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Interrupts Microcontroller

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Interrupts Microcontroller

8051 Timer Programming in Assembly and C Microcontroller

    8051 Timer Programming in Assembly and C

    Objectives:

    At the end of this chapter, we will be able to:

    ; List the timers of 8051 and their associated registers

    ; Describe the various modes of the 8051 timers.

    ; Program the 8051 timers in assembly and C and

    ; Program the 8051 counters in assembly and C.

    Programming 8051 Timers:

    The 8051 has two timers/counters; they can be used either as Timers to generate a time delay or as event counters to count events happening outside the microcontroller. Basic Timers of 8051:

    Both Timer 0 and Timer 1 are 16 bits wide. Since 8051 has an 8-bit architecture, each 16-bits timer is accessed as two separate registers of low byte and high byte. The low byte register is called TL0/TL1 and the high byte register is called TH0/TH1. These registers can be accessed like any other register. For example:

    ; MOV TL0,#4FH

    ; MOV R5,TH0

    Figure 1: Timer 0 and Timer1 register

    TMOD (timer mode) Register:

    Both timers 0 and 1 use the same register, called TMOD (timer mode), to set the various timer operation modes. TMOD is an 8-bit register. The lower 4 bits are for Timer 0 and the upper 4 bits are for Timer 1. In each case, the lower 2 bits are used to set the timer mode the upper 2 bits to specify the operation. The TMOD register is as shown in figure 2 below:

Prof. Roopa Kulkarni, GIT, Belgaum Page 1

8051 Timer Programming in Assembly and C Microcontroller

    Timers of 8051 do starting and stopping by either software or hardware control. In using software to start and stop the timer where GATE=0. The start and stop of the timer are controlled by way of software by the TR (timer start) bits TR0 and TR1. The SETB instruction starts it, and it is stopped by the CLR instruction. These instructions start and stop the timers as long as GATE=0 in the TMOD register. The hardware way of starting and stopping the timer by an external source is achieved by making GATE=1 in the TMOD register.

    Mode 1 Programming:

    The following are the characteristics and operations of mode 1:

    1. It is a 16-bit timer; therefore, it allows value of 0000 to FFFFH to be loaded into the timer’s register TL and TH.

    2. After TH and TL are loaded with a 16-bit initial value, the timer must be started. This is done by SETB TR0 for timer 0 and SETB TR1 for timer 1.

    Prof. Roopa Kulkarni, GIT, Belgaum Page 2

8051 Timer Programming in Assembly and C Microcontroller

    3. After the timer is started, it starts to count up. It counts up until it reaches its limit of FFFFH. When it rolls over from FFFFH to 0000, it sets high a flag bit called TF (timer flag). Each timer has its own timer flag: TF0 for timer 0 and TF1 for timer 1. This timer flag can be monitored. When this timer flag is raised, one option would be to stop the timer with the instructions CLR TR0 or CLR TR1, for timer 0 and timer 1, respectively.

    4. After the timer reaches its limit and rolls over, in order to repeat the process. TH and TL must be reloaded with the original value, and TF must be reloaded to 0.

Steps to program in mode 1:

    To generate a time delay, using timer in mode 1, following are the steps: 1. Load the TMOD value register indicating which timer (timer 0 or timer 1) is to be used and which timer mode (0 or 1) is selected.

    2. Load registers TL and TH with initial count value.

    3. Start the timer.

    4. Keep monitoring the timer flag (TF) with the JNB TFx, target instruction to see if it is raised. Get out of the loop when TF becomes high.

    5. Stop the timer.

    6. Clear the TF flag for the next round.

    7. Go back to Step 2 to load TH and TL again.

     Example 1 In the following program, we create a square wave of 50% duty cycle (with equal portions high and low) on the P1.5 bit. Timer 0 is used to generate the time delay. Analyze the program. Also calculate the delay generated. Assume XTAL=11.0592MHz. Program: MOV TMOD,#01 ;Timer 0, mode 1(16-bit mode) HERE: MOV TL0,#0F2H ;TL0=F2H, the low byte

    MOV TH0,#0FFH ;TH0=FFH, the high byte CPL P1.5 ;toggle P1.5

    ACALL DELAY SJMP HERE

     DELAY:

    SETB TR0 ;start the timer 0

     AGAIN: JNB TF0,AGAIN ;monitor timer flag 0 until it rolls over

    CLR TR0 ;stop timer 0

     CLR TF0 ;clear timer 0 flag

    RET

(a)In the above program notice the following step. 1. TMOD is loaded.

