--- In 6..., "Jefferson Smith" wrote:
>
> --- In 6..., "theobee00" wrote:
> > Sorry for that, I did not want to imply that you are a hacker, I
> > know better.
> >
> > It was just a small dig at you when you failed to explain why the
> > particular chip you use could work without ground when you did your
> > test.
>
> Hey Theo, thanks. I was hoping you'd respond about the hacker thing
> ;-) However I see nothing left out in my explanations, except that I
> don't use the same terminology you do. So it is you who failed to
> understand my explanation.

That would be bloody right, sme again.

> It's quite intriquing though, isn't it, how
> connecting one or two random wires between two otherwise isolated
> circuits makes a new common ground for the new circuit including them
> both, isn't it?

Old principle, same drawback and problems as before, high impedance ground, lots of common mode noise, I must admit this Philips chip allows for some plenty of headroom.

> > As I said at the time it either was possible because the bus was
> > driving protective devices or it relied on active circuits to
> > achieve this, I observed at the time there was an amazingly small
> > difference between the "soft" grounds.
> >
> > You now forced me to delve into the specific device you used which I
> > tried to avoid since I prefer spending my time arguing theory
> >
> > It turns out that the specific Philips device you use? has a
> > reference generator build in, it says it sets the differential
> > outputs to half rail voltage, I.e 2.5 Volts, since the bus drivers
> > are 180 degrees out of phase, the average is always 2.5 Volts, the
> > receivers in turn have a resistor to the ref Voltage, I quote;
>
> I did catch your previous comments about protective or active circuits
> and was trying to get back to that because of course that is what I
> was trying to say that it is... which means of course that OP may not
> need to send the common Gnd with it. In fact, I think it was a
> Freescale SR where they told me that I did not need the Gnd
> connections on the CAN cable. I avoided saying that because I don't
> want to look it up to verify it, but I think it was (I have terrible
> memory except for certain technical details that I think important).
>
> So anyway, victory! My whole plot was to force you to look at the
> specific hardware and not just talk general theory :)

Caught again, got me consulting for free again:-)
Do take into consideration that it is designed for cars where the whole body acts as a grounded plane with very little DC ofsets, the problems you get would be mainly RF related.

> I don't know if you noticed, but I believe the original question was
> from a less experienced (probably newbie) asking about the Gnd pins on
> their CAN header on either a Freescale evaluation board, or similar
> clone implementation (Dragon12, Techarts, etc). I believe the poster
> did not specify the 82C250 because of the assumption that it was the
> only way to implement a CAN bus. Also note that it is not only
> Phillips. I heard that their proprietary rights to this nifty CAN
> driver have expired.

Can't see how they can have serious rights to a line driver, these have been around since I was a kid, around the same era as Methusalem.

Now back to that soup recipy...
Cheers,

Theo

--- In 6..., "theobee00" wrote:
> Sorry for that, I did not want to imply that you are a hacker, I
> know better.
>
> It was just a small dig at you when you failed to explain why the
> particular chip you use could work without ground when you did your
> test.

Hey Theo, thanks. I was hoping you'd respond about the hacker thing
;-) However I see nothing left out in my explanations, except that I
don't use the same terminology you do. So it is you who failed to
understand my explanation. It's quite intriquing though, isn't it, how
connecting one or two random wires between two otherwise isolated
circuits makes a new common ground for the new circuit including them
both, isn't it?

> As I said at the time it either was possible because the bus was
> driving protective devices or it relied on active circuits to
> achieve this, I observed at the time there was an amazingly small
> difference between the "soft" grounds.
>
> You now forced me to delve into the specific device you used which I
> tried to avoid since I prefer spending my time arguing theory
>
> It turns out that the specific Philips device you use? has a
> reference generator build in, it says it sets the differential
> outputs to half rail voltage, I.e 2.5 Volts, since the bus drivers
> are 180 degrees out of phase, the average is always 2.5 Volts, the
> receivers in turn have a resistor to the ref Voltage, I quote;

I did catch your previous comments about protective or active circuits
and was trying to get back to that because of course that is what I
was trying to say that it is... which means of course that OP may not
need to send the common Gnd with it. In fact, I think it was a
Freescale SR where they told me that I did not need the Gnd
connections on the CAN cable. I avoided saying that because I don't
want to look it up to verify it, but I think it was (I have terrible
memory except for certain technical details that I think important).

