1. Chapter 9 Counter/Timer Programming in the 8051

2. Objective 8051 有 2 個 Timer/Counter 的機制，我們在這一章裡要瞭解：

3. Sections
9.1 Programming 8051 Timers
9.2 Counter Programming

4. Section 9.1 Programming 8051 Timers

5. Figure 4-1. 8051 Pin Diagram 8051 (8031) PDIP/Cerdip 1 2 3 4 5 6 7 8 910 11 12 13 14 15 16 17 18 19 20 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21
P1.0
P1.1
P1.2
P1.3
P1.4
P1.5
P1.6
P1.7
RST
(RXD)P3.0
(TXD)P3.1
(T0)P3.4
(T1)P3.5
XTAL2
XTAL1
GND
(INT0)P3.2
(INT1)P3.3
(RD)P3.7
(WR)P3.6
Vcc
P0.0(AD0)
P0.1(AD1)
P0.2(AD2)
P0.3(AD3)
P0.4(AD4)
P0.5(AD5)
P0.6(AD6)
P0.7(AD7)
EA/VPP
ALE/PROG
PSEN
P2.7(A15)
P2.6(A14)
P2.5(A13)
P2.4(A12)
P2.3(A11)
P2.2(A10)
P2.1(A9)
P2.0(A8)
8051
(8031)

6. Inside Architecture of 8051
External interrupts
On-chip ROM for program code
Timer/Counter
Interrupt Control
On-chip RAM
Timer 1
Counter Inputs
Timer 0
CPU
Serial Port
Bus Control
4 I/O Ports
OSC
P0 P1 P2 P3
TxD RxD
Address/Data
Figure 1-2. Inside the 8051 Microcontroller Block Diagram

7. Timers /Counters
The 8051 has 2 timers/counters: timer/counter 0 and timer/counter 1. They can be used as
The timer is used as a time delay generator.
The clock source is the internal crystal frequency of the 8051.
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.

8. 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
LCD
Set
Timer 0
TH0
TL0

9. 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
External input from Tx input pin.
We use Tx to denote T0 or T1.
8051
TH0 P1 to
LCD
TL0
P3.4
a switch T0

10. Registers Used in Timer/Counter
TH0, TL0, TH1, TL1
TMOD (Timer mode register)
TCON (Timer control register)
You can see Appendix H (pages ) for details.
Since 8052 has 3 timers/counters, the formats of these control registers are different.
T2CON (Timer 2 control register), TH2 and TL2 used for 8052 only.

11. Basic Registers of the Timer
Both timer 0 and timer 1 are 16 bits wide.
These registers stores
the time delay as a timer
the number of events as a counter
Timer 0: TH0 & TL0
Timer 0 high byte, timer 0 low byte
Timer 1: TH1 & TL1
Timer 1 high byte, timer 1 low byte
Each 16-bit timer can be accessed as two separate registers of low byte and high byte.

12. Timer Registers TH0 TL0 TH1 TL1 Timer 0 Timer 1 D15 D8 D9 D10 D11 D12

13. TMOD Register Timer mode register: TMOD GATE C/T M1 M0 Timer 1 Timer 0
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)
Set upper 4 bits for Timer (Set to 0000 if not used)
Not bit-addressable
GATE
C/T M1 M0
Timer 1
Timer 0
(MSB)
(LSB)

14. Figure 9-3. TMOD Register
GATE Gating control when set. Timer/counter is enabled only while the INTx pin is high and the TRx control pin is set. When cleared, the timer is enabled whenever the TRx control bit is set.
C/T Timer or counter selected cleared for timer operation (input from internal system clock). Set for counter operation (input from Tx input pin).
M Mode bit 1
M Mode bit 0
GATE
C/T M1 M0
Timer 1
Timer 0
(MSB)
(LSB)

15. C/T (Clock/Timer)
This bit is used to decide whether the timer is used as a delay generator or an event counter.
C/T = 0 : timer
C/T = 1 : counter

16. Gate Every timer has a mean of starting and stopping. GATE=0 GATE=1
Internal control
The start and stop of the timer are controlled by way of software.
Set/clear the TR for start/stop timer.
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).

