First, you prove that a prescaler is there (which I doubt).  You can=20
do that by getting CCS to output a lst file, and you analyze what the=20
compiler is doing before it gets to main().  You check to see if it is=20
doing anything with the clock to add a prescaler.  You can also look at=20
the disassembly to sanity check the values passed to the delay()=20
function (it's probably a macro).
My guess is that a floating input pin RBx is causing interrupts and the=20
isr is using up most of the processors time.  So, you check to see if=20
the compiler put any isr vectors in there.  And you prevent portb=20
interrupt on change on floating inputs by enabling portb pullups.
If you have debugger capabilities, it is easier to do some of these=20
things by setting breakpoints, examining register contents, and single=20
stepping.   Inserting dummy code in main() before the loop will let you=20
set a breakpoint.  The dummy code should do something that doesn't get=20
optimized out, like "unsigned char n; n =3D PORTB;"
If it's the watchdog timer, or something else causing a reset, you will=20
find that if you set a breakpoint before the loop, the breakpoint will=20
be hit more than once.  You can also set a breakpoint at the isr vector=20
to see if it is a robue interrupt.

Joe W


On 6/18/2011 5:03 AM, geetansh wrote:
> how can i remove that prescaler then?
>
>
>
> RussellMc wrote:
>>> when the delay between on off  was set
>>> 10ms the leds took 2.3sec
>> 2300 mS / 10 mS ~~~=3D 256
>>
>> Delay MAY be out by a factor of 256, indicating that there is possibly a
>> prescaler in the clock that is not being accounted for.
>>
>>
>>                     Russell
>> --=20
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>>


--=20
Joe Wronski
Stillwater Embedded Engineering
www.stillwatereng.net

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