The document discusses different approaches to using timers/counters in 8051 microcontrollers. It describes software-based, hardware-based, and combined approaches. It provides details on registers used for timers like TH0, TL0, TMOD, and TCON. It explains how to use these registers to initialize and control timers, and provides examples of generating square waves using timers.

1. 8051 counters/timers
N.Bhaskar
12MT06PED007

2. Introduction :
• To generate precise time delay and to count the
number of pulses timer / counter plays an
important role
Timer operations are basically three
types
1.software based approach
2. hardware based approach
3. combined software and hardware
approach
SJCET

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start
Set P1.0 to 1
Initialize counter
register(CR)
CR←CR-1
is
CR=0
CPL P1.0
N
Y
SETB P1.0
Loop: MOV R6,#255
Here: DJNZ R6,Here
CPL P1.0
SJMP Loop

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5. CCoommbbiinneedd SS//WW aanndd HH//WW aapppprrooaacchh :
Registers Used in Timers/Counters:
– TH0, TL0, TH1, TL1
– TMOD (Timer mode register)
– TCON (Timer control register)
SJCET

6. TTCCOONN RReeggiisstteerr
Timer control register: TCON
• Upper nibble for timer control, lower nibble for
interrupts
TR (run control bit)
• TR0 for Timer 0 control; TR1 for Timer 1 control.
• TR is set by programmer to turn timer/counter
on/off.
• TR=0: off (stop)
• TR=1: on (start)
(MSB) (LSB)
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt

7. TF (timer overflow flag bit)
TF0 is a timer 0 overflow flag bit; TF1 is a
timer 1 overflow flag bit.
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
(MSB
)
(LSB)
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt

8. Equivalent Instructions ffoorr tthhee TTiimmeerr CCoonnttrrooll
RReeggiisstteerr::
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9. TTMMOODD RReeggiisstteerr
Timer mode register: TMOD
• An 8-bit register
• Set the usage mode for two timers
• Set lower 4 bits for Timer 0 (Set to 0000
if not used)
• 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

11. VNIT

13. SStteeppss FFOORR TTIIMMEERRSS OOPPEERRAATTIIOONN
1. Choose mode 1 timer 0
MOV TMOD,#01H
1. Set the initiat value to TH0 and TL0.
MOV TH0,#FFH
MOV TL0,#FCH
1. Better to clear the flag to monitor: TF0=0.
CLR TF0
1. Start the timer.
SETB TR0

14. 5.The 8051 starts to count up by incrementing the TH0-TL0.
TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000H

15. 6. When TH0-TL0 rolls over from FFFFH to
0000, the 8051 set TF0=1.
Now TF0=1
6. Keep monitoring the timer flag (TF) to see
if it is raised.
AGAIN: JNB TF0, AGAIN
6. Clear TR0 to stop the process.
CLR TR0
6. Clear the TF flag for the next round.
CLR TF0

16. Timer Delay CCaallccuullaattiioonn ffoorr XXTTAALL ==
1111..00559922 MMHHzz
(a) in hex
• (FFFF – YYXX ) × 1.085 ms
• where YYXX are TH, TL initial values
respectively.
• Notice that values YYXX are in hex.
(b) in decimal
• Convert YYXX values of the TH, TL register to
decimal to get a NNNNN decimal number
• then (65536 – NNNNN) × 1.085 ms

17. • 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 ms and get n.
2. Perform 65536 –n
3. Convert the result of Step 2 to hex ( yyxx)
4. Set TH = yy and TL = xx.

18. Example 1:
• 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 ms = 9216
4. 65536 – 9216 = 56320 in decimal = DC00H in
hex.
5. TL1 = 00H and TH1 = DCH.

19. MOV TMOD,#10H ;timer 1, mode 1
AGAIN: MOV TL1,#00 ;Timer value= DC00H
MOV TH1,#0DCH
SETB TR1 ;start
BACK: JNB TF1,BACK
CLR TR1 ;stop
CPL P2.3
CLR TF1 ;clear timer flag 1
SJMP AGAIN ;reload timer

20. Example 2:
square wave of 50% duty on P1.5
Timer 0 is used
;each loop is a half clock
MOV TMOD,#01 ;Timer 0,mode 1(16-bit)
HERE: MOV TL0,#0F2H ;Timer value = FFF2H
MOV TH0,#0FFH
CPL P1.5
ACALL DELAY
SJMP HERE
50% 50%
whole clock
P1.5

21. ;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
FFF2 FFF3 FFF4 FFFF 0000
TF0 =0 TF0 =0 TF0 =0 TF0 =0 TF0 =1

22. QUERIES
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23. THANK YOU
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