17. M1, M0 M0 and M1 select the timer mode for timers 0 & 1.
M1 M0 Mode Operating Mode
bit timer mode
8-bit THx + 5-bit TLx (x= 0 or 1)
bit timer mode
8-bit THx + 8-bit TLx
bit auto reload
8-bit auto reload timer/counter;
THx holds a value which is to be reloaded into
TLx each time it overflows.
Split timer mode

18. Example 9-1
Indicate which mode and which timer are selected for each of the
following.
(a) MOV TMOD,#01H (b) MOV TMOD,#20H
(c) MOV TMOD,#12H
Solution:
(a) TMOD = , mode 1 of timer 0 is selected.
(b) TMOD = , mode 2 of timer 1 is selected.
(c) TMOD = , mode 2 of timer 0,
and mode 1 of timer 1 are selected.
timer timer 0

19. Example 9-2
Find the timer’s clock frequency and its period for various 8051-
based systems, with the following crystal frequencies.
(a) 12 MHz (b) 16 MHz (c) MHz
Solution:
XTAL
oscillator
÷ 12
(a) 1/12 × 12 MHz = 1 MHz and T = 1/1 MHz = 1 s
(b) 1/12 × 16 MHz = MHz and
T = 1/1.333 MHz = .75 s
(c) 1/12 × MHz = KHz;
T = 1/921.6 KHz = s

20. Example 9-3
Find the value for TMOD if we want to program timer 0 in mode 2,
use 8051 XTAL for the clock source, and use instructions to start
and stop the timer.
Solution:
TMOD= Timer 1 is not used.
Timer 0, mode 2,
C/T = 0 to use XTAL clock source (timer)
gate = 0 to use internal (software)
start and stop method.

21. TCON Register (1/2) Timer control register: TMOD TR (run control bit)
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)
TF1
TR1
TF0
TR0
IE1
IT1
IE0
IT0
Timer Timer0
for Interrupt
(MSB)
(LSB)

22. TCON Register (2/2) TF (timer flag, control flag) TF1 TR1 TF0 TR0 IE1
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.
TF1
TR1
TF0
TR0
IE1
IT1
IE0
IT0
Timer Timer0
for Interrupt
(MSB)
(LSB)

23. Table 9-2: Equivalent Instructions for the Timer Control Register
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
SETB TF1
SETB TCON.7
CLR TF1
CLR TCON.7
TCON: Timer/Counter Control Register
TF1
IT0
IE0
IT1
IE1
TR0
TF0
TR1

24. 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.
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.

25. Steps of Mode 1 (1/3) Chose mode 1 timer 0
MOV TMOD,#01H
Set the original value to TH0 and TL0.
MOV TH0,#FFH
MOV TL0,#FCH
You had better to clear the flag to monitor: TF0=0.
CLR TF0
Start the timer.
SETB TR0

26. Steps of Mode 1 (2/3)
The 8051 starts to count up by incrementing the TH0-TL0.
TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000H
TR0=1
TR0=0
TH0
TL0
Start timer
Stop timer
FFFC
FFFD
FFFE
FFFF
0000
TF = 0
TF = 0
TF = 0
TF = 0
TF = 1 TF
Monitor TF until TF=1

27. Steps of Mode 1 (3/3)
When TH0-TL0 rolls over from FFFFH to 0000, the 8051 set TF0=1.
TH0-TL0= FFFEH, FFFFH, 0000H (Now TF0=1)
Keep monitoring the timer flag (TF) to see if it is raised.
AGAIN: JNB TF0, AGAIN
Clear TR0 to stop the process.
CLR TR0
Clear the TF flag for the next round.
CLR TF0

28. Mode 1 Programming ÷ 12 XTAL oscillator TR TH TL TF
TF goes high when FFFF
overflow flag
C/T = 0 MC
Start timer

29. Initial Count Values The initial count value = FFFC.
The number of counts = FFFFH-FFFCH+1 = 4
we add one to 3 because of the extra clock needed when it rolls over from FFFF to 0 and raises the TF flag.
The delay = 4 MCs
If MC=1.085 ms, then the delay = 4.34 ms
Figure 9-4 show a formula for delay calculations using mode 1 of the timer for a crystal frequency of XTAL= MHz.
Examples 9-4 to 9-9 show how to calculations the delay generated by timer.

