interrupts programming

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1Week 9 Vocational Training Council, Hong Kong.

Timer Programming and InterruptsTimer Programming and Interrupts

EEE3410 Microcontroller ApplicationsDepartment of Electrical Engineering

Lecture 8


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EEE3410 Microcontroller Applications

Timer Programming

Contrast and compare interrupt versus polling

Interrupt Handling

Interrupts of the 8051

Purpose of the interrupt vector table

Enable or disable interrupts

Timers using interrupts

Interrupt priority

In this LectureIn this Lecture ..


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Timers Programming (1/10)

The 8051 has two timers: Timer 0 and Timer 1

They can be used either as timers to generate a time delay of

as counter to count events happening outside the

microcontroller

Both Timer 0 and Timer 1 are 16 bits wide

They are accessed as two separate registers, low byte and

high byte. (TL0 & TH0 for Time 0 and TL1 & TH1 for

timer 1)


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There are three Special Function Registers for timer settings

1. Timer Registers (Timer 0 & Timer 1) store the starting valuesof the Timer 0 & Timer 1. Each timer is 16-bit register which is

split into two bytes (THx & TLx).

D0D2D4D6D8D10D12D14 D1D3D5D7D9D11D13D15

TL0TH0

Timer 0 register (16 Bit)

D0D2D4D6D8D10D12D14 D1D3D5D7D9D11D13D15

TL1TH1

Timer 1 register (16 Bit)


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2. Timer Control Register (TCOD) use to turn ON/OFF of the timers

and timer interrupt control. It is a 8-bit register and bit-addressable,

and only the upper 4-bit refers to timer control.

Timer ON/OFF control bits

IT0IE0IT1IE1TR0TF0TR1TF1

(LSB)(MSB)

TR1 Timer 1 run control bit. Set = Timer ON and Clear = Timer OFF

TF1 Timer flag which is set when the Timer 1 rolls over from FFFFH to 0000H.

TR0 Timer 0 run control bit. Set = Timer ON and Clear = Timer OFF

TF0 Timer flag which is set when Timer 0 rolls over from FFFFH to 0000H.

IE1 and IT1 Set the trigger mode of external interrupt 1

IE0 and IT0 Set the trigger mode of external interrupt 1

Trigger mode of external interrupt control bit





3. Timer Mode Register (TMOD) use to set the various timer operation

modes. It is a 8-bit register and bit-addressable.

Timer 1

M0M1C/TGATEM0M1C/TGATE

(LSB)(MSB)

GATE Gating control when set. The timer/counter is enable only while the INTx

pin is high and the TRx control pin is set. When cleared, the timer is enabledwhenever the TRx control bit is set.

C/T Timer or counter selected. Cleared for timer operation (input from internal

system clock) and Set for counter operation (input from Tx input pin)

M1 & M0 Mode bits

Timer 0





M1 & M0 Mode bits

Split timer mode311

8-bit auto reload timer/counter mode

THx holds a value that is to be reloaded into TLx each

time it overflows.

201

16-bit timer/counter mode

Timer value range from 0000H to FFFFH in TH - TL

110

13-bit timer/counter mode

Timer value range from 0000H to 1FFFH in TH - TL

000

Operating ModeModeM0M1

Modes 1 & 2 are used most widely.




Example 8Example 8--11

Solution:

Convert the values from hex to binary:

(a) TMOD = 0000 0001, mode 1 of Timer 0 is selected

(b)TMOD = 0010 0000, mode 2 of Timer 1 is selected

(c) TMOD = 0001 0010, mode 2 of Timer 0, and mode 1 of Timer 1are

selected

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:

(1)

MOV TMOD, #02H ; Timer 0 and mode 1 set, C/T = 0 to use XTAL

; clock source, Gate = 0 to use software ON/OFF

MOV TL0, #0ABH ; TL0 =ABH

MOV TH0, #0F0H ; TH0 =F0H

SETB TR0 ; Start Timer 0

1. Write instructions to do the followings:

a. Set Timer 0 in mode 1, use 8051 XTAL for the clock source,

instructions to start and stop the timer,b. Set value F0ABH to Timer 0.

c. Start Timer 0

2. Determine the time for Timer 0 rolling over if XTAL = 12 MHz.






Solution:

(2)

Counts for Timer rolling over = Counts from F0ABH to FFFFHplus rolling over to 0

Timer clock cycles = (FFFFH F0ABH + 1) = 0F55H

= 3925 in decimal

Time for Timer 0 rolls over = 3925 x 1 s = 3925 s #

Time for 1 Timer clock = 1 machine cycle = s1s12x10x12

16

=

Assume 12 MHz clock :

3410 i ll li i





Solution:The period of the square wave, T = 1/50 Hz = 20 ms

of it for the high and low portions of the pulse = 10 ms

10 ms/ 1s = 10000 timer cycles are needed for each pulse.

