1. Chapter 9 8051 Timer Programming in Assembly and C

2. Objective Embedded Systems with Timer/Counter
紅綠燈、電子鐘
碼錶(count up)、計時器(count down)
計數器、計步器、跑步機等運動器材。
8051 有 2 個 Timer/Counter 的機制，我們在這一章裡要瞭解：
什麼是 Timer，什麼是 Counter。
如何來使用 Timer。
如何來使用 Counter。

3. Sections 9.1 Programming 8051 timers 9.2 Counter programming
9.3 Programming timers 0 and 1 in 8051 C

4. Section 9.1 Programming 8051 Timers

5. 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
再三強調 timer/counter 是兩個獨立於 CPU 之外的硬體, 有自己的線路與運作, 但 CPU 要控制它們, 以達到目的.
P0 P1 P2 P3
TxD RxD
Address/Data
Figure 1-2. Inside the 8051 Microcontroller Block Diagram

6. 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.

7. Timer
8051 timers use 1/12 of XTAL frequency as the input of timers, regardless of machine cycle.
Because the input of timer is a regular, fixed-periodic square wave, we can count the number of pulses and calculate the time delay.
8051
XTAL
oscillator
÷ 12
Timer P1 to
LCD
TH0
Set
Timer 0
TL0

8. Counter Count the number of events
External input from Tx input pin (x=0 or 1).
We use Tx to denote T0 or T1.
External input from T0 input pin (P3.4) for Counter 0
External input from T1 input pin (P3.5) for Counter 1
8051
TH0 P1 to
LCD
TL0
Vcc
P3.4
a switch T0

9. 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)

10. Figure 9-8: Timer/Counter 0
timer input
XTAL
oscillator
÷ 12
TR0
INT0 Pin
Pin 3.2
C/T = 0
Gate
T0 Pin
Pin 3.4
C/T = 1
TH0
TL0
counter input
1:start
0:stop
TF0
1. monitor by JNB
2. interrupt
hardware control
Sec 9.2 

11. Figure 9-9: Timer/Counter 1
timer input
XTAL
oscillator
÷ 12
TR1
INT1 Pin
Pin 3.3
C/T = 0
Gate
T1 Pin
Pin 3.5
C/T = 1
TH1
TL1
counter input
1:start
0:stop
TF1
1. monitor by JNB
2. interrupt
hardware control

12. Registers Used in Timer/Counter
TH0, TL0 (Timer 0 registers)
TH1, TL1 (Timer 1 registers)
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.

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

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

15. TCON Register (1/2) Timer control register: TCON 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.
TRx is set by programmer to turn timer/counter on/off.
TRx=0: off (stop)
TRx=1: on (start)
TF1
TR1
TF0
TR0
IE1
IT1
IE0
IT0
Timer Timer0
for Interrupt
(MSB)
(LSB)

16. TCON Register (2/2) TF (timer flag, control flag) TF1 TR1 TF0 TR0 IE1
TF0 for timer/counter 0; TF1 for timer/counter 1.
TFx is like a carry. Originally, TFx=0. When TH-TL roll over to 0000 from FFFFH, the TFx is set to 1.
TFx=0 : not reach
TFx=1: reach
If we enable interrupt, TFx=1 will trigger ISR.
TF1
TR1
TF0
TR0
IE1
IT1
IE0
IT0
Timer Timer0
for Interrupt
(MSB)
(LSB)

17. 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

18. 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)

19. 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)

20. 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

21. 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 software.
Set/clear the TR0 (or TR1) 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 INT0 (or INT1) pin has an 1 to 0 transition and the TR0 (or TR1) control pin is set.
INT0: P3.2, pin 12; INT1: P3.3, pin 13.
GATE=0 表示只要 TR0/TR1 被設成 1, timer/counter 就會開始運作.
GATE=1 表示除了 TR0/TR1 要被設成 1 之外, INT0/INT1 也要被設成0, timer/counter 就會開始運作.
the 1 to 0 transition on INT0 (P3.2)

22. 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 (x= 0 or 1)
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
在mode 3 中, counter/timer 0 的 TL0 為 8 bit 的 timer/counter; TH0 為 8 bit 的 timer. 至於 Timer/counter 1 則停止動作.

