (Bit 2) is set to 1 to allow counter mode operation.

2. M1 : M0 bits (bits 1 : 0) in TMOD register are set to 01 to select mode 1.

3. When Gate bit (Bit 3) in TMOD register is cleared to 0, TR0 bit (bit 4 of TCON) is set to 1 to start the counter.

4. When Gate bit (Bit 3) in TMOD register is set to 1, counter will run only if TR0 is set to 1 and the logic signal on external interrupt pin  is high.

Counter 1 in Mode 1 : The Fig. 16.5.3 (a) shows the block diagram of counter 1 in mode 1 and the Fig. 16.5.3 (b) shows the block diagram of counter 1 in mode 1 when GATE = 0 and  = 1.

The operation of counter 1 in mode 1 is same as counter 0 operation. The only difference is that here registers for counter 1 are programmed instead of counter 0.

Counter in Mode 2 : In this mode, counter is used in auto-reload mode instead of 16-bit counter. Rest of the operation is exactly same as that of Mode 1. The Fig. 16.5.4 (a) shows the block diagram of counter 0 in Mode 2 and the Fig. 16.5.4 (b) shows the block diagram of counter 0 in Mode 2 when GATE = 0 and  = 1

Example 16.5.1 Write a program for counter 1 in mode 2 to count the pulses and display the state of TL1 count on port 2. Assume that clock input is connected to T1 pin (P 3.5).

AU : Dec.-12, Marks 8

Solution :

MOV TMOD, #01100000B ; Initialize counter 1 in Mode 2,  = 1

MOV TH1, #0 ; Clear TH1

SETB P3.5 ; Make T1 input

START : SETB TR1 ; Start the counter

BACK : MOV A, TL1 ; Get the count from TL1

MOV P2, A ; Sent it to port 2

JNB TF1, BACK ; If TF1 = 0 repeat

CLR TR1 ; Otherwise stop counter 1

CLRTF1 ; Make TF1=0

SJMP START ; Repeat

Note When 8051 is powered up ports are configured as input ports. To make them work as output port we have to send high output on it. Therefore, to behave T1 as input P 3.5 is set.

Example 16.5.2 Write a program to display counter 0 on 7-segment LEDs. Assume that clock input is connected to pin (P 3.4).

Solution :

MOV TMOD, #00000110 ; Initialize counter 0 in Mode 2, C/T=1

MOV THO, #00H ; Reset counter value

SETB P3.4 ; Make T0 as input

START: SETB TR0 ; Start counter 0

BACK: MOV A, TL0 ; Get the count value

ACALL CONVBCD

MOV P2, A ; Send count value in BCD on port 2

JNB TF0, BACK ; If TF0 = 0 repeat

CLR TR0 ; Otherwise stop counter 0

CLR TF0 ; Make TF0=0

SJMP START ; Repeat

CONVBCD: ADD A, #00H ; [Use DAA for

DA A ; BCD conversion]

RET ; Return to main program

Example 16.5.3 Assume that XTAL = 11.0592 MHz. Write a program to generate a square wave of 2 kHz frequency on pin P1.5.

Solution : T = 1 /f = l/2 kHz = 500 µs is period of square wave. 1/2 of it for high and low portion of the pulse is 250 µs. 250 µs/1.085 µs = 230 and 65536 - 230 = 65306 which in hex is FF1AH.

ஃ TL = 1AH and TH = FFH

Program is as follows

MOV TMOD, #10H ; timer 1, mode 1

AGAIN : MOV TL1, #1AH ; low byte of timer

MOV TH1, #0FFH ; high byte of timer

SETB TR1 ; Start timer 1

BACK : JNB TF1, BACK ; Stay until timer rolls over

CLR TR1 ;  Stop timer 1

CPL P1.5 ; Complement P1.5

CLR TF1 ; Clear timer flag

SJMP AGAIN ; reload timer

Example 16.5.4 Generate a square wave of frequency 1 kHz using timer 1 in mode 1, on Pin P1.2. Explain the TMOD word used to configure the timer 1 for this application. Show the necessary calculations to find the value of count to be loaded into TH1 and TL1 registers. Assume XTAL frequency - 11.0592 MHz.

