microcontroller 8051- 2.ppt

1

• Section 2

Microprocessors course

Dr. S.O.Fatemi

By: Mahdi Hassanpour

2

Contents:

I/O Programming; Bit ManipulationTime delay Generation and calculationTimer/Counter Programming

-Timers- Counters

Interrupts ProgrammingSerial Communication

3

I/O Programming; Bit Manipulation

• To toggle every bit of P1 continuously, 3 ways exists:• Way 1: Send data to Port 1 through ACC ：

BACK: MOV A,#55H ;A=01010101B MOV P1,A ACALL DELAY MOV A,#0AAH ;A=10101010B MOV P1,A ACALL DELAY SJMP BACK

• Way 2: Access Port 1 directly ：BACK: MOV P1,#55H ;P1=01010101B ACALL DELAY MOV P1,#0AAH ;P1=10101010B ACALL DELAY SJMP BACK

• Read-modify-write feature ： MOV P1,#55H ;P1=01010101B AGAIN: XRL P1,#0FFH ACALL DELAY SJMP AGAIN

– The instruction XRL P1,#0FFH do EX-OR P1 and FFH （ That is, to toggle P1. ）

4

Bit Manipulation

• Sometimes we need to access only 1 or 2 bits of the port instead of the entire 8 bits.• This table shows how to name each pin for each I/O port. • Example:

Write a program to perform the following.(a) Keep monitoring the P1.2 bit until it becomes high,(b) When P1.2 becomes high, write value 45H to port 0, and(c) Send a high-to-low (H-to-L) pulse to P2.3.

Solution: SETB P1.2 ;make P1.2 an input MOV A,#45H ;A=45HAGAIN:JNB P1.2,AGAIN;get out when P.2=1 MOV P0,A ;issue A to P0 SETB P2.3 ;make P2.3 high CLR P2.3 ;make P2.3 low for H-to-LNote ：1. JNB: jump if no bit （ jump if P1.2 = 0 ） 2. a H-to-L pulse by the sequence of instructions SETB and CLR.

5

Single-Bit Addressability of Ports

D7P3.7P2.7P1.7P0.7

D6P3.6P2.6P1.6P0.6

D5P3.5P2.5P1.5P0.5

D4P3.4P2.4P1.4P0.4

D3P3.3P2.3P1.3P0.3

D2P3.2P2.2P1.2P0.2

D1P3.1P2.1P1.1P0.1

D0P3.0P2.0P1.0P0.0

Port BitP3P2P1P0

6

Time delay Generation and calculation

• Machine cycle

• For the CPU to execute an instruction takes a certain number of block cycles. In the 8051 family, these clock cycles are referred to as machine cycles.

• The frequency of the crystal connected to the 8051 family ca vary from 4MHz to 30 MHz, depending on the chip rating and manufacturer. Very often the 11.0592 MHz crystal oscillator is used to make the 8051-based system compatible with the serial port of the IBM PC.

• In the 8051, one machine cycle lasts 12 oscillator periods.

8

Example:Find the time delay for the following subroutine, assuming a crystal frequency of 11.0592 MHz

DELAY: MOV R3,#250 ; 1 MCHERE: NOP ; 1 MC

NOP ; 1 MCNOP ; 1 MCNOP ; 1 MCDJNZ R3,HERE ; 2 MCRET ; 1 MC

Solution:250x(1+1+1+1+2)+2x1.085 us=1627.5 us

9

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.

10

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

toLCD

P1

8051

TL0

TH0

P2SetTimer 0

11

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

toLCD

P3.4

P1

8051

a switch

TL0

TH0

12

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.

13

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.

14

Timer Registers

D15 D8D9D10D11D12D13D14 D7 D0D1D2D3D4D5D6

TH0 TL0

D15 D8D9D10D11D12D13D14 D7 D0D1D2D3D4D5D6

TH1 TL1

Timer 0

Timer 1

15

TMOD Register

• Timer mode register: TMODMOV 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 M0Timer 1 Timer 0

(MSB) (LSB)

16

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)

17

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

18

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).

