Sure enough, went back to the data sheet looking for info regarding
reads/writes.  It's there in section 10.4.  You must write the high byte
first because that goes in to a buffer that is written when you write the
low byte.  Original code works like a charm after flipping the order of the
high and low byte writes.  Thank you Scott and everyone else that took the
time to respond.



On 11/15/06, Scott Dattalo <scott@dattalo.com> wrote:
>
> On Tue, 2006-11-14 at 18:07 -0800, Shawn Wilton wrote:
>
>
> > Does anyone see anything wrong with this code:
> >
> > //Set page 103 for details regarding the +2
> >     tmr0l = (unsigned char)65000;
> >     tmr0h = (unsigned char)(65000 >> 8);  //Right shift so the top byte
> > becomes the bottom byte
>
> Others may or may not be correct about the default signedness of
> chars, but I believe your problem is even more fundamental. When
> writing to a 16-bit timer, you must write the high byte first. The
> write to the low byte will initiate a full 16-bit to real
> hardware. Similarly, to read a 16-bit timer, read the low byte
> first. This will initiate an access of the high byte too which you may
> read on the subsequent instruction.
>
> Scott
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-- 

Shawn Wilton (b9 Systems)
http://b9Systems.com  <- New web page
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