30. Figure 9-4. Timer Delay Calculation for XTAL = 11.0592 MHz
(a) in hex
(FFFF – YYXX + 1) ×
1.085 s 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 s

31. Example 9-4 (1/4)
In the following program, we are creating 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.
;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
P1.5
50%
50%
whole clock

32. Example 9-4 (2/4) ;generate delay using timer 0 DELAY:
SETB TR ;start the timer 0
AGAIN:JNB TF0,AGAIN
CLR TR ;stop timer 0
CLR TF ;clear timer 0 flag
RET
FFF2
FFF3
FFF4
FFFF
0000
TF0 = 0
TF0 = 1

33. Example 9-4 (3/4) Solution:
In the above program notice the following steps.
1. TMOD = 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.
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.

34. Example 9-4 (4/4)
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).

35. Example 9-5
In Example 9-4, calculate the amount of time delay in the DELAY
subroutine generated by the timer. Assume that XTAL =
MHz.
Solution:
The timer works with the internal system clock.
frequency of internal system clock = /12 = KHz
machine cycle = 1 /921.6 KHz = s (microsecond)
The number of counts = FFFFH – FFF2H +1 = 14 (decimal).
The delay = number of counts × s = 14 × s = s for half the clock.
For the entire period of a clock, it is T = 2 × s = s as the time delay generated by the timer.

36. Example 9-6 (1/2)
In Example 9-5, calculate the frequency of the square wave
generated on pin P1.5.
Solution:
In the time delay calculation of Example 9-5, we did not include
the overhead due to instructions in the loop.
To get a more accurate timing, we need to add clock cycles from Table A-1 in Appendix A, as shown below.

37. Example 9-6 (2/2) HERE: MOV TL0,#0F2H 2 MOV TH0,#0FFH 2 CPL P1.5 1
ACALL DELAY
SJMP HERE
; delay using timer 0
DELAY:
SETB TR
AGAIN: JNB TF0,AGAIN
CLR TR
CLR TF
RET
Total
T = 2 × 28 × s = s and F = Hz.

38. Example 9-7 (1/2)
Find the delay generated by timer 0 in the following code, using
both of the methods of Figure 9-4. Do not include the overhead due
to instructions.
CLR P2.3
MOV TMOD,#01 ;Timer 0,mode 1(16-bit)
HERE: MOV TL0,#3EH ;Timer value=B83EH
MOV TH0,#0B8H
SETB P2.3
SETB TR ;start the timer 0
AGAIN:JNB TF0,AGAIN
CLR TR0
CLR TF0
P2.3

39. Example 9-7 (2/2) Solution:
(a) (FFFF – B83E + 1) = 47C2H = in decimal and ×
1.085 s = ms.
(b) Since TH – TL = B83EH = (in decimal ) we have
65536 – = This means that the timer counts from
B83EH to FFFF. This plus rolling over to 0 goes through a
total of clock cycles, where each clock is s in
duration. Therefore, we have × s = ms
as the width of the pulse.

40. Example 9-8 (1/2)
Modify TL and TH in Example 9-7 to get the largest time delay
possible. Find the delay in ms. In your calculation, exclude the
overhead due to the instructions in the loop.
Solution:
TH0=TL0=0 means that
the timer will count from 0000 to FFFF, and then roll over to raise the TF0 flag.
As a result, it goes through a total of states. Therefore, we have delay = (65536 – 0) × s = ms.