Solution :

T of square wave = 1/f = 1/2 kHz = 500 µs

ஃ Count to be loaded in TH1 and TL1 = 65536 - 230 = 65306 = FF1AH

ஃ TH1 = FF and TL1 = 1AH

Program

MOV TMOD, #10H ; Timer 1, mode 1 (16-bit)

AGAIN : MOV TL1, #1AH ; Load lower byte of timer

MOV TH1, #0FFH ; Load higher byte of timer

SETB TR1 ; Start timer 1

BACK : JNB TF1, BACK ; Wait for timer rolls over

CLR TR1 ; Stop timer 1

CPL P1.2 ; Complement P1.2

CLR TF1 ; Clear timer flag 1

SJMP AGAIN ; Reload timer 1 and continue

Example 16.5.5 Explain the steps to program timers in model and write an 8051 program to generate a square wave of 50 %        duty cycle on the pin P1.5.

Solution : Square wave

MOV TMOD, #01 ; Timer 0 mode 1

Here : MOV TL0, #0F2H ; Load TL0 - 0F2

MOV TH0, #0FFH ; Load TH0 - 0FF

CPL P1.5 ; toggle input on P1.5

ACALL Delay

SJMP Here

Delay : SETB TR0 ; Start Timer 0

again : JNB TF0, again ; Monitor Timer 0 flag until it rolls over

CLR TRO

CLR TF0

RET

Example 16.5.6 Write an ALP to generate square wave on pin Pl.5 of 500 Hz (approximately) with a subroutine to provide a time delay of 30.38 is using timer 0. Assume that crystal frequency of 8051 is 11.0592 Hz.

Solution :

65536 - 28 = 65508 = FFE4H

ஃ To get a delay of 30.38 we have to load TH0 = FFH and TL0 = E4H

For square wave T = 1/500 = 2 ms

ஃ T_ON = T_OFF = T/2 = 1 ms

Thus we have call delay routine (1 ms/30.38 µs) = 33 = 21H times

Program

MOV TMOD, #01 ; Timer0, mode1

HERE : MOV TL0, #E4H ; Load TL0 = E4H

MOV TH0, #FFH ; Load TH0 = FFH

CPL P1.5 ; toggle P1.5

MOV R0, #21H ; Load count in R0

BACK : ACALL Dealy ; wait for 30.38 µs × 33 = 1ms

DJNZ R0, BACK

SJMP HERE

DELAY : SETB T0 ; start timer 0

AGAIN : JNB TF0, AGAIN ; Wait for TF0 to roll over

CLR TR0 ; stop timer 0

CLR TF0 ; Clear TF0

RET ; Return

Example 16.5.7 Find the delay generated by timer 0 in the following code. Calculate the delay generated excluding the instruction overhead. What count has to be loaded in TL0 and TH0 if delay has to be increased to 25 msec ?

CLR P2.3

HERE : MOV TMOD, #01

MOV TL0, #3Eh

MOV TH0, #0B8h

SETB TF0

AGAIN: JNB TF0, AGAIN

CLR TF0

CLR TR0

CLR P2.3

Solution :

Timer count = B83E

(FFFF - B83E + 1) = 47C2H = (18370)₁₀

Assuming XTAL = 11.0592 MHz

T = 12 / 11.0592 × 10⁶ = 1.085 × 10^-6

ஃ Delay = 1.085 × 10^{- 6} × 18370 = 19.93 ms

For 25 msec delay

Decimal count =  25 ms / 1.085 × 10^-6 = 23041

= 5A01 H

ஃ TL0 = 01H and TH0 = 5AH

Example 16.5.8 Find out Hex. number to be loaded in TH0, to produce delay of 4.096 msec in mode 'O' operation. Assume clock frequency of 12 MHz.

Solution : Timer clock frequency = Crystal frequency + 12 =12 MHz + 12

 = 1 MHz

ஃ Timer clock period = 1 µs

Maximum count in mode 0 is 1FFFH (8191) and we have to count for 4096 counts to get a delay of 4.096 ms

ஃ Count to be loaded in timer is 1FFFH

(8191 in decimal) - 4096 = FFFH (4095 in decimal)

Therefore, we have to load 0FH in TH0 and FFH in TL0.

Example 16.5.9 Using autoreload mode of timer 0 in 8051, generate a frequency of 10 kHz on pin P1.0. Write assembly language program for it.

