DELAY and the 8051


Hi All
I've obtained the "The 8051 Micro Controller and Embedded Systems Using
Assembly and C-2nd-Ed -", and in this manual the DELAY routine is described
as below, which really confuses me a lot
######################################
In original 8051, one machine cycle lasts 12 oscillator periods
So, 1 oscilator period for a 12Mhz Crystal will be 12.0 Mhz / 12 = 1Mhz
Therefore one machine cycle is is 1/1Mz = 1us (1 micro second)
For 12Mhz the delay is as follows for this example:
DELAY : MOV R3 ,#200 ;= 1 Machine Cycle
REPEAT: DJNZ R3,REPEAT ;= 2 Machine Cycles
RET ;= 2 Machine Cycles
END
Therefore the above routine equals to a delay of 403us
"#200" is equal to a delay of , [(200x2)+1+2]x1us = 403us (403
Microseconds)
######################################
If the above is 100% correct, how will a 1 Second delay look like ?
I see a few examples showing a 1 sec delay as " MOV R1, #0FFH" - but don't
seem to "reverse engineer" the "#0FFH" back to 1 Second using the above
information
Can someone please help me to understand this, I don't just want to use this
without knowing what I do
Kind Regards
Lodewicus Maas
Reply to
Lodewicus Maas
Loading thread data ...
mov r3,#0ff mov r4,#0ff mov r5,#08 delay: djnz r3,delay 256x2uS ( 2 instructions)= approx 512uS djnz r4,delay 256xx512uS = 131 mS approx djnz r5,delay 8 x 131 mS = appex 1 sec ret sort of thing... Much better to use a timer counter set to time at say 1mS and do the 'counting ' in the RTC interrupt routine...
Reply to
TTman
The delay routine timing you posted is correct. The information about the 1 sec delay is, obviously, missing some information. The 'standard' way of making longer delays is to make a fixed delay and call it multiple times, think of nested loops, or calling a fixed delay in a loop.
Reply to
WangoTango
Hi TTman
You're completely lossing me here - Hex FF is equal to 255 - is this 0 to 255 which is in fact 256 ?:
In line 1) you set delay=512us(2us x ff), and then in line 2) you multiply the delay set in line 1) with 256 (ff), and then in line 3) you multiple the new delay value set in line 2) with 8
Do I understand this correctly ?
Can I build up a few "DELAY" routines to obtain different delay timings like DELAY_1, DELAY_2 ........ DELAY_N, everytime using its own r3,r4, .........rn value , or is there any limit to the total DELAY routines which you can declare ?
mov r3,#0ff mov r4,#0ff mov r5,#08 1) delay: djnz r3,delay 256x2uS ( 2 instructions)= approx 512uS 2) djnz r4,delay 256xx512uS = 131 mS approx 3) djnz r5,delay 8 x 131 mS = appex 1 sec
Lodewicus Maas
Reply to
Lodewicus Maas
YES....
YES, ABSOLUTELY, BUT U MAY NEED R5 AS WELL
Reply to
TTman
Thank you TTman - much appreciated
Reply to
Lodewicus Maas
Setting the register to #0ff only is 255 If you start with #00 then it will be 256
Reply to
Eddards
ed
t
his
Now that others have shown you how to do it, you should also know that you should almost never waste processor cycles in long delays, at least in "real" (as opposed to educational or hobby) applications. For example, in your case of a 1 second delay your processor could have done a million clock cycles worth of useful work.
So go ahead now and understand how to nest delays, but also understand that in real life a 1 second delay would usually be achieved in some other, more efficient manner (typically involving timers and interrupts).
Mike
Reply to
snarflemom

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