need timer interrupt using assembly language.

i am attaching u my program can u guide me how i multiply 3 digit no 132. so kindly look into this

well 132 is a 8-bit number and ADC values is also 8-bit so just multiply both using MUL AB instruction in 8051.

put ADC value in A
132 in B

AxB = BA => B(MSB), A(LSB)

a 16-bit value.. you can easily manipulate the result for display.

THANK YOU FOR UR HELP BUT NOW I AM FINDING DIFFICULTY IN DIVIDING 16 BIT BY 8 BIT HOPE U GIVE ME A EXPLANATION. COZ AS U DO IT USING SUBTRACTION I GONE THRU UR PROGRAM BUT HOPE U GIVE ME EXAMPLE OF ATLEAST ONE CYCLE MEANS IF I WANNA CONVERT FFH TO 336V.
THANK YOU.

i think you can find 16 bit division in code section..

check this link:
http://www.8051projects.net/downloads35.html

I want timer interrupt program of 5 sec in assembly language help me

what value xtal are you using ? direct 5 sec delay is not possible, you'd have to first do a delay for 50millisec and then a counter for 100 times , 50 millisec X 100 times = 5000 msec or 5 seconds.


Arun

i want an example of timer interrupt in assembly language if u can coz i have completed with display to show 336 vdc on 7 segment display.

here is a small example of timer 0 interrupt

org 0H
sjmp main

org 0BH ;ISR location for Timer 0 interrupt
sjmp T0_ISR

;Main routine
main:
mov TH0,#00 ;delay value
mov TL0,#00
clr TF0 ;clear timer flag
setb ET0 ;enable timer0 interrupt
setb EA ;enable global interrupt
setb TR0 ; enable timer
sjmp $

T0_ISR:
clr TF0 ;clear interrupt
;do something in ISR
RETI

end