PIC interrupt priorities

nobody wrote on Tuesday, November 16, 2004:

Just getting started looking at the PIC 18 port to see if it will fit my needs. Thanks for all the work by the way. Very cool.

I ran across something that I am a "bit" confused on. (Pardon the pun…) The UART RX and TX interrupts are set as low priority interrupts (TXIP and RXIP bits = 0). And the ISR function is appropriately named "prvLowInterrupt" but the pragma that surrounds the ISR locates this code at 0x08, which according to my datasheet is the vector for the high priority interrupt. Have I missed something? I am not too new to PICs but my past applications were simple enough to employ polling. So I am new to interrupts in the PIC. Has my datasheet led me astray or might there be an error (perhaps in my head) here.

rtel wrote on Tuesday, November 16, 2004:

The PIC18 has a compatibility mode which effectively switches off this interrupt prioritisation mechanism - this is the default.  When interrupt priorities are not enabled the chip operates as per the mid range devices, and all interrupt vector to 0x08.

See the description of the bit IPEN (RCON<7>) and the interrupt section of the PIC18 manual for more details.

Thanks for you comments.

Regards.

nobody wrote on Wednesday, November 17, 2004:

Doh. Didn’t catch that on my first read through the interrupt section. Thanks for setting me straight.

I plan to add the PortB OnChange Interrupt. Should I just add it to the low priority interrupts, or would there be any advantage to handling it as a low priority and having the tick interrupt at a high priority? Seems like I would have to implement nested interrupts to use both priority levels.

Thank again.

rtel wrote on Wednesday, November 17, 2004:

Keeping it as a low priority would be the easiest - nesting the interrupts would require a larger stack (stating the obvious I know).

Provided the new ISR is kept short it is unlikely to cause a problem.  If the interrupts occur at the same time you may have a small jitter in the tick period, but as the tick runs from a free running timer any timing error will not accumulate over time  the next tick will still occur at the correct time.

Regards.