...is the best search term I've come up with, and still no relevant info...
So here's the deal: say I've got a multi-voltage DC power-supply ~10ft away from my equipment... the equipment consists of a [single-board] computer and various peripherals, all powered from that same supply.
Now, say the computer runs off 5V, it's connected to a hard disk which runs off 5V and 12V, and it's connected to the computer via USB.
Forget "ground loops", here we're talking *several* factors which could affect data, or even inject weird/negative voltages where they don't belong.
There's the obvious loop on the ground, around 25ft [10ft from the power supply to the computer, 10ft from the supply to the drive, 5ft between the drive and computer]. But, again, who cares about the audio-nuts' much-loathed "ground-loop" "hum" in a digital system?
Forget that for a second, let's look at the devices. The computer uses 3A max, but varies dramatically depending on CPU load. The hard disk's 5V is used only for digital stuff [I'm presuming], whereas the 12V is used for motors. Let's look only at the 5V system for now, and also presume that the disk's 5V load stays relatively small and somewhat constant.
Now, again, looking at the "ground" [or maybe more appropriately "common"] signal from *just* the perspective of the 5V loads, there will be different voltage-drops on those 10ft cables to the compy and drive. Say the drive uses half an amp at 5V, and say there's 1ohm in its 10ft connection to "common"... that leaves the drive's "common" [at the drive] at +0.5V relative to the power supply.
Now say the compy is asleep, and drawing only 0.1A, its "common" rests at 0.1V WRT the supply. Compy and disk are attached via 5ft USB cable; their "commons" also.
What happens, now? A USB cable, with its comparatively-thin wires [unless both sides' shields are connected to common, which is often the case], say 1ohm, has 0.4V across its common wire... it *should* not carry current [except in the case of a USB-powered device, but that's another matter], but V=IR, I=V/R=0.4/1=0.4A.
That's nearly *all* of the disk's circuitry current, travelling not through the disk's power wiring, but through the USB cable and back to the power-supply through compy's power cable.
Now, obviously this can't be the case... because those calcs rely on compy's common carrying 0.1A, and our result is that it's carrying 0.5A, etc. So, some balance must be achieved.
Let's assume, for a moment, somehow that balances to 0 current through the USB's common. [Note that I can't quite visualize how, in consideration of such scenario].
Now, consider the transients; a load changes suddenly, maybe compy's awakened to draw 2A, it will take time for the system to balance out, again. There will be a moment, however brief, where current [plausibly a lot] will flow through the USB cable's common. A surge of current [nevermind the voltage difference] through a long wire parallel to several others which are coupled inductively... a transformer, sending a surge of current through data lines.
Now, sure, maybe CRC-errors will result in data retransmission... but this is looking kinda ugly already, and we haven't even considered the 12V system of noisy power-hungry motors, etc. sharing that same "common," nor other large/varying loads such as a USB-powered DVD drive deriving its power through compy's power cables. Further, we haven't considered nearly identical effects on the 5V "loop," nor external interference/"hum" commonly-associated with "ground" loops [that could just-as-well occur on a 5V loop].
So far we have reasonable potential for data-voltages that, at least, aren't within high/low voltage tolerances for data transmission. It's also plausible such signals may *exceed* input-specs. I.E. If the drive's 5V circuitry's "common" is at 0.4V higher than compy's, and it outputs a 5V data signal, compy will see a 5.4V input. And the reverse, compy might output a 0V data signal, which will be -0.4V into the drive's circuitry.
And all this is only in consideration of *inter*-system data communication. What about the effects of a 5V digital circuit's *power rails* fluctuating so dramatically?
And, of course, there's plenty of room for our 5V source wires to also drop those same voltages due to the same length and wire-resistance... 0.5V on the drive's circuitry's "common" *and* 0.5V on its 5V wires, resulting in only 4V for the circuitry. [This is where remote-voltage-sense is appropriate, but can't do that with multiple endpoints!]
These values may seem extreme, but we are talking *amps* of current, non-negligible load fluctuations, and long wires.
It's almost a wonder a typical full-tower PC can even keep things straight with various drives and multiple endpoints, floppy and IDE cables with interconnected commons looping all over the place, motor-currents [however briefly] surging through ribbon cables with alternating data wires.
I'm a bit baffled as to how to best go about powering these devices. Almost thinking it best to just send 12V to each and have a local 5V regulator. But even still the long "common" wires, USB cables, voltage offsets, etc.
Oh, and I meant to go further into thoughts on the overall "balance" of the system. I'm almost certain it involves current through the USB cable's power wires, maybe even a lot, even in steady-state. Am somewhat convinced it may wind-up that the two long power cables both carry pretty much the same current, regardless of the endpoints' different needs. Then the USB cable must handle that balancing, so compy running full-tilt at 3A while the disk is sleeping means 1.5A through a measly usb-cable? And, further, reversed in direction... 1.5A *leaving* compy's "common," through the USB cable, *to* the device... I'm not liking this.