19

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

20

Example 9-3

Find the value for TMOD if we want to program timer 0 in mode 2,

use 8051 XTAL for the clock source, and use instructions to start

and stop the timer.

Solution:

TMOD= 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

21

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 IT0Timer 1 Timer0 for Interrupt

(MSB) (LSB)

22

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.

TF1 TR1 TF0 TR0 IE1 IT1 IE0 IT0Timer 1 Timer0 for Interrupt

(MSB) (LSB)

23

Equivalent Instructions for the Timer Control Register

For timer 0SETB TR0 = SETB TCON.4CLR TR0 = CLR TCON.4

SETB TF0 = SETB TCON.5CLR TF0 = CLR TCON.5

For timer 1SETB TR1 = SETB TCON.6CLR TR1 = CLR TCON.6

SETB TF1 = SETB TCON.7CLR TF1 = CLR TCON.7

TF1 IT0IE0IT1IE1TR0TF0TR1

TCON: Timer/Counter Control Register

24

Timer Mode 1

• In following, we all use timer 0 as an example.• 16-bit timer (TH0 and TL0)• TH0-TL0 is incremented continuously when TR0 is set

to 1. And the 8051 stops to increment TH0-TL0 when TR0 is cleared.

• The timer works with the internal system clock. In other words, the timer counts up each machine cycle.

• When the timer (TH0-TL0) reaches its maximum of FFFFH, it rolls over to 0000, and TF0 is raised.

• Programmer should check TF0 and stop the timer 0.

25

Steps of Mode 1 (1/3)

1. Chose 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

26


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 TL0Start timerStop timer

Monitor TF until TF=1

TR0=1 TR0=0

TF

27


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

28

Mode 1 Programming

XTALoscillator ÷ 12

TR

TH TL TF

overflow flag

C/T = 0

TF goes high when FFFF 0

29

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

30

Example 9-4 (1/3)

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

31

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

32

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).

33

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 1AGAIN:MOV TL1,#34H ;timer value=3476H MOV TH1,#76H SETB TR1 ;startBACK: JNB TF1,BACK CLR TR1 ;stop CPL P1.5 ;next half clock CLR TF1 ;clear timer flag 1 SJMP AGAIN ;reload timer1

34

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.

35

Find Timer Values

• Assume that we know the amount of timer delay and XTAL = 11.0592 MHz .

• How to find the inter values needed for the TH, TL?– Divide the desired time delay by 1.085 s.

– Perform 65536 –n, where n is the decimal value we got in Step 1.

– Convert th result of Step 2 to hex, where yyxx is the initial hex value to be loaded into the timer’s registers.

– Set TH = yy and TL = xx.

• Example 9-10

36

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.

37

Example 9-12 (2/2)

MOV TMOD,#10H ;timer 1, mode 1AGAIN: MOV TL1,#00 ;Timer value = DC00H

MOV TH1,#0DCH SETB TR1 ;startBACK: 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

38

Generate a Large Time Delay

• The size of the time delay depends on two factors:– They crystal frequency

– The timer’s 16-bit register, TH & TL

• The largest time delay is achieved by making TH=TL=0. What if that is not enough?

• Example 9-13 show how to achieve large time delay.

39

Example 9-13Examine 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 msTotal delay = 200 × 70.820 ms = 14.164024 seconds

40

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.

41

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.• In the following example, we want to generate a

delay with 200 MCs on timer 0.• See Examples 9-14 to 9-16

42


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

43


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. – 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.

44

Timer 1 Mode 2 with External Input


TR1

TL1

TH1

TF1overflow flag

reload

TF goes high when FF 0

C/T = 0

45

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 AGAINDELAY:SETB TR0 ;startBACK: 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.

46

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

47

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 tat 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.

48

Example 9-17 (2/2)

MOV TMOD,#2H ;Timer 0,mode 2 MOV TH0,#-150 ;Count=150AGAIN:SETB P1.3 ACALL DELAY ACALL DELAY CLR P1.3 ACALL DEALY SJMP AGAIN

DELAY:SETB TR0 ;startBACK: JNB TF0,BACK CLR TR0 ;stop CLR TF0 ;clear TF RET

high period

low period

49

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)

50

Port 3 Pins Used For Timers 0 and 1

Pin Port Pin Function Description

14 P3.4 T0 Timer/Counter 0 external input15 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)

51

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).