41. Example 9-8 (2/2) MOV TMOD,#01 ;Timer 0,mode1(16-bit)
CLR P ;clear P2.3
MOV TMOD,#01 ;Timer 0,mode1(16-bit)
HERE: MOV TL0,#0 ;TL0=0, the low byte
MOV TH0,#0 ;TH0=0, the high byte
SETB P ;SET high P2.3
SETB TR ;start timer 0
AGAIN: JNB TF0,AGAIN ;monitor timer Flag 0
CLR TR ;stop timer 0
CLR TF ;clear timer 0 flag
CLR P2.3

42. Example 9-9 (1/2)
The following program generates a square wave on pin P1.5
continuously using timer 1 for a time delay. Find the frequency of
the square wave if XTAL = MHz. In your calculation do
not include the overhead due to instructions in the loop.
MOV TMOD,#10H ;timer 1, mode 1
AGAIN:MOV TL1,#34H ;timer value=3476H
MOV TH1,#76H
SETB TR ;start
BACK: JNB TF1,BACK
CLR TR ;stop
CPL P ;next half clock
CLR TF ;clear timer flag 1
SJMP AGAIN ;reload timer1

43. Example 9-9 (2/2) Solution:
In mode 1, the program must reload the TH1, TL1 register every timer if we want to have a continuous wave.
FFFFH – 7634H + 1 = 89CCH = clock count
Half period = × s = ms
Whole period = 2 × ms = ms
Frequency = 1/ ms = Hz.
Also notice that the high portion and low portion of the square wave are equal.
In the above calculation, the overhead due to all the instructions in the loop is not included.

44. Find Timer Values
Assume that we know the amount of timer delay and XTAL = MHz .
How to find the inter values needed for the TH, TL?
Divide the desired time delay by s.
Perform –n, where n is the decimal value we got in Step 1.
Convert th result of Step 2 to hex, where yyxx is the initial hex value to be loaded into the timer’s registers.
Set TH = yy and TL = xx.
Example 9-10

45. Example 9-10 (1/2) Assume that XTAL = 11.0592 MHz.
What value do we need to load into the timer’s registers 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 = MHz  MC = s.
5 ms / s = 4608 MCs.
To achieve that we need to load into TL and TH the value – 4608 = = EE00H.
Therefore, we have TH = EE and TL = 00.

46. Example 9-10 (2/2) MOV TMOD,#01 ;Timer 0,mode 1 HERE: MOV TL0,#0
CLR P2.3
MOV TMOD,#01 ;Timer 0,mode 1
HERE: MOV TL0,#0
MOV TH0,#0EEH
SETB P ;SET high P2.3
SETB TR ;start
AGAIN: JNB TF0,AGAIN
CLR TR ;stop
CLR TF0
5ms
P2.3

47. Example 9-11 (1/2)
Assuming that XTAL = 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.
The period of square wave = 1 / frequency
= 1 / 2 kHz = 500 s.
(b) The half period = 500 s /2 = 250 s.
(c) 250 s / s = 230
65536 – 230 = = FF1AH.
(d) TL1 = 1AH and TH1 = FFH

48. Example 9-11 (2/2) AGAIN:MOV TL1,#1AH ;timer value = FF1AH
MOV TMOD,#10H ;timer 1, mode 1
AGAIN:MOV TL1,#1AH ;timer value = FF1AH
MOV TH1,#0FFH
SETB TR ;start
BACK: JNB TF1,BACK
CLR TR ;stop
CPL P1.5
CLR TF ;clear timer flag 1
SJMP AGAIN

49. Example 9-12 (1/2)
Assuming XTAL = MHz, write a program to generate a
square wave of 50 Hz frequency on pin P2.3.
Solution:
Look at the following steps.
(a) The period of the square wave = 1 / 50 Hz = 20 ms.
(b) The high or low portion of the square wave = 10 ms.
(c) 10 ms / s = 9216
65536 – 9216 = in decimal = DC00H in hex.
(d) TL1 = 00H and TH1 = DCH.

50. Example 9-12 (2/2) MOV TMOD,#10H ;timer 1, mode 1
AGAIN: MOV TL1,#00 ;Timer value = DC00H
MOV TH1,#0DCH
SETB TR ;start
BACK: JNB TF1,BACK
CLR TR ;stop
CPL P2.3
CLR TF ;clear timer flag 1
SJMP AGAIN ;reload timer since
;mode 1 is not
;auto-reload

51. Generate a Large Time Delay
The size of the time delay depends on two factors:
They 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 9-13 show how to achieve large time delay.

