2
\$\begingroup\$

I'm interfacing a product that operates on a kind of CAN bus. Specifically, it uses the CANL line only.

I'd like to use a transceiver I've used before, the MCP2562, and I suspect there's not likely to be any problem transmitting, but will it receive with nothing connected to the CANH line?

How should I manage termination in this instance?

The bus operates at 1Mbit and my logic analyzer measures levels of 0.9V dominant and 1.8V recessive. I realise these are not standard CAN levels. The frame format is CAN, but I'm happy to use an alternative transceiver type if I can achieve the same levels and it manages the line correctly.

Any help much appreciated !

\$\endgroup\$
8
  • 1
    \$\begingroup\$ To use single-ended Rx, CANH needs to be biased at mid-point voltage and loss of SNR due to any CM noise is expected. Pull-up/down to Thevenin voltage and impedance to match bus and Tx impedance reduces reflection errors if distance/slew rate is greater than risetime. Otherwise matched impedance is best practice. \$\endgroup\$ Commented Feb 13, 2020 at 10:32
  • 2
    \$\begingroup\$ The bus operates at 1mbit That's probably nonesense so please state something meaningful here. \$\endgroup\$
    – Andy aka
    Commented Feb 13, 2020 at 11:06
  • \$\begingroup\$ This is high speed CAN on a bus no longer than a shoelace. The measured speed on the logic analyser is indeed 1mbit/second and the length of each bit and frame corroborates this. \$\endgroup\$
    – quiver
    Commented Feb 13, 2020 at 11:34
  • \$\begingroup\$ @quiver 1millibit is nonsense. That is what Andy means. You did not measure a speed of 1millibit on your logic analyzer because it is not a speed. You also did not measure a speed of 1 millibit/second unless you were waiting hours for each message to transmit. Correctly type out what you really mean. Don't say 0.001 when what you mean is 1,000,000 \$\endgroup\$
    – DKNguyen
    Commented Feb 13, 2020 at 14:24
  • 5
    \$\begingroup\$ @DKNguyen you're right. I did not measure a speed of 1 millibit, because it is not a speed. And you could have reasonably assumed I meant 1Mbit. \$\endgroup\$
    – quiver
    Commented Feb 13, 2020 at 21:54

1 Answer 1

1
\$\begingroup\$

I've just discovered that there are heaps of CAN transceivers that operate at 3V3, and thus have a 1.9V recessive voltage (yes, they can only speak to other CAN devices that operate at 3V3). This solves the levels part of my question, I'll use the tip provided by @'Tony Stewart Sunnyskyguy EE75' to try and hold CANH steady and see where I land.

Thanks!

\$\endgroup\$
3
  • \$\begingroup\$ No these aren't CAN transceivers, they are something-else transceivers. CAN has standardized voltage levels. The vast majority of 3V3 transceivers are 3V3 tolerant on the tx rx lines towards the MCU, which is something else entirely. Most often they are supplied with 5V. There are also transceivers that support 3V3 supply + signal levels without going out of spec on the CANH and CANL lines, such as SN65HVD232. You can use this one to speak with CAN standard nodes. Briefly put, 1.7V to 3.3V is still within spec range. TI has a good app note. \$\endgroup\$
    – Lundin
    Commented Feb 14, 2020 at 11:56
  • \$\begingroup\$ Direct link to that app note: ti.com/lit/pdf/slla337 \$\endgroup\$
    – Lundin
    Commented Feb 14, 2020 at 11:57
  • 3
    \$\begingroup\$ From p31 of the SN65HVD232 datasheet: "Most implementers of CAN buses recognize that ISO 11898 does not sufficiently specify the electrical layer and that strict standard compliance alone does not ensure full interchangeability." The levels are different. TI have worked hard to ensure their devices are broadly compatible.. but there are other products including the Maxim 33054E which do indeed have levels out of spec (by ISO 11898) yet are still billed as CAN transceivers. The Maxim 33054E gets me where I need to be to interface this product. \$\endgroup\$
    – quiver
    Commented Feb 16, 2020 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.