ahh .. I had the wrong field when inverting to get sec/intr in the
spreadsheet

Thanks..

-R

-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Mike Harrison
Sent: Thursday, March 13, 2003 2:15 PM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [PIC] question about timing..


On Thu, 13 Mar 2003 14:02:45 -0800, you 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.
>
>-R
No - your maths are wrong.
at 20MHz, you do 5M instructions/sec, 200ns per instruction, 5 instruction
times per microsecond,
5000 instruction times per millisecond

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