"Russell C. Hay" wrote:
>
> 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.

20MHz = 50nS per cycle.

4 cycles per instruction = 200nS.

1uS delay = 5 cycles
1mS       = 5000 cycles
1S        = 5 million cycles

etc

--
Best regards

Tony

mICros
http://www.bubblesoftonline.com
mailto:sales@bubblesoftonline.com

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