Microcontroller 8051 Timer Counter Interrrupt

hsabaghianb @ kashanu.ac.ir
hsabaghianb @ kashanu.ac.ir Microprocessors
Microprocessors 1-
1-1
1
8051
timer/counter

Microprocessors 1-
1-2
2
Timers /Counters Programming
 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.
2. 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.

Microprocessors 1-
1-3
3

Microprocessors 1-
1-4
4
Timer
 Set the initial value of registers
 Start the timer and then the 8051 counts up.
 Input from internal system clock (machine
cycle)
 When the registers equal to 0 and the 8051
sets a bit to denote time out
to
LCD
P1
8051
TL0
TH0
P2
Set
Timer 0

Microprocessors 1-
1-5
5
Counter
 Count the number of events
 Show the number of events on registers
 External input from T0 input pin (P3.4) for Counter 0
 External input from T1 input pin (P3.5) for Counter 1
 External input from Tx input pin.
 We use Tx to denote T0 or T1.
T0
to
LCD
P3.4
P1
8051
a switch
TL0
TH0

Microprocessors 1-
1-6
6
Registers Used in Timer/Counter
TH0, TL0, TH1, TL1
TMOD (Timer mode register)
TCON (Timer control register)
You can see Appendix H (pages 413-415) 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.

Microprocessors 1-
1-7
7
Basic Registers of the Timer
Both timer 0 and timer 1 are 16 bits wide.
These registers stores
the time delay as a timer
the number of events as a counter
Timer 0: TH0 & TL0
Timer 0 high byte, timer 0 low byte
Timer 1: TH1 & TL1
Timer 1 high byte, timer 1 low byte
Each 16-bit timer can be accessed as two
separate registers of low byte and high byte.

Microprocessors 1-
1-8
8
Timer Registers
D15 D8
D9
D10
D11
D12
D13
D14 D7 D0
D1
D2
D3
D4
D5
D6
TH0 TL0
D15 D8
D9
D10
D11
D12
D13
D14 D7 D0
D1
D2
D3
D4
D5
D6
TH1 TL1
Timer 0
Timer 1

Microprocessors 1-
1-9
9
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)

Microprocessors 1-
1-10
10
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
GATE C/T M1 M0 GATE C/T M1 M0
Timer 1 Timer 0
(MSB) (LSB)

Microprocessors 1-
1-11
11
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

Microprocessors 1-
1-12
12
Gate
 Every timer has a mean of starting and stopping.
 GATE=0
Internal control
The start and stop of the timer are controlled by way of
software.
Set/clear the TR for start/stop timer.
 GATE=1
External control
The hardware way of starting and stopping the timer by
software and an external source.
Timer/counter is enabled only while the INT pin is high and
the TR control pin is set (TR).

Microprocessors 1-
1-13
13
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
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

Microprocessors 1-
1-14
14
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

Microprocessors 1-
1-15
15
Timer modes

Microprocessors 1-
1-16
16
TCON Register (1/2)
Timer control register: TMOD
Upper nibble for timer/counter, lower nibble for
interrupts
TR (run control bit)
TR0 for Timer/counter 0; TR1 for Timer/counter 1.
TR is set by programmer to turn timer/counter on/off.
TR=0: off (stop)
TR=1: on (start)
TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0
Timer 1 Timer0 for Interrupt
(MSB) (LSB)

Microprocessors 1-
1-17
17
TCON Register (2/2)
TF (timer flag, control flag)
TF0 for timer/counter 0; TF1 for timer/counter 1.
TF is like a carry. Originally, TF=0. When TH-TL roll
over to 0000 from FFFFH, the TF is set to 1.
TF=0 : not reach
TF=1: reach
If we enable interrupt, TF=1 will trigger ISR.
(MSB) (LSB)

Microprocessors 1-
1-18
18
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 IT0
IE0
IT1
IE1
TR0
TF0
TR1
TCON: Timer/Counter Control Register

Microprocessors 1-
1-19
19
Timer Mode 1
 In following, we all use timer 0 as an example.
 16-bit timer (TH0 and TL0)
 TH0-TL0 is incremented continuously when TR0 is set
to 1. And the 8051 stops to increment TH0-TL0 when
TR0 is cleared.
 The timer works with the internal system clock. In
other words, the timer counts up each machine cycle.
 When the timer (TH0-TL0) reaches its maximum of
FFFFH, it rolls over to 0000, and TF0 is raised.
 Programmer should check TF0 and stop the timer 0.