52. Example 9-13
Examine the following program and find the time delay in seconds.
Exclude the overhead due to the instructions in the loop.
MOV TMOD,#10H
MOV R3,#200
AGAIN: MOV TL1,#08
MOV TH1,#01
SETB TR1
BACK: JNB TF1,BACK
CLR TR1
CLR TF1
DJNZ R3,AGAIN
Solution:
TH – TL = 0108H = 264 in decimal
65536 – 264 =
One of the timer delay = × s = ms
Total delay = 200 × ms = seconds

53. Timer Mode 0
Mode 0 is exactly like mode 1 except that it is a 13-bit timer instead of 16-bit.
8-bit TH0 + 5-bit TL0
The counter can hold values between 0000 to 1FFF in TH0-TL0.
213-1= 2000H-1=1FFFH
We set the initial values TH0-TL0 to count up.
When the timer reaches its maximum of 1FFFH, it rolls over to 0000, and TF0 is raised.

54. Timer Mode 2 8-bit timer. Auto-reloading
It allows only values of 00 to FFH to be loaded into TH0.
Auto-reloading
TL0 is incremented continuously when TR0=1.
In the following example, we want to generate a delay with 200 MCs on timer 0.
See Examples 9-14 to 9-16

55. Steps of Mode 2 (1/2) Chose mode 2 timer 0
MOV TMOD,#02H
Set the original value to TH0.
MOV TH0,#38H
Clear the flag to TF0=0.
CLR TF0
After TH0 is loaded with the 8-bit value, the 8051 gives a copy of it to TL0.
TL0=TH0=38H
Start the timer.
SETB TR0

56. Steps of Mode 2 (2/2)
The 8051 starts to count up by incrementing the TL0.
TL0= 38H, 39H, 3AH,....
When TL0 rolls over from FFH to 00, the 8051 set TF0=1. Also, TL0 is reloaded automatically with the value kept by the TH0.
TL0= FEH, FFH, 00H (Now TF0=1)
The 8051 auto reload TL0=TH0=38H.
Go to Step 6 (i.e., TL0 is incrementing continuously).
Note that we must clear TF0 when TL0 rolls over. Thus, we can monitor TF0 in next process.
Clear TR0 to stop the process.

57. Mode 2 programming ÷ 12 XTAL oscillator TR TH TF
TF goes high when FF
overflow flag
C/T = 0
reload

58. Example 9-14 (1/2) MOV TMOD,#20H ;Timer 1,mode 2
Assuming that XTAL = MHz, find
(a) the frequency of the square wave generated on pin P1.0 in the following program
(b) the smallest frequency achievable in this program, and the TH value to do that.
MOV TMOD,#20H ;Timer 1,mode 2
MOV TH1,#5 ;not load TH1 again
SETB TR ;start (no stop TR1=0)
BACK:JNB TF1,BACK
CPL P1.0
CLR TF ;clear timer flag 1
SJMP BACK ;mode 2 is auto-reload

59. Example 9-14 (2/2) Solution:
(a) First notice that target address of SJMP. In mode 2 we do not
need to reload TH since it is auto-reload.
Half period = (FFH – 05 +1) × s = s
Total period = 2 × s = s
Frequency = kHz.
(b) To get the smallest frequency, we need the largest period and that is achieved when TH = 00.
Total period = 2 × 256 × s = s
Frequency = 1.8kHz.

60. Example 9-15
Find the frequency of a square wave generated on pin P1.0.
Solution:
MOV TMOD,#2H ;Timer 0,mode 2
MOV TH0,#0
AGAIN:MOV R5,#250 ;count 250 times
ACALL DELAY
CPL P1.0
SJMP AGAIN
DELAY:SETB TR ;start
BACK: JNB TF0,BACK
CLR TR ;stop
CLR TF ;clear TF
DJNZ R5,DELAY ;timer 2: auto-reload
RET
T = 2 (250 × 256 × s) = ms, and frequency = 72 Hz.

