Delay doesn't seem to work - Help! - Electronic Circuits Projects Diagrams Free

bsodmike · 14th June 2004, 05:01 AM

Code:

Delay   movlw	d'39'		; load this to the timer

Loop	btfss	_intcon,2	; if tmr0 overflow is not set
	goto	Loop		; loop, else skip

	bcf	_intcon,2	; clear the timer interrupt flag

	decfsz	w,w		; loop until the wreg is zero
	goto 	Loop		; it isn't zero so loop

	return			; zero, so we return

Not sure why this doesn't seem to work. My thinking is as follows:

Clock speed of 4000000Hz/4 = 1000000hz

This 1/4th value is further divided by my pre-scaler of choice /256: 1000000hz/256 = 3906.25hz

3906.25 pulses take 1 second - so to count 10ms delay we need to count when this has tripped over 39 times. To prove this 1/3906.25 = 2.56x10-4 * 39 = 9.98ms.

Right or wrong? Is there anything else needed to get the TMR0 overflow to work?

Thanks,

Mike

bsodmike · 14th June 2004, 06:01 AM

Any ideas?

**Nigel Goodwin** · 14th June 2004, 06:17 AM

Quote:

Originally Posted by bsodmike

Any ideas?

My main idea is 'why bother'.

Why not simply use the delay code generator on the PICList to generate delay code for you? - it's far more versatile!.

Are you sure you are setting everything up correctly?, you've only posted a small piece of the code, so there's no indication of how it's set up at all.

bsodmike · 14th June 2004, 06:20 AM

I move to RPO and have set this:

movlw b'10100000'
movwf _intcon

however by using this delay:

Delay clrf _tmr0
tloop movlw d'39'
subwf _tmr0,w
btfss _status,0
goto tloop
return

I have got it to work - but it alters tmr0. I could also count 10000 lines of code (big nested loop...) and it'll work...

I will settle for the way I've posted above...thanks

Mike

bsodmike · 14th June 2004, 06:30 AM

Code:

; Delay = 0.01 seconds
; Clock frequency = 4 MHz

; Actual delay = 0.01 seconds = 10000 cycles
; Error = 0 %

	cblock
	cnt1
	cnt1_1
	endc

Delay
			;9993 cycles
	movlw	0xCE
	movwf	cnt1
	movlw	0x08
	movwf	cnt1_1
Delay_0
	decfsz	cnt1, f
	goto	$+2
	decfsz	cnt1_1, f
	goto	Delay_0

			;3 cycles
	goto	$+1
	nop

			;4 cycles (including call)
	return

This is what the generator gives me. CE = 206 and 8, so wouldn't this mean 206 * 8 ??? How is it 9993 ?!?!

**Nigel Goodwin** · 14th June 2004, 08:17 AM

Quote:

Originally Posted by bsodmike

This is what the generator gives me. CE = 206 and 8, so wouldn't this mean 206 * 8 ??? How is it 9993 ?!?!

It's not as simple as that, the inner loop (cnt1) loops round to 255 after it reaches zero.

Don't worry about it, just use it!.

motion · 14th June 2004, 09:47 AM

I took a quick glance at your code. You can't use the ff. sequence because the W register is not a regular file register.

Code:

   decfsz   w,w      ; loop until the wreg is zero 
   goto    Loop      ; it isn't zero so loop

It will compile without error but because the value of W is defined as "0", the register that will be accessed will be the INDF register. You have to use a real file register instead of W.

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14th June 2004, 05:01 AM	(permalink (permalink))
bsodmike	Delay doesn't seem to work - Help! Code: Delay movlw d'39' ; load this to the timer Loop btfss _intcon,2 ; if tmr0 overflow is not set goto Loop ; loop, else skip bcf _intcon,2 ; clear the timer interrupt flag decfsz w,w ; loop until the wreg is zero goto Loop ; it isn't zero so loop return ; zero, so we return Not sure why this doesn't seem to work. My thinking is as follows: Clock speed of 4000000Hz/4 = 1000000hz This 1/4th value is further divided by my pre-scaler of choice /256: 1000000hz/256 = 3906.25hz 3906.25 pulses take 1 second - so to count 10ms delay we need to count when this has tripped over 39 times. To prove this 1/3906.25 = 2.56x10-4 * 39 = 9.98ms. Right or wrong? Is there anything else needed to get the TMR0 overflow to work? Thanks, Mike __________________ "Ambition is the last refuge of failure..." --Oscar Wilde "Success is not final, failure is not fatal: it is the courage to continue that counts." -- Winston Churchill

14th June 2004, 06:01 AM	(permalink (permalink))
bsodmike	Any ideas? __________________ "Ambition is the last refuge of failure..." --Oscar Wilde "Success is not final, failure is not fatal: it is the courage to continue that counts." -- Winston Churchill

14th June 2004, 06:20 AM	(permalink (permalink))
bsodmike	I move to RPO and have set this: movlw b'10100000' movwf _intcon however by using this delay: Delay clrf _tmr0 tloop movlw d'39' subwf _tmr0,w btfss _status,0 goto tloop return I have got it to work - but it alters tmr0. I could also count 10000 lines of code (big nested loop...) and it'll work... I will settle for the way I've posted above...thanks Mike __________________ "Ambition is the last refuge of failure..." --Oscar Wilde "Success is not final, failure is not fatal: it is the courage to continue that counts." -- Winston Churchill

14th June 2004, 06:30 AM	(permalink (permalink))
bsodmike	Code: ; Delay = 0.01 seconds ; Clock frequency = 4 MHz ; Actual delay = 0.01 seconds = 10000 cycles ; Error = 0 % cblock cnt1 cnt1_1 endc Delay ;9993 cycles movlw 0xCE movwf cnt1 movlw 0x08 movwf cnt1_1 Delay_0 decfsz cnt1, f goto $+2 decfsz cnt1_1, f goto Delay_0 ;3 cycles goto $+1 nop ;4 cycles (including call) return This is what the generator gives me. CE = 206 and 8, so wouldn't this mean 206 * 8 ??? How is it 9993 ?!?! __________________ "Ambition is the last refuge of failure..." --Oscar Wilde "Success is not final, failure is not fatal: it is the courage to continue that counts." -- Winston Churchill

14th June 2004, 09:47 AM	(permalink (permalink))
motion	I took a quick glance at your code. You can't use the ff. sequence because the W register is not a regular file register. Code: decfsz w,w ; loop until the wreg is zero goto Loop ; it isn't zero so loop It will compile without error but because the value of W is defined as "0", the register that will be accessed will be the INDF register. You have to use a real file register instead of W. __________________ "Having to do with Motion Control"