Microprocessors 1-
1-20
20
Steps of Mode 1 (1/3)
1. Choose mode 1 timer 0
 MOV TMOD,#01H
2. Set the original value to TH0 and TL0.
 MOV TH0,#FFH
 MOV TL0,#FCH
3. You had better to clear the flag to monitor:
TF0=0.
 CLR TF0
4. Start the timer.
 SETB TR0

Microprocessors 1-
1-21
21
5.The 8051 starts to count up by incrementing the
TH0-TL0.
TH0-TL0= FFFCH,FFFDH,FFFEH,FFFFH,0000H
FFFC FFFD FFFE FFFF 0000
TF = 0 TF = 0 TF = 0 TF = 0 TF = 1
TH0 TL0
Start timer
Stop timer
Monitor TF until TF=1
TR0=1 TR0=0
TF

Microprocessors 1-
1-22
22
6. When TH0-TL0 rolls over from FFFFH to 0000,
the 8051 set TF0=1.
TH0-TL0= FFFEH, FFFFH, 0000H (Now TF0=1)
7. Keep monitoring the timer flag (TF) to see if it is
raised.
AGAIN: JNB TF0, AGAIN
8. Clear TR0 to stop the process.
CLR TR0
9. Clear the TF flag for the next round.
CLR TF0

Microprocessors 1-
1-23
23
Mode 1 Programming
XTAL
oscillator ÷
12
TR
TH TL TF
Timer
overflow
flag
C/T = 0
TF goes high when FFFF 0

Microprocessors 1-
1-24
24
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

Microprocessors 1-
1-25
25
Example 9-4 (1/3)
 square wave of 50% duty on P1.5
 Timer 0 is used
;each loop is a half clock
MOV TMOD,#01 ;Timer 0,mode 1(16-bit)
HERE: MOV TL0,#0F2H ;Timer value = FFF2H
MOV TH0,#0FFH
CPL P1.5
ACALL DELAY
SJMP HERE
50% 50%
whole clock
P1.5

Microprocessors 1-
1-26
26
Example 9-4 (2/3)
;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

Microprocessors 1-
1-27
27
Example 9-4 (3/3)
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.
7. Timer 0 is stopped by the instruction “CLR TR0”. The DELAY subroutine
ends, and the process is repeated.
Notice that to repeat the process, we must reload the TL and TH
registers, and start the timer again (in the main program).

Microprocessors 1-
1-28
28
Example 9-9 (1/2)
 This program generates a square wave on pin P1.5 Using timer 1
 Find the frequency.(dont include the overhead of instruction delay)
 XTAL = 11.0592 MHz
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

Microprocessors 1-
1-29
29
Example 9-9 (2/2)
Solution:
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.
Note
Mode 1 is not auto reload then the program must reload
the TH1, TL1 register every timer overflow if we want to
have a continuous wave.

Microprocessors 1-
1-30
30
Find Timer Values
 Assume that XTAL = 11.0592 MHz .
 And we know desired delay
 how to find the values for the TH,TL ?
1. Divide the delay by 1.085 s and get n.
2. Perform 65536 –n
3. Convert the result of Step 2 to hex (yyxx )
4. Set TH = yy and TL = xx.

Microprocessors 1-
1-31
31
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:
1. The period of the square wave = 1 / 50 Hz = 20 ms.
2. The high or low portion of the square wave = 10 ms.
3. 10 ms / 1.085 s = 9216
4. 65536 – 9216 = 56320 in decimal = DC00H in hex.
5. TL1 = 00H and TH1 = DCH.

