8051 ch9-950217

1
Chapter 9
8051 Timer Programming in
Assembly and C

2
Objective
• Embedded Systems with Timer/Counter lights, clock
stopwatch (count up), timer (count down) counters,
pedometers, treadmills and other exercise equipment.
8051 2 Timer/Counter mechanism, in this chapter, we
want to know: what is the Timer, what is the Counter.
How to use a Timer. How to use

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
CPU
On-chip
RAM
On-chip
ROM for
program
code
4 I/O Ports
Timer 0
Serial
Port
Figure 1-2. Inside the 8051 Microcontroller Block Diagram
OSC
Interrupt
Control
External interrupts
Timer 1
Timer/Counter
Bus
Control
TxD RxDP0 P1 P2 P3
Address/Data
Counter
Inputs

6
Timers /Counters
• The 8051 has 2 timers/counters: timer/counter 0 and
timer/counter 1. They can be used as
1. The timer is used as a time delay generator.
– The clock source is the internal crystal frequency of the
8051.
1. 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.
to
LCD
P1
8051
TL0
TH0
Set
Timer 0
XTAL
oscillator
÷ 12 Timer

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
T0
to
LCD
P3.4
P1
8051
a switch
TL0
TH0
Vcc

9
Figure 4-1. 8051 Pin Diagram
PDIP/Cerdip
1
2
3
4
5
6
7
8
9
10
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)

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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
TH0 TL0
TF0
1. monitor by JNB
2. interrupt
timer input
hardware control
1:start
0:stop
counter input
Sec 9.2 

11
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
TH1 TL1
TF1
1. monitor by JNB
2. interrupt
1:start
0:stop
timer input
counter input
hardware control

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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.

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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

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Timer Registers
D15 D8D9D10D11D12D13D14 D7 D0D1D2D3D4D5D6
TH0 TL0
TH1 TL1
Timer 0
Timer 1

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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.
• TRx=0: off (stop)
• TRx=1: on (start)
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt
(MSB) (LSB)

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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, 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 1 Timer0 for Interrupt
(MSB) (LSB)

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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
TF1 IT0IE0IT1IE1TR0TF0TR1
TCON: Timer/Counter Control Register

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

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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
Timer 1 Timer 0
(MSB) (LSB)

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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

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Gate
• Every timer has a mean of starting and stopping.
– GATE=0
• 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.

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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

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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
Timer 1 Timer 0
(MSB) (LSB)

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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) 11.0592 MHz
Solution:
(a) 1/12 × 12 MHz = 1 MHz and T = 1/1 MHz = 1
µs
(b) 1/12 × 16 MHz = 1.333 MHz and
T = 1/1.333 MHz = 0.75 µs
(c) 1/12 × 11.0592 MHz = 921.6 KHz;
T = 1/921.6 KHz = 1.085 µs
XTAL
oscillator
÷ 12

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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.
Timer 1 Timer 0
(MSB) (LSB)

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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.

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

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

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5. The 8051 starts to count up by incrementing the TH0-
TL0.
– TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000H
FFFC FFFD FFFE FFFF 0000
TF0=0 TF0=0 TF0=0 TF0=0 TF0=1
TH0 TL0Start timer
Stop timer
Monitor TF01 until TF0=1
TR0=1 TR0=0
TF0
Roll over

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6. When TH0-TL0 rolls over from FFFFH to 0000,
the 8051 set TF0=1.
– TH0-TL0= FFFEH, FFFFH, 0000H (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

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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 µs, then the delay = 4.34 µs
• Figure 9-4 show a formula for delay calculations
using mode 1 of the timer for a crystal frequency of
XTAL=11.0592 MHz.
• Examples 9-4 to 9-9 show how to calculations the
delay generated by timer.

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

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Find Timer Values
• Assume XTAL = 11.0592 MHz .
• How to find the inter values needed for the TH, TL to get
delay of 20 ms?
– Divide the desired time delay by 1.085 µs.
20ms ÷ 1.085 µs = 18433
– Perform 65536 –n, where n is the decimal value we got in Step 1.
65536-18433=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: Delay of 200 ms Then time delay in µs 200,000÷1.085=184332.
65536-184332=47203=B863H, TH=B8H, TL=63H

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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
50% 50%
whole clock
P1.5

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Example 9-4 (2/4)
;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

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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.

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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).

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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 µs (microsecond)
The number of counts = FFFFH – FFF2H +1 = 14 (decimal).
The delay = number of counts × 1.085 µs = 14 × 1.085 µs = 15.19
µs for half the clock.
For the entire period of a clock, it is T = 2 × 15.19 µs = 30.38 µs
as the time delay generated by the timer.

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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.

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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 µs = 60.76 µs, and F = 16458.196 Hz.

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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 TR0 ;start the timer 0
AGAIN:JNB TF0,AGAIN
CLR TR0
CLR TF0
CLR P2.3
P2.3

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Example 9-7 (2/2)
Solution:
(a) or (b) is OK.
(a) (FFFF – B83E + 1) = 47C2H = 18370 in decimal and 18370 ×
1.085 µs = 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 µs in
duration. Therefore, we have 18370 × 1.085 µs = 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:
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 65536 states. Therefore, we
have delay = (65536 – 0) × 1.085 µs = 71.1065 ms.