    2. FFF2H is loaded into TH0-TL0. Prof. Roopa Kulkarni, GIT, Belgaum Page 3

    3. P1.5 is toggled for the high and low portions of the pulse.

     4. The DELAY subroutine using the timer is called.

    5. In the DELAY subroutine, timer 0 is started by the SETB TR0 instruction.

    6. Timer 0 counts up with the passing of each clock, which is provided by the crystal

    8051 Timer Programming in Assembly and C Microcontroller

     Example 2: Find the delay generated by timer 0 in the following code, using hex as well as decimal method. Do not include the overhead due to instruction.

     Program:

    CLR P2.3 ;Clear P2.3

     MOV TMOD,#01 ;Timer 0, 16-bitmode

    HERE: MOV TL0,#3EH ;TL0=3Eh, the low byte

    MOV TH0,#0B8H ;TH0=B8H, the high byte

    SETB P2.3 ;SET high timer 0

    SETB TR0 ;Start the timer 0

    AGAIN: JNB TF0,AGAIN ;Monitor timer flag 0

    CLR TR0 ;Stop the timer 0

    CLR TF0 ;Clear TF0 for next round

    CLR P2.3

    Solution: (a) (FFFFH B83E + 1) = 47C2H = 18370 in decimal and 18370 × 1.085 us = 19.93145 ms (b) Since TH TL = B83EH = 47166 (in decimal) we have 65536 47166 = 18370. This means that the timer counts from B38EH to FFFF. This plus Rolling over to 0 goes through a total of 18370 clock cycles, where each clock is 1.085µs in duration. Therefore, we have 18370 × 1.085 us = 19.93145 ms as the width of the pulse.

    Finding values to be loaded into the timer: Prof. Roopa Kulkarni, GIT, Belgaum Page 4

8051 Timer Programming in Assembly and C Microcontroller

To calculate the values to be loaded into the TL and TH registers, look at the following steps.

    Assume XTAL = 11.0592 MHz, we can use the following steps for finding the TH, TL

    registers’ values:

    1. Divide the desired time delay by 1.085µs. 2. Perform 65536 n, where n is the decimal value we got in Step1. 3. Convert the result of Step2 to hex, where yyxx is the initial hex value to be loaded into the

    timer’s register and

    4. Set TL = xx and TH = yy.

     Example 3: Assume that XTAL = 11.0592 MHz. What value do we need to load the

    timer’s register if we want to have a time delay of 5 ms ? Show the program for timer 0

     to create a pulse width of 5 ms on P2.3.

    Solution:

     Since XTAL = 11.0592 MHz, the counter counts up every 1.085 us. This means that out of many 1.085 us intervals we must make a 5 ms pulse. To get that, we divide one by the other. We need 5 ms / 1.085µs = 4608 clocks. To Achieve that we need to load into TL

    and TH the value 65536 4608 = EE00H. Therefore, we have TH = EE and TL = 00.

    Program:

    CLR P2.3 ;Clear P2.3

    MOV TMOD,#01 ;Timer 0, 16-bitmode

    HERE: MOV TL0,#0 ;TL0=0, the low byte

    MOV TH0,#0EEH ;TH0=EE, the high byte

    SETB P2.3 ;SET high P2.3

    SETB TR0 ;Start timer 0 AGAIN: JNB TF0,AGAIN ;Monitor timer flag 0

    CLR TR0 ;Stop the timer 0 CLR TF0 ;Clear timer 0 flag

     Example 4: Assume that XTAL = 11.0592 MHz, write a program to generate a square wave of 2 kHz frequency on pin P1.5. Solution:

    This is similar to Example 9-10, except that we must toggle the bit to generate the square wave. Look at the following steps.

    (a) T = 1 / f = 1 / 2 kHz = 500 us the period of square wave. (b) 1 / 2 of it for the high and low portion of the pulse is 250 us.