Microprocessors 1-
1-32
32
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

Microprocessors 1-
1-33
33
Generate a Large Time Delay
 The size of the time delay depends on two factors:
They crystal frequency
The timer’s 16-bit register, TH & TL
 The largest time delay is achieved by making TH=TL=0.
 What if that is not enough?
 Next Example show how to achieve large time delay

Microprocessors 1-
1-34
34
Example 9-13
Examine the following program and find the time delay in seconds.
Exclude the overhead due to the instructions in the loop.
MOV TMOD,#10H
MOV R3,#200
AGAIN: MOV TL1,#08
MOV TH1,#01
SETB TR1
BACK: JNB TF1,BACK
CLR TR1
CLR TF1
DJNZ R3,AGAIN
Solution:
TH – TL = 0108H = 264 in decimal
65536 – 264 = 65272.
One of the timer delay = 65272 × 1.085 s = 70.820 ms
Total delay = 200 × 70.820 ms = 14.164024 seconds

Microprocessors 1-
1-35
35
Timer Mode 0
Mode 0 is exactly like mode 1 except that it is a
13-bit timer instead of 16-bit.
8-bit TH0
5-bit TL0
The counter can hold values between 0000 to
1FFF in TH0-TL0.
213
-1= 2000H-1=1FFFH
We set the initial values TH0-TL0 to count up.
When the timer reaches its maximum of 1FFFH, it
rolls over to 0000, and TF0 is raised.

Microprocessors 1-
1-36
36
Timer Mode 2
8-bit timer.
 It allows only values of 00 to FFH to be loaded into TH0.
Auto-reloading
TL0 is incremented continuously when TR0=1.
next example: 200 MCs delay on timer 0.
See Examples 9-14 to 9-16

Microprocessors 1-
1-37
37
1. Chose mode 2 timer 0
MOV TMOD,#02H
2. Set the original value to TH0.
MOV TH0,#38H
3. Clear the flag to TF0=0.
CLR TF0
4. After TH0 is loaded with the 8-bit value,
the 8051 gives a copy of it to TL0.
TL0=TH0=38H
5. Start the timer.
SETB TR0

Microprocessors 1-
1-38
38
6. The 8051 starts to count up by incrementing the TL0.
 TL0= 38H, 39H, 3AH,....
7. When TL0 rolls over from FFH to 00, the 8051 set
TF0=1. Also, TL0 is reloaded automatically with the
value kept by the TH0.
 TL0= FEH, FFH, 00H (Now TF0=1)
 The 8051 auto reload TL0=TH0=38H.
 Clr TF0
 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.
 Clr TR0

Microprocessors 1-
1-39
39
Timer 1 Mode 2 with internal Input
XTAL
oscillator ÷
12
TR1
TL1
TH1
TF1
overflow flag
reload
TF goes high when FF 0
C/T = 0

Microprocessors 1-
1-40
40
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 TR0 ;start
BACK: JNB TF0,BACK ;wait until TL0 ovrflw auto-reload
CLR TR0 ;stop
CLR TF0 ;clear TF
DJNZ R5,DELAY
RET
T = 2 (250 × 256 × 1.085 s) = 138.88 ms, and frequency = 72 Hz.

Microprocessors 1-
1-41
41
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

Microprocessors 1-
1-42
42
Example 9-17 (1/2)
Find
(a) the frequency of the square wave generated in the following code
(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 is twice tat of the low portion (66% duty cycle).
The total period = high portion + low portion
T= 325.5 s + 162.25 s = 488.25 s
Frequency = 2.048 kHz.

Microprocessors 1-
1-43
43
Example 9-17 (2/2)
MOV TMOD,#2H ;Timer 0,mode 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

Microprocessors 1-
1-44
44
Counter
These timers can also be used as counters
counting events happening outside the
8051.
When the timer is used as a counter, it is a
pulse outside of the 8051 that increments
the TH, TL.
When C/T=1, the counter counts up as
pulses are fed from
T0: timer 0 input (Pin 14, P3.4)
T1: timer 1 input (Pin 15, P3.5)

Microprocessors 1-
1-45
45
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)

Microprocessors 1-
1-46
46
Timer/Counter selection

Microprocessors 1-
1-47
47
Counter Mode 1
 16-bit counter (TH0 and TL0)
 TH0-TL0 is incremented when TR0 is set to 1 and an
external pulse (in T0) occurs.
 When the counter (TH0-TL0) reaches its maximum of
FFFFH, it rolls over to 0000, and TF0 is raised.
 Programmers should monitor TF0 continuously and stop
the counter 0.
 Programmers can set the initial value of TH0-TL0 and
let TF0=1 as an indicator to show a special condition.
(ex: 100 people have come).

