Is it necessary to separate DC power cables and data cables?

Question

I have had this discussion with my colleague earlier. A DC power supply is non-alternating, so the magnetic field a DC power wire generates is constant (is it?). Now, I know the rule is to separate power cables and data cables, but I'm assuming that's when it comes to AC power. Is it the same rule when it comes to a regulated DC power supply?

We are using CAN bus twisted pair wires next to regulated DC power cables (12V and GND). I understand that CAN is immune to noise, but if you had a different data cable (let's say UART aka serial, or Ethernet), would the DC power cables have any impact? If so, why?

Answer 1

The answer is "It all depends".

• What is the load on the DC? If it's very noisy inductive loads, you're going to have noise on the DC line, and it might be considerably more than you'd think
• What is the signalling rate on the data lines? Faster rates are vastly much more sensitive
• [EDIT] What line encoding have you got? Anything differential, such as RS-485, is going to be a lot more robust than something voltage-based, such as RS-232

• What coding scheme are you using? If you have got any scheme with error detection, perhaps it will be ok

• What happens if there are errors on the line? If it is updating a clock display, with the effect of a little time skew that's different to dropping heavy machinery onto workers.

Having said all that, it is quite common to have signal and DC power adjacent. I have quite a lot of underwater telemetry where we use specially-made DC-power and twisted pair cable for 24 VDC and 250 Kbit/s RS-485. In another much more noisy environment we use 9600 bit/sec. Per commenters, of course power-over-ethernet is one of the best examples of high-speed, long-distance, high-power DC and data in the same cable. (Long and high compared to for example USB or a bus on a PCB. 100 metres, 12 Watts.)

In short: it's perfectly doable, but pay good attention.

Answer 2

The current drawn over a DC power supply is usually not constant. Changing current results in a changing magnetic field.

So it might be necessary to separate power and data, it might not be. In USB or PoE Power and data are not separate. In SATA it is.

So you might need to take measurements and either separate the cables or get a better shielding between power and data.

Answer 3

Honestly AC vs DC isn't really very relavent.

There are two reasons to seperate power from data lines.

The first is safety. Voltages above 50V or so can be a shock risk. Currents over a handful of amps can be a fire risk. For this reason electrical regulations often require either a certain seperation between mains and communication circuits or extra precuations to be taken (such as earthed metal barriers or mains rated insulation on both the power and communication lines, exactly what is and isn't allowable will depend on what standards you are working to).

The second is interference. As you say constant DC isn't going to couple into your communications lines. If you have gone for a half-decent twisted pair for the data lines then 50Hz is unlikely to be much of a problem either. The real problem is the transients and interference that all too often end up superimposed on power wiring. How bad this is will depend very much on the characteristics of your supply and loads.

Answer 4

For 12V CAN bus there is usually no good reason to separate data lines from device's own power lines.

Any certified CAN device has to pass the test for immunity to coupled transient noise (ISO 7637 or similar), which specifies quite harsh conditions such as repetitive high frequency disturbance (e.g from a relay arcing under load). Arguably this is much worse than the noise of the power lines of your own device, so if you manage to certify your device for car use, it will have enough immunity so that your own 12V power cable in the vicinity will not be a problem.

UART will likely not work in an environment where CAN is used.

Answer 5

The reason to separate AC power from comms is the electrical code.

The reason in Code is the risk of power wires taking damage and shorting AC distribution voltages (100-277V) onto the comms wires, creating arcing/fire and shock hazards where they would be least expected.

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There's one exception to the Code. If the comms circuit, from stem to stern, end to end all the way, is insulated to Mains (Class 1) wiring standards, including the equipment at the points of use, then yeah, that mains-insulated comms wiring can intermix with mains. Some examples:

• 3-way smart switches that reuse an old messenger wire as their comms wire.
• office lighting with RR7 latching relays, where they send 24V to all the switch locations, which then make momentary contact to send 24V back to the relays. That wiring is typically run in conduit rated for mains.
• Ethernet wiring between SCADA control equipment, where all that equipment is housed inside enclosures rated for enclosing mains wiring. All that equipment is rated for having the ethernet card take a 277V hit.

What you can't do is run an ethernet cable in the conduit with mains wires, and then have the LV/comms circuit exit the mains wiring via an Ethernet cover plate, common cable and plug into a PC. That "exit" is the thing you cannot do.

Answer 6

Its your responsibility to have local-charge-storage, with dampening to prevent ringing.

That keeps the high-frequency current fluctuations to a minimum.

Let's run an example: high frequency Power Wiring trash (0.1 ampere peak, at 100MHz Fring, thus the dI/dT is 0.1amp * d/dt(100MHz) == 63 Million amp/sec To simplify the math, assume the power RETURN wire is some distance away, so we will assume a SINGLE power wire, with irksomely fast ringing.

Assume the victim is 1 meter of wire for the data, and the data_return is 1mm away, and not twisted pair.

Assume 1mm distance between power and data wires.

Vinduce = [ MU0 * MUr * Area / (2 * pi * Distance)] * dI/dT

For MU0 = 4 * pi * e-7, MUr = 1 (air, copper, aluminum, FR-4), the

Vinduce = 2e-7 * Area/Distance * dI/dT [we ignore a weak natural_log coeff]

And we plug in the numbers

Vinduce = 2e-7 * 1meter * 1mm /1mm * 63Million amp/second

Vinduce = 2e-7 * 1 * 0.63e+7 = 1.26 volts interference

solution: use twisted-pairs, with different turns/inch hot ppwer versus data signals

So what to do, if ONE bundle of wires is the only choice? Use "Local Batteries".

^{simulate this circuit – Schematic created using CircuitLab}