Programming 8051 Timers

Parul Institute Of Engineering & Technology

Subject code:- 151001
Name of subject:- Microcontroller and Interfacing

Name of unit:- 8051 Timer Programming In Assembly And C

(Topic):-Programming 8051 Timers

Name of faculty:- Ms.Srushti Soni mam

Name of students:- 1) Vivek Patel (Roll.no- 146)

Programming 8051 Timers
 The 8051 has 2 timers/counters:
 timer/counter 0
 timer/counter 1

They can be used as
1. The timer is used as a time delay generator.
 The clock source is the internal crystal
frequency of the 8051.

1. An event counter.
 External input from input pin to count the
number of events on registers.
 These clock pulses cold represent the
number of people passing through an
entrance, or the number of wheel
rotations, or any other event that can be
converted to pulses.

 Timer:
 Set the initial value of registers
 Start the timer and then the 8051 counts up.
 Input from internal system clock (machine
cycle)

 When the registers equal to 0 and the 8051
sets a bit to denote time out
8051

P2 P1
to
Set LCD
Timer 0 TH0

TL0

 Counter :
 Count the number of events
 Show the number of events on registers

 External input from T0 input pin (P3.4) for
Counter 0
 External input from T1 input pin (P3.5) for
Counter 1

 Externalinput from Tx input pin.
 We use Tx to denote T0 or T1.

8051
TH0
P1
TL0 to
LCD
P3.4
a switch T0

 Register used in 8051

 TH0, TL0, TH1, TL1
 TMOD (Timer mode register)
 TCON (Timer control register)

 TMOD Register:
 Timer mode register: TMOD
MOV TMOD,#21H
 An 8-bit register

 Set the usage mode for two timers

 Set lower 4 bits for Timer 0 (Set to 0000

if not used)

TMOD Register:
 Set
upper 4 bits for Timer 1 (Set to
0000 if not used)
 Not bit-addressable

(MSB) (LSB)
GATE C/T M1 M0 GATE C/T M1 M0
Timer 1 Timer 0

TMOD
 GATE=1
 External control
 The hardware way of starting and
stopping the timer by software and an
external source.
 Timer/counter is enabled only while the
INT pin is high and the TR control pin is
set (TR).

TMOD
 Gate:
 Every timer has a mean of starting and stopping.
 GATE=0
 Internal control

 The start and stop of the timer are controlled
by way of software.
 Set/clear the TR for start/stop timer.

SETB TR0
CLR TR0

TMOD
 Timer/counter is enabled only while the
INT pin is high and the TR control pin is
set (TR).

(MSB) (LSB)
GATE C/T M1 M0 GATE C/T M1 M0
Timer 1 Timer 0

TMOD
 C/T

 Timer or counter selected cleared for timer
operation (input from internal system clock).
Set for counter operation (input from Tx input
pin).

TMOD
 M1 M0 Mode Operating Mode

 0 0 0 13-bit timer mode 8-bit
THx + 5-bit TLx (x= 0 or 1)
 0 1 1 16-bit timer mode 8-bit
THx + 8-bit TLx
 1 0 2 8-bit auto reload 8-bit
auto reload timer/counter;
 THx holds a
value which is to be reloaded into
 TLx each
time it overflows.

TCON REGISTERS:
 Timer control register: TMOD
 Upper nibble for timer/counter, lower
nibble for interrupts
 TR (run control bit)
 TR0 for Timer/counter 0; TR1 for
Timer/counter 1.
 TR is set by programmer to turn
timer/counter on/off.
 TR=0: off (stop)

 TR=1: on (start)

TCON REGISTERS:

(MSB) (LSB)
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt

TCON REGISTERS:
 TF (timer flag, control flag)
 TF0 for timer/counter 0; TF1 for
timer/counter 1.
 TF is like a carry. Originally, TF=0. When TH-
TL roll over to 0000 from FFFFH, the TF is
set to 1.
 TF=0 : not reach

 TF=1: reach

 If we enable interrupt, TF=1 will

trigger ISR.

TCON REGISTERS
For timer 0
SETB TR0 = SETB TCON.4
CLR TR0 = CLR TCON.4

SETB TF0 = SETB TCON.5
CLR TF0 = CLR TCON.5

For timer 1
SETB TR1 = SETB TCON.6
CLR TR1 = CLR TCON.6

TCON REGISTERS

TCON: Timer/Counter Control Register
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0

Timer Mode 1:
 In following, we all use timer 0 as an example.
 16-bit timer (TH0 and TL0)
 TH0-TL0 is incremented continuously when
TR0 is set to 1. And the 8051 stops to
increment TH0-TL0 when TR0 is cleared.
 The timer works with the internal system
clock. In other words, the timer counts up
each machine cycle.

Timer Mode 1:
 When the timer (TH0-TL0) reaches its
maximum of FFFFH, it rolls over to 0000,
and TF0 is raised.
 Programmer should check TF0 and stop
the timer 0.

Step of Timer Mode 1:
1. Choose mode 1 timer 0
 MOV TMOD,#01H

2. Set the original value to TH0 and TL0.
 MOV TH0,#FFH

 MOV TL0,#FCH

3. You had better to clear the flag to monitor:
TF0=0.
 CLR TF0

4. Start the timer.
 SETB TR0

Step of Timer Mode 1:
 TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000H
 The 8051 starts to count up by incrementing
the TH0-TL0.

TR0=1 TR0=0
Start timer TH0 TL0
Stop timer

FFFC FFFD FFFE FFFF 0000

TF = 0 TF = 0 TF = 0 TF = 0 TF = 1

Timer Mode 0
generate delay using timer 0
DELAY:
SETB TR0 ;start the timer 0
AGAIN:JNB TF0,AGAIN
CLR TR0 ;stop timer 0
CLR TF0 ;clear timer 0 flag
RET

Timer Mode 0

FFF2 FFF3 FFF4 FFFF 0000

TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 0 TF0 = 1

Timer Mode 0

In the above program notice the following
steps.
1. TMOD = 0000 0001 is loaded.
2. FFF2H is loaded into TH0 – TL0.
3. P1.5 is toggled for the high and low
portions of the pulse.
4. The DELAY subroutine using the timer is
called.

Timer Mode 0
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
oscillator.
As the timer counts up, it goes through the
states of FFF3, FFF4, FFF5, FFF6, FFF7, FFF8,
FFF9, FFFA, FFFB, FFFC, FFFFD, FFFE, FFFFH.
One more clock rolls it to 0, raising the timer
flag (TF0 = 1). At that point, the JNB instruction
falls through.

Timer Mode 0
7. Timer 0 is stopped by the instruction
“CLR TR0”. The DELAY subroutine ends,
and the process is repeated.

Notice that to repeat the process, we must
reload the TL and TH
registers, and start the timer again (in the
main program).

Find Timer Values
 Assume that XTAL = 11.0592 MHz .
 And we know desired delay
 how to find the values for the TH,TL ?
1. Divide the delay by 1.085 µs and get n.

2. Perform 65536 –n

3. Convert the result of Step 2 to hex (yyxx )

4. Set TH = yy and TL = xx.

 Assuming XTAL = 11.0592 MHz,
 write a program to generate a square wave of
50 Hz frequency on pin P2.3.

Solution:
1. The period of the square wave = 1 / 50 Hz = 20
ms.
2. The high or low portion of the square wave =
10 ms.
3. 10 ms / 1.085 µs = 9216

4. 65536 – 9216 = 56320 in decimal = DC00H in
hex.
5. TL1 = 00H and TH1 = DCH.

Programming 8051 Timers

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