Microprocessors 1-
1-48
48
Timer 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

Microprocessors 1-
1-49
49
Counter Mode 2
8-bit counter.
It allows only values of 00 to FFH to be loaded
into TH0.
Auto-reloading
• TL0 is incremented if TR0=1 and external
pulse occurs.
See Figure 9.6, 9.7 for logic view
• See Examples 9-18, 9-19

Microprocessors 1-
1-50
50
Example 9-18 (1/2)
Assuming that clock pulses are fed into pin T1, write a program for
counter 1 in mode 2 to count the pulses and display the state of the
TL 1 count on P2.
Solution:
MOV TMOD,#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

Microprocessors 1-
1-51
51
Example 9-18 (2/2)
P2 is connected to 8 LEDs
and input T1 to pulse.
T1
to
LEDs
P3.5
P2
8051
 Timer 1 as an event counter fed into pin3.5.
 “SETB P3.5” make P3.5 an input port by making it high

Microprocessors 1-
1-52
52
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

Microprocessors 1-
1-53
53
Example 9-19 (2/3)
ACALL LCD_SET_UP ;initialize the LCD
MOV TMOD,#00000110B ;Counter 0,mode2
MOV TH0,#-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
ACALL DISPLY ;display on LCD
JNB TF0,BACK ;loop if TF0=0
CLR TR0 ;stop
CLR TF0
SJMP AGAIN

Microprocessors 1-
1-54
54
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

Microprocessors 1-
1-55
55
A Digital Clock
 Example 9-19 shows a simple digital clock.
 If we feed an external square wave of 60 Hz frequency
into the timer/counter, we can generate the second, the
minute, and the hour out of this input frequency and
display the result on an LCD.
 You might think that the use of the instruction
“JNB TF0,target” to monitor the raising of the
TF0 flag is a waste of the microcontroller’s time.
 The solution is the use of interrupt. See Chapter 11.
 In using interrupts we can do other things with the 8051.
 When the TF flag is raised it will inform us.

Microprocessors 1-
1-56
56
GATE=1 in TMOD
All discuss so far has assumed that GATE=0.
The timer is stared with instructions “SETB TR0”
and “SETB TR1” for timers 0 and 1, respectively.
If GATE=1, we can use hardware to control the
start and stop of the timers.
INT0 (P3.2, pin 12) starts and stops timer 0
INT1 (P3.3, pin 13) starts and stops timer 1
This allows us to start or stop the timer externally
at any time via a simple switch.

Microprocessors 1-
1-57
57
GATE (external control)
Timer 0 must be turned on by “SETB TR0”
If GATE=1 count up if
 INT0 input is high
TR0=1
If GATE=0 count up if
TR0=1

Microprocessors 1-
1-58
58

Microprocessors 1-
1-59
59

Microprocessors 1-
1-60
60

Microprocessors 1-
1-61
61

Microprocessors 1-
1-62
62
8051
Interrupts

Microprocessors 1-
1-63
63
Interrupts Programming
 An interrupt is an external or internal event that
interrupts the microcontroller to inform it that a
device needs its service.
Interrupts vs. Polling
 A single microcontroller can serve several devices.
 There are two ways to do that:
 interrupts
 polling.
 The program which is associated with the interrupt
is called the interrupt service routine (ISR) or
interrupt handler.

