How did you get to 12.5x10-6 seconds?

John Dammeyer



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> -----Original Message-----
> From: pic microcontroller discussion list
> [mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of Russell C. Hay
> Sent: Thursday, March 13, 2003 2:03 PM
> To: PICLIST@MITVMA.MIT.EDU
> Subject: [PIC] question about timing..
>
>
> I just want to make sure that I'm thinking about this
> correctly.  Please
> correct me if I'm wrong in this.
>
> So assuming a 20Mhz PIC (in this case, it's going to be a
> 16f876), that
> means we have 20x10^6 clock cycles per second (duh, right?).  Each
> instruction cycle is 4 clock cycles, so now we have 5x10^6
> instructions per
> second, which is basically 12.5x10-6 seconds (12.5 nanoseconds) per
> instruction.  So, to get a 1 microsecond delay, I'd need 80
> instruction
> cycles, which in turn means that I'd need 80,000 instruction
> cycles to get a
> 1 milisecond delay.  Is this right?
> I think my logic (and math) is sound, but since the number is
> so big, I just
> want to ask someone who knows a bit more than me if I'm correct.
>
> -R
>
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