I think I CAN, I think I CAN...

[EV_Builder]

Can you see it???

Txs for sharing this. Assumptions are the mother of all...

[Gregski]

so can I fix it? will you help me? if I was a Diode on this logic board, what would I look like? and where would I live?

please and thank you

[celeron55]

You don't really need more than a multimeter with a diode check function to check the polarity of any HV power component.

If you get a diode voltage between 0.3V and 1.0V, that's a diode - no matter whether it's a real diode or just looks like one, it acts like one - don't connect your HV this way, there will be infinite current through the diode. In reality the "diode" is the reverse diodes inside the transistors that form a half bridge. In order to know more, type that into Google image search (not sponsored).

Then you switch your probes, and will see the diode check voltage rise from 0.0 to 1.0V and once it goes above, the multimeter will generally show its "open circuit" display. This means the multimeter just charged the capacitor inside the component to above its diode check voltage range and it's also possible to charge it more without issues.

The probe colors are now showing you the way you should connect the HV.

If the measurement does not act in this exact way, then you need to find another way, maybe a more high power one. There might be a bleed down resistor inside the component, or something weird going on.

And you of course should do this at the end you're about to connect to the battery, so that you notice any wiring errors (or have two errors that cancel themselves out... and 2 errors = 0 errors, obviously. Just be wrong enough, right?)

[EV_Builder]

so can I fix it? will you help me? if I was a Diode on this logic board, what would I look like? and where would I live?

please and thank you

Post pictures we might get an idea. Fixing car stuff is not always easy since sometimes parts are not clearly marked etc...

[Gregski]

so can I fix it? will you help me? if I was a Diode on this logic board, what would I look like? and where would I live?

please and thank you

Post pictures we might get an idea. Fixing car stuff is not always easy since sometimes parts are not clearly marked etc...

thank you, did couple threads above

[arber333]

I would advise you to continue working with CAN bus in the meantime.
Try to connect twisted pair with Canalyst, Microchip, Arduino DUE and Volt charger connected together. This will allow you to spy on your line with one device while you will be able to train to command with others.

Also as far as i know your battery BMS is also transmitting on its CAN bus. I just cant say at what speed.

[EV_Builder]

so can I fix it? will you help me? if I was a Diode on this logic board, what would I look like? and where would I live?

please and thank you

Check the 3 partnumbers with the 3 legged back to two. (one is looks cut).

Those could be diodes. In one of your pictures I also see 12volts on the HV side? That's not to bad I would guess..

[marengo]

Thanks for the documentation of your efforts. I have been wanting to dive into the CAN but so far am paralyzed. For instance, I'm not sure what files to download from https://github.com/collin80/SavvyCAN/re ... continuous and what to do with them. I use a windows computer primarily and know nothing about "compiling"

The CAN I want to reverse engineer lives in my Alta motorcycle. I have the OEM laptop (ubuntu) and dongle and can see the CAN data running through the Alta proprietary diagnostic software. I suppose I should use this computer going forward as it already set up for this. The Alta dongle is Lawicel based, which may or may not be important per my understanding.

[Gregski]

Thanks for the documentation of your efforts. I have been wanting to dive into the CAN but so far am paralyzed. For instance, I'm not sure what files to download from https://github.com/collin80/SavvyCAN/re ... continuous and what to do with them. I use a windows computer primarily and know nothing about "compiling"

The CAN I want to reverse engineer lives in my Alta motorcycle. I have the OEM laptop (ubuntu) and dongle and can see the CAN data running through the Alta proprietary diagnostic software. I suppose I should use this computer going forward as it already set up for this. The Alta dongle is Lawicel based, which may or may not be important per my understanding.

I can only comment on the Windows environment and in this environment the SavvyCAN software is a stand alone piece of code meaning nothing needs to be compiled nor installed, that's right you don't even need to install it or run setup or anything like that. You just download it from here:

https://github.com/collin80/SavvyCAN/re ... yCAN32.zip

then unzip or "extract it" and it will launch, now it won't work yet but the program will open up

what it needs beneath it is called GVRET. GVRET is not so much an application as it is firmware a lower level code that controls the hardware, this is the stuff you need Arduino IDE to compile onto a GVRET compatible hardware such as the Arduino DUE

hope this helps a little bit

[Jack Bauer]

https://vimeo.com/manage/videos/511514530

[asavage]

This is not addressed to Pete specifically, but rather to add to this thread's conversation around CAN termination resistors in general.

The ideal is to have a termination at each end but as long as there is a termination somewhere you should be fine on a small network (few m of cable). Too many terminations reduce the signal level, possibly to the point where messages aren't received (but that would need quite a few extra terminations).