52

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

53

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

54

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 portAGAIN:SETB TR1 ;startBACK: 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

55

Example 9-18 (2/2)

We use timer 1 as an event counter where it counts up as clock pulses are fed into pin3.5.

Notice in the above program the role of the instruction “SETB

P3.5”. Since ports are set up for output when the 8051 is powered

up , we must make P3.5 an input port by making it high.

P2 is connected to 8 LEDs and input T1 to pulse.

T1

toLEDs

P3.5

P2

8051

56

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

toLCD

P3.4

P1

8051

1 Hz clock

57

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 inputAGAIN:SETB TR0 ;starts the counterBACK: 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

58

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

59

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.

60

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.

61

Example for GATE=1

• The 8051 is used in a product to sound an alarm every second using timer 0.

• Timer 0 is turned on by the software method of using the “SETB TR0” instruction and is beyond the control of the user of that product.

• However, a switch connected to pin P3.2 can be used to turn on and off the timer, thereby shutting down the alarm.

62

Timer/Counter 0


TR0

INT0 Pin Pin 3.2

C/T = 0

Gate

T0 Pin Pin 3.4

C/T = 1

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. That are two ways to do that: interrupts or polling.

• The program which is associated with the interrupt is called the interrupt service routine (ISR) or interrupt handler.

64

Steps in executing an interrupt:

• it finishes the instruction it is executing and serves the address of the next instruction (PC) on the stack.

• It also saves the current status of all the interrupts internally (i.e. not on the stack)

• It jumps to a fixed location in memory called the interrupt vector table that holds the address of the interrupt service routine.

• The microcontroller gets the address of the ISR from the interrupt vector table and jumps to it. It starts to execute the interrupt service routine until it reaches the last instruction of the subroutine which is RETI (return from interrupt)

• Upon executing the RETI instruction, the microcontroller returns to the place where it was interrupted. First, it gets the program counter (PC) address from the stack by popping the top two bytes of the stack into the PC. Then it starts to execute from that address.

65

SJMP FIRST

ORG 13H ;TSR FOR INT1MOV A,P1 ;read dataACALL ;MUL39;R3R2R1=A*39MOV A,R3FCAL PUTHMOV A,R2FCAL PUTHMOV A,R1FCAL PUTH ;DISP VOLTAGE IN mVSETB P3.0 ;RD=1 FOR NEXT

CLR P3.1 ;WR=0SETB P3.1 ;WR=1 ,start conversion

ORG 30HFIRST:

…END

Six interrupts in 8051

66

Enabling and disabling an interrupt:

Example:Write a program using interrupts to simultaneously create 7

kHz and 500 Hz square waves on P1.7 and P1.6.

67

71s

143s

1ms

2ms

P1.7

P1.6

8051

Solution:

ORG 0LJMP MAINORG 000BHLJMP T0ISRORG 001BHLJMP T1ISRORG 0030H

MAIN: MOV TMOD,#12HMOV TH0,#-71SETB TR0SETB TF1MOV IE,#8AHMOV IE,#8AHSJMP $

T0ISR: CLR P1.7RETI

T1ISR: CLR TR1MOV TH1,#HIGH(-1000)MOV TL1,#LOW(-1000)SETB TR1CPL P1.6RETIEND

68

External Interrupts:

IE0 (TCON.3)

0003

INT0(Pin 3.2) 0

12

IT0

Edge-triggered

Level-triggered (default)

IE1 (TCON.3)

INT0(Pin 3.3) 0

12

IT1

Edge-triggered

Level-triggered (default)

0013

69

Exercise

• We have a motor that send pulses to micro proportional to it’s r.p.m. write a program that if the number of pulses per 10-second are less than 100, send 1 to P1.0, and if more than 200, send 1 to P1.1

• Write a program and design hardware that connect key-pad to micro and identifies which key is pressed.

70

Serial Communication

71

Basics of serial communication

•Baud Rate

73

Start and stop bits

74

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

75

MAX232

microcontroller 8051- 2.ppt

Documents