So anyway, victory! My whole plot was to force you to look at the
specific hardware and not just talk general theory :)

I don't know if you noticed, but I believe the original question was
from a less experienced (probably newbie) asking about the Gnd pins on
their CAN header on either a Freescale evaluation board, or similar
clone implementation (Dragon12, Techarts, etc). I believe the poster
did not specify the 82C250 because of the assumption that it was the
only way to implement a CAN bus. Also note that it is not only
Phillips. I heard that their proprietary rights to this nifty CAN
driver have expired.

--- In 6..., "Darren Moore" wrote:
> The data will pass without the ground, this is not the
> full CAN spec, which includes power down and other control
> modes used with the lines not in the normal common mode
> operation. So it depends on what you want from the data
> link as to if it will do all you need and at what data
> rates.
>
> The full CAN spec from Bosch has all this information in
> it.
>
> Collision detection is done by two nodes pulling the bus
> active at the same time, who ever releases the bus first
> loses and tries again later, there is no data loss for
> the other talker.

> All the drivers on the bus are OR'd
> together as the drivers are open collector outputs.

Yes but there is a twist, the CANH and CANL are driven in opposite directions, i.e. one is driven from a source the other from a sink.
The result is that they are driven together in opposite directions to develop the differential voltage aross the terminating resistor.

Somewhat different from the standard differential pair where the lines are driven symmetrically.

Cheers,
Theo

> If this is so - then how come when I set the two transceivers at a
> different ground potential of 40 volts (either way) it did not affect the
> can signals or anything else at all ??????

Without going back to the spec sheet I recall the max. allowable voltage on the bus is about 40V.
Cant recall if this was also specified as the common mode voltage range, these devices typically stay operational as long as the inputs are in the common mode range, common mode of course means this voltage is applied to both inputs, the differential being your signal.

> (outputs were not opto-isolated). Is the spec for the transceiver higher
> than this ? 40 volts is a lot for an IC. I bet I could do more but that
> is as high as my supply went. Remember - I am talking about the grounds of
> two networks floating at different potentials.I see AN96116 mentions a
> "Maximum transient bus terminal voltage" of up to 200 volts but I am nnot
> sure what this is.

It looks like the device can safely handle up to 40V continuous on the CAN lines without cooking and up to 200V transients without devices arcing over, a pretty solid spec indeed considering the supply is 5V.

> I think the CAN OUTPUT signals are referenced to each other and NOT to
> ground and this is how this can work in my experiment.
> It is the voltage
> difference between them that is important in a system - not their absolute
> value above ground. ( except of course inside the transceiver chip
> itself). Otherwise, my 40 volts would likely have upset my little network.
> Are there any other explanations ?

For sure, it is a differential pair, i.e. the actual signal is developed from the difference between CANH and CANL, OTOH the difference in ground potential shows up on both lines as the same value, i.e. as a common mode voltage.

> Why is this important ?? Vehicles can have fairly larget ground
> potentials due to heavy curents flowing through the chassis. I can show
> you a place on my car that two grounds develop 1.4 volts when I put the
> brake lights on. Steel isn't -that- good of a conductor as copper and
> silver. So, if in a truck trailer - if there is 2 volts drop (let's say)
> in the positive wire from the front to the lamp at the back - is there a
> zero drop along the chassis ? Do we not have two series resistor plus the
> lamp ????

Yup, it is for that self same reason the CAN devices are designed to handle large common mode voltages.

> BTW - it appears the Philips (and others) are not compliant to ISO15765 in
> that they can't tolerate a short of CAN_Hi to ground or one of Hi or Lo
> open (with a good ground attached).

You will find you need a more complicated input network than a single terminating resistor between the lines, obviously with just a single resistor there is no differential voltage developed when one line goes open circuit, however with two resistors on each line the remaining line will still be active and develop a voltage.

