• State
  • Locked Locked
  • Replies 2 replies
  • Subscribers 82 subscribers
  • Views 167 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TMDSCNCD28388D: Sample Code

May
Genius 9880 points
Part Number: TMDSCNCD28388D
Other Parts Discussed in Thread: C2000WARE

Hi,

Customer is looking for a sample code regarding the device, below is the details.

"

I have read referenced links of previous posts prior to contacting TI CSC. However, I would appreciate more if you could provide sample code related to SCI-A to SCI-B link (SCI duplex configuration), or perhaps there is a simple way (e.g. by commenting out few lines of code) to obtain it from the SCI-A loopback demo?

Also, what is the maximal baud-rate that reliable SCI link (8-N-1) can sustain, assuming galvanically isolated communication channel interfaces and shielded cables are employed (e.g. baud-rates over 1Mbps), as well as 20MHz or 50MHz crystal oscillator frequency ?

Thank you in advance.

Regards,
Maynard

  • 0
    TI__Mastermind 22810 points

    Hi Maynard,

    We do not have a sample program that showcases communication between SCI-A and SCI-B. We have an example for SPI that could be good to look at to understand the flow between communication (SCI will have a similar flow). (C:\ti\c2000\C2000Ware_version\driverlib\f2838x\examples\c28x\spi\spi_ex3_external_loopback_fifo_interrupts.c)

    You'll need to set one of the up one of the modules as a master which the SCI-A loopback example already does and then configure the other module as a slave. Make sure to setup the correct GPIOs for SCI-B. 

    The maximum baud rate SCI can have is LSPCLK/16. The LSPCLK is derived from SYSCLK. Supposing SYSCLK is configured to be the highest frequency (200MHz), LSPCLK can equal 200MHz as well. Therefore, the max baud rate would be 200MHz/16 = 12.5Mbps. 

    Best Regards,

    Marlyn

  • 0 in reply to Marlyn Rosales Castaneda20
    Intellectual 440 points

    Hi Maynard,

    Thanks a lot to you and your colleagues for providing me the answer which enables significant upgrade of our service ports.

    Best regards

    Nenad Težak