I know that the most robust way of transferring an analog signal over long distances is via differential pairs which are twisted together and then shielded (shield grounded at one end only).

But sometimes, we have to make do with single ended signals and I have seen other engineers twist a bunch of single ended signals with their very own GND line - all connected to the same net: AO0+GND, AO1+GND, AO3+GND, etc. Then shield each pair individually and connect one end of the shield to the same GND.

In my opinion, the very objective of twisting is to make both the signal and the return be affected by the same induction from the radiating environment. But if they don't have the same impedance, they can't be affected the same way right? I don't see how GND could have the same impedance as any of those signals, so in fact I think GND will see the noise picked up by all lines which is bad for all signals and even other parts of the circuit.

Assume DC signals, if that makes a difference (potential offset errors do matter).

In a nutshell: a) is it really any better? b) is it actually worse?

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    \$\begingroup\$ When you twist, you minimize the loop area for coupling as well. \$\endgroup\$
    – vir
    Jun 23, 2022 at 16:45
  • \$\begingroup\$ I think you are right that it should not be done blindly. It can indeed do more harm than good (for example increase cross-talk), but it depends. \$\endgroup\$
    – Eugene Sh.
    Jun 23, 2022 at 16:51
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    \$\begingroup\$ Twisting a signal with a ground does not make it a balanced differential pair. For one thing the source and destination impedances are different for the signal and the ground wire, so this by itself makes in unbalanced. \$\endgroup\$
    – SteveSh
    Jun 24, 2022 at 1:55
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    \$\begingroup\$ I don't know where you heard about that strange practice, but shields must by bonded to ground at both ends to work properly. However, in internal twisted return conductor (e.g. at GND potential) should be driven only from one side, same as the signal wire. \$\endgroup\$
    – tobalt
    Jun 24, 2022 at 3:33
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    \$\begingroup\$ @user42875 Please read my answer. I explain that that ground loops don't automatically couple to the signal. They only do, if you abuse the shielding as signal return. \$\endgroup\$
    – tobalt
    Jun 27, 2022 at 14:38

3 Answers 3


Yes, it is better.

Basic Twisted Pair Model

Model the cable as two transmission lines: one representing the differential mode (signal to GND wire), stacked on top of one representing the common mode (GND wire over GND plane). For coax over GND, this is a reasonably direct model; for twisted pair, use ideal transformers at either end to place the CM mode at the midpoint of the DM ports (this version is shown above).

The signal couples into CM corresponding to wavelength versus electrical length, exhibiting notches in the frequency response at harmonics, corresponding to maximum coupling and thus CM emissions/susceptibility (which aren't necessarily an overall problem, for example the cable or system can be shielded overall; but if it is exposed, then it's a prime suspect).

So in short, it's good when the signal bandwidth (signal bandwidth or pulse risetime) is short or on par with cable length, i.e. at frequencies where the coupling is not excessive.

Note there's always some kind of "plane" environment which defines the CM or GND-wire-over-GND-plane impedance. Even if the distance to that plane approaches infinity (~free space). In the extreme case, a system might look like, for example, two boards connected by a loose bit of spaghetti: an end-loaded dipole, with the feedpoint inbetween being driven by coupling from the unbalanced pair. The CM impedance is still well defined in this case (antenna impedance).

As for particular cases, consider a 3.3V CMOS logic signal for example. Maybe the 2ns edges couple a good ~20% of amplitude into the gnd-GND loop, or to each other (which is fairly typical for multiconductor with alternating GND, whether as a round cable bundle, or ribbon style). The leading edge of a square pulse is attenuated by about this much, and rises more gradually afterwards (maybe with lots of ringing due to various kinds of reflections). Well, that's no problem because the input threshold is 30-70% (typical for CMOS; some are 20-80% or even worse though). So the receiver still triggers on the leading edge, and it's fine.

Meanwhile, that 20% signal loss corresponds to a massive some-100s mV CM signal, which will blow the hell out of your EMC test if it's open to the environment (unshielded). (Typical limits are in the <1mV range, ballpark of course, depending on what and how you're measuring it.) Conversely, that remaining 10% noise threshold is easily violated by a couple volts of noise from external sources -- commercial testing levels of 3V (conducted) or 3V/m (radiated) will easily induce bit errors.

If the alternative is very few GNDs at all, then the signals in a multiconductor cable simply couple into each other, basically looking like grounds with respect to a given driven wire. Which easily blows out the noise margin on the other wires, causing chatter at every receiver when an output changes state. The cable acts as a differentiator, coupling sharp rising edges between lines. Again, it's entirely dependent on bandwidth and length: if you filter all outputs (or enable slew rating reduction options, as many MCUs have these days), you can still get successful communication through such media -- assuming you don't need the bandwidth (bitrate), of course.

For example, this allows RS-232 to be carried on, well, it's not even bad with wet string if it's salty. RS-232 is slew rate limited to a bit under a microsecond, and current limited to some 10s mA, so it's quite modest on emissions, and tolerant of unterminated cables. Likewise, it doesn't take much filtering to avoid external noise sources. (And just to improve things even further, it's usually wired to a shielded D-sub (DE-9, etc.) connector, providing even more immunity.)