Microprocessors 1-
1-64
64
Steps in executing an interrupt
 Finish current instruction and saves the PC on stack.
 Jumps to a fixed location in memory depend on type
of interrupt
 Starts to execute the interrupt service routine until
RETI (return from interrupt)
 Upon executing the RETI the microcontroller returns
to the place where it was interrupted. Get pop PC
from stack

Microprocessors 1-
1-65
65
Interrupt Sources
Original 8051 has 6 sources of interrupts
 Reset
 Timer 0 overflow
 Timer 1 overflow
 External Interrupt 0
 External Interrupt 1
 Serial Port events (buffer full, buffer empty, etc)
Enhanced version has 22 sources
 More timers, programmable counter array, ADC, more
external interrupts, another serial port (UART)

Microprocessors 1-
1-66
66
Interrupt Vectors
Each interrupt has a specific place in code memory where
program execution (interrupt service routine) begins.
External Interrupt 0: 0003h
Timer 0 overflow: 000Bh
External Interrupt 1: 0013h
Timer 1 overflow: 001Bh
Serial : 0023h
Timer 2 overflow(8052+) 002bh
Note: that there are
only 8 memory
locations between
vectors.

Microprocessors 1-
1-67
67
SJMP main
ORG 03H
ljmp int0sr
ORG 0BH
ljmp t0sr
ORG 13H
ljmp int1sr
ORG 1BH
ljmp t1sr
ORG 23H
ljmp serialsr
ORG 30H
main:
…
END
ISRs and Main Program in 8051

Microprocessors 1-
1-68
68
Interrupt Enable (IE) register
All interrupt are disabled after reset
We can enable and disable them bye IE

Microprocessors 1-
1-69
69
Enabling and disabling an
interrupt
by bit operation
Recommended in the middle of program
SETB EA ;Enable All
SETB ET0 ;Enable Timer0 ovrf
SETB ET1 ;Enable Timer1 ovrf
SETB EX0 ;Enable INT0
SETB EX1 ;Enable INT1
SETB ES ;Enable Serial port
by mov instruction
Recommended in the first of program
MOV IE, #10010110B
SETB IE.7
SETB IE.1
SETB IE.3
SETB IE.0
SETB IE.2
SETB IE.4

Microprocessors 1-
1-70
70
Example
A 10khz square wave with 50% duty cycle
ORG 0 ;Reset entry poit
LJMP MAIN ;Jump above interrupt
ORG 000BH ;Timer 0 interrupt vector
T0ISR:CPL P1.0 ;Toggle port bit
RETI ;Return from ISR to Main program
ORG 0030H ;Main Program entry point
MAIN: MOV TMOD,#02H ;Timer 0, mode 2
MOV TH0,#-50 ;50 us delay
SETB TR0 ;Start timer
MOV IE,#82H ;Enable timer 0 interrupt
SJMP $ ;Do nothing just wait
END

Microprocessors 1-
1-71
71
Example
Write a program using interrupts to
simultaneously create 7 kHz and 500 Hz
square waves on P1.7 and P1.6.
71s
143s
1ms
2ms
P1.7
P1.6
8051

Microprocessors 1-
1-72
72
71s
143s
1ms
2ms
P1.7
P1.6
8051
Solution
ORG 0
LJMP MAIN
ORG 000BH
LJMP T0ISR
ORG 001BH
LJMP T1ISR
ORG 0030H
MAIN: MOV TMOD,#12H
MOV TH0,#-71
SETB TR0
SETB TF1
MOV IE,#8AH
MOV IE,#8AH
SJMP $
T0ISR: CPL P1.7
RETI
T1ISR: CLR TR1
MOV TH1,#HIGH(-1000)
MOV TL1,#LOW(-1000)
SETB TR1
CPL P1.6
RETI
END

Microprocessors 1-
1-73
73
Timer ISR
Notice that
There is no need for a “CLR TFx” instruction in
timer ISR
8051 clears the TF internally upon jumping to ISR
Notice that
We must reload timer in mode 1
There is no need on mode 2 (timer auto reload)