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Example 9-8 (2/2)
CLR P2.3 ;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 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

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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
AGAIN:MOV TL1,#34H ;timer value=3476H
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

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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 µs = 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 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 = 11.0592 MHz ⇒ MC = 1.085 µs.
5 ms / 1.085 µs = 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 P2.3
5ms

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.
(a) The period of square wave = 1 / frequency
= 1 / 2 kHz = 500 µs.
(b) The half period = 500 µs /2 = 250 µs.
(c) 250 µs / 1.085 µs = 230
65536 – 230 = 65360 = FF1AH.
(d) TL1 = 1AH and TH1 = FFH

50
Example 9-11 (2/2) ×
AGAIN:MOV TL1,#1AH ;timer value = FF1AH
MOV TH1,#0FFH
SETB TR1 ;start
BACK: JNB TF1,BACK
CLR TR1 ;stop
CPL P1.5
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 µs = 9216
65536 – 9216 = 56320 in decimal = DC00H in hex.
(d) TL1 = 00H and TH1 = DCH.

52
Example 9-12 (2/2)×
AGAIN: MOV TL1,#00 ;Timer value = DC00H
MOV TH1,#0DCH
SETB TR1 ;start
BACK: JNB TF1,BACK
CLR TR1 ;stop
CPL P2.3
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 µs = 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-1=1FFFH
• When the timer reaches its maximum of 1FFFH, it
rolls over to 0000, and TF0 is raised.
TH0 TL0
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

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Timer Mode 2
• 8-bit timer
– 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
XTAL
oscillator ÷ 12
TR
TL
TH
TF
TF goes high
when FF 0
overflow
flag
C/T = 0
reload

58
1. Chose mode 2 timer 0
– MOV TMOD,#02H
1. Set the original value to TH0.
– MOV TH0,#FCH
1. Clear the flag to TF0=0.
– CLR TF0
1. Start the timer.
– SETB TR0
– Note that the instruction SETB TR0 dose not load TH0 to
TL0. So TL0 still is 00H.

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

60
6. 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)
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 TR1=0)
BACK:JNB TF1,BACK
CPL P1.0
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 µs = 272.33 µs
Total period = 2 × 272.33 µs = 544.67 µs
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 µs = 555.52 µs
Frequency = 1.8kHz.

63
Example 9-15 ×
Find the frequency of a square wave generated on pin P1.0.
Solution:
MOV TH0,#0
AGAIN:MOV R5,#250 ;count 250 times
ACALL DELAY
CPL P1.0
SJMP AGAIN
DELAY:SETB 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 µs) = 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 ⇒ 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 × 1.085 µs = 162 µs.
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 µs + 162.25 µs = 488.25 µs
Frequency = 2.048 kHz.

67
Example 9-17 (2/2)
MOV TH0,#-150 ;Count=150
AGAIN:SETB P1.3
ACALL DELAY
ACALL DELAY
CLR P1.3
ACALL DEALY
SJMP AGAIN
DELAY:SETB TR0 ;start
BACK: JNB TF0,BACK
CLR TR0 ;stop
CLR TF0 ;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)
T0
to
LCD
P3.4
P1
8051
a switch
TL0
TH0
Vcc
Counter 0


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.
T1
to
LCD
P3.5
P1
8051
a switch
TL1
TH1
Vcc
Counter 1

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
GATE C/T=1 M1 M0 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 0
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 0
overflow
flag
C/T = 1

75
Counter Mode 2
• 8-bit counter.
– 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 0
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 0
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.
P2 is connected to 8 LEDs
and input T1 to pulse.
T1
to
LEDs
P3.5
P2
8051

79
Example 9-18 (2/2)
MOV TMOD,#01100000B ;mode 2, counter 1
MOV TH1,#0
SETB P3.5 ;make T1 input port
AGAIN:SETB TR1 ;start
BACK: MOV A,TL1
MOV P2,A ;display in P2
JNB TF1,BACK ;overflow
CLR TR1 ;stop
CLR TF1 ;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,#00000110B ;Counter 0,mode2
MOV TH0,#C4H ;C4H=-60
SETB P3.4 ;make T0 as input
AGAIN:SETB TR0 ;starts the counter
BACK: MOV A,TL0 ; every 60 events
ACALL CONV ;convert in R2,R3,R4
JNB TF0,BACK ;loop if TF0=0
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 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.

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
1.085µs ×51968 ≈ 56.384ms
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=11.0592MHz 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
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) {
if (c==0) { TL0=0x67; TH0=0xFC };
// FFFFH-FC67H+1=921, about 999.285µs, 500 Hz
else { TL0=0x9A; TH0=0xFD };
// FFFFH-FD9AH+1=614, about 666.19µs, 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) {
mybit=~mybit;
T0Delay(); }}

95
Example 9-21 (2/3)
(a) Assume XTML=11.0592MHz for the AT89C51
FFFFH-4BFDH+1=B403H=46083
1.085µs ×46083 ≈ 50ms
void T0Delay() {
TL0=0xFD; TH0=0x4B; //initial value
TR0=1; //turn on T0
while (TF0==0); //BACK: JNZ TF0,BACK

96
Example 9-21 (3/3)
(b) Assume XTML=11.0592MHz for the DS89C4x0
FFFFH-4BFDH+1=B403H=46083
1.085µs ×46083 ≈ 50ms
void T0Delay() {
TL0=0xFD; TH0=0x4B; //initial value
TR0=1; //turn on T0
while (TF0==0); //BACK: JNZ TF0,BACK

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.
T1
P3.5
P1
1Hz
TL1
TH1
LEDs

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
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 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
5 7 2 1 0 4 3
dial tone
start
timer
short
timer
long
timer

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 µs ），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
µs ）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

8051 ch9-950217

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