Chapter 9 8051 Timer Programming in Assembly and C

1
Chapter 9 8051 Timer Programming in Assembly and
C
2
Objective

• Embedded Systems with Timer/Counter
• ???????
• ??(count up)????(count down)
• ?????????????????
• 8051 ? 2 ? Timer/Counter ???,??????????
• ??? Timer,??? Counter?
• ????? Timer?
• ????? Counter?

3
Sections

• 9.1 Programming 8051 timers
• 9.2 Counter programming
• 9.3 Programming timers 0 and 1 in 8051 C

4
Section 9.1Programming 8051 Timers
5
Inside Architecture of 8051
External interrupts
On-chip ROM for program code
Timer/Counter
Interrupt Control
Timer 1
On-chip RAM
Counter Inputs
Timer 0
CPU
Serial Port
Bus Control
4 I/O Ports
OSC
TxD RxD
P0 P1 P2 P3
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 (x0 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
PDIP/Cerdip
10
Figure 9-8 Timer/Counter 0
timer input
TH0
TL0
counter input
1start 0stop
TF0
1. monitor by JNB 2. interrupt
hardware control
Sec 9.2 ?
11
Figure 9-9 Timer/Counter 1
timer input
TH1
TL1
counter input
1start 0stop
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 607-611) 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
Timer 0
Timer 1
15
TCON Register (1/2)

• Timer control register TCON
• 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.
• TRx0 off (stop)
• TRx1 on (start)

16
TCON Register (2/2)

• TF (timer flag, control flag)
• TF0 for timer/counter 0 TF1 for timer/counter 1.
• TFx is like a carry. Originally, TFx0. When
  TH-TL roll over to 0000 from FFFFH, the TFx is
  set to 1.
• TFx0 not reach
• TFx1 reach
• If we enable interrupt, TFx1 will trigger ISR.

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
18
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)
• Set upper 4 bits for Timer 1 (Set to 0000 if
  not used)
• Not bit-addressable

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).
• M1 Mode bit 1
• M0 Mode bit 0

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.
• GATE0
• Internal control
• The start and stop of the timer are controlled by
  software.
• Set/clear the TR0 (or TR1) for start/stop timer.
• GATE1
• 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.

22
M1, M0

• M0 and M1 select the timer mode for timers 0 1.
• 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 (x 0 or 1)
• 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

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 00000001, mode 1 of timer 0 is
  selected.
• (b) TMOD 00100000, mode 2 of timer 1 is
  selected.
• (c) TMOD 00010010
• mode 2 of timer 0, and mode 1 of timer
  1 are selected.

timer 1 timer 0
24
Example 9-2 ?

• Find the timers clock frequency and its period
  for various 8051-
• based systems, with the following crystal
  frequencies.
• (a) 12 MHz (b) 16 MHz
  (c) 11.0592 MHz
• Solution

• (a) 1/12 12 MHz 1 MHz and T 1/1 MHz 1 ?s
• (b) 1/12 16 MHz 1.333 MHz and
• T 1/1.333 MHz 0.75 ?s
• (c) 1/12 11.0592 MHz 921.6 KHz
• T 1/921.6 KHz 1.085 ?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 0000 0010 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.

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
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
  TF00.
• 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

TR01
TR00
TH0
TL0
Start timer
Stop timer
Roll over
TF00
TF00
TF00
TF00
TF01
Monitor TF01 until TF01
TF0
30
Steps of Mode 1 (3/3)

• When TH0-TL0 rolls over from FFFFH to 0000, the
  8051 set TF01.
• TH0-TL0 FFFEH, FFFFH, 0000H (Now TF01)
• 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

31
Initial Count Values

• The initial count value FFFC.
• The number of counts FFFFH-FFFCH1 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 MC1.085 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 XTAL11.0592 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 1.085 ?s.
• 20ms 1.085 ms 18433
• Perform 65536 n, where n is the decimal value we
  got in Step 1.
• 65536-1843347103B7FFH
• Convert the result of Step 2 to hex, where yyxx
  is the initial hex value to be loaded into the
  timers registers.
• Set TH yy and TL xx.
• THB7H, TLFFH
• Example 9-10

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 TR0 start the timer 0
• AGAINJNB TF0,AGAIN
• CLR TR0 stop timer 0
• CLR TF0 clear timer 0 flag
• RET

36
Example 9-4 (3/4)

• Solution
• 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.
• 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 11.0592
• MHz.
• Solution
• The timer works with the internal system clock.
• frequency of internal system clock
  11.0592/12 921.6 KHz
• machine cycle 1 /921.6 KHz 1.085 ?s
  (microsecond)
• The number of counts FFFFH FFF2H 1 14
  (decimal).
• The delay number of counts 1.085 ?s 14
  1.085 ?s 15.19 ?s for half the clock.
• For the entire period of a clock, it is T 2
  15.19 ?s 30.38 ?s as the time delay generated
  by the timer.