Solution : Assume XTAL frequency = 12 MHz. To generate a frequency of 10 kHz, one half of the cycle is of period 0.05 ms. Since XTAL frequency = 12 MHz we have cycle period is Ips. Therefore, we have to decrement count equal to 50. Thus the initial value to be loaded in TH0 = (256 - 50) = 206.

Program

MOV TMOD, #2H ; Timer 0, mode 2 (8-bit auto reload)

MOV TH0 #206 ; TH0 = 206

SETB TR0 ; Start timer 0

BACK : JNB TF1 BACK ; Stay till timer rolls over

CPL P1.0 ; Complement P1.0

CLR TF0 ; Clear timer flag 0

SJMP BACK ; Continue

Example 16.5.10 Write an assembly language program to make LED ON and OFF connected to P 1.0 continuously with ON time 20 msec and off time 40 msec.

Solution : Assume crystal frequency 12 MHz

Time for one instruction = (1/12MHz)* 12 = 1 µsec.

Count to be loaded in TH0:TL0 = >20 µsec / 1 p sec. = 20000 (Decimal)

65536 - 20000 = 45536 = B1E0H

ORG 0000H

START : MOV TMOD,#01H

MOV TH0,#0B1H

MOV TL0,#0E0H

SETB P1.0

ACALL DELAY

CLR P1.0

ACALL DELAY

ACALL DELAY

SJMP START

DELAY20MS : SETB TR0

HERE : JNB TF0, HERE

CLR TF0

CLR TF0

RET

Example 16.5.11 Write an assembly language program to generate a square wave of frequency 5 kHz on pin P3.0 using auto reload mode of timer 0 in 8051.

Solution : For 5 kHz T = 1/5 kHz = 200µs and T/2 = 100 µs

Assume clock frequency 12 MHz so time for one cycle is 1 µs. So the value of TH0 and TL0 is 255 - 100 = 155

ORG 0000H

MOV TMOD, #02H ; Timer 0 in mode 2

MOV TH0,#155 ; TH0 = 155

SETB TR0 ; Start timer 0

HERE : JNB TF0, HERE

CPL P3.0

ORG 001BH

CLR TF0

SJMP HERE

1. Programming Timers in 8051 C

The general purpose registers of 8051, such as R0-R7, A and B are under control of the C compiler and are not accessed directly by C statement. However, in case of SFRs entire RAM space of 80H-FFH is directly accessible to 8051 C statements. In this section we discuss the accessing of timers using C statements.

Accessing Timer Registers in C

In 8051 C, we can access the timer registers TH, TL and TMOD directly with the inclusion reg51.h file in the program. We can also access TR and TF bits directly. This is illustrated in example.

Example 16.5.12 Write an 8051 C program to toggle all bits of port P0 continuously. Use timer 0 to generate delay of 1 sec between each toggle.

Solution :

#include <reg51.h>

void DELAY (void);

void main (void)

{

while(1) /* Repeat forever */

{

P0=0x00 ; /* Make P0 bits all zero */

DELAY();/* Wait for 1 sec */

P0 = 0xFF; /* Make P0 bits all one */

DELAY();/* Wait for 1 sec */

}

}

void DELAY()

{

unsigned char i;

for(i=0; i<20; i++)

{

TMOD = 0x01; /* Configure timer 0 in mode 1 */

TL0 = 0x28; /* Load count in TL0*/

TH0 = 0x29; /* load count in TH0 */

TR0 = 1; /* Turn on T0 */

while(TF0 = = 0); /* Wait for TF0 to rollover */

TR0 = 0; / turn off T0 */

TF0 = 0; /* clear TF0 */

}

}

Timer Count Calculations

Assume crystal frequency = 12 MHz

ஃ  T = 12 / 12 × 10^-6 = 1 µs

Let us determine the count to get a delay of 50 ms

ஃ We need 50 ms / 1 µs = 50000 clocks.

ஃ Count = 65536 - 50000 = 10536 = 2928 H

To get a delay of 1 sec we have to repeat the delay of timerO 20 (1/50 msec) times.

Example 16.5.13 Write an 8051 C program to toggle only pin P1.0 continuously every 500 ms.

Solution :

Let us use mode 2 of timer0 to create the delay. Mode 2 is an 8-bit auto reload mode.