23. 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.
GATE
C/T M1 M0
Timer 1
Timer 0
(MSB)
(LSB)
timer timer 0

24. 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 = 0.75 s
(c) 1/12 × MHz = KHz;
T = 1/921.6 KHz = s

25. 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.
GATE
C/T M1 M0
Timer 1
Timer 0
(MSB)
(LSB)
Ask student how to set TMOD for the following conditions:
(1) Timer 1, Mode 1, software control => MOV TMOD, #10H
(2) Timer 1, Mode 2, software + hardware control, Counter 0, Mode 2, software control => TMOD= , MOV TMOD, #84H

26. 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 every 12 clocks from XTAL.
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.

27. Mode 1 Programming ÷ 12 XTAL oscillator TR TH TL TF
TF goes high when FFFF
overflow flag
C/T = 0
like MC for 89C51
Start timer

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

29. 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
Roll over
TF0=0
TF0=0
TF0=0
TF0=0
TF0=1
TF0
Monitor TF01 until TF0=1

30. 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
要特別注意, JNB 需要 2 個 MCs. 所以每執行一次 JNB, TH0-TL0 會加 2.
在執行完 AGAIN 迴圈, JNB 發現 TF0 != 0, 離開 AGAIN, 將要執行 CLR TR0, 此時 TH0-TL0 = 0002.

31. 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 for 89C51
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.

32. 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

33. Find Timer Values Assume XTAL = 11.0592 MHz .
How to find the inter values needed for the TH, TL?
Divide the desired time delay by s.
20ms ÷ ms = 18433
Perform –n, where n is the decimal value we got in Step 1.
=47103=B7FFH
Convert the 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.
TH=B7H, TL=FFH
Example 9-10
Ex: 想要產生 200 ms 的 time delay. 200,000÷1.085= =47203=B863H, TH=B8H, TL=63H

34. 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

35. 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

36. 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.

37. Example 9-4 (4/4)
7. Timer 0 is stopped by the instruction “CLR TR0”. The DELAY subroutine ends, and the process is repeated.
8. Remember to clear TF0 by the instruction “CLR TF0”.
Notice that to repeat the process, we must reload the TL and TH
registers, and start the timer again (in the main program).

38. 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.
Ask whether the answer is precise enough or not. 是一點點不準或很不準?

39. Example 9-6 (1/2)
In Example 9-5, calculate the accurate 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.

40. 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.

41. 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
這不是 square wave, 只是在求 level 1 的時間長度.
P2.3

42. Example 9-7 (2/2) Solution: (a) or (b) is OK.
(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.

43. 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.

44. 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

45. 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

46. 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.

47. 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 TL0 and TH0 the value – 4608 = = EE00H.
Therefore, we have TH0 = EE and TL0 = 00.

48. 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

49. 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

50. 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

51. 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.

52. 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

53. 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:
TH1-TL1 = 0108H = 264 in decimal
65536 – 264 =
One of the timer delay = × s = ms
Total delay = 200 × ms = seconds
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.

54.    Timer Mode 0 Mode 0 is used to compatible with MSC-48
Mode 0 is exactly like mode 1 except that it is a 13-bit timer instead of 16-bit.
The counter can hold values between 0000 to 1FFF
213-1= 2000H-1=1FFFH
When the timer reaches its maximum of 1FFFH, it rolls over to 0000, and TF0 is raised.
D15 D8 D9
D10
D11
D12
D13
D14 D7 D0 D1 D2 D3 D4 D5 D6
TH0
TL0
  
Timer 0

55. Timer Mode 3 For Timer/Counter 0 For Timer/Counter 1 As a Timer:
TH0 and TL0 can be 8-bit timers.
As a Counter:
TL0 can be an 8-bit counter.
For Timer/Counter 1
Not available

56. Timer Mode 2 8-bit timer Auto-reloading
TL0 is incremented (or roll over) continuously when TR0=1.
Auto-reloading
TH0 is loaded into TL0 automatically when TL0=00H.
TH0 is not changed.
You need to clear TF0 after TL0 rolls over.
In the following example, we want to generate a delay with 200 MCs on timer 0.
See Examples 9-14 to 9-16

