SE PAI Unit 5_Timer Programming in 8051 microcontroller_Part 2

1
Unit 5: 8051 Timers/Counters Programming
Part 2
Subject: Processor Architecture & Interfacing
(SPPU, Pune 2015 Course of Information Technology)
Class: SEIT Semester: II
Prepared By,
Ms. K. D. Patil, Assistant Professor
Department of Information Technology
(NBA accredited)
Sanjivani College of Engineering, Kopargaon-423603.
Maharashtra, India.
(An Autonomous Institute, Affiliated with SPPU, Pune.)
NAAC ‘A’ Grade Accredited, ISO 9001:2015 certified

Contents
Prepared By: Ms. K. D. Patil, Dept. of Information
Technology, Sanjivani COE, Kopargaon2
 Part 1
 Introduction
 Special Function Registers
 Timer 0 Register
 Timer 1 Register
 TMOD Register
 TCON Register
 How to UseTimer
 Operating Modes ofTimer
 Part 2
 Timer Programming
 Mode 1 Programming
 Mode 2 Programming
 Introduction to Counters
 Counter Programming

Steps in Mode 1 Programming
1. Load theTMOD 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 registersTL andTH with initial count value
3. Start the timer (SETBTRx)
4. Keep monitoring the timer flag (TF) with JNBTFx, target
instruction to see if it is raised. Get out of the loop whenTF
becomes high
5. Stop the timer (CLRTRx)
6. Clear theTF flag for the next round (CLRTFx)
7. Go back to Step 2 to loadTH andTL again

Steps in Mode 2 Programming
1. Load theTMOD value register indicating which timer
(timer 0 or timer 1) is to be used and which timer mode
(2) is selected
2. Load registers TH with initial count value
3. Start the timer (SETBTRx)
4. Keep monitoring the timer flag (TF) with JNBTFx, target
instruction to see if it is raised. Get out of the loop when
TF becomes high
5. Clear theTF flag for the next round (CLRTFx)
6. Go back to Step 4, since mode 2 is auto reloaded

Clock Source for Timer
 Timer needs a clock pulse to ticks
 if C/T = 0, the crystal frequency attached to the 8051 is the source
of the clock for the timer
 Size of Crystal frequency also decides the speed at which 8051
timer ticks
 8051 based systems have XTAL frequency of 10MHz to 40MHz
 XTAL = 11.0592 MHz allows 8051 system to communicate with
IBM PC with no errors
 The frequency for the timer is always 1/12th the frequency of the
crystal attached to the 8051

Clock Source for Timer
 When , XTAL =11.0592MHz
 1 Machine cycle of 8051 consists of 12 OSC periods (see fig.)
 Timer Frequency
= (11.0592/12) = 921.6 KHz
 Time period of each machine cycle
= 1/921.6 KHz= 1.085 us
 1 instruction takes 1.085 us for execution

Finding Values to be loaded in TH-TL
 If XTAL = 11.0592 MHz
 Timer Frequency = (11.0592/12) = 921.6 KHz
 We know that each clock hasTime period of each machine
cycle = 1/921.6 KHz= 1.085 us
1. To get the number of counts, Divide the desired time delay
by 1.085 us, we will get n
2. Perform 65536 – n, where n is the decimal value we got in
step 1
3. Convert the result of Step2 to hex, whereYYXX is the
initial hex value to be loaded into the timer’s register
4. SetTL = XX andTH =YY

Calculate the amount of Time Delay
 If XTAL = 11.0592 MHz
 Timer Frequency = (11.0592/12) = 921.6 KHz
 We know that each clock hasTime period of each machine cycle
= 1/921.6 KHz= 1.085 uSec
 Means,Timer 0 counts up each 1.085 us resulting in
delay = number of counts × 1.085us.
 The number of counts for the roll over is
FFFF H –YYXX H = Result
 However, we add one to Result because of the extra clock needed
when it rolls over from FFFF to 0 and raise theTF flag
 This gives (Result + 1) × 1.085us = ___us for half the pulse
 For the entire period it isT = 2 × ___us = ___us as the time delay
generated by the timer

Calculate the amount of Time Delay
 In Hex
(FFFF –YYXX + 1) × 1.085 us,
whereYYXX areTH,TL initial
values respectively.
 Notice that value YYXX are in
hex.
will give you delay for half
pulse
 In Decimal
 ConvertYYXX values of
theTH,TL register to
decimal to get a NNNN
decimal, then
(65536 - NNNN) × 1.085 us
will give you delay for half
pulse

Example No.1 (Mode 1 Programming)
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
(milliseconds)? Show the program for timer 0 to create a pulse
width of 5 ms on P2.3.
Solution:
XTAL = 11.0592 MHz
Timer Frequency = (11.0592/12) = 921.6 KHz
Each clock hasTime period = 1/921.6 KHz= 1.085 uSec
No. of clocks = 5000 us /1.085us = 4608
To achieve that we need to load the values inTH-TL
= 65536 – 4608 = 60928 in decimal
60928 D = EE00 H so, TH = EEH and TL = 00H

