Super capacitors vs batteries for distributed power?

[e100]

Apologies if this sounds vague as it's just an idea at this stage.
Essentially my plan is to have a number of microcontrollers spread out along the length of 100 meters of CAT5 that provides power and uses CAN for communication. Each will have some kind of motion sensor and a 1w LED light that stays on for a minute or two. CAT5 is only rated to about 1 amp per core so each node will need to have some kind of local power storage so that multiple lights can switch on at the same time without overloading wires or causing the bus voltage to sag.

Does this sound like a job for supercapacitors trickle charged from the bus power?
In theory each microcontroller could simply monitor the voltage level and halt charging when the voltage limit is reached. When a light is switched on voltage decays exponentially so I'll need to do some PWM manipulation to give the illusion of constant brightness.
Charging lithium batteries is usually done with dedicated charger IC which adds another level of complexity to the design.

[fourtytwo42]

Seems to me the simple answer to your problem is increase the voltage, e.g 48V, using two pairs for power this could easily deliver 48W and probably more.
There are loads of dc-dc converters for such input voltages designed for Power Over Ethernet applications.
Please don't be tempted to use lethal voltages or non-isolated sources.

[Psi]

i was just typing that, but you bet me too it

Cat5 is rated for up to 57V.
And you could parallel some of the wires to get more than 1A.

[Doctorandus_P]

Agree with the above.
Add a SMPS module to each node.

Also keep an eye on the common mode voltage for your CAN bus.

If you use for example 50V for power and you have a 10V drop over the 100m cable, then the CAN bus will be 5V below GND, because the GND level is lifted at the end.

[e100]

Agree with the above.
Add a SMPS module to each node.

Also keep an eye on the common mode voltage for your CAN bus.

If you use for example 50V for power and you have a 10V drop over the 100m cable, then the CAN bus will be 5V below GND, because the GND level is lifted at the end.

I'm not sure I understand the transceiver specs regrading negative voltage limits.
Looking at voltages received at the far end, if CAN L goes below 0 v it looks like it doesn't matter (providing it stays within the electrical limit of the silicon)?
But if CAN H goes below 0 v then the transceiver doesn't see a difference between H and L and therefore cannot determine the state?
So does that mean the higher the ground lift the worse the noise margin as CAN L sinks below 0v and you are now effectively measuring the voltage between CAN H and ground?

So for some realistic numbers, if the transmitted peak CAN H signal from the near end is 3.5v does that mean the bus will cease to function if the voltage boost in the ground wire at the far end exceeds 3.5v?

Coming back the other way, for signals transmitted from the far end, does the ground lift effectively raise the CAN L and H voltages seen at the near end?
If so does that mean you can calculate the ground voltage drop by seeing how far the peak voltage is above the nominal 3.5v CAN H signal level?

[drvtech]

One solution to the ground lift problem would be to use two separate grounds. You've got 4 pairs available so a pair for CAN then two pairs for power+ and power- and another pair for logic ground. Your logic ground is connected to power- at the far end only and passes very little current so is roughly the same voltage at both ends. Your LED switching transistors go down to power- and the odd volt here and there can be allowed for by biassing resistors on the gates. The power for the microcontrollers comes from an smps at each node.

[Doctorandus_P]

There is nothing special (at low frequencies) for the cable. It's just a bunch of resistors.

Take one end as GND reference, and a 50V power supply (and a CAN transceiver) and you then measure 40V at the other end of the cable, GND is lifted by half that, so 5V.

If the CAN wires are between 0V and 3V3, then at the other end of the cable the voltages will be practically the same because the current through the CAN bus wires is low.
Because of the 5V lifted GND reference at the other end though, the CAN transceiver on that other end will have it's signal lines between -5V and -1.7V, compared to it's local GND.

For RS485 this is a part of the specification. The signal levels can be anywhere between -7V and +12V (which is 5V plus the same 7V margin). I don't use CAN much, but different transceivers may have different specifications, and some may work just fine if GND gets lifted.

[e100]

There is nothing special (at low frequencies) for the cable. It's just a bunch of resistors.

Take one end as GND reference, and a 50V power supply (and a CAN transceiver) and you then measure 40V at the other end of the cable, GND is lifted by half that, so 5V.

If the CAN wires are between 0V and 3V3, then at the other end of the cable the voltages will be practically the same because the current through the CAN bus wires is low.
Because of the 5V lifted GND reference at the other end though, the CAN transceiver on that other end will have it's signal lines between -5V and -1.7V, compared to it's local GND.

For RS485 this is a part of the specification. The signal levels can be anywhere between -7V and +12V (which is 5V plus the same 7V margin). I don't use CAN much, but different transceivers may have different specifications, and some may work just fine if GND gets lifted.

My mistake, I had a simplistic view of how the receiver chips operated, then I found this page which shows the resistive divider and bias system used to deal with incoming signals below ground level.
https://e2e.ti.com/blogs_/b/analogwire/archive/2016/02/09/rs-485-basics-the-rs-485-receiver

[e100]

Please correct me if I'm wrong.
The 24 gauge wire used in CAT 5 has a DC resistance of about 10 ohms per 100m.
So with a CAN receiver common mode limit of -7v, the max allowable current per 24 gauge ground wire is about 0.7 amps.
So using 2 ground wires and a 48v supply I'll have approximately 66 watts available?

Correction, with a 14v out and back drop it'll effectively be a 34v supply giving 47 watts if the load is concentrated at the far end.

[e100]

One solution to the ground lift problem would be to use two separate grounds. You've got 4 pairs available so a pair for CAN then two pairs for power+ and power- and another pair for logic ground. Your logic ground is connected to power- at the far end only and passes very little current so is roughly the same voltage at both ends. Your LED switching transistors go down to power- and the odd volt here and there can be allowed for by biassing resistors on the gates. The power for the microcontrollers comes from an smps at each node.

Does this still work for intermediate nodes half way along the cable as in my original plan?

[Doctorandus_P]

With all the amps through the whole cable you have a worst case scenario.

If you have 10 nodes (or some other number) divided equally over the cable, then the part of the cable near the end will have less current, and therefore also less voltage drop.

To do a more thorough calculation, first draw a schematic with all the partial cable resistances between your nodes, and then start adding in numbers for currents and voltages.

But many things depend on implementation details.
Best results is probably when you use all 8 wires available. Then use 2 for the bus, 3 for +Power and 3 for GND. You can split the GND for a "clean" GND for the logic and communication, and a "dirty" ground for the LED current, which will then have more voltage drop over the cable.

10 Ohms for 100m of wire sounds about right.
Note that dissipation is quadratic with current, At around 1A through one wire pair I noticed some warming but it was manageable. 1A through all 6 of the wires may be too much for CAT5. I have not punched in any numbers for calculation.

[drvtech]

One solution to the ground lift problem would be to use two separate grounds. You've got 4 pairs available so a pair for CAN then two pairs for power+ and power- and another pair for logic ground. Your logic ground is connected to power- at the far end only and passes very little current so is roughly the same voltage at both ends. Your LED switching transistors go down to power- and the odd volt here and there can be allowed for by biassing resistors on the gates. The power for the microcontrollers comes from an smps at each node.

Does this still work for intermediate nodes half way along the cable as in my original plan?

I think it should, although I haven't actually drawn it out. The main thing is to keep the low current, 'logic ground' separate from the high current 'power -' everywhere except the far end and only use the logic ground as a ground rail for the microcontrollers (and wherever your CANBUS signalling is coming from).
As a previous poster suggested, draw it out with resistor symbols to represent the resistance of each of the long bits of wire and it should become clear.