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

58. Steps of Mode 2 (1/3) Chose mode 2 timer 0
MOV TMOD,#02H
Set the original value to TH0.
MOV TH0,#FCH
Clear the flag to TF0=0.
CLR TF0
Start the timer.
SETB TR0
Note that the instruction SETB TR0 dose not load TH0 to TL0. So TL0 still is 00H.
在 timer/counter mode 2, 只有當 TL 從 FFH rolls over 成00H 時, TH 會被載入 TL. 對於 timer 第一次啟動時, 因為沒有任何 roll over 的動作, 所以 TL 的值為 00H (if no initial value for TL), TL 會從 00H 開始往上數.
直到第一次 TL rolls over (FFH -> 00H), TL 才會被設成 TH, 以後的運作才會按照 TH 的設定來產生 time delay.

59. Steps of Mode 2 (2/3)
The 8051 starts to count up by incrementing the TL0.
TL0= ..., FCH,FDH,FEH,FFH,FCH
TH0=FCH
TL0=FCH
TF0=10
TR0=1
auto reload
Clear TF0
Start timer 00 01 FE FF FC FD
...
roll over
00H
TF0 = 0
TF0 = 0
TF0 = 0
TF0 = 0
TF0 = 1
TF0 = 1
TF0 =0
TH0=FCH
roll over
00H
TL0=00H FE FF FC FD
TF0 = 0
TF0 = 0
TF0 = 1
TF0 = 1
auto reload: TL=-FCH immediately

60. Steps of Mode 2 (3/3)
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= FCH, FDH, FEH, FFH, FCH(Now TF0=1)
The 8051 auto reload TL0=TH0=FCH.
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.
00H

61. 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

62. 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 TH1 = 00.
Total period = 2 × 256 × s = s
Frequency = 1.8kHz.

63. 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.

64. 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 (AT89C51), 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

65. 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

66. 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 that of the low portion (66% duty cycle).
The total period = high portion + low portion
= s s = s
Frequency = kHz.

67. 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

68. Section 9.2 Counter Programming

69. Counter (1/2)
These timers can also be used as counters counting events happening outside the 8051 by setting 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) 
8051 P1
Counter 0
TH0 to
LCD
TL0
Vcc
P3.4 T0
a switch

70. Counter (2/2)
When the timer is used as a counter, it is a pulse outside of the 8051 that increments
TH0 & TL0 for counter 0.
TH1 & TL1 for counter 1.
8051 P1
Counter 1
TH1 to
LCD
TL1
Vcc
P3.5 T1
a switch

71. 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)

72. 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 as an indicator to show a special condition. (ex: 100 people have come).

73. Figure 9-5. (a) Counter 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

74. Figure 9-5. (b) Counter 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

75. Counter Mode 2 8-bit counter. Auto-reloading
TL0 is incremented if TR0=1 and external pulse occurs.
Auto-reloading
TH0 is loaded into TL0 when TF0=1.
It allows only values of 00 to FFH to be loaded into TH0.
You need to clear TF0 after TL0 rolls over.
See Figure 9.6, 9.7 for logic view
See Examples 9-18, 9-19

76. Figure 9.6: Counter 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

77. Figure 9.7: Counter 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

78. 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
TL1 count on P2.
Solution:
We use timer 1 as an event counter where it counts up as clock pulses are fed into pin3.5. to
LEDs
P3.5 P2
8051 T1
P2 is connected to 8 LEDs and input T1 to pulse.

79. Example 9-18 (2/2) MOV TMOD,#01100000B ;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
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.

80. 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 R4 R3 R2

81. Example 9-19 (2/3) ACALL LCD_SET_UP ;initialize the LCD
MOV TMOD,# B ;Counter 0,mode2
MOV TH0,#C4H ;C4H=-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

82. 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
A 的值介在 196 – 255 之間, 為一個 triple figure ( ), 所以可以用 C4H=196 不斷除以10的方式, 轉成十進位變成 百位數 (the hundreds digit)=1, 十位數(the tens digit)=9, 個位數(the units digit)=6 為範例來說明.