Microprocessors 1-
1-74
74
External interrupt type control
 By low nibble of Timer control register TCON
 IE0 (IE1): External interrupt 0(1) edge flag.
 set by CPU when external interrupt edge (H-to-L) is detected.
 Does not affected by H-to-L while ISR is executed(no int on int)
 Cleared by CPU when RETI executed.
 does not latch low-level triggered interrupt
 IT0 (IT1): interrupt 0 (1) type control bit.
 Set/cleared by software
 IT=1 edge trigger
 IT=0 low-level trigger
(MSB) (LSB)

Microprocessors 1-
1-75
75
External Interrupts
IE0 (TCON.3)
0003
INT0
(
Pin 3.2
) 0
1
2
IT0
Edge-triggered
Level-triggered (default)
IE1 (TCON.3)
INT0
(
Pin 3.3
) 0
1
2
IT1
Edge-triggered
Level-triggered (default)
0013

Microprocessors 1-
1-76
76
Example of external interuupt
ORG 0000H
LJMP MAIN
;
;interrupt service routine (ISR)
;for hardware external interrupt INT1
;
ORG 0013H
SETB P1.1
MOV R0,200
WAIT: DJNZ R0,WAIT
CLR P1.1
RETI
;
;main program for initialization
;
ORG 30H
MAIN: SETB IT1 ;on negative edge of INT1
MOV IE,#10000100B
WAIT2: SJMP WAIT2
END

Microprocessors 1-
1-77
77

Microprocessors 1-
1-78
78
Org 0000h
Ljmp main
Org 0003h
x0isr: clr p1.7
Reti
Org 0013h
x1isr: setb p1.7
Reti
Org 0030h
Main: mov ie,#85h
Setb it0
Setb it1
Setb p1.7
Jb p3.2,skip
Clr p1.7
Skip: Sjmp $
end

Microprocessors 1-
1-79
79
Interrupt Priorities
 What if two interrupt sources interrupt at the same
time?
 The interrupt with the highest PRIORITY gets
serviced first.
 All interrupts have a power on default priority order.
1. External interrupt 0 (INT0)
2. Timer interrupt0 (TF0)
3. External interrupt 1 (INT1)
4. Timer interrupt1 (TF1)
5. Serial communication (RI+TI)
 Priority can also be set to “high” or “low” by IP reg.

Microprocessors 1-
1-80
80
Interrupt Priorities (IP) Register
IP.7: reserved
IP.6: reserved
IP.5: timer 2 interrupt priority bit(8052 only)
IP.4: serial port interrupt priority bit
IP.3: timer 1 interrupt priority bit
IP.2: external interrupt 1 priority bit
IP.1: timer 0 interrupt priority bit
IP.0: external interrupt 0 priority bit
--- PX0
PT0
PX1
PT1
PS
PT2
---

Microprocessors 1-
1-81
81
Interrupt Priorities Example
 MOV IP , #00000100B or SETB IP.2 gives priority order
1. Int1
2. Int0
3. Timer0
4. Timer1
5. Serial
 MOV IP , #00001100B gives priority order
1. Int1
2. Timer1
3. Int0
4. Timer0
5. Serial
--- PX0
PT0
PX1
PT1
PS
PT2
---

Microprocessors 1-
1-82
82
Interrupt inside an interrupt
--- PX0
PT0
PX1
PT1
PS
PT2
---
 A high-priority interrupt can interrupt a low-priority
interrupy
 All interrupt are latched internally
 Low-priority interrupt wait until 8051 has finished
servicing the high-priority interrupt

Microprocessors 1-
1-83
83
Serial
Communication

Microprocessors 1-
1-84
84
Basics of serial communication
Parallel: expensive - short distance – fast
Serial :cheaper– long (two different cities by modem)-slow

Microprocessors 1-
1-85
85
Basics of serial communication

Microprocessors 1-
1-86
86
Start and stop bits
When there is no transfer the signal is high
Transmission begins with a start (low) bit
LSB first
Finally 1 stop bit (high)
Data transfer rate (baud rate) is stated in bps
bps: bit per second

