> > > an > > > > > idle > > > > > > > > bus. > > > > > > > > > > > > > > Technically what's needed is an idle condition on the bus as specified > > > > > > > in the ISO. i.e. 11 consecutive sampled recessive bits after a full > > > > > > > frame (if one is currently in transmission). > > > > > > > > > > > > Right. What happens if another CAN frames comes before there are > 11 > > > > > > consecutive sampled recessive bits? The IFS for CAN is 3 bits? > > > > > > > > > > Not quite. Between the Frames is an IFS that's correct but the IFS > consists > > > of > > > > > an Intermission which is 3 bits long + a bus idle condition of 11 bits. > Regular > > > > > frames have to wait for the IFS to elapse BUT there's an exception for > error > > > > > frames and overload frames. EF may be up to 12 bit, OF are 8 dominant > + 8 > > > > > recessive bits, browsing through the spec I think only 2 OFs can happen > > > > > consecutively. Adding another 32 bits to the formula should cover this. > > > > > > > > Re-reading the spec again I noticed that the part where I wrote > > > > that there's an "idle condition" after the intermission is > > > > wrong. What follows the intermission is just "bus idle", > > > > defined two paragraphs later as "The period of bus idle may be > > > > of arbitrary length." The 11 recessive bits can be removed from > > > > the formula again. The longest period (under the assumption > > > > there aren't multiple/continuous errors on the bus) will be > > > > Frame + Error Frame (12 bit) + 2 x Overload Frame. > > > ^^^^^^^^^^^^^^^^^^^^ > > > > > > How did you find that a error frame is 12 bits? From section 10.4.4 > > > "Specification of EF", I can read: > > > > > > The EF shall consist of two different fields. The first field > > > shall be given by the superposition of error flags contributed > > > from different nodes. The second field shall be the error > > > delimiter. > > > > > > Then: > > > > > > Two forms of error flag may be used, the active error flag and > > > the passive error flag, where > > > > > > - the active error flag shall consist of 6 consecutive > > > dominant bits, and > > > > > > - the passive error flag shall consist of 6 consecutive > > > recessive bits unless it is overwritten by dominant bits > > > from other nodes. > > > > > > Finally: > > > > > > The error delimiter shall consist of 8 recessive bits. > > > > > > So the error frame should be 14 bits (6 + 8), not 12, right? > > That was imprecise, you're right - I meant an Error Flag, not Frame and hence > the 8 recessive bits were missing. There's an active error flag + passive error > flag though which can be 6 bits long each. In section 10.4.4.2 The total length > of this sequence may vary between a minimum of 6 bit and a maximum of 12 > bit. > > The active error flag and the passive error flag may both occur, but > in that case, they occur as a superposition (c.f. above quotation). > This also means that the passive error flag is seen if and only if all > the nodes send a passive error flag. As long as one node sends an > active error flag, the superposition will hide any other passive error > flag. > > So, I think that the error flag is always 6. I think if one node detects a bit error (that's not on the bus but inside the receiving node) on the second bit of the intermission it will start sending the 6 dominant bits, starting on the third bit of the intermission. Now, according to 10.4.2.2 " A node sampling a dominant bit during its suspend transmission time or at the third bit of intermission shall accept it as SOF.", treating the next 5 dominant bits as part of the ID of a new DF and only then sending its own error flag, giving a total of 12 dominant bits followed by 8 recessive bits EF delimiter etc. Am I missing something here? For this (I believe worst case when only considering 1 EF you could deduct one bittime - the third bit in the intermission) > > > For the great total, if you want the absolute worst case, you should > > > consider that: > > > > > > - The error frame may start at any point. For example, you may > > > observe the first two bits of the intermission and have it > > > broken by an error frame. It may also appear in the middle of > > > a data frame. So you may consider cases such as: partial > > > frame + error frame + intermission + frame + ... > > > > > > - The overload frame is required to "destroy the fixed form of > > > the intermission field". So even if not explicitly specified, > > > I think that overload frame may start after the second > > > recessive bits of the intermission, e.g. frame + 2 bits of > > > intermission + overload frame + 2 bits of intermission + > > > overload frame + full 3 bits intermission > > > > > > - An error frame or an overload frame may trigger another error > > > frame if a node does not correctly receive one of the bits in > > > the fixed form delimiter. The only exception is the last bit > > > of that delimiter (c.f. section 10.11 "Error detection" > > > paragraph d) "Form error"). In other words, you can have > > > the two overload frames, then an error frame, then two > > > overload frames again. > > Correcting myself: that last example was invalid. There are at maximum > two overload frames before the next data or remote frame. So the > presence of an error frame should not reset the overload frame count. > > > > This is to say that the worst case scenario is just not something > > > worth consideration. > > > > ACK > > > The assumption that only one error frame occurs is pretty arbitrary. I > > > would suggest making it simpler and simply ignore error and overload > > > frames. As long as frame + intermission works well in empirical tests, > > > I would suggest to stay with that and call it a day. > > > > Correct, that's why I wrote: (under the assumption there aren't > multiple/continuous errors on the bus) but I agree that's a pretty arbitrary > limit. So I'm fine with changing it to one or two full-sized frames + > intermission) > > ACK. At the end, you have the hardware, so you should decide what is > the best delay to fix the issue. I just hope that my message was > helpful to simplify the formula. I agree a simple formula will be the best way forward as the rest is academic only anyway. Best Regards, Thomas
