What is the total number of data frames needed to transfer an entire 1024 bytes block?

Also, what is the minimum amount of time needed to complete this method?

  • \$\begingroup\$ CAN has a maximum payload of 8 bytes per message. \$\endgroup\$
    – Martin
    Dec 5 '14 at 17:24
  • \$\begingroup\$ smells like homework to me. \$\endgroup\$
    – akohlsmith
    Dec 7 '14 at 5:24

You need to define the time cycle for your communication. It depends on the length of your frame (it depends on the identifier, 29 bits or 11 bits). Usually the maximum length is 126 bits for a data frame (CAN 2.0), and the bite rate in your case (1 Mbit/s). You also need to introduce the bit stuffing. You must also define the maximum distance of your CAN installation.

  • \$\begingroup\$ 11 bit identifier it is \$\endgroup\$ Dec 5 '14 at 16:10
  • \$\begingroup\$ I dont have any more information apart from this \$\endgroup\$ Dec 5 '14 at 16:13
  • \$\begingroup\$ in your case you have Tcycle = N*(Lmax/C+r) where C=1Mbps and Lmax is the maximum frame lenth (depend on the payload) and N numbre of frames and r is the frame latency (Rx+Tx) from to node. \$\endgroup\$
    – R Djorane
    Dec 5 '14 at 18:30
  • \$\begingroup\$ if you take the formula without the latency than assumes that the max frame lenth is 15 bytes than you have to transmit at N=1024/15 with N is integer so you have N= 69 frames so you have T= 69*(15*8/1Mbps)= 9 ms \$\endgroup\$
    – R Djorane
    Dec 5 '14 at 18:42
  • \$\begingroup\$ Isn't the maximum data field length in a CAN frame 8 bytes? \$\endgroup\$
    – markt
    Dec 5 '14 at 23:59

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