INTERRUPT_RTCC2.SXB question

Rsadeika

I am considering modifying the program to run at an interrupt rate of returnint 104, so this value slows the interrupt down to 26 uS. Now, in order to make the program work, do I change the B19K2 CON 4 to B19K2 CON 8. My thinking is, if you slow the interrupt down , you have to increase the sampling rate. Is this a correct assumption? I would like to have this program work at returnint 104, but I am not sure what is really involved.

Any ideas?

Thanks

Ray

Bean

If you half the interrupt rate, you need to half the sample period. So you would need to change "B19K2 CON 4" to "B19K2 CON 2". And two sample periods might not be enough to get reliable data.

Bean.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Cheap used 4-digit LED display with driver IC·www.hc4led.com

Low power SD Data Logger www.sddatalogger.com
SX-Video Display Modules www.sxvm.com

There are only two guaranteed ways to become weathy.
Spend less than you make.
Make more than you spend.
·

Rsadeika

Thanks Bean,

So, if I use a 9600 baud selection, the B9600 con 8 would be B9600 con 4, which would be the same value as the original B19K2 con, so that should do the trick, right? And, I guess I would have to change tht BitTm, and BitTm15 values to the new baud rate. I guess in these types of examples it would be nice if their were two, one at one baud rate, and the other at a different baud rate. That way I guess one could look at the code and see what is changing at the different baud rates, and in the end learn some general concepts about serial coding.

Thanks

Ray