Semaphore_post() Function about 1000 cycles?

h.web

Other Parts Discussed in Thread: TMS320C28346

Hi,

i want to use a semaphore to send the event from IRQ to the task. But the Semaphore_post() call needs around 3,5us at 300Mhz which means around 1000 Cycles to post the semaphore. I thinks this is a quite long time?! Why is there so much time needed and is there a way to reduze event generation?

I need a software cycle time of 50us. But now the semaphore alone consumes 3,5us (7%) of the time.

Best regards

Holger

Sys/Bios 6.33.04.39

TMS320C28346

over 13 years ago

0 MarkGrosen over 13 years ago

TI__Expert 4125 points

Holger,

That does not look right - see the benchmarks with this release:

http://software-dl.ti.com/dsps/dsps_public_sw/sdo_sb/targetcontent/bios/sysbios/6_33_04_39/exports/bios_6_33_04_39/packages/ti/sysbios/benchmarks/doc-files/Timings_C28_large.html

There the time is 304 cycles. Perhaps there is a problem with wait states with the memory containing the code?

Mark

0 h.web over 13 years ago in reply to MarkGrosen

Prodigy 110 points

Hi Mark,

your answer give me the right hint. I changed the SYS/BIOS library type to "non-instrumented" and the sem_post call reduced to 660ns. So with the "instrumented" view there is much time needed to save the debug data to memory.

Thank you for your fast response.

Best regards

Holger

Processors

Processors forum

Semaphore_post() Function about 1000 cycles?