Microprocessors 1-
1-87
87
How to communicate 8051 to PC
 Connect TXD to RXD and RXD to TXD from pc to 8051
 Use max232 to transform signal from TTL level to RS232 level
 The baud rate of the 8051 must matched the baud rate of the pc
 PC standard baud rate
 2400-4800-9600-14400-19200-28800-33600-57600
 Serial mode 1 is used
 Timer 1 is used
 The 8051 UART divides the machine cycle frequency by 32
 Machine cycle is 1/12 XTAL frequency
 We use timer1 in mode 2 (auto reload)
 See example 10-1

Microprocessors 1-
1-88
88
RxD and TxD pins in the 8051
TxD pin 11 of the 8051 (P3.1)
RxD pin 10 of the 8051 (P3.0)
SBUF register
MOV SBUF,#’D’ ;load SBUF=44H, ASCII for ‘D’
MOV SBUF,A ;copy accumulator into SBUF
MOV A,SBUF ;copy SBUF into accumulator

Microprocessors 1-
1-89
89
MAX232

Microprocessors 1-
1-90
90
Serial port block diagram

Microprocessors 1-
1-91
91

Microprocessors 1-
1-92
92

Microprocessors 1-
1-93
93
Serial control (SCON) Register
SM0 (SCON.7) : mode specifier
SM1 (SCON.6) : mode specifier
SM2 (SCON.5) : used for multi processor communication
REN (SCON.4) : receive enable (by software enable/disable)
TB8 (SCON.3) : transmit bit8
RB8 (SCON.2) : receive bit 8
TI (SCON.1) : transmit interrupt flag set by HW clear by SW
RI (SCON.0) : receive interrupt flag set by HW clear by SW
SM0 RI
TI
RB8
TB8
REN
SM2
SM1

Microprocessors 1-
1-94
94
Mode of operation
SM0 SM1 MODE operation transmit rate
0 0 0 shift register fixed (xtal/12)
0 1 1 8 bit UART variable (timer1)
1 0 2 9 bit UART fixed (xtal/32 or
xtal/64)
1 1 3 9 bit UART variable (timer1)

Microprocessors 1-
1-95
95
Mode of operation
 Mode 0 :
 Serial data enters and exits through RxD
 TxD outputs the shift clock.
 8 bits are transmitted/received(LSB first)
 The baud rate is fixed a 1/12 the oscillator frequency.
 Application
 Port expansion
8051
TXD
RXD Shift register
clk
data

Microprocessors 1-
1-96
96
Mode of operation
 Mode 1
 Ten bits are transmitted (through TxD) or received (through RxD)
 A start bit (0), 8 data bits (LSB first), and a stop bit (1)
 On receive, the stop bit goes into RB8 in SCON
 the baud rate is determined by the Timer 1 overflow rate.
 Timer1 clock is 1/32 machine cycle (MC=1/12 XTAL)
 Timer clock can be programmed as 1/16 of machine cycle
 Transmission is initiated by any instruction that uses SBUF as a destination register

Microprocessors 1-
1-97
97
Mode of operation

Microprocessors 1-
1-98
98
Mode of operation
 Mode 2 :
 Eleven bits are transmitted (through TxD), received (through RxD)
 A start bit (0)
 8 data bits (LSB first)
 A programmable 9th data bit
 and a stop bit (1)
 On transmit, the 9th bit (TB8) can be assigned 0 or 1.
 On receive, the 9the data bit goes into RB8 in SCON.
 the 9th
can be parity bit
 The baud rate is programmable to 1/32 or 1/64 the oscillator frequency in Mode 2 by
SMOD bit in PCON register
 Mode 3
 Same as mode 2
 But may have a variable baud rate generated from Timer 1.