Program:
ORG 00H
CLR P2.3 ;Clear P2.3
MOV TMOD,#01 ;Timer 0, 16-bitmode
HERE: MOV TL0,#00H ;TL0=00, the low byte
MOVTH0,#0EEH ;TH0=EE, the high byte
SETB P2.3 ;SET high P2.3
SETBTR0 ;Start timer 0
AGAIN: JNBTF0,AGAIN ;Monitor timer flag 0
CLRTR0 ;Stop the timer 0
CLRTF0 ;Clear timer 0 flag
RET
END

 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 = 1 / 2 kHz = 500 us the period of square wave.
 1 / 2 of it for the high and low portion of the pulse is 250
us.
 250 us / 1.085 us = 230 and 65536 – 230 = 65306 which
in hex is FF1AH.
 TL = 1A andTH = FF, all in hex.

Program:
ORG 00H
MOV P1, # 0000 0000 B
MOV TMOD,#0000 0001 B ;Timer 0, 16-bitmode
MAIN:
SETB P1.5 ; make P1.5 as high
ACALL DELAY
CLR P1.5 ; make P1.5 as low
ACALL DELAY
SJMP MAIN ;Reload timer
DELAY: MOV TL1,#1AH ;TL1=1A,low byte of timer
MOV TH1,#0FFH ;TH1=FF, the high byte
SETB TR0 ;Start timer 0
BACK: JNB TF0, BACK ;until timer rolls over
CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
SETB P1.5
RET
END

 Assume XTAL = 11.0592 MHz, find the
frequency of the square wave generated on
pin P1.0 in the following program
MOVTMOD,#20H ;T1/8-bit/auto reload
MOVTH1,#5 ;TH1 = 5
SETBTR1 ;start the timer 1
BACK: JNB TF1,BACK ;till timer rolls over
CPL P1.0 ;P1.0 to high, low
CLRTF1 ;clearTimer 1 flag
SJMP BACK ;mode 2 is auto-reload
Solution:
 In mode 2 we do not need to
reloadTH since it is auto-reload.
 Delay
= (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 periodT is twice that
 T = 2 × 272.33 us = 544.67 us
 frequency = 1.83597 kHz

 Assume that XTAL = 22 MHz, write a program to generate a
square wave of 1 kHz frequency on pin P1.2.
Solution:
 XTAL = 22MHz
 Timer Frequency = 22/12 = 1.83 us
 Clock Period = 1 / 1.83us = 0.546 us
 T = 1 / f = 1 / 1 kHz = 1000 us the period of square wave.
 1 / 2 of it for the high and low portion of the pulse is 500 us.
 No. of Cycles in 5 ms = 500 us / 0.546 us = 915
 No. of Cycles in 1 ms = 915/5 = 183 cycles
 Value loaded toTH = FF H – B7H = 48 H

Program:
ORG 00H
MOV TMOD, #02 H ;Timer 0, mode 2
MAIN: CPL P1.2 ; complement P1.2
MOV R6, #05 ; ; count for multiple delay
DELAY: MOV TH0, #48H ;TH0=48 H
SETB TR0 ;Start timer 0
BACK: JNB TF0, BACK ;until timer rolls over
CLR TR0 ;Stop the timer 0
CLR TF0 ;Clear timer 0 flag
DJNZ R0, DELAY ; repeat until R0= 0
SJMP MAIN ; repeat to get train of pulses
END

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 theTH,TL registers
 TMOD andTH,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

Counter Programming
 The C/T bit in theTMOD 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 calledT0 (timer 0 input) andT1 (timer 1 input)

Counter Programming
 GATE Bit inTMOD
 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 to start or stop the timer externally at
any time via a simple switch

Example No. 5 (Counter Programming)
 Assuming that clock pulses are fed into pinT1, write a program for counter 1 in
mode 2 to count the pulses and display the state of theTL1 count on P2, which
connects to 8 LEDs.
Solution:
MOVTM0D,#01100000B ;counter 1, mode 2, ;C/T=1 external pulses
MOVTH1,#0 ;clearTH1
SETB P3.5 ;makeT1 input
AGAIN: SETBTR1 ;start the counter
BACK: MOV A,TL1 ;get copy ofTL
MOV P2,A ;display it on port 2
JNBTF1,Back ;keep doing, ifTF = 0
CLRTR1 ;stop the counter 1
CLRTF1 ;makeTF=0
SJMPAGAIN ;keep doing it

Reference
 “The 8051 Microcontroller and Embedded Systems using
Assembly and C” , Muhammad Ali Mazidi, Janice Gillispie
Mazidi, Rolin D. McKinlay, Second Edition, Pearson
publication

SE PAI Unit 5_Timer Programming in 8051 microcontroller_Part 2

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