    (c) 250 us / 1.085 us = 230 and 65536 230 = 65306 which in hex is FF1AH. (d) TL = 1A and TH = FF, all in hex. The program is as follow.

    MOV TMOD,#01 ;Timer 0, 16-bitmode AGAIN: MOV TL1,#1AH ;TL1=1A, low byte of timer

    MOV TH1,#0FFH ;TH1=FF, the high byte

     SETB TR1 ;Start timer 1

    BACK: JNB TF1,BACK ;until timer rolls over

     CLR TR1 ;Stop the timer 1

    CLR P1.5 ;Clear timer flag 1

     CLR TF1 ;Clear timer 1 flag

    SJMP AGAIN ;Reload timer

Prof. Roopa Kulkarni, GIT, Belgaum Page 5

8051 Timer Programming in Assembly and C Microcontroller

Example 5: Assume XTAL = 11.0592 MHz, write a program to generate a square wave of 50 kHz frequency on pin P2.3.

    Solution: Look at the following steps.

    (a) T = 1 / 50 = 20 ms, the period of square wave. (b) 1 / 2 of it for the high and low portion of the pulse is 10 ms.

    (c) 10 ms / 1.085 us = 9216 and 65536 9216 = 56320 in decimal, and in hex it is DC00H.

    (d) TL = 00 and TH = DC (hex).

    Program:

     MOV TMOD,#10H ;Timer 1, mod 1

    AGAIN: MOV TL1,#00 ;TL1=00,low byte of timer

     MOV TH1,#0DCH ;TH1=DC, the high byte

     SETB TR1 ;Start timer 1 BACK: JNB TF1,BACK ;until timer rolls over

     CLR TR1 ;Stop the timer 1 CLR P2.3 ;Comp. p2.3 to get high and low

     SJMP AGAIN ;Reload timer

    ;mode 1 isn’t auto-reload

Mode 2 Programming:

    The following are the characteristics and operations of mode 2:

    1. It is an 8-bit timer; therefore, it allows only values of 00 to FFH to be loaded into the timer’s register TH

    2. After TH is loaded with the 8-bit value, the 8051 gives a copy of it to TL

    ; Then the timer must be started

    ; This is done by the instruction SETB TR0 for timer 0 and SETB TR1 for timer 1 3. After the timer is started, it starts to count up by incrementing the TL register

    ; It counts up until it reaches its limit of FFH

    ; When it rolls over from FFH to 00, it sets high the TF (timer flag)

    4. When the TL register rolls from FFH to 0 and TF is set to 1, TL is reloaded automatically with the original value kept by the TH register

    ; To repeat the process, we must simply clear TF and let it go without any need by the

    programmer to reload the original value

    ; This makes mode 2 an auto-reload, in contrast with mode 1 in which the programmer

    has to reload TH and TL

Prof. Roopa Kulkarni, GIT, Belgaum Page 6

8051 Timer Programming in Assembly and C Microcontroller

Steps to program in mode 2:

    To generate a time delay

    1. Load the TMOD value register indicating which timer (timer 0 or timer 1) is to be used,

    and the timer mode (mode 2) is selected. 2. Load the TH registers with the initial count value. 3. Start timer.

    4. Keep monitoring the timer flag (TF) with the JNB TFx, target instruction to see whether it

    is raised. Get out of the loop when TF goes high. 5. Clear the TF flag and

    6. Go back to Step 4, since mode 2 is auto reload.

     Example 6: Assume XTAL = 11.0592 MHz, find the frequency of the square wave generated on pin P1.0 in the following program Program:

    MOV TMOD, #20H ; T1/8-bit/auto reload

    MOV TH1, #5 ; TH1 = 5

    SETB TR1 ; start the timer 1

    BACK: JNB TF1, BACK ; till timer rolls over

     CPL P1.0 ; P1.0 to high, low

    CLR TF1 ; clear Timer 1 flag

     SJMP BACK ; mode 2 is auto-reload

     Solution:

    First notice the target address of SJMP. In mode 2 we do not need to reload TH since it is auto-reload. Now (256 - 05) × 1.085 us = 251 × 1.085 us = 272.33 us is the high portion of

    the pulse. Since it is a 50% duty cycle square wave, the period T is twice that; as a result T = 2 × 272.33 us = 544.67 us and the frequency = 1.83597 kHz

