Lab # 9

Task # 1

OBJECT

Write a program for generating a pulse which has a low portion for 10ms and a high portion for 5ms.Use timer 0 in mode 1 for generating delay.

PROGRAM COMMENTS

org 00h ; originate the code from memory location 00h

main: ;main label

here: setb p1.1 ;here label, making p1.1 high

acall delay ;calling delay subroutine

clr p1.1 ;making p1.1 low

acall delay ;calling delay subroutine

acall delay ;calling delay subroutine

sjmp here ;unconditional jump tohere label

delay: ;delay subroutine

mov TMOD,#01h ;Selecting timer 0 mode 1

mov Th0,#0eeh ;Th0 = EEH

mov Tl0,#00h ;Tlo = 00h

setb tr0 ;Starting timer 0

h1:jnb Tf0,h1 ;jump to label h1 until Tf0 becomes high, H1 label

clr tr0 ;Stopping timer 0

clr tf0 ;clearing TF0 flag

ret ; return back to normal routine

end ;end directive, to end program

DELAY CALCULATION

For 5 ms

5x10^(-3)/1.085x10^(-6) = 4608

65536-4608=60927 60927 = EE00H

RESULT

[pic]

Task # 2

OBJECT

Repeat task # 1 using timer 0 in mode 2.

PROGRAM COMMENTS

org 00h ; originate the code from memory location 00h

main: ;main label

here:setb p1.2 ;here label, making p1.2 high

acall delay ;calling delay subroutine

clr p1.2 ;making p1.2 low

acall delay ;calling delay subroutine

acall delay ;calling delay subroutine

sjmp here ;unconditional jump to here label

delay: ;delay subroutine

mov TMOD,#02h ; Selecting timer 0 mode 2

mov R0,#20 ;Moving value of 20 in R0

mov th0,#1Ah ;Th0 = 1AH

Here3: ;Here3 label

setb tr0 ;Starting timer 0

back:jnb tf0,back ;jump to label back until Tf0 becomes high, back label

clr tr0 ;Stopping timer 0

clr tf0 ;clearing TF0 flag

djnz R0,Here3 ;decrement in R0 and jump to Here3 label until R0=0

ret ; return back to normal routine

end ;end directive, to end program

DELAY CALCULATION

FFh – 1Ah+1=E6h

E6h=230

230 x (1.085x10^(-6))=2.4955x10^(-4)

Since loop runs 20 times therefore

(2.4955x10^(-4)) x 20 = 5x10^(-3)

RESULT

[pic]

Task # 3

OBJECT

Write a program code for generating a pulse of 75% duty cycle using timer 1 mode 2.Take 1 sec for complete time period of single pulse.

PROGRAM COMMENTS

org 00h ; originate the code from memory location 00h

here1: ;here1 label

setb p1.3 ;making p1.3 high

acall delay ;calling delay subroutine

acall delay calling delay subroutine

acall delay ;calling delay subroutine

clr p1.3 ;making p1.3 low

acall delay ;calling delay subroutine

sjmp here1 ;unconditional jump to here1 label

delay: ;delay subroutine

mov tmod,#20h ; Selecting timer 1 mode 2

mov th1,#8dh ;Th0 = 1AH

mov r0,#10 ;Moving value of 10 in R0

here2:mov r1,#200 ;Moving value of 200 in R0

here:setb tr1 ;Starting timer 1

back:jnb tf1,back ;jump to label back until Tf0 becomes high, back label

clr tr1 ;Stopping timer 0

clr tf1 ;clearing TF0 flag

djnz r1,here ;decrement in R0 and jump to here label until R1=0

djnz r0,here2 ;decrement in R0 and jump to here label until R0=0

ret ; return back to normal routine

end ;end directive, to end program

DELAY CALCULATION

FFh – 8Dh+1=73h

73h=115

115 x (1.085x10^(-6))= 1.24775x10^(-4)

Since loop runs 200x10 times therefore

(2.4955x10^(-4)) x 200x10 = 0.25 sec

RESULT

[pic]

Task # 4

OBJECT

Write a program code for program counter 0 in which you have in count the no.s of pulses provide to input bit p1.6 and also keeping value in r2 register.

PROGRAM COMMENTS

org 00h ; originate the code from memory location 00h

mov TMOD,#06h ; Selecting counter 0 ,mode 1

setb p1.6 ;make port 1.6 as input

mov Th0,#0h ;clear Th0

setb tr0 ;Starting the counter

back: ;Back label

mov A,TL0 ;Get copy of count from TLO

mov r2,TL0 ;send it to re register

jnb TF0,back ;jump tp label back if Tf0 = 0

end ;end directive, to end program