Changing the frequency of a PIC inside a program.

[mkstevo]

Is there a way of changing the oscillator frequency 'on the fly'?

Thanks to Chris Roper, the white noise source is running well. The 'noise' generated sounds rather different at different ocsillator frequencies and I thought it might be useful to be able to change this using a button to select the different oscillator frequencies.

Is this possible with GCB, or is the #Chip compiler directive the only way of setting the frequency?

[Anobium]

Yes, you can. With ease but... you will impact the delay timing and other timing calcs.

So, set the chip frequency as normal. Then, when you change simply proportionally change the delays. Remember, the USART would change also - so, just recalc and be aware of the proportional change.

How?
I would create a program with the different frequencie(s). Compile. Open the ASM. Lift the top piece of INITSYS and paste into your main program as a new Sub.

;Source: system.h (99)
INITSYS
;asm showdebug OSCCON type is 105 'Bit(SPLLEN) Or Bit(IRCF3) And NoBit(INTSRC) and ifdef Bit(IRCF3)
;osccon type is 105
;OSCCON = OSCCON AND b'10000111'
;OSCCON = OSCCON OR b'01100000'
;Set IRCF3 On
  banksel OSCCON
  bsf OSCCON,IRCF3
;Set IRCF2 On
  bsf OSCCON,IRCF2
;Set IRCF1 Off
  bcf OSCCON,IRCF1
;Set IRCF0 Off
  bcf OSCCON,IRCF0
;Set SPLLEN Off
  bcf OSCCON,SPLLEN
;Ensure all ports are set for digital I/O and, turn off A/D

So, the code you needs for this is the Great Cow BASIC code.

This code is for ,2

Set IRCF3 On
Set IRCF2 On
Set IRCF1 Off
Set IRCF0 Off
Set SPLLEN Off

Then, I create new code with ,8. If you look carefully Set IRCF1 On

Set IRCF3 On
Set IRCF2 On
Set IRCF1 On
Set IRCF0 Off
Set SPLLEN Off

Now create all that you need. Take the code from INITSYS and stop at the line ;Ensure all ports are set for digital I/O and, turn off A/D

[mkstevo]

Thanks.
As I'm not using any timing or other bits and pieces, I shouldn't need to do much else.

[mkstevo]

Here is my code for adjusting the frequency.
A button on PortA.3 is tested for being pressed for a period of time which then increments the frequency. The button must be released for a period of time before the next change can be made.

#Chip 12F1840, 32

'Inspired by the Practical Electronics White Noise project, September 2019
'White Noise source
'Using a 31 bit shift register random generator XOR Q31 and Q28

'Most of this work was done by Chris Roper
'of the GCB forum for which I am most grateful.

#Define Button  PortA.3
Dir Button      In

#Define Pressed 0

Dim Noise       As Long  ' Create a 32bit shift register
Let Noise     = Random   ' Seed the LSFR
Dim PressTime   As Word
Dim ReleaseTime As Word
Dim TimeOut     As Word

Dim LastFreq    As Byte
Dim Freq        As Byte

Dim LS          As Bit
'This allows us to the port on, or off. Without it the port goes
'momentarily low in each cycle, even if it is intended to stay high

'You end up with this:

'____|‾|‾|‾|____|‾|‾|____

'Instead of this:

'____|‾‾‾‾‾|____|‾‾‾|____

Let PressTime   = 0
Let ReleaseTime = 0
Let LastFrq     = 0 '0 + 1 = 1 = 2MHz
Let Freq        = Frq2


#Define  Frq01  5
#Define  Frq05  7
#Define  Frq1   11
#Define  Frq2   12
#Define  Frq4   13
#Define  Frq8   14
#Define  Frq16  15
#Define  Frq32  30

Wait 100 mS
ChangeFrequency

Do

  If Button=Pressed Then
    If PressTime > TimeOut Then
      If ReleaseTime = 0 Then
        ChangeFrequency
        Let ReleaseTime = PressTime
      End If
    Else
      Let PressTime = PressTime + 1
    End If
  Else
    If ReleaseTime > 0 Then
      Let ReleaseTime = ReleaseTime - 1
      If ReleaseTime > TimeOut Then
        Let ReleaseTime = TimeOut
      End If
    End If
    If ReleaseTime = 0 Then
      Let PressTime   = 0
    End If
  End If