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 2
• SJMP HERE 2
• ----------delay using timer 0
• DELAY
• SETB TR0 1
• AGAIN JNB TF0,AGAIN 14
• CLR TR0 1
• CLR TF0 1
• RET 2 0
• 
  Total 28
• T 2 28 1.085 ?s 60.76 ?s, and F
  16458.196 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 valueB83EH
• MOV TH0,0B8H
• SETB P2.3
• SETB TR0 start the timer 0
• AGAINJNB TF0,AGAIN
• CLR TR0
• CLR TF0
• CLR P2.3

P2.3
42
Example 9-7 (2/2)

• Solution
• (a) or (b) is OK.
• (a) (FFFF B83E 1) 47C2H 18370 in decimal
  and 18370
• 1.085 ?s 19.93145 ms.
• (b) Since TH-TL B83EH 47166 (in decimal ) we
  have
• 65536 47166 18370. This means that the
  timer counts from
• B83EH to FFFF. This plus rolling over to 0
  goes through a
• total of 18370 clock cycles, where each
  clock is 1.085 ?s in
• duration. Therefore, we have 18370 1.085
  ?s 19.93145 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
• TH0TL00 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 65536
  states. Therefore, we have delay (65536 0)
  1.085 ?s 71.1065 ms.

44
Example 9-8 (2/2)

• CLR P2.3 clear P2.3
• MOV TMOD,01 Timer 0,mode1(16-bit)
• HERE MOV TL0,0 TL00, the low byte
• MOV TH0,0 TH00, the high byte
• SETB P2.3 SET high P2.3
• SETB TR0 start timer 0
• AGAIN JNB TF0,AGAIN monitor timer Flag 0
• CLR TR0 stop timer 0
• CLR TF0 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 11.0592 MHz. In your
  calculation do
• not include the overhead due to instructions in
  the loop.
• MOV TMOD,10H timer 1, mode 1
• AGAINMOV TL1,34H timer value3476H
• MOV TH1,76H
• SETB TR1 start
• BACK JNB TF1,BACK
• CLR TR1 stop
• CPL P1.5 next half clock
• CLR TF1 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 35276 clock count
• Half period 35276 1.085 ?s 38.274 ms
• Whole period 2 38.274 ms 76.548 ms
• Frequency 1/ 76.548 ms 13.064 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 timers
  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 11.0592 MHz ? MC 1.085 ?s.
• 5 ms / 1.085 ?s 4608 MCs.
• To achieve that we need to load into TL0 and TH0
  the value 65536 4608 60928 EE00H.
• Therefore, we have TH0 EE and TL0 00.

48
Example 9-10 (2/2) ?

• CLR P2.3
• MOV TMOD,01 Timer 0,mode 1
• HERE MOV TL0,0
• MOV TH0,0EEH
• SETB P2.3 SET high P2.3
• SETB TR0 start
• AGAIN JNB TF0,AGAIN
• CLR TR0 stop
• CLR TF0
• CLR P2.3

5ms
P2.3
49
Example 9-11 (1/2) ?

• Assuming that XTAL 11.0592 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 / 1.085 ?s 230
• 65536 230 65360 FF1AH.
• (d) TL1 1AH and TH1 FFH

50
Example 9-11 (2/2) ?

• MOV TMOD,10H timer 1, mode 1
• AGAINMOV TL1,1AH timer value FF1AH
• MOV TH1,0FFH
• SETB TR1 start
• BACK JNB TF1,BACK
• CLR TR1 stop
• CPL P1.5
• CLR TF1 clear timer flag 1
• SJMP AGAIN

51
Example 9-12 (1/2) ?

• Assuming XTAL 11.0592 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 / 1.085 ?s 9216
• 65536 9216 56320 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 TR1 start
• BACK JNB TF1,BACK
• CLR TR1 stop
• CPL P2.3
• CLR TF1 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 65272.
• One of the timer delay 65272 1.085 ?s
  70.820 ms
• Total delay 200 70.820 ms 14.164024 seconds

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-11FFFH
• When the timer reaches its maximum of 1FFFH, it
  rolls over to 0000, and TF0 is raised.