Also obvious is that when you implement such an H bridge you also decouple the two lines from each other, thereby preventing shorts on one line affecting the performance of the remaining line.
It could simply be that what you observe is the result of the short on one line killing the other line by overloading it.

BTW, I seem to recall from my quick visit to the data sheet that the Philips device converts the differential signal to a single ended one to present to the CPUs CAN receiver, an easy spot to check if you still have a signal under specific fault conditions.

> BTW2 - LIN is a single wire 0-12 volt UART based network operating at up to
> 20 Kbps (LIN 2.0) or about 10 Kbps (J2602) and more usually 9600 baud.
> The physical layer is the same as ISO 9141 which can be found as the main
> network and diagnostics (older) in many European and Japanese cars. It is
> obviously not a transmission line as CAN is. LIN is a very current
> protocol and there is a lot of development happening right now...at least
> in North America and Europe. It is cheap to make (buy the parts) but
> fairly expensive in engineering time to develop.

It appears to be intended as a local communication system, i.e. the can acts as the main long haul transmission and from there you can cheaply farm out to some devices in the vicinity, not sure why such a system would be expensive to develop, too many software layers?

> BTW3 - in my informal tests - a network needs at least one 120 ohm resistor
> to work. It as if the chips need a little bias to work at all.
>It might
> work with a higher resistor - never tried this. I have built small systems
> up to about 10 nodes, 500 Kbps and about 30 feet long (scrap wire) using
> just one resistor with no problems. After about 20 nodes, it gets flakey
> and downright cantankerous and as soon as I put in the second termination
> resistor, the network works perfectly.

Indeed, short distance at somewhat lower speed, you get away with a bit more
Go through the previous post on the subject, been discussed quite extensively.
Basically these types of busses are current driven, it needs a resistor to drive into and at higher speeds this also acts as a terminator to stop reflections, the actual value is important.

> At a bus company in NYC - they didn't believe me until one of them yanked
> one of the resistors out of the bus and there was no change. The bus ran
> fine.

If I would find a resistor interfering with my bus I would call the cops.

> I don't recommend making systems with no ground or one termination resistor
> - but this shows how CAN really works and how tough it is.
>
> I don't know how this discussion got here and not on the CAN board.

The HC12 is a very popular chip indeed in cars, that would make this board probably the home of the CAN CAN.

Cheers,

Theo

> As far as CANH and CANL are not isolated, is much better a
> "good and known" common ground reference than floating !!!
> (or leave the nodes create undesired ground loops)

The data will pass without the ground, this is not the
full CAN spec, which includes power down and other control
modes used with the lines not in the normal common mode
operation. So it depends on what you want from the data
link as to if it will do all you need and at what data
rates.

The full CAN spec from Bosch has all this information in
it.

Collision detection is done by two nodes pulling the bus
active at the same time, who ever releases the bus first
loses and tries again later, there is no data loss for
the other talker. All the drivers on the bus are OR'd
together as the drivers are open collector outputs.

The address in the first part of the packet and its set
up that the node that has the highest priority has an
address that releases the bus last, hence it wins.

It's a very cleaver system.

Regards,
Darren

Totally agree.

As far as CANH and CANL are not isolated, is much better a "good and known" common ground reference than floating !!!
(or leave the nodes create undesired ground loops)