83. 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, interrupt service routine starts.

84. 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.
“SETB TR0” and “SETB TR1” for timers 0 and 1 are necessary.
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.

85. 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.
The timer is turned ON/OFF by the 1 to 0 transition on INT0 (P3.2) when TR0 =1 (hardware control).
However, a switch connected to pin P3.2 can be used to turn on and off the timer, thereby shutting down the alarm.
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. 表示 TR0 的優先權比 user control (INTx) 大.

86. Section 9.3 Programming Timers 0 and 1 in 8051 C

87. 8051 Timers in C
In 8051 C we can access the timer registers TH, TL, and TMOD directly using the reg51.h header file.
See Example 9-20
Timers 0 and 1 delay using mode 1
See Example 9-22 and Example 9-25
Timers 0 and 1 delay using mode 2
See Examples 9-23 and 9-24
Look by yourself

88. Example 9-20 (1/2)
Write an 8051 C program to toggle bits of P1 continuously with some time delay. Use Timer 0, 16-bit mode.
Solution:
#include <reg51.h>
void T0Delay(void);
void main(void) {
while(1) { //repeat forever
P1=0x55;
T0Delay(); //time delay
P1=0xAA;
T0Delay(); }}

89. Example 9-20 (2/2) Assume XTML=11.0592MHz for the AT89C51
FFFFH-3500H+1=CB00H=51968
1.085ms 51968  ms
void T0Delay() {
TMOD=0x01; //Timer 0, Mode 1
TL0=0x00; TH0=0x35; //initial value
TR0=1; //turn on T0
while (TF0==0); //text TF to roll over
TR0=0; //turn off T0
TF0=0; } //clear TF0

90. Example 9-25 (1/3)
A switch is connected to P1.7. Write an 8051 C program to monitor SW and create the following frequencies on P1.5.
SW=0: 500 Hz; SW=1: 750 Hz. Using Timer 0, mode 1.
Assume XTML= MHz for the AT89C51
Solution:
#include <reg51.h>
sbit mybit=P1^5;
sbit SW=P1^7;
void T0Delay(void);

91. Example 9-25 (2/3) void main(void) { SW=1; //make P1.7 an input pin
while(1) { //repeat forever
mybit=~mybit; //toggle
if (SW==0)
T0Delay(0); //500Hz time delay
else
T0Delay(1); //750Hz time delay }}

92. Example 9-25 (3/3) void T0Delay(unsigned char c) {
TMOD=0x01; //Timer 0, Mode 1
if (c==0) { TL0=0x67; TH0=0xFC };
// FFFFH-FC67H+1=921, about ms, 500 Hz
else { TL0=0x9A; TH0=0xFD };
// FFFFH-FD9AH+1=614, about ms, 750 Hz
TR0=1;
while (TF==0);
TR0=0;
TF0=0; }

93. Time Delay with Different Chips
Although the numbers of clocks per machine cycle are vary with different versions, the frequency for the timer is always 1/12 the frequency of the crystal.
To maintain compatibility with the original 8051
The same codes put on AT89C51 and DS89C420 will generate different time delay.
The factors of C compiler and MC of others instructions
The C compiler is a factor in the delay size since various 8051 C compilers generate different hex code sizes. So, we still have approximate delay only.
See Examples 9-21

94. Example 9-21 (1/3)
Write an 8051 C program to toggle bits P1.5 continuously every 50 ms. Use Timer 0, mode 1 for (a) AT89C51 and
(b) DS89C420.
Solution:
#include <reg51.h>
void T0Delay(void);
sbit mybit=P1^5;
void main(void) {
while(1) { //repeat forever
mybit=~mybit;
T0Delay(); }}

95. Example 9-21 (2/3) (a) Assume XTML=11.0592MHz for the AT89C51
FFFFH-4BFDH+1=B403H=46083
1.085ms 46083  50ms
void T0Delay() {
TMOD=0x01; //Timer 0, Mode 1
TL0=0xFD; TH0=0x4B; //initial value
TR0=1; //turn on T0
while (TF0==0); //BACK: JNZ TF0,BACK
TR0=0; //turn off T0
TF0=0; } //clear TF0