Timer 1 value to be set = 65536 10000 = 55536 in decimal = D8F0H

i.e. TH1 = D8H and TL1 = F0H

MOV TMOD, #10H ; Timer 1, mode 1

AGAIN: MOV TL1, #0F0H ; TL1 =F0H

MOV TH1, #0D8H ; TH1 =D8HSETB TR1 ; Start Timer 1

BACK: JNB TF1, BACK ; Stay until timer rolls over

CLR TR1 ; Stop Timer 1

CPL P1.3 ; Complement P1.3 to set Hi, Low

CLR TF1 ; Clear Timer flagSJMP AGAIN

Assume XTAL = 12 MHz, write a program to generate a square wave of 50 Hz

frequency on pin P1.3 by using timer 1 as time control.


EEE3410 Mic ocont olle Applications





; Time Delay subroutine

MOV TMOD, #10H ; Timer 1, mode 1

MOV R3, #200AGAIN: MOV TL1, #08H ; TL1 =08H

MOV TH1, #01H ; TH1 =01H

SETB TR1 ; Start Timer 1

BACK: JNB TF1, BACK ; Stay until timer rolls over

CLR TR1 ; Stop Timer 1

CLR TF1 ; Clear Timer flag

DJNZ R3, AGAIN ; Loop until R3 = 0

RET

Examine the following subroutine and find the time delay for it. (12 MHz clock)


Solution :

Timer value = 0108H = 264 in decimal, counts to rolls over = 65536 264 = 65272 in decimal

Timer cycle of Timer = 65272

Total machine cycles of the subroutine = [2 + 1 + (5 + 65272 + 4) x 200 + 2] = 13056205Total delay time 13056205 x 1s = 13056.2 ms

Machine cycles

2

12

2

1

2

1

1

2

2





A single microcontroller always connects to

serve several peripheral devices through itsI/O ports

There are two ways for the peripheral devicesto request service from microcontroller

Polling

Interrupt

Interrupt & PollingInterrupt & Polling



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Microcontroller continuously monitors the status

of a certain number of devices in sequence

Services to a device if preset condition met

After the service, the microcontroller will move

on to monitor the status of another device until

all devices are servicedThe operation described above is called polling

Programmed I/O (Polling)Programmed I/O (Polling)



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Whenever any device needs the service, it notifies

the microcontroller by sending it an interruptrequest (IR) signal while the microcontroller isdoing other work.

The microcontroller suspends its work to service

the device at once.

Note that each IR is associated with an interruptservice routine (ISR)

Interrupt I/O (Interrupt)Interrupt I/O (Interrupt)



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pp

Comparison between Interrupt and PollingComparison between Interrupt and Polling

NoYesPriority setting

MoreLessNeed of MCUtime

SlowerFasterResponse time

MCU continuously

monitors devices to

determine whether

they need service

Devices notify MCU

by sending it an

interrupt signals

while the MCU is

doing another work

Method

PollingInterrupt



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Steps in Handling an Interrupt Request

The current instruction will be finished

The PC is saved on the stack

The current interrupt status is saved internally The PC is loaded with the vector address of the ISR

from the interrupt vector table (Jump to execute ISR)

The ISR is executed and will be finished with a RETIinstruction (return from interrupt)

Return to main program by popping the PC from the

stack

When an interrupt activates and is accepted by the MCU, the main

program is interrupted. The following actions occurs:



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Program Execution with Interrupts

Main Main Main

ISR ISR

stack

PC saved

on stack

PC popped

from stack

The program that deals with an interrupt is called an interrupt

service routine (ISR)

ISR executes in response to the interrupt and generally performs I/O

operation to a device



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There are 6 interrupts in the 8051

Types of Interrupt in the 8051Types of Interrupt in the 8051

Reset when the reset pin is activated, the 8051 will reset all

registers and ports and jumps to address location 0000H starting upexecution.

2 external interrupts Hardware external interrupts (INT0 andINT1) at pins 12 & 13 are used to receive interrupt signals fromexternal devices.

2 timer interrupts They are Timer 0 and Timer 1 which will giveout interrupt signal when the timers count to zero.

1 serial port interrupt - It is used for data reception andtransmission during serial communication.

Apart from the Reset, only the 5 interrupts are available to the user



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Interrupt Vector Table for the 8051

Serial port

Timer 1

External 1

(INT1)

Timer 0

External 0

(INT0)

Reset

Interrupt

RI or TI

TF1

IE1

TF0

IE0

RST

Flag

---0023H6

---001BH5

P3.3 (13)0013H4

---000BH3

P3.2 (12)0003H2

90000H1

PinROM locationPriority

interrupt vector table holds the addresses of ISR



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Enabling and Disabling an Interrupt

Upon reset, all interrupts are disabled

The interrupts must be enabled by software

A register called IE (interrupt enable) register, which is bit-

addressable, is responsible for enabling and disabling the interrupts

Bit IE. 7 must be set high to allow the rest of register to take effect

EA = 1 ; Global enable interrupt

EA = 0 ; Global disable interrupt

EX0ET0EX1ET1ESET2--EAIE.0IE.1IE.2IE.3IE.4IE.5IE.6IE.7

IE (interrupt enable) register



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Enabling and Disabling an Interrupt