Jordi Costa

>
> > The problem is that your setup is what I call a "lab setup". In the
> > real world you have other electrical influences on the wires and a
> > ground reference that may be at different voltage levels, especially
> > in a factory environment where the factory ground structure isn't
> > very good and you have high power machines. If you don't connect the
> > ground wire here and you have a long distance between the nodes and
> > the nodes are connected to the ground structure, the only connection
> > between the nodes are the communication wires. If there is a
> > difference in the grounds for some reason, say a power surge caused
> > by a high power machine, the current will flow from one ground to
> > the other through the communication wires then there could be more
> > than 40V difference between the CANH, CANL and the circuit ground.
> > Even though the difference between CANH and CANL is only a couple of
> > volts. A ground wire here would let the temporary surge current flow
> > through the ground wire instead since it has a much lower impedance
> > to ground. This may in turn cause other problems (ground current
> > loops) but it doesn't destroy the circuits.
> >
> > If you, on the other hand, have the nodes isolated from ground, you
> > have a DC insulation between your circuit and the ground structure.
> > You still have a capacitive coupling to the different grounds which
> > also could cause HF currents to flow between the nodes. This
> > capacitive coupling is not to be negligated, especially at higher
> > frequencies. A shield connected to the ground structure (not 0V) at
> > both ends helps here. This may also cause current loops but the
> > connection in one end can be done through a capacitor to block the
> > DC and low frequency current flow. The HF is almost always a bigger
> > problem than the low frequency current loop though.
>
> Your confusion is apparently that although you finally admitted this
> is a "differential" signal, you still do not seem to understand that I
> am suggesting that the differential part is between CANL (the 'ground'
> reference, whatever 'ground' means) and CANH.
>
> You, and maybe Theo, are only disagreeing with me by claiming that if
> there is only CANL and CANH (no common Gnd), that it cannot work at
> all. I don't agree, but you obviously don't agree with yourself
> either. So... get a grip on yourself! :)
>
> Great, somebody make a larger implementation... I'd like to know how
> far theirs goes without the Gnd wire. No, mine is not a "lab setup".
> It is my product. I've got a 50 foot cable, but really... Your point
> in this case is my point exactly. What I have been trying to say is
> that in order to engineer the specific implementation, we need to
> consider the environment that we are facing. The OP asking the
> original question did not specify any of this, therefore we cannot
> give the answer. You are probably thinking of some permenant structure
> that uses earth ground on both ends. Certainly you have different
> things to consider than me. But do you even have an implementation? I
> did not rememeber if you said you implement anything, or if you are
> only talking abstract theory, like Theo is.
>
> My product is portable, fully battery powered (sealed lead-acid). I
> have none of those considerations to worry about, like
> lightning strikes.
>

Isn't this all about the difference between "robust" (real world) and
"special" (like battery operated nodes) usage?
And that the "special" variant may work in theory but not always well in the
real world because:

The prerequisites are:

- The CANH and CANL pins can both sink and source current
- The circuitry works ok as long as the common mode voltage is within limits

The first means that a node without ground connection will float at a
voltage determined by the levels of CANH and CANL.

The common mode voltage can be relative high (like +/12 or +/- 40, this is
no problem to achieve, check the INA117 amplifier or similar).

So far the groundless system seems to be ok.

The "real world" problems would then be that:

A: The node without ground will only work (due to the common mode rule
above) as long as the node is completely isolated DC and AC which is hard to
guarantee, it is by design very loosely coupled groundwise and therefore
easy to disturb by static, electric fields, coupling, leakage etc. It may
stop working just by placing it close to other power equipment or even by
holding it in the hand(!).

So the groundless connection is normally not wanted but could theoretically
work if the environment is very clean (which probably could be more
expensive than adding the ground conductor).

B: In practice, there always exists some unbalance in the CANH and CANL
lines, the input resistances or capacitances are not exactly equal. This
means that the groundless node will not float at a nice constant DC voltage
but instead jump up and down with the frequency of the data which makes the
hole connection between the nodes to act as one great antenna. In the
grounded connection this will not occur because the ground wire will return
all this unbalance current.

If you want a clean and quiet (and legal?) groundless system you may
therefore need to shield the complete system, again the ground wire seems to
be more optimal?

Regards,
Anders

You now forced me to delve into the specific device you used which I tried
to avoid since I prefer spending my time arguing theory

It turns out that the specific Philips device you use? has a reference
generator build in, it says it sets the differential outputs to half rail
voltage, I.e 2.5 Volts, since the bus drivers are 180 degrees out of phase,
the average is always 2.5 Volts, the receivers in turn have a resistor to
the ref Voltage, I quote;

------------------
the CANH and CANL inputs are biased to a voltage level of 2.5 V nominal via
receiver input networks with an internal impedance of 17 kOhm typical.
----------------------

So the "soft" ground you generated comes from the 2.5 V reference with an
impedance of around 8.5 kOhm to the power supply ground (when receiving).