Microprocessors 1-
1-99
99
What is SMOD
 Bit 7 of PCON register
 If SMOD=1 double baud rate
 PCON is not bit addressable
 How to set SMOD
Mov a, pcon
Setb acc.7
Mov pcon,a

Microprocessors 1-
1-100
100
Serial example(1)
An example of sending a message.
;initialization
MOV TMOD,#20H
MOV TH1,#-12
MOV SCON,#52H
;begin to trnasmit
SETB TR1
AGAIN1: MOV A,#'B'
CALL TRANSS
MOV A,#'A'
CALL TRANSS
MOV A,#'L'
CALL TRANSS
MOV A,#'E'
CALL TRANSS
SJMP AGAIN1
;seial transmiting subroutine
TRANSS: MOV SBUF,A
AGAIN2: JNB TI,AGAIN2
CLR TI
RET
END

Microprocessors 1-
1-101
101
Serial example(2)
An example for serial port interrupt
ORG 0000H
LJMP MAIN
;jump to serial ISR
ORG 23H
LJMP ISR
;main program
ORG 30H
;1-initializtion
MAIN: MOV P0,#0FFH
MOV TMOD,#20H
MOV TH1,#-13
MOV SCON,#50H
MOV IE,#90H
;2-begin
SETB TR1
AGAIN: MOV A,P0
MOV P1,A
SJMP AGAIN
;
;ISR for reading from serial port
ISR: PUSH ACC
JB TI,TRANSM
MOV A,SBUF
MOV P2,A
CLR RI
SJMP ISREND
TRANSM: CLR TI
ISREND: POP ACC
RETI
END

Microprocessors 1-
1-102
102
Serial example(3)
an example for serial port interrupt
;for transmitting
ORG 0000H
LJMP MAIN
;jump to serial ISR
ORG 23H
LJMP ISR
;main program
ORG 30H
;initializtion
MAIN: MOV P0,#0FFH
MOV TMOD,#20H
MOV TH1,#-13
MOV SCON,#50H
MOV IE,#90H
;2-begin
SETB TR1
AGAIN: SJMP AGAIN
;ISR for receive from serial to p0
;transmitting to serial from p1
ISR: JB TI,TRANSM
MOV A,SBUF
mov P0,A
CLR RI
RETI
TRANSM: MOV A,P1
MOV SBUF,A
CLR TI
RETI
END

Microprocessors 1-
1-103
103
Serial example(4)
ORG 0000
;Initialize serial port & timer
INIT: MOV SCON,#52H ;Serial port mode 1
MOV TMOD,#20H ;Timer 1, mode 2
MOV TH1,#-13 ;Reload count for 2400 baud
SETB TR1 ;Start timer 1
;move character 'B' to accumulator for transmitting
MOV A,#'B'
;Transmit characters by serial port
OUTCHR: MOV C,P ;Put parity bit in C flag
CPL C ;Change to odd parity
MOV ACC.7,C ;Add to character code
AGAIN: JNB TI,AGAIN ;Buffer empty? no:check again
CLR TI ;Yes:clear falg and
MOV SBUF,A ;send character
CLR ACC.7 ;Strip off parity bit
JMP $
END

Microprocessors 1-
1-104
104
Power control register

Microprocessors 1-
1-105
105
Power control
A standard for applications where power
consumption is critical
two power reducing modes
 Idle
 Power down

Microprocessors 1-
1-106
106
Idle mode
 An instruction that sets PCON.0 causes Idle mode
 Last instruction executed before going into the Idle mode
 the internal CPU clock is gated off
 Interrupt, Timer, and Serial Port functions act normally.
 All of registers , ports and internal RAM maintain their data
during Idle
 ALE and PSEN hold at logic high levels
 Any interrupt
 will cause PCON.0 to be cleared by HW (terminate Idle mode)
 then execute ISR
 with RETI return and execute next instruction after Idle
instruction.
 RST signal clears the IDL bit directly

Microprocessors 1-
1-107
107
Power-Down Mode
 An instruction that sets PCON.1 causes power dowm
mode
 Last instruction executed before going into the power
down mode
 the on-chip oscillator is stopped.
 all functions are stopped,the contents of the on-chip
RAM and Special Function Registers are maintained.
 The ALE and PSEN output are held low
 The reset that terminates Power Down

Microprocessors 1-
1-108
108
Power control example
Org 0000h
Ljmp main
Org 0003h
Orl pcon,#02h ;power down mode
Reti
Org 0030h
Main:
……
……
……
Orl pcon,#01h ;Idle mode
end

Microprocessors 1-
1-109
109
example

Microcontroller 8051 Timer Counter Interrrupt

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