? ? ?
Timer 0
55
Timer Mode 3

• For Timer/Counter 0
• 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
• TL0 is incremented (or roll over) continuously
  when TR01.
• Auto-reloading
• TH0 is loaded into TL0 automatically when
  TL000H.
• 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
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 TF00.
• CLR TF0
• Start the timer.
• SETB TR0
• Note that the instruction SETB TR0 dose not load
  TH0 to TL0. So TL0 still is 00H.

59
Steps of Mode 2 (2/3)

• The 8051 starts to count up by incrementing the
  TL0.
• TL0 ..., FCH,FDH,FEH,FFH,FCH

TF010
TH0FCH
TL0FCH
TR01
auto reload
Clear TF0
Start timer
...
roll over
00H
TF0 0
TF0 0
TF0 0
TF0 0
TF0 1
TF0 1
TF0 0
TH0FCH
roll over
00H
TL000H
TF0 0
TF0 1
TF0 1
TF0 0
auto reload TL-FCH immediately
60
Steps of Mode 2 (3/3)

• When TL0 rolls over from FFH to 00, the 8051 set
  TF01. Also, TL0 is reloaded automatically with
  the value kept by the TH0.
• TL0 FCH, FDH, FEH, FFH, FCH(Now TF01)
• The 8051 auto reload TL0TH0FCH.
• 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)

• Assuming that XTAL 11.0592 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 TR1 start (no stop TR10)
• BACKJNB TF1,BACK
• CPL P1.0
• CLR TF1 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) 1.085 ?s
  272.33 ?s
• Total period 2 272.33 ?s 544.67 ?s
• Frequency 1.83597 kHz.
• (b) To get the smallest frequency, we need the
  largest period and that is achieved when TH1
  00.
• Total period 2 256 1.085 ?s 555.52 ?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
• AGAINMOV R5,250 count 250 times
• ACALL DELAY
• CPL P1.0
• SJMP AGAIN
• DELAYSETB TR0 start
• BACK JNB TF0,BACK
• CLR TR0 stop
• CLR TF0 clear TF
• DJNZ R5,DELAY timer 2 auto-reload
• RET
• T 2 (250 256 1.085 ?s) 138.88 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 1 256-200 56 38H
• Way 2 -200 -C8H ? 2s 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 ? 2s complement of 200 100H C8H
  38 H

Decimal 2s 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 1.085 ?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
• 325.5 ?s 162.25 ?s 488.25 ?s
• Frequency 2.048 kHz.

67
Example 9-17 (2/2)

• MOV TMOD,2H Timer 0,mode 2
• MOV TH0,-150 Count150
• AGAINSETB P1.3
• ACALL DELAY
• ACALL DELAY
• CLR P1.3
• ACALL DEALY
• SJMP AGAIN
• DELAYSETB TR0 start
• BACK JNB TF0,BACK
• CLR TR0 stop
• CLR TF0 clear TF
• RET

high period
low period
68
Section 9.2Counter Programming
69
Counter (1/2)

• These timers can also be used as counters
  counting events happening outside the 8051 by
  setting C/T1.
• 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
P3.4
Vcc
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
P3.5
Vcc
T1
a switch
71
Table 9-1 Port 3 Pins Used For Timers 0 and 1
Pin Port Pin Function Description
14 P3.4 T0 Timer/Counter 0 external input
15 P3.5 T1 Timer/Counter 1 external input
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)
74
Figure 9-5. (b) Counter 1 with External Input
(Mode 1)
75
Counter Mode 2

• 8-bit counter.
• TL0 is incremented if TR01 and external pulse
  occurs.
• Auto-reloading
• TH0 is loaded into TL0 when TF01.
• 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)
77
Figure 9.7 Counter 1 with External Input (Mode 2)
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.
  

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 P3.5 make T1 input port
• AGAINSETB TR1 start
• BACK MOV A,TL1
• MOV P2,A display in P2
• JNB TF1,BACK overflow
• CLR TR1 stop
• CLR TF1 make TF0
• 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, TL00. To
  solve this problem, load TH0 with -60 at the
  beginning of the program.