If you sketch out this set-up it will be immediately obvious any difference
in ground potential between sender and receiver will appear across the
input of the receiver, this is no problem as long as you stay in the
specified voltage range of the receiver, however if there is noise or a DC
offset between the grounds you will find your transmission line is a lot
better of with an actual ground connection somewhere.

Cheers,

Theo

***************************************************************************
**************************************************************
If this is so - then how come when I set the two transceivers at a
different ground potential of 40 volts (either way) it did not affect the
can signals or anything else at all ??????
(outputs were not opto-isolated). Is the spec for the transceiver higher
than this ? 40 volts is a lot for an IC. I bet I could do more but that
is as high as my supply went. Remember - I am talking about the grounds of
two networks floating at different potentials.I see AN96116 mentions a
"Maximum transient bus terminal voltage" of up to 200 volts but I am nnot
sure what this is.

I think the CAN OUTPUT signals are referenced to each other and NOT to
ground and this is how this can work in my experiment. It is the voltage
difference between them that is important in a system - not their absolute
value above ground. ( except of course inside the transceiver chip
itself). Otherwise, my 40 volts would likely have upset my little network.
Are there any other explanations ?

Why is this important ?? Vehicles can have fairly larget ground
potentials due to heavy curents flowing through the chassis. I can show
you a place on my car that two grounds develop 1.4 volts when I put the
brake lights on. Steel isn't -that- good of a conductor as copper and
silver. So, if in a truck trailer - if there is 2 volts drop (let's say)
in the positive wire from the front to the lamp at the back - is there a
zero drop along the chassis ? Do we not have two series resistor plus the
lamp ????

BTW - it appears the Philips (and others) are not compliant to ISO15765 in
that they can't tolerate a short of CAN_Hi to ground or one of Hi or Lo
open (with a good ground attached).

BTW2 - LIN is a single wire 0-12 volt UART based network operating at up to
20 Kbps (LIN 2.0) or about 10 Kbps (J2602) and more usually 9600 baud.
The physical layer is the same as ISO 9141 which can be found as the main
network and diagnostics (older) in many European and Japanese cars. It is
obviously not a transmission line as CAN is. LIN is a very current
protocol and there is a lot of development happening right now...at least
in North America and Europe. It is cheap to make (buy the parts) but
fairly expensive in engineering time to develop.

BTW3 - in my informal tests - a network needs at least one 120 ohm resistor
to work. It as if the chips need a little bias to work at all. It might
work with a higher resistor - never tried this. I have built small systems
up to about 10 nodes, 500 Kbps and about 30 feet long (scrap wire) using
just one resistor with no problems. After about 20 nodes, it gets flakey
and downright cantankerous and as soon as I put in the second termination
resistor, the network works perfectly.

At a bus company in NYC - they didn't believe me until one of them yanked
one of the resistors out of the bus and there was no change. The bus ran
fine.

I don't recommend making systems with no ground or one termination resistor
- but this shows how CAN really works and how tough it is.

I don't know how this discussion got here and not on the CAN board.