#include <reg51.h>

void Delay(void);

sbit portbit = P1 ^ 0;

void main(void)

{

unsigned char i,j;

while(1)

{

portbit = ~ portbit; /* toggle P1.0 */

for(i=0; i<500; i++)

for(j=0; j<40; j++)

Delay();

}

}

void Delay(void)

{

TMOD=0x02; /* Time 0, mode 2(8-bit auto-reload) */

TH0=-25; /* load TH0 (auto-reload value) */

TR0=l; /* turn on TO */

while (TF0==0); /* wait for TF0 to roll over */

TR0=0; /* turn off TO */

TF0=0; /* clear TF0 */

}

Delay calculations : Assume XTAL frequency = 12 MHz.

T = 12/12 MHz = 1 µs

256 - 25 = 231

25 × 1.0 µs = 25 µs

Total delay = 25 µs × 500 × 40 = 500 ms

Note Due to inclusion of for loop of C in delay generation, the delay may be slightly more than expected. To get a correct delay we can adjust delay loop count by observing frequency of port 1.0 on oscilloscope.

Example 16.5.1 Write an 8051 C program to create a frequency of 2 kHz on pin P2.0.

Solution : Let us use mode 2 of timerl to create the delay. Mode 2 is an 8-bit auto reload mode.

#include <reg51.h>

void Delay (void);

sbit portbit=P2 ^ 0;

void main(void)

{

while(1)

{

portbit=~portbit; /* toggle P2.0 */

Delay();

}

}

void Delay(void)

{

TMOD=0x20; /* Timer 1, mode 2(8-bit auto-reload) */

TH1=-250; /* load Thl (auto-reload value) */

TR1 = 1; /* turn on T1 */

while(TF1 = =0); /* wait for TF1 to roll over */

TR1=0; /* turn off T1 */

TF1=0; /* clear TF1 */

}

Delay calculations : Assume XTAL frequency = 12 MHz

T = 12/12 MHz = 1 µs

1/2 kHz = 0.5 ms

0.5 ms/2 = 0.25 ms

0.25 ms / 1 µs = 250

Example 16.5.15 Two switches are connected to pin P1.0 and Pl.l, respectively. Write an 8051 C program to monitor switches and create the following frequencies on pin P2.0 according switch positions.

Solution :

#include <reg51.h>

sbit sw0=P1^0;

sbit sw1= P1^1;

sbit portbit=P2 ^ 0;

void Delay (unsigned char);

void main(void)

{

SW0=1; /* Make P1.0 an input */

SW1 = 1; /* make P1.1 an input */

while(1)

{

portbit=~portbit; /* toggle P2.0 */

if(sw1= =0&sw0= =0) /* check switch */

Delay(1);

if(sw1 = =0&sw0= = l)

Delay(2);

if(sw1= = 1&sw0==0)

Delay(3);

if(sw1 = = 1&SW0= = 1)

Delay (4);

}

}

void Delay(unsigned char c)

{ TMOD=0x01;

do

{

TL0=0x78;

TH0=0xEC;

TR0=l;

while (TF0==0);

TR0=0;

TF0=0;

c=c-l;

}

while(c!=0);

}

EC78H = 60536

65536 - 60536 = 5000

5000 × 1 µs = 5 ms

1 / (5 ms × 2) = 100 Hz

C Programming of Timers 0 and 1 as Counters

Let us see how to use timers 0 and 1 as event counters. A timer can be used as a counter if we provide external clock instead of using the frequency of the crystal oscillator as the clock source. By feeding pulses to the TO (P3.4) and T1 (P3.5) pins, we can use timer 0 and timer 1 as counter 0 and counter 1, respectively. Following examples show us how timers 0 and 1 are programmed as counters using the C language.

Example 16.5.16 Assume that a 1-Hz external clock is being fed into pin Tl. Write a C program for counter 1 in mode 2 (8-bit auto reload) to count up and display the state of the TL1 count on P0. Start the count from 00H.

Solution :

 #include <reg51.h>

Sbit T1 = P3^5;

Void main(void)

{

T1=1; /* make T1 an input */

TMOD=0x60;

TH1=0; /* set count to 0 */

While(1) /* repeat forever */

{

TR1=1 /* start timer */

do

{

P0=TL1; /* place value on port P0 */

}

while (TF1 = = 0); /* wait for TF1 to rollver */

TR1= 0; /* stop timer */

TF1 = 0; /* clear flag */

}

}

Example 16.5.17 Assume that a 1 Hz external clock is being fed into pin TO. Write a C program for counter 0 in mode 1 (16 bit) to count the pulses and display the TH0 and TL0 registers on P1 and P0, respectively.