    Example 7: Write an ALP to generate a square wave of frequency 72Hz on pin P1.0. Solution: Assume XTAL=11.0592MHz. With TH=00, the delay generated is 256 x 1.085 µs = 277.76 µs. therefore to generate a delay of (1 / 72) = 138.88ms, the count

    to be loaded is 250 x 2=500. That is

T = 2 (250 × 256 × 1.085 µs) = 138.88ms, and frequency = 72 Hz

     Program:

     MOV TMOD, #2H ; Timer 0, mod 2;(8-bit, auto reload) MOV TH0, #0

    AGAIN: MOV R5, #250 ; multiple delay count

     ACALL DELAY

     CPL P1.0

     SJMP AGAIN

     DELAY: SETB TR0 ; start the timer 0

    BACK: JNB TF0,BACK ; stay timer rolls over

     CLR TR0 ; stop timer

     CLR TF0 ; clear TF for next round

     DJNZ R5,DELAY

     RET

    Prof. Roopa Kulkarni, GIT, Belgaum Page 7

8051 Timer Programming in Assembly and C Microcontroller

Assemblers and Negative values:

     Example 8: Assuming that we are programming the timers for mode 2, find the value (in hex) loaded into TH for each of the following cases.

    (a) MOV TH1,#-200 (b) MOV TH0,#-60

    (c) MOV TH1,#-3 (d) MOV TH1,#-12

    (e) MOV TH0,#-48

    Solution:

    You can use the Windows scientific calculator to verify the result provided by the

     assembler. In Windows calculator, select decimal and enter 200. Then select hex, then +/-

    to get the TH value. Remember that we only use the right two digits and ignore the rest

     since our data is an 8-bit data.

    Decimal 2’s complement (TH value) -3 FDH

    -12 F4H -48 D0H

    -60 C4H -200 38H

Counter Programming:

    Timers can also be used as counters, counting events happening outside the 8051. When it is used as a counter, it is a pulse outside of the 8051 that increments the TH, TL register. TMOD and TH, TL registers are the same as for the timer discussed previously. Programming the timer in the last section also applies to programming it as a counter, except the source of the frequency.

C/T bit in TMOD register

    The C/T bit in the TMOD registers decides the source of the clock for the timer. When C/T = 1, the timer is used as a counter and gets its pulses from outside the 8051. The counter counts up as pulses are fed from pins 14 and 15, these pins are called T0 (timer 0 input) and T1 (timer 1 input).

Prof. Roopa Kulkarni, GIT, Belgaum Page 8

    8051 Timer Programming in Assembly and C Microcontroller

Example 9: Assuming that clock pulses are fed into pin T1, write a program for counter 1 in mode 2 to count the pulses and display the state of the TL1 count on P2, which connects

    to 8 LEDs.

    Program:

    MOV TM0D,#01100000B ;counter 1, mode 2,C/T=1 external pulses

    MOV TH1,#0 ;clear TH1

     SETB P3.5 ;make T1 input

    AGAIN: SETB TR1 ;start the counter

     BACK: MOV A,TL1 ;get copy of TL

    MOV P2,A ;display it on port 2

     JNB TF1,Back ;keep doing, if TF = 0

    CLR TR1 ;stop the counter 1 CLR TF1 ;make TF=0

    SJMP AGAIN ;keep doing it Solution:

    Notice in the above program the role of the instruction SETB P3.5. Since ports are set up for output when the 8051 is powered up, we make P3.5 an input port

    by making it high. In other words, we must configure (set high) the T1 pin (pin P3.5) to

    allow pulses to be fed into it.

    TCON (timer control) register: TCON is an 8- bit register. It is a bit addressable register.

Prof. Roopa Kulkarni, GIT, Belgaum Page 9

8051 Timer Programming in Assembly and C Microcontroller

     Figure 3: TCON register

    If GATE = 1, the start and stop of the timer are done externally through pins P3.2 and P3.3 for timers 0 and 1, respectively. This hardware way allows starting or stopping the timer externally at any time via a simple switch.

Prof. Roopa Kulkarni, GIT, Belgaum Page 10

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