61. Assemblers and Negative Values
Since the timer is 8-bit in mode 2, we can let the assembler calculate the value for TH.
For example, if we want to generate a time delay with 200 MCs, then we can use
MOV TH1,#38H or
MOV TH1,#-200
Way 1: = 56 = 38H
Way 2: -200 = -C8H  2’s complement of –200 = 100H – C8H = 38 H

62. Example 9-16
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:
Some 8051 assemblers provide this way.
-200 = -C8H  2’s complement of –200 = 100H – C8H = 38 H
Decimal
2’s complement (TH value)
-200 = - C8H
38H
- 60 = - 3CH
C4H
- 3
FDH
- 12
F4H
- 48
D0H

63. Example 9-17 (1/2)
Find (a) the frequency of the square wave generated in the
following code, and (b) the duty cycle of this wave.
Solution:
“MOV TH0,#-150” uses 150 clocks.
The DELAY subroutine = 150 × s = 162 s.
The high portion of the pulse is twice tat of the low portion (66% duty cycle).
The total period = high portion + low portion
= s s = s
Frequency = kHz.

64. Example 9-17 (2/2) MOV TH0,#-150 ;Count=150 AGAIN:SETB P1.3
MOV TMOD,#2H ;Timer 0,mode 2
MOV TH0,#-150 ;Count=150
AGAIN:SETB P1.3
ACALL DELAY
CLR P1.3
ACALL DEALY
SJMP AGAIN
DELAY:SETB TR ;start
BACK: JNB TF0,BACK
CLR TR ;stop
CLR TF ;clear TF
RET
high period
low period

65. Section 9.2 Counter Programming

66. Counter
These timers can also be used as counters counting events happening outside the 8051.
When the timer is used as a counter, it is a pulse outside of the 8051 that increments the TH, TL.
When C/T=1, the counter counts up as pulses are fed from
T0: timer 0 input (Pin 14, P3.4)
T1: timer 1 input (Pin 15, P3.5)

67. Table 9-1: Port 3 Pins Used For Timers 0 and 1
Function
Description 14
P3.4 T0
Timer/Counter 0 external input 15
P3.5 T1
Timer/Counter 1 external input
GATE
C/T=1 M1 M0
Timer 1
Timer 0
(MSB)
(LSB)

68. Counter Mode 1 16-bit counter (TH0 and TL0)
TH0-TL0 is incremented when TR0 is set to 1 and an external pulse (in T0) occurs.
When the counter (TH0-TL0) reaches its maximum of FFFFH, it rolls over to 0000, and TF0 is raised.
Programmers should monitor TF0 continuously and stop the counter 0.
Programmers can set the initial value of TH0-TL0 and let TF0=1 as an indicator to show a special condition. (ex: 100 people have come).

69. Figure 9-5. (a) Timer 0 with External Input (Mode 1)
Timer 0 external input Pin 3.4
TR0
TH0
TL0
TF0
TF0 goes high when FFFF
overflow flag
C/T = 1

70. Figure 9-5. (b) Timer 1 with External Input (Mode 1)
Timer 1 external input Pin 3.5
TR1
TH1
TL1
TF1
TF1 goes high when FFFF
overflow flag
C/T = 1

71. Counter Mode 2 8-bit counter. Auto-reloading
It allows only values of 00 to FFH to be loaded into TH0.
Auto-reloading
TL0 is incremented if TR0=1 and external pulse occurs.
See Figure 9.6, 9.7 for logic view
See Examples 9-18, 9-19

72. Figure 9.6: Timer 0 with External Input (Mode 2)
Timer 0 external input Pin 3.4
TR0
TL0
TH0
TF0
TF0 goes high when FF
overflow flag
C/T = 1
reload

73. Figure 9.7: Timer 1 with External Input (Mode 2)
Timer 1 external input Pin 3.5
TR1
TL1
TH1
TF1
TF1 goes high when FF
overflow flag
C/T = 1
reload

74. Example 9-18 (1/2)
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
TL 1 count on P2.
Solution:
MOV TMOD,# B ;mode 2, counter 1
MOV TH1,#0
SETB P ;make T1 input port
AGAIN:SETB TR ;start
BACK: MOV A,TL1
MOV P2,A ;display in P2
JNB TF1,Back ;overflow
CLR TR ;stop
CLR TF ;make TF=0
SJMP AGAIN ;keep doing it

75. Example 9-18 (2/2)
We use timer 1 as an event counter where it counts up as clock pulses are fed into pin3.5.
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 must make P3.5 an input port by making it high. T1 to
LEDs
P3.5 P2
8051
P2 is connected to 8 LEDs and input T1 to pulse.