Bob Boys
San Mateo, California

--- In 6..., "Jefferson Smith" wrote:
>
> --- In 6..., "Ruben Jsson" wrote:
> > > > That is also where the confusion stems from, the drivers would
> > > > need to supply enough current to drive the termination resistor,
> > > > i.e. an output to supply current, but it is not what is known as
> > > > a current loop, these assume low freq. or DC levels.
> > >
> > > Hmmm, does it even matter whether this bus is called
> > > "current loop" or is just a loop of current?
> >
> > Because it isn't a current loop.
> >
> > Current loop needs no common ground. Reference voltage is
> > derived from the current loop itself. Just two wires. In
> > the receiver a voltage over a resistor is derived from
> > the current.
>
> Before I go on, let me make one thing clear: I am not arguing at ALL
> that having an extra Gnd would not improve distance. I am merely
> saying, "yes, it works without the extra Gnd wire". Shoot, docs only
> said you could expect about a mile, and not at the full 1 Mbps. So how
> far do you need yours to go?
>
> OK I hope by now you and Theo figure out what I mean by "it does not
> matter". Honest, guys, it does not matter. We are not trying to
> discuss what current loop means. I did not need to know whether it was
> officailly called a "current loop" or not. The fact that this is not
> called a current loop, just does not change the question that was
> asked about needing the extra Gnd wire. It simply does not change the
> fact that the CAN communication works fine without the extra Gnd wire.
> Just drop the "current loop" thing, because it honestly does not
> matter. We are officially NOT talking about a "current loop". OK good.
> Now let's try to figure out what the heck you think you're talking
> about when you say that CAN needs a "common ground" in ADDITION to
> CANH and CANL.
>
> > A differential transmission line, on the other hand, works with
> > voltages. Voltages referenced to a common ground. The driver does
> > not adjust (regulate) the voltages and currents depending on
> > different loads. It is true that the logic level on the bus is
> > derived from the voltage difference between the two lines BUT the
> > voltage is not derived using the termination resistor, it is coming
> > directly from the transmitter circuit.
>
> I am following your explanation except for the part about "voltage is
> not derived using the termination resistor". I've never heard of
> something that derives voltage using the termination resistor except
> that it is derived by Ohm's law. Surely you understand Ohm's law, yet
> you telling me that you do not. If there were zero ohms terminating
> (i.e. no resistance), there would be no voltage to be received. The
> voltage that will be received is ACROSS the terminating resistance,
> which sould of course be large enough to read. As both you and Theo
> have agreed with me, that resistor improves reliability (such as
> reducing 'reflections'). Sheeshe, if I confuse you guys this much, you
> should just start ignoring me and let the guys listen to me who
> understand me :-/ I am certain I am speaking sensibly (I go back and
> read my words again).
>
> > The termination resistor is
> > there to "terminate" the line, impedance wise. Sort of to make a
> > smoth ending for the signal which would otherwise bounce and cause
> > ripple on the line when the voltage is swithed fast.
>
> Honestly, guys, where have I mislead you to make you think that I was
> saying anything different?
>
> > The problem is that your setup is what I call a "lab setup". In the
> > real world you have other electrical influences on the wires and a
> > ground reference that may be at different voltage levels, especially
> > in a factory environment where the factory ground structure isn't
> > very good and you have high power machines. If you don't connect the
> > ground wire here and you have a long distance between the nodes and
> > the nodes are connected to the ground structure, the only connection
> > between the nodes are the communication wires. If there is a
> > difference in the grounds for some reason, say a power surge caused
> > by a high power machine, the current will flow from one ground to
> > the other through the communication wires then there could be more
> > than 40V difference between the CANH, CANL and the circuit ground.
> > Even though the difference between CANH and CANL is only a couple of
> > volts. A ground wire here would let the temporary surge current flow
> > through the ground wire instead since it has a much lower impedance
> > to ground. This may in turn cause other problems (ground current
> > loops) but it doesn't destroy the circuits.
> >
> > If you, on the other hand, have the nodes isolated from ground, you
> > have a DC insulation between your circuit and the ground structure.
> > You still have a capacitive coupling to the different grounds which
> > also could cause HF currents to flow between the nodes. This
> > capacitive coupling is not to be negligated, especially at higher
> > frequencies. A shield connected to the ground structure (not 0V) at
> > both ends helps here. This may also cause current loops but the
> > connection in one end can be done through a capacitor to block the
> > DC and low frequency current flow. The HF is almost always a bigger
> > problem than the low frequency current loop though.
>
> Your confusion is apparently that although you finally admitted this
> is a "differential" signal, you still do not seem to understand that I
> am suggesting that the differential part is between CANL (the 'ground'
> reference, whatever 'ground' means) and CANH.
>
> You, and maybe Theo, are only disagreeing with me by claiming that if
> there is only CANL and CANH (no common Gnd), that it cannot work at
> all. I don't agree, but you obviously don't agree with yourself
> either. So... get a grip on yourself! :)
>
> Great, somebody make a larger implementation... I'd like to know how
> far theirs goes without the Gnd wire. No, mine is not a "lab setup".
> It is my product. I've got a 50 foot cable, but really... Your point
> in this case is my point exactly. What I have been trying to say is
> that in order to engineer the specific implementation, we need to
> consider the environment that we are facing. The OP asking the
> original question did not specify any of this, therefore we cannot
> give the answer. You are probably thinking of some permenant structure
> that uses earth ground on both ends. Certainly you have different
> things to consider than me. But do you even have an implementation? I
> did not rememeber if you said you implement anything, or if you are
> only talking abstract theory, like Theo is.
>
> My product is portable, fully battery powered (sealed lead-acid). I
> have none of those considerations to worry about, like lightning strikes.