Solution :

#include <reg51.h>

void main(void)

{       

T0=l; /* make T0 an input */

TMOD=0x05;

TL0=0;       /* set count to 0 */

TH0=0;       /* set count to 0 */

while(1)      /* repeat forever */

{        TR0=1;       /* start timer */

do

{

P0=TL0; /* place value of TL0 on port 0 */

P1=TH0; /* place value of TH0 on port 1 */

}

while(TF0==0); /* wait here */

TR0=0;       /* stop timer */

TF0=0;       /*clear flag */

}

}

Example 16.5.18 Assume that a 2 Hz external clock is being fed into pin T1. Write a C program for counter 0 in mode 2 (8-bit auto reload) to display the count in ASCII. The 8-bit binary count must be converted to ASCII. Display the ASCII digits (in binary) on PO serially. Display least significant digit first.

Solution : To display the TL1 count we must convert 8 bit binary data to ASCII.

#include <reg51.h>

void BinToASCn(unsigned char);

void main()

{       

unsigned char value;

T1=l;

TMOD=0x06;

TH0=0;

while(1)

{

do

{

TR0=1;

value=TL0;

BinToASCII(value);

}

while (TF0==0);

TR0=0;

TF0=0;

}

}       

void BinToASCII (unsigned char value)

unsigned char Abyte, i,Q;

unsigned char R[3];

Abyte = 0XFB;

i=0;

do

{

Q=Abyte/10; /* divide by 10 */

R[i]=Abyte%10; /* find remainder and save it */

Abyte=Q; /* save quotient as a number */

i = i+1;

}

while(Q!= 0) /* Repeat until quotient = 0 */

for(; i>0 ; i- -)

{

P0=R[i-1]+0×30; /* Make binary to ASCII */

}

}

Example 16.5.19 Assume that a 100 Hz external clock is being fed into pin TO. Write a C program for counter 0 in mode 2 (8-bit auto-reload) to display the seconds and minutes on P0 and P1, respectively.

Solution :

#include <reg51.h>

void Time(unsigned char);

void main(void)

{

unsigned char val=0;

T0=l;

TMOD=0x06;  /* TO, mode 2, counter */

TH0=100; /* sec=100 pulses */

while(1)

{ TR0=1; /* start timer */

while (TF0==0);

Time(val);

val+ + ;

TR0=0; /* stop timer */

TF0=0; /* clear flag */

}

}

void Time(unsigned char value)

{

unsigned char sec.min;

min=value/60;

sec=value % 60;

P0=sec;

P1=min;

}

Example 16.5.20 Write an 8051 C program to generate a rectangular wave of 2 kHz with 60 % duty cycle in pin P1.2. Assume crystal frequency as 11.0592 MHz. Use Timer - 0 in mode-1 operation. Show delay calculations.

Solution : For 2 kHz rectangular wave,

Count for T_OFF = 184

Count to be loaded for T_ON = 65536 - 276 = 65260 = FEECH

TH0 = FEH, TL0 = ECH

Count to be loaded for T_OFF = 65536 - 184 = 65352 = FF48H

TH0 = FFH, TL0 = 48H

#include <reg51.h>

sbit mybit = P2^1;

void T0M1D (unsigned char) ;

void main (void)

{

while (1)

{

mybit = 1; /*Make P2.1 High*/

T0M1D (1); /* Wait for 0.3 msec */

mybit = 0; /*Make P2.1 Low*/

T0M1D = (0); /*Wait for 0.2 msec*/

}

}

void T0M ID (unsigned char i)

{

TMOD = 0 × 01; /*Timer 0, mode 1*/

if (i = = 0)

{

TL0 = ECH; /*Load FEECH*/

TH0 = FEH;

}

else

{

TL0 = 48H; /*Load FF48H*/

TH0 = FFH;

}

TR0 = 1; /*Turn ON Timer 0*/

while (TF0 = = 0); /Wait for TF0 to roll over */

TR0 = 0; /*Turn OFF Timer 0 */

TF0 = 0; /*Clear TF0*/

}

Review Question

1. Write a note on counter programming of 8051.