  LSFR

Loop

Sub ChangeFrequency

  Let LastFreq = LastFreq + 1
  If LastFreq > 8 Then 
    Let LastFreq = 1
  End If


  Select Case LastFreq
    Case 1
      Let Freq = Frq2
      Let TimeOut = 400
    Case 2
      Let Freq = Frq4
      Let TimeOut = 800
    Case 3
      Let Freq = Frq8
      Let TimeOut = 1600
    Case 4
      Let Freq = Frq16
      Let TimeOut = 3200
    Case 5
      Let Freq = Frq32
      Let TimeOut = 6400
    Case 6
      Let Freq = Frq01
      Let TimeOut = 25
    Case 7
      Let Freq = Frq05
      Let TimeOut = 50
    Case 8
      Let Freq = Frq1
      Let TimeOut = 200
  End Select

  If Freq.0 = 1 Then
    Set IRCF0 On
  Else
    Set IRCF0 Off
  End If

  If Freq.1 = 1 Then
    Set IRCF1 On
  Else
    Set IRCF1 Off
  End If

  If Freq.2 = 1 Then
    Set IRCF2 On
  Else
    Set IRCF2 Off
  End If

  If Freq.3 = 1 Then
    Set IRCF3 On
  Else
    Set IRCF3 Off
  End If

  If Freq.4 = 1 Then
    Set SPLLEN On
  Else
    Set SPLLEN Off
  End If

End Sub

Sub LSFR
'This line causes an error:
'Error: GCASM: Symbol 3.0 has not been defined
  'Let C = Noise.28 XOR Noise.31 ' Test and set Carry flag

'Replaced with this If...Then
  If Noise.27 <> Noise.30 Then
    Let C = 1
  Else
    Let C = 0
  End If

  Rotate Noise Left             ' Rotate all 32 bits
  If Noise.31 <> LS Then
    Let LS = Noise.31
    Let PortA.2 = LS            ' Output bit 31
  End If
End Sub

[Chris Roper]

You shouldn't be running the Processor at 32MHz.
You are hearing a strange pitch because you are running 4 times faster than the original code was designed for.
The ASM in the article and the GCBASIC version both expect a Clock frequency of 8MHz.
Change your Chip definition to:

#Chip 12F1840,8

and I think you will find that your pitch issues are reduced if not eliminated with no need to have a Clock adjust button.

[mkstevo]

This is intended to go into a unit that will connect to headphones (with appropriate attenuation) to create "noise" to mask external sounds (snoring!) that prevent me from sleeping. The actual frequency of the "noise" isn't important to me, but the effectiveness of it helping me to sleep is. I wanted the option of changing the pitch (if noise has any pitch at all) to give me a few things to try.

If anything, the very low frequencies seemed to have most potential. I was using the lowest (125kHz) when debugging as I tried to decode why the XOR wasn't working, eventually shifting all zeros into the register.

[mkstevo]

The orignal device from P.E. was a 12F617 (I originally mistakenly said it was a 12F675). The source code comments suggest it was running at 2MHz?

; 2 to the power of 31 -1 results = 2,147,483,647
; clock is 2MHz so 13 cycles is 153.8451kHz generating max frequency of 76.923kHz
; minimum frequency is 76.923kHz/ 2,147,483,647 = 35.83uHz
; sequence repeats after 13us x 2,147,483,647 = 27,917,280s or 7.75hours

[Chris Roper]

The PIC has a 4 phase Clock so that the Fetch, Execute, Store and Branch portions of the instruction execution can overlap. If you read any of the Datasheets you will see mention of FSC/4 which is Front Side Clock or Clock Speed divided by 4. For that reason the Clock frequency of the PIC is divided by 4 to get that actual instruction Cycle time.In this case an 8Mhz Clock speed would result in a 2MHz Instruction cycle time hence 2Mhz being used to calculate the bandwidth in the article.