96. Example 9-21 (3/3) (b) Assume XTML=11.0592MHz for the DS89C4x0
FFFFH-4BFDH+1=B403H=46083
1.085ms 46083  50ms
void T0Delay() {
TMOD=0x01; //Timer 0, Mode 1
TL0=0xFD; TH0=0x4B; //initial value
TR0=1; //turn on T0
while (TF0==0); //BACK: JNZ TF0,BACK
TR0=0; //turn off T0
TF0=0; } //clear TF0

97. 8051 Counters in C External pulses to T0 (P3.4) and T1 (P3.5).
See Examples 9-26 to 9-29.

98. Example 9-26 (1/2)
Assume that a 1-Hz external clock is being fed into T1 (P3.5). Write a C program for counter 1 in mode 2 to count up and display the state of the TL1 count on P1. Start the count at 0H.
Solution:
P1 is connected to 8 LEDs.
T1 is connected to 1Hz external clock.
TH1 P1
LEDs
TL1
P3.5
1Hz T1

99. Example 9-26 (2/2) #include <reg51.h> sbit T1=P3^5;
void main(void) {
T1=1; //make T1 an input pin
TMOD=0x60; //counter 1, mode 2
TH1=0; //reload value
while(1) { //repeat forever
do {TR1=1; P1=TL1;} while (TF1==0);
TR1=0; TF1=0; //clear flags }}

100. 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 in Assembly and C to generate time delays
Program the 8051 counters in Assembly and C as event counters

101. DigitMap Descriptor A dialing plan A start timer, to start
a digit map (11x|080xxxxxx|03xxxxxxx|002x.)
A start timer, to start
A short timer, when more digits are needed
A long timer, to differentiate different routing
Timers are applied by default values.
off-hook
Digital map 是一種表示電話號碼規劃 (dialing plan) 的方式.
Digital map 會透過 MEGACO 命令由 MGC 送給 MG, 然後存在 MG 中.
Digital map 的目的是讓 MG 瞭解會從 caller 收到甚麼型式的 digit strings. 當 gateway 收集到所有輸入的電話號碼, 會將整個號碼送給 MGC 做進一步的處理.
也可以讓 trunking gateway 去收集一些資訊, 如信用卡號等資料.
Ex: (11x|080xxxxxx|03xxxxxxx|002x.) 就是一個 digit map.
The symbol "x" which matches any digit ("0" to "9")
MEGACO 的 digit map 中有三個 symbols, 用以代表設定的 timers.
Start timer (T) 代表從 dial tone 到使用者輸入第一個 digit, MG 所能容忍的最長的時間間隔.
Short timer (S) 代表兩個連續輸入的 digits 間最大的間隔.
Long timer (L) 代表已收到一個完整的數值, 但如果使用者繼續輸入 digits, 則會改變 routing 的方式. 因此在得到完整電話號碼後, 會等待 long timer 的時間, 以確定這是最終的電話號碼. 5 7 2 1 4 3
start timer
short timer
long timer
dial tone

102. CSMA/CD 通訊協定 1. 先監聽網路上是否有其他電腦正在傳送frame資料，一直監聽到沒有訊號。
2. 若沒有聽到網路上有frame，將自己的 frame送出；若有則繼續監聽，直到訊號消失，則將frame送出。
3. 繼續監聽網路（51.2 ms），若發生frame的碰撞（collision），則送出擾亂訊號（Jamming Signal）。
4. 以二元指數退後演算法設定一段時間（r 51.2 ms）後再重新開始步驟 1。

103. 資料遺失與重傳 設定重傳的 timeout 時間
這是收一筆message就送回一個ack的範例. 若 timeout 仍未收到 ack, 便會重傳.
然而TCP不見得要收到上一個ack才能送第二個message. 因為常常一端會大量送資料, 另一端可能都沒有甚麼資料要送, 就會是 send many message, 然後才 1 個 ack message.