If you find a network is really sensitive to terminations there are probably other problems (such as high electrical noise) too.

In general if you are connecting to an existing network (e.g. to debug it) don't add a terminator on the dongle and keep the cables as short as possible. If you are creating a new network with the dongle at one end then add a termination on the dongle.

As a differential voltage signalling scheme, and unlike some other data buses that use termination primarily to damp signal reflections from the endpoints of the bus, the (high-speed, ISO 11898-2) CAN bus uses the 60 ohm bus impedance (2 * 120 ohm termination resistors, in parallel) to return both CAN lines to their nominal/passive/undriven state (typ. 2.5v) which equates to logical 1 and is referred to as Recessive state

When the CAN lines are being actively driven, CANH is driven ~1v higher toward 3.5v, AND CANL is driven ~1v lower toward 1.5v, with a total differential voltage of approximately 2.0v.

When all CAN transceivers stop driving the lines (eg when changing from a 0/Dominant state to a 1/Recessive state), both CAN lines have to transition to as close to zero differential voltage as practical, and they have to make that transition within a defined timeframe. This is accomplished via the termination resistors!

If there is too much termination resistance -- eg only one termination resistor installed = 120 ohms across the CANL & CANH -- the bus cannot "reset" to differential voltage of 0v fast enough (or at all, if there is no termination at all), and data errors will occur because the next bit will be read as 0/Dominant when it should be 1/Recessive, which will eventually fail a CRC or other check (there is a CRC on EVERY frame). It affects the trailing-edge slew rate.

If there is too little termination resistance -- eg too many termination resistors (installed in parallel) -- then the CAN hardware drivers struggle to drive the differential voltage high enough, against the too-low impedance, it might take too long to rise to spec, and the next bit could/will be read as 1/Recessive instead of 0/Dominant. It affects the leading-edge slew rate.

In a nutshell, this is why having 60 ohms resistance across the CANL & CANH lines on a high-speed CAN bus is important. Ideally, this resistance is spread across two termination resistors, at the ends of the bus, but it's more important to have at or near the correct 60 ohms across the bus than the physical location of that resistance (though both are important design considerations). IOW, worry more about the value than the location of the CAN bus resistance.

Ref. https://en.wikipedia.org/wiki/CAN_bus#P ... ganization



Ref. https://www.can-cia.org/can-knowledge/c ... nsmission/




---

Having said the above . . . below is mere speculation . . .

Just noticed from your pictures that it's reporting an "error passive".. I think this means that it's not getting an ack back from the charger (same problem I had), is it definitely set to the right baud rate and are the can bus wires the right way round?

I believe so because it reads messages just fine, which makes me think the high and low wires are connected properly and the termination is on, this guy has software controlled resistor terminator thingie and I tried it with both on and off, it is on in this case

Different CAN transceivers have slightly different internal specs for deciding when a differential voltage on the CAN lines is Dominant or Recessive, and it's possible that the Microchip unit is OK to read with your setup but not to transmit; with insufficient resistors (too little resistance) it might be reading some 0s when it is waiting for a 1 -- the CAN bus arbitration is predicated on only being able to transmit when someone else of higher priority is NOT transmitting, and that is defined by the Msg. ID (lower Msg. ID = higher priority). It's a bit dense to relate, but at the transceiver level, when there's a "window" in which a new frame can be transmitted (ie the bus traffic is at a sort of idle state momentarily), any transceiver can start transmitting. The way the prioritization scheme works is that the transceiver is required to monitor the bus as it's transmitting, and if it reads the wrong bit on the bus, it either loses priority arbitration, or it goes to error state (I'm simplifying here).

So if Greg's Microchip black box tries to TX and drives the CAN bus to 0/Dominant but reads the resulting voltage as not quite high enough, it'll think it lost bus arbitration and will wait a few cycles for the bus to settle down, then try again. If the box tries to "send" a 1/Recessive (which involves not driving the bus to 2v differential voltage), but instead "reads" a 0, same thing. If this happens enough times, the transceiver reports an error, which SavvyCan presumably echoes back.

If I was a better writer, I could probably have said the above more concisely, but the Wikipedia CAN reference is pretty good, though it does require a time commitment to read through.

[Pete9008]

The ideal is to have a termination at each end but as long as there is a termination somewhere you should be fine on a small network (few m of cable). Too many terminations reduce the signal level, possibly to the point where messages aren't received (but that would need quite a few extra terminations).

If you find a network is really sensitive to terminations there are probably other problems (such as high electrical noise) too.