76. Example 9-19 (1/3)
Assume that a 1-Hz frequency pulse is connected to input pin 3.4.
Write a program to display counter 0 on an LCD. Set the initial
value of TH0 to -60.
Solution:
Note that on the first round, it starts from 0 and counts 256 events, since on RESET, TL0=0. To solve this problem, load TH0 with -60 at the beginning of the program. T0 to
LCD
P3.4 P1
8051
1 Hz clock

77. Example 9-19 (2/3) ACALL LCD_SET_UP ;initialize the LCD
MOV TMOD,# B ;Counter 0,mode2
MOV TH0,#-60
SETB P ;make T0 as input
AGAIN:SETB TR ;starts the counter
BACK: MOV A,TL ; every 60 events
ACALL CONV ;convert in R2,R3,R4
JNB TF0,BACK ;loop if TF0=0
CLR TR ;stop
CLR TF0
SJMP AGAIN

78. Example 9-19 (3/3) ;converting 8-bit binary to ASCII
CONV: MOV B,#10 ;divide by 10
DIV AB
MOV R2,B ;save low digit
MOV B,#10 ;divide by 10 once more
ORL A,#30H ;make it ASCII
MOV R4,A
MOV A,B
ORL A,#30H
MOV R3,A
MOV A,R2
MOV R2,A ;ACALL LCD_DISPLAY here
RET R4 R3 R2

79. A Digital Clock Example 9-19 shows a simple digital clock.
If we feed an external square wave of 60 Hz frequency into the timer/counter, we can generate the second, the minute, and the hour out of this input frequency and display the result on an LCD.
You might think that the use of the instruction “JNB TF0,target” to monitor the raising of the TF0 flag is a waste of the microcontroller’s time.
The solution is the use of interrupt. See Chapter 11.
In using interrupts we can do other things with the 8051.
When the TF flag is raised it will inform us.

80. GATE=1 in TMOD All discuss so far has assumed that GATE=0.
The timer is stared with instructions “SETB TR0” and “SETB TR1” for timers 0 and 1, respectively.
If GATE=1, we can use hardware to control the start and stop of the timers.
INT0 (P3.2, pin 12) starts and stops timer 0
INT1 (P3.3, pin 13) starts and stops timer 1
This allows us to start or stop the timer externally at any time via a simple switch.

81. Example for GATE=1
The 8051 is used in a product to sound an alarm every second using timer 0.
Timer 0 is turned on by the software method of using the “SETB TR0” instruction and is beyond the control of the user of that product.
However, a switch connected to pin P3.2 can be used to turn on and off the timer, thereby shutting down the alarm.

82. Figure 9-8: Timer/Counter 0
XTAL
oscillator
÷ 12
TR0
INT0 Pin
Pin 3.2
C/T = 0
Gate
T0 Pin
Pin 3.4
C/T = 1

83. Figure 9-9: Timer/Counter 1
XTAL
oscillator
÷ 12
TR1
INT1 Pin
Pin 3.3
C/T = 0
Gate
T1 Pin
Pin 3.5
C/T = 1

84. Timer/Counter Mode 3 For Timer/Counter 0 For Timer/Counter 1
As a Timer:
Two 8-bit timers
As a Counter:
8-bit counter.
For Timer/Counter 1
Not available

86. You are able to
List the timers of the 8051 and their associated registers
Describe the various modes of the 8051 timers
Program the 8051 timers to generate time delays
Program the 8051 counters as event counters

87. Homework
Chapter 9 Problems：9,20,22,26,31,36,44,47,48