81
Example 9-19 (2/3)

• ACALL LCD_SET_UP initialize the LCD
• MOV TMOD,00000110B Counter 0,mode2
• MOV TH0,C4H C4H-60
• SETB P3.4 make T0 as input
• AGAINSETB TR0 starts the counter
• BACK MOV A,TL0 every 60 events
• ACALL CONV convert in R2,R3,R4
• JNB TF0,BACK loop if TF00
• CLR TR0 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
• DIV AB
• ORL A,30H make it ASCII
• MOV R4,A
• MOV A,B
• ORL A,30H
• MOV R3,A
• MOV A,R2
• ORL A,30H
• MOV R2,A ACALL LCD_DISPLAY here
• RET

R4
R3
R2
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 microcontrollers
  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
GATE1 in TMOD

• All discuss so far has assumed that GATE0.
• The timer is stared with instructions SETB TR0
  and SETB TR1 for timers 0 and 1, respectively.
• If GATE1, 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 GATE1

• 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.

86
Section 9.3Programming 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 ltreg51.hgt
• void T0Delay(void)
• void main(void)
• while(1) //repeat forever
• P10x55
• T0Delay() //time delay
• P10xAA
• T0Delay()

89
Example 9-20 (2/2)

• Assume XTML11.0592MHz for the AT89C51
• FFFFH-3500H1CB00H51968
• 1.085ms ?51968 ? 56.384ms
• void T0Delay()
• TMOD0x01 //Timer 0, Mode 1
• TL00x00 TH00x35 //initial value
• TR01 //turn on T0
• while (TF00) //text TF to roll over
• TR00 //turn off T0
• TF00 //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.
• SW0 500 Hz SW1 750 Hz. Using Timer 0, mode
  1.
• Assume XTML11.0592MHz for the AT89C51
• Solution
• include ltreg51.hgt
• sbit mybitP15
• sbit SWP17
• void T0Delay(void)

91
Example 9-25 (2/3)

• void main(void)
• SW1 //make P1.7 an input pin
• while(1) //repeat forever
• mybitmybit //toggle
• if (SW0)
• T0Delay(0) //500Hz time delay
• else
• T0Delay(1) //750Hz time delay

92
Example 9-25 (3/3)

• void T0Delay(unsigned char c)
• TMOD0x01 //Timer 0, Mode 1
• if (c0) TL00x67 TH00xFC
• // FFFFH-FC67H1921, about 999.285ms, 500 Hz
• else TL00x9A TH00xFD
• // FFFFH-FD9AH1614, about 666.19ms, 750 Hz
• TR01
• while (TF0)
• TR00
• TF00

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 ltreg51.hgt
• void T0Delay(void)
• sbit mybitP15
• void main(void)
• while(1) //repeat forever
• mybitmybit
• T0Delay()

95
Example 9-21 (2/3)

• (a) Assume XTML11.0592MHz for the AT89C51
• FFFFH-4BFDH1B403H46083
• 1.085ms ?46083 ? 50ms
• void T0Delay()
• TMOD0x01 //Timer 0, Mode 1
• TL00xFD TH00x4B //initial value
• TR01 //turn on T0
• while (TF00) //BACK JNZ TF0,BACK
• TR00 //turn off T0
• TF00 //clear TF0

96
Example 9-21 (3/3)

• (b) Assume XTML11.0592MHz for the DS89C4x0
• FFFFH-4BFDH1B403H46083
• 1.085ms ?46083 ? 50ms
• void T0Delay()
• TMOD0x01 //Timer 0, Mode 1
• TL00xFD TH00x4B //initial value
• TR01 //turn on T0
• while (TF00) //BACK JNZ TF0,BACK
• TR00 //turn off T0
• TF00 //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
T1
1Hz
99
Example 9-26 (2/2)

• include ltreg51.hgt
• sbit T1P35
• void main(void)
• T11 //make T1 an input pin
• TMOD0x60 //counter 1, mode 2
• TH10 //reload value
• while(1) //repeat forever
• do TR11 P1TL1 while (TF10)
• TR10 TF10 //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 digit map (11x080xxxxxx03xxxxxxx002x.)
• 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
5
7
2
1
0
4
3
start timer
short timer
long timer
dial tone
102
CSMA/CD ????

• 1. ?????????????????frame??,??????????
• 2. ?????????frame,???? frame?????????,??????,??fr
  ame???
• 3. ??????(51.2 ms),???frame???(collision),???????(
  Jamming Signal)?
• 4. ????????????????(r ?51.2 ms)???????? 1?

103
???????
????? timeout ??