time delay

Comments

• kwinn Posts: 8,697

  2013-01-20 11:10 edited 2013-01-20 11:10

  The longest delay waitcnt can provide is about 56 seconds (2^32 clocks) at 80MHz. What you need to do is put a shorter wait in a repeat loop. I would suggest:
  

     repeat 300
     waitcnt(clkfreq + cnt) ' one second delay
  

  
  This will be very close to a 5 minute delay, but slightly more since it does not take execution time into account.
• mikea Posts: 283

  2013-01-20 11:13 edited 2013-01-20 11:13

  Understood, thank you kwinn-mike
• Mike Green Posts: 23,101

  2013-01-20 11:19 edited 2013-01-20 11:19

  A more accurate scheme would be:

  PRI delay5Minutes | time
     time := cnt
     repeat 300
        waitcnt(time += clkfreq)   ' delay 1 second
  

  This works by waiting for one second beyond the current time. Since all time references are relative to the initial reference to cnt, any overhead in the loop is absorbed into the wait.
• Mike G Posts: 2,702

  2013-01-20 11:22 edited 2013-01-20 11:22

  Here's another way using a cog as a timer and a global variable
  

  CON
    _clkmode = xtal1 + pll16x     
    _xinfreq = 5_000_000
  
  
  OBJ
    pst             : "Parallax Serial Terminal"
  
  DAT
    seconds   long  $00
  
  VAR
    long stack[20]
  
  PUB Main
      pst.Start(115_200)
      pause(500)
  
      cognew(SecondCounter, @stack)
  
      repeat
        pst.dec(seconds)
        pst.char(13)
        if(seconds == 20)
          'Do something
          SetSeconds(0)
        pause(1000)
      
  
  PUB GetSeconds
    return seconds
  
  PUB SetSeconds(value)
    seconds := value
  
  PUB SecondCounter
    repeat
      pause(1_000)
      seconds++
    
  PRI pause(Duration)  
    waitcnt(((clkfreq / 1_000 * Duration - 3932) #> 381) + cnt)
    return