He, he, excellent rant that.

The fact of the matter is that it might or might not work in a given situation, see my previous post on the whyfor, it remains a poor design practise that some poor beginner might try to emulate and fall flat.

I have no doubt that relying on a relatively high impedance to remain in the operating range of the receivers is not recommended in any situation, even at ten meters distance, even if it works for you.
Cheers,
Theo

--- In 6..., "Jefferson Smith" wrote:

> > I have always seen the CAN bus depicted as a single line with
> > terminating
> > resistors at each end. Can the bus be used in a star configuration?
> > (probably) If so, what does that do to the terminating resistors?

I think there is a LIN? Bus working of a single wire at low frequencies, intended for short distance work where a differential bus isn't needed, either system should work in star at short distances and low freqs.

> I recently found a doc you probably have, the Phillips Semiconductors
> Application Note AN96116. Good information there, for this thread.

It gives a good run down on the why fors and limits of high freq differential transmission lines, but these issues are pretty basic.

It also does explain the issue of phase differentials, not mentioned here so far.
An issue only arising when there is bi-directional transmission, return transmissions have to fall in the expected opportunity slot, lightspeed limitations will create a phase shift along a transmission line, what can I say, read all about it:-).

> From what I read, it's best to do it in a single bus, i.e. the line.
> Maybe run that in a circle or "chain".
> But we're talking some
> distance. If some nodes are just a few feet away from the main bus, it
> seems fine to me. Of course if you want more bitrate you need to do
> some engineering work, analyze the phase delays, etc. There's good
> variables in configuring each node that can help with that, which is
> why you can't just convigure each node with a simple "bitrate value".
>
> Personally, I like Theo's suggestion that I am just a hacker who
> doesn't understand theory (though that should obviously be not true),
> but I recommend understanding the theory in the adjustments.

Sorry for that, I did not want to imply that you are a hacker, I know better.

It was just a small dig at you when you failed to explain why the particular chip you use could work without ground when you did your test.

As I said at the time it either was possible because the bus was driving protective devices or it relied on active circuits to achieve this, I observed at the time there was an amazingly small difference between the "soft" grounds.

You now forced me to delve into the specific device you used which I tried to avoid since I prefer spending my time arguing theory

It turns out that the specific Philips device you use? has a reference generator build in, it says it sets the differential outputs to half rail voltage, I.e 2.5 Volts, since the bus drivers are 180 degrees out of phase, the average is always 2.5 Volts, the receivers in turn have a resistor to the ref Voltage, I quote;

------------------
the CANH and CANL inputs are biased to a voltage level of 2.5 V nominal via
receiver input networks with an internal impedance of 17 kOhm typical.
----------------------

So the "soft" ground you generated comes from the 2.5 V reference with an impedance of around 8.5 kOhm to the power supply ground (when receiving).

If you sketch out this set-up it will be immediately obvious any difference in ground potential between sender and receiver will appear across the input of the receiver, this is no problem as long as you stay in the specified voltage range of the receiver, however if there is noise or a DC offset between the grounds you will find your transmission line is a lot better of with an actual ground connection somewhere.

Cheers,

Theo