Enable/disable external interrupt 0EX0IE.0

Enable/disable timer 0 interruptET0IE.1

Enable/disable external interrupt 1EX1IE.2

Enable/disable timer 1 interruptET1IE.3

Enable/disable serial port interruptESIE.4

Not use for 8051 (8052 only)ET2IE.5

Not implemented, reserved for future use--IE.6

Global Enable/disableEAIE.7

Description (1=Enable, 0=Disable)SymbolBit



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Solution:

(a) MOV IE, #10010110B ; enable serial, Timer 0, EX1 interruptsor

SETB IE.7 ; EA=1, Global enable

SETB IE.4 ; enable serial interrupt

SETB IE.1 ; enable Timer 0 interruptSETB IE.2 ; enable EX1 interrupt

(b) CLR IE.1 ; disable Timer 0 interrupt

(c) CLR IE.7 ; disable all interrupts

Write instructions to

(a) Enable serial interrupt, Timer 0 interrupt and external interrupt 1, an

(b) Disable Timer 0 interrupt only, then

(c) Disable all the interrupt with a single instruction



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Timer Interrupt

When timer rolls over, its timer flag (TF) is set.

If the timer interrupt in the IE register is enable, whenever the TF isset, the microcontroller is interrupted and jumps to the interruptvector table to service the ISR.

With timer interrupt is enabled, microcontroller can do other thingsand no need to monitor the TF for rolling over.

000BHJumps to1

Timer 0 Interrupt VectorTF0

001BHJumps to1

Timer 1 Interrupt VectorTF1


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Interrupt Priority

Upon reset, the priorities of interrupt source are assigned from

top to bottom as in the following Table 8 , i.e. if INT0 and

INT1 are activated at the same time, INT0 is first responded.

(RI+TI)Serial Communication

(TF1)Timer Interrupt 1(INT1)External Interrupt 1

(TF0)Timer Interrupt 0

(INT0)External Interrupt 0

Table 8: 8051 Interrupt Priority Upon Reset

HighestPriority

Lowest

Priority



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Setting Interrupt Priority with the IP register

The sequence of Table 8 can be changed by assigning a

higher priority to any one of the interrupts.

It is done by setting high at the corresponding bit in the IP(interrupt priority) register.

PX0PT0PX1PT1PSPT2----

IP.0IP.1IP.2IP.3IP.4IP.5IP.6IP.7

IE (interrupt priority) register (Bit-addressable)

Priority bit = 1 (assign high priority) Priority bit = 0 (assign low priority)

PT2, PT1 & PT0 Timer 2 (8052 only), Timer 1 & Timer 0 interrupts

PX1 & PX0 External interrupts 1 & 0, PS Serial port interrupt



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Example 8.7 :

(a) Program the IP register to assign the highest priority to INT1,

(b) Discuss what happens if INT0, INT1 and TF0 are activated at thesame time.

Solution :(a) MOV IP, #00000100B ; set IP.2=1 INT1 has the highest priority

or SETB IP.2

(b) Priority of interrupt will be changed to INT1 > INT0 > TF0The 8051 will services INT1 first and then INT0 and TF0.

(As the INT0 and TF0 bits in IP register are 0, their priorities follow the

sequence in Table 8.1)



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Example 8.8 :

The IP register is set by the instruction MOV IP, #00001100B

after reset. Discuss the sequence in which the interrupts are serviced.

Solution :

MOV IP, #00001100B instruction sets INT1 & TF1 to a higherpriority level compared with the rest of the interrupts.

Priority of interrupt will then beINT1 > TF1 > INT0 > TF0 > Serial



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Interrupt inside an interrupt

Higher-priority interrupt can interrupt a low-priority interrupt.

An interrupt cannot be interrupted by a low-priority interrupt

No low priority interrupt can get the immediate attention of the

CPU until the 8051 has finished servicing the high-priority

interrupts

Main Main Main

ISR ISR

ISR

ISR

Low ISR is

interrupted by a

higher priority ISR

higher ISRlower Lower ISR canninterrupt a higher

priority ISR


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Example 8.9 - Solution

ORG 0000HLJMP MAIN ;by-pass interrupt vector table

; ISR for INT1

ORG 0013HLJMP ISR_INT1 ; jump to ISR_INT1

; Main ProgramORG 30H

MAIN: MOV IE,#10000100B ;enable external INT 1HERE: SJMP HERE ;stay here until get interrupted

; INT1 ISR

ISR_INT1: SETB P1.3 ;turn on LEDMOV R3,#255BACK: DJNZ R3, BACK ;keep LED on for a while

CLR P1.3 ;turn off LEDRETI

END


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The 8051 Microcontroller and Embedded Systems -

Using Assembly and C, Mazidi

Chapter 9 P.239 P.255

Chapter 11 P.317 P.339

Read referenceRead reference


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EEE3410 Microcontroller ApplicationsDepartment of Electrical Engineering

END of Lecture 8 Timer Programming and InterruptsTimer Programming and Interrupts

interrupts programming

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