In general if you are connecting to an existing network (e.g. to debug it) don't add a terminator on the dongle and keep the cables as short as possible. If you are creating a new network with the dongle at one end then add a termination on the dongle.

As a differential voltage signalling scheme, and unlike some other data buses that use termination primarily to damp signal reflections from the endpoints of the bus, the (high-speed, ISO 11898-2) CAN bus uses the 60 ohm bus impedance (2 * 120 ohm termination resistors, in parallel) to return both CAN lines to their nominal/passive/undriven state (typ. 2.5v) which equates to logical 1 and is referred to as Recessive state

When the CAN lines are being actively driven, CANH is driven ~1v higher toward 3.5v, AND CANL is driven ~1v lower toward 1.5v, with a total differential voltage of approximately 2.0v.

When all CAN transceivers stop driving the lines (eg when changing from a 0/Dominant state to a 1/Recessive state), both CAN lines have to transition to as close to zero differential voltage as practical, and they have to make that transition within a defined timeframe. This is accomplished via the termination resistors!

If there is too much termination resistance -- eg only one termination resistor installed = 120 ohms across the CANL & CANH -- the bus cannot "reset" to differential voltage of 0v fast enough (or at all, if there is no termination at all), and data errors will occur because the next bit will be read as 0/Dominant when it should be 1/Recessive, which will eventually fail a CRC or other check (there is a CRC on EVERY frame). It affects the trailing-edge slew rate.

If there is too little termination resistance -- eg too many termination resistors (installed in parallel) -- then the CAN hardware drivers struggle to drive the differential voltage high enough, against the too-low impedance, it might take too long to rise to spec, and the next bit could/will be read as 1/Recessive instead of 0/Dominant. It affects the leading-edge slew rate.

In a nutshell, this is why having 60 ohms resistance across the CANL & CANH lines on a high-speed CAN bus is important. Ideally, this resistance is spread across two termination resistors, at the ends of the bus, but it's more important to have at or near the correct 60 ohms across the bus than the physical location of that resistance (though both are important design considerations). IOW, worry more about the value than the location of the CAN bus resistance.

Great explanation, completely agree

I was more trying to make the point (probably fairly badly) that on a small network the termination is unlikely to be the problem so it makes more sense to concentrate on other things first.

I was just curious what the resistance limits might be so did a few sums:

Assuming a small network, lets say 5m. Now for reflections the worst case is transmission to the far end and back, so 5m x 2 x 5ns = 50ns. A bit time at 500kbps is 2us so with this length cable the first reflection will occur a 40th of a bit period in. Even allowing for multiple loop reflections we can pretty much ignore this it so accurately terminating the line is a nicety rather than a necessity. Also the location of the terminators is not important just the value.

As you say the other main reason the terminator is there is to provide a return to the recessive state in an acceptable time so the question is what is the maximum value of resistance that can do this. If we assume a cable capacitance of 60pF/m (screened twisted pair) then the total cable capacitance for a 5m length is 300pF. To allow for PCB trace capacitance, connectors, transceiver, etc lets call it a round 0.5nF. So how much resistance do we need to restore the recessive level in say 1/10th of a bit period. Bit period is 2us so 1/10th is 200ns so R needs to be no more than 400R (just calculated from t=RC so a 66% return to 0v state).

On the lower end most transceivers are rated to drive a 40R load, so in theory you could have 4 terminators on the bus and it still work - definitely not recommended though.

So on a 5m network it should work with a bus resistance of anywhere between 40R and 400R. In no way ideal but it should still work!

I actually have CAN bus running on my home BMS system which breaks all the rules, multiple 10-20m cable runs on a star topology and not a single 120R terminator anywhere (but I do have slew rate limited transceivers, lower data rates and distributed higher value terminators) - it works fine

To sum up on proper installs always do it right, 120R at each end on a daisy chain install will give the best system performance. On a small test install don't get too hung up on terminators, as long as you have at least one 120R on the network it should work - check other things first.

Edit - just for completeness, if the bus length is 10m the max R would be 200R and at 40m it would be 50R (pretty much the 60R for two 120R terminators). At this length the return trip propagation time is 40 x 2 x 5ns = 400ns which is getting close to the 1/4 bit period so we can no longer ignore transmission line effects. The transmission line effects mean that the transceivers do not 'see' the termination impedance/line capacitance any more, instead they 'see' the cable impedance. This means that further increases in length do not increase the capacitive loading on the transceiver, they just continue to see the cable impedance. There is therefore no change to the maximum value of the terminator as the line length continues to increase and the optimal terminator value is one which matches the characteristic impedance of the twisted pair (or 120R at each end!).