PIC16F84 ASM Delay program

[Samji]

For an assignment at college, I have to write an ASM program for the PIC16F84 microcontroller which:

Outputs number 2h to port B,
delays for 0.5 second and then outputs 3h to port B.

I don't think I have done this task correctly, so I have posted here to ask for anyone who understands PIC16F84 ASM to evaluate this code and tell me what needs changed.

; Written by      SAMUEL SAINT-PETTERSEN
; Date            19-04-2006
; File saved      delay.asm
; Device          PIC16F84
; Oscillator      XT (4 MHz)
;
; Function        Output number 2h to port B,
;                 delay for 0.5 second and then output
;                 3h to port B.
;
; ------------------------EQUATES--------------------------
  portb      EQU       0x06       ; portb equates to 0x06
  counter    EQU       0x01       ; countera equates to 0x01
; ---------------------------------------------------------
             ORG       0x00       ; origin is 0x00
start       clrw                 ; reset W register
             movlw     0x02       ; move 2h into W
             tris      portb      ; copy W tristate, portb outputs
             clrw                 ; reset W register

             movlw     0x05       ; move 5h into W
delay      movwf     counter    ; move 5h into countera
             goto      decrm      ; loop to decrm
             
decrm     decfsz    counter, 1 ; decrement countera if > 0
             goto      delay      ; loop to delay and loop if so

             clrw                 ; reset W register
             movlw     0x03       ; move 3h into W
             tris      portb      ; copy w trisate, portb outputs
             goto      start      ; loop back to start
             end                  ; end the program

Thanks in advance.

[Wooloomooloo]

I don't think you program is correct. The way it is now, I think it'll just loop forever, because you're always overwriting the value in counter with whatever is in W (I'm guessing 5h) at the beginning of the loop. Try putting the decremented value in W instead of placing it back into counter.
Also, with a 4MHz clock, you'll need 0.5 million non-branching instructions (each taking 4 clock cycles or 1 microsecond to execute) to get a 0.5 second delay. You need to count to way more than 5 .

[phreak]

edit: nevermind... I can't read

[Samji]

Originally Posted by Wooloomooloo

I don't think you program is correct. The way it is now, I think it'll just loop forever, because you're always overwriting the value in counter with whatever is in W (I'm guessing 5h) at the beginning of the loop. Try putting the decremented value in W instead of placing it back into counter.
Also, with a 4MHz clock, you'll need 0.5 million non-branching instructions (each taking 4 clock cycles or 1 microsecond to execute) to get a 0.5 second delay. You need to count to way more than 5 .

Thanks for your help. So decrementing a value by one each time would take 1 microsecond each time, so therefore decrementing 5 to 0 would take 5 microseconds.

So if a microsecond is 10^-6 of a second?

[Wooloomooloo]

1 microsecond is 10e-6 seconds. As it is now, your loop has 4 instructions:

delay    movwf    counter
         goto     decrm
decrm    decfsz   counter, 1
         goto     delay

Instructions 1 and 3 take 1 microsecond each; instructions 2 and 4 (goto) takes 2 microseconds, so one iteration of the loop takes 6 microseconds to execute. You're counting to 5 which gives you a delay of about 30 microseconds (the last iteration will be a bit shorter since the branch will be taken, skipping "goto delay").

[Samji]

Thanks for your help. Our tutor said that we had to have a delay of 0.5 second, but he didn't explain this.

[Wooloomooloo]

No problem. You should really take a look at the datasheet for the PIC if you haven't already, especially the chapter about the instruction set.