• The 8051 has 2 timers/counters:
• timer/counter 0
• timer/counter 1
They can be used as
1.
The timer to generate time delay.
– The clock source is the internal crystal frequency of the 8051.
2.
An event counter .
– External pulse input from input pin to count the number of events on registers.
– These clock pulses could 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 .

• TH0, TL0 : timer/counter register of timer 0
• TH1, TL1 : timer/counter register of timer 1
• TMOD : Mode Select register
• TCON : Control Register

• Accessed as lower byte and higher byte
 The lower byte register is TL0 / TL1
 The higher byte register is TH0 / TH1
 Accessed like any other register
• mov TL0, #4Fh
• mov R5, TH0

• TR (Timer run control bit)
– TR0 for Timer/counter 0; TR1 for Timer/counter 1.
– TR is set by programmer to turn timer/counter on/off.
• CLR TR0 : off (stop)
• SETB TR0 : on (start)
• TF (timer flag bit)
– TF0 for timer/counter 0; TF1 for timer/counter 1.
– TF is like a carry. Originally TF=0. When TH-TL rolls over from FFFFh to
0000, the 8051 sets TF to 1. TF should be cleared by software in polling method.
• TF=0 : not reach
• TF=1: reach

timer clock

• 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
– 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).

M1 M0 Mode Operation
0 0 0
0 1 1
1 3-bit timer mode 8-bit THx + 5-bit TLx (x= 0 or 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.
1 1 3 Split timer mode

Assume XTAL frequency = 12MHz
 Clock frequency of the timer = 12MHz/12 = 1MHz
 Clock period = 1/1MHz = 1uS
 Devide the desired time delay by 1uS = decimal value n
 Perform 65536 n (for 16bit mode)
 convert the result to hex , where yyxx is the initial value to be loaded into the timer’s register
 Set TH = yy and TL = xx.

Example
• Generate 500us time delay. Let crystal frequency = 12MHz
 Timer clock period ( 12 / Xtal freq. ) = 1us
 The required Delay time = 500 us
 No. of counts to be counted by timer = 500us/1us = 500
 The value in TH0 & TL0 = 65536-500 = 65036 = FE0Ch
 TH0 = FEh, TL0 = 0Ch

1.
Set the timer & mode in TMOD register
2.
Load 16 bit register with the initial count value
3.
Start the timer SETB TRX
4.
Keep monitoring TFx JNB TFx, step4
TFX = 0 until timer reaches FFFFh
TFX = 1 when timer rolls over from FFFFh to 0000h
5.
Stop the process CLR TRX
6.
Clear the TFx flag for the next round
7.
Go back to step 2 to load TH and TL

Example
• Assuming XTAL = 12MHz write a program to generate a square wave of 1khz at P2.0 with 50% duty cycle
 The required ON/OFF time or Delay time = ( 1 /1KHz)/2 = 500 µS
 Machine cycle period = 12MHz / 12 = 1 µS
 No. of counts to be counted by timer = 500us/1 µS = 500
 The value in TH0 & TL0 = 64536 - 500 = 64036 = FE0Ch

Example cont..
MAIN:
Again:
Here:
ORG 0 ljmp MAIN
;Reset entry point
;Jump above interrupt
ORG 0030H ;Main Program entry point after vector table space mov TMOD,#02H ;Timer 0, mode 1 mov TH0, #0FAh mov TL0, #024h setb TR0
JNB TF0, Here clr tr0
; TH0 = FEh
; TL0 = 0Ch
;Start timer
;keep monitoring TF0
;Stop timer cpl p2.0
clr TF0 sjmp Again
END
;toggle bit 0 of port2
;clear timer 0 flag bit
;go for next round

Example
• Assuming XTAL = 12MHz write a program to generate a square wave of 10khz at P1.0 with 50% duty cycle
 The required ON/OFF time or Delay time = ( 1 /10KHz)/2 = 50 µS
 Machine cycle period = 12MHz / 12 = 1 µS
 No. of counts to be counted by timer = 50us/1 µS = 50
 The value in TH0 & TL0 = 256 - 50 = 206 = CEh

Example Cont…
MAIN:
Again:
Here:
ORG 0 sjmp MAIN
ORG 0030H mov TMOD,#02H
;Reset entry point
;Jump above interrupt
;Timer 0, mode 2 mov TH0, #0CEh mov TL0, #0CEh
SETB TR0
; TH0 = CEh
; TL0 = CEh
;Start timer
JNB TF0, Here clr tr0 cpl p1.0
clr TF0 sjmp Again
END
;keep monitoring TF0
;stop the process
;toggle bit 0 of port1
;clear timer 0 flag bit
;go for next round

• The size of the time delay depends on two factors:
– The crystal frequency
– The timer’s 16-bit register, TH & TL
• The largest time delay is achieved by making TH=TL=0.
• What if that is not enough?

Example
Examine the following program and find the time delay in seconds.
Exclude the overhead due to the instructions in the loop.
org 0 again: mov TMOD,#10H mov R3, #200 mov TL1,#08 mov TH1,#01 setb TR1
;2
;2
;1 back: jnb TF1,back clr TR1 clr TF1
;65272
;1
;1 djnz R3, again ;2 end
Solution:
TH – TL = 0108H = 264 in decimal; 65536 – 264 = 65272.
One of the timer delay = 65272 X 1
 s = 65.272 ms
Total delay = 200 X 65.272 ms = 13.054400 seconds

 Timer = internal frequency
 Counter = pulses from external input pins (T0/1)
 C/T bit in the TMOD decides the source frequency

• 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
– External input from Tx input pin. a switch
8051
TH0
P1
T0
TL0
P3.4
to
LCD

Example
Assume that clock pulses are fed into pin T1, write a program for counter
1 in mode 2 to count and display the state of the TL1 count on P2.
Org 00h
Start:
Sjmp start
Org 030h mov TMOD, #60h mov th1,#00h setb P3.5
Count_again: setb tr1
Back: mov a, tl1 mov p2, a jnb tf1, back clr tr1 clr tf1 sjmp Count_again end
; Counter 1 mode 2,
;clear th1
;make T1 as input
;start the counter
;get copy of TL1
;display it on port 2
;keep monitoring timer1 flag bit
;stop timer
;make tf1 =0 (polling mode)