I've extended discussion to this webpage on my website: https://www.seventransistorlabs.com/Articles/CableModel.html

  • \$\begingroup\$ Side note, RS-232 is slew rate limited to 30V per microsecond. In other words, the signal can fully switch from +5 to -5 volts in only 333ns. \$\endgroup\$
    – Justme
    Jun 23, 2022 at 19:45
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    \$\begingroup\$ One common mistake is trying to reduce the number of "ground loops" in a system. If you can get that number down to zero, that's good for reducing crosstalk. But in a complex system, that's often very difficult. In general, while zero is good, a small number of ground loops is very bad. If you can't get rid of all the ground loops, "let ground abound" is usually the best policy. If you have a lot of ground conductors, magnetic fields in the system can't induce much EMF because the web of grounds effectively short circuits it. \$\endgroup\$
    – John Doty
    Jun 25, 2022 at 12:34
  • \$\begingroup\$ Thanks for the reply. I +1'd all answers here, but can't decide which to accept. I'm tempted to accept this one but I'm having a hard time understanding it entirely. Would you mind adding schematics to illustrate what you are saying? \$\endgroup\$
    – user42875
    Jun 27, 2022 at 14:36
  • \$\begingroup\$ @user42875 how's this? \$\endgroup\$ Jun 27, 2022 at 19:38

It strongly helps when done right. Allow me to make a small story arc to explain my answer.

The problem

Ground planes are good dump-and-forget nodes for return currents at both high and low frequency because they are present everywhere on the board. But the issue comes when using it as return for precision single-ended signals, because any local change in the plane's potential will add to the measured signal amplitude at the receiver - bad.

The solution

Make the single-ended signal differential by adding a second dedicated return conductor to it, which can be also at ground potential


simulate this circuit – Schematic created using CircuitLab

The trick herein is that all low-frequency signal return currents for this signal will flow in the new conductor and all other return currents will not flow in this conductor. By adding Rreturn, one can even match the impedance of this wire to be similar to the signal sender's.

Potential noise from the GND plane and other sources would impact both traces similarly, highlighting the benefit of the differential signaling.

These benefits in signal integrity only come to fruitation when the receiver is differential. If you have a single-ended receiver, referenced to its local ground, then the signal integrity will not improve as much vs. omitting the ground wire. However, even in this case, having the ground wire close to the signal wire strongly helps with EMC because the return current loop size is essentially zero.

Differential receiver

The receiver doesn't need a reference to its local ground, i.e. it can be floating. Still, it must be assured, that the differential signal is still within the receiver's valid common-mode input voltage range.

Below are two examples of differential-to-single-ended receivers. The transformer approach on the left has very large input common-mode voltage range, limited only by its isolation strength. It can be used for HF signalling because then the transformer doesn't need to be so big.

The op-amp approach on the right is more common for low-frequency precision signalling because it works down to DC. The common-mode voltage range is limited by the supplies (or a multiple of them depending on gain). Therefore, the op-amp supplies must be referenced to the same ground as the sender, even if only by large 1M resistor to prevent the sender and receiver from floating too far apart.


simulate this circuit

Interboard transmission & shielding

When laid out on a PCB the two traces can be made as a neighboring trace pair, but between boards, one would typically use a twisted pair arrangement for these two wires. Optionally, one might opt to provide additional shielding around the twisted pair in harsh environments, but this shield must bond to a low impedance chassis/ground node at both the receiving and sending end to properly fulfill its function. This shield would encapsulate both wires used for transmission. If you transmit between two metallic enclosures, the shield bonds to their chassis. Lacking metallic enclosures, shields can be bonded to the respective circuit grounds instead.

This will typically form "Ground loops", but these are never a problem for a shield. They are only a problem for "single-ended precision" signals, which - as I wrote above - don't exist anyway because aiming for high precision in single-ended common-return signals is futile. The proper way is to use shields as intended (both ends bonded), and use differential signalling when dealing with a precision signal.

  • \$\begingroup\$ Thanks for your answer. I don't see any argument supporting the fact that ground loops are not a problem when bonding a shield at both ends. Nor why we should bond it at both ends. Care to expand? \$\endgroup\$
    – user42875
    Jun 27, 2022 at 14:41
  • \$\begingroup\$ @user42875 If you bond a shield only at one end, the dangling end will essentially float at high enough frequencies (depending on cable length, this becomes serious >10..100 MHz). As a result, the cable will readily pick up common-mode noise, failing EMI. And why the grounding is not an issue is explained in the paragraph "The solution". That said, ground loop are an issue, if you don't provide dedicated return conductors. \$\endgroup\$
    – tobalt
    Jun 27, 2022 at 14:45
  • \$\begingroup\$ In that case could you make it explicit how you ground the receiver and the shield on the diagram? Just to confirm? \$\endgroup\$
    – user42875
    Jun 27, 2022 at 14:50
  • \$\begingroup\$ @user42875 I will update the sketch tomorrow maybe. But to clarify: the shield is not shown at all in the sketch, as IMO, it is optional..It would additionally enclose both wires and be bonded to GND at both ends directly. The receiver is differential..It might use ground-referenced supplies or might even use GND as its negative supply for example. But the signal is not compared to ground by the receiver. The receiver can also convert it back into a single-ended signal (referenced to its local receiver ground) after evaluating it differentially. \$\endgroup\$
    – tobalt
    Jun 27, 2022 at 14:59
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    \$\begingroup\$ @user42875 I have updated/expanded the answer. Thank you for your feedback. \$\endgroup\$
    – tobalt
    Jun 28, 2022 at 6:47

We can distinguish two kinds of coupling, inductive and capacitative. Using balanced twisted pairs helps reduce inductive coupling. But the question is about unbalanced twisted pairs.

It turns out that if there are many signal wires in a cable, twisting the signal wires with ground wires can reduce capacitative coupling between the signal wires. Each wire is capacitatively coupled with every other wire, but is more strongly coupled with wires that are (on average) closer. Therefore, having the closest wire to a signal wire be a ground wire will mean that other signal wires will have less coupling with that wire. The capacitative coupling to the ground will attenuate the signal somewhat, but the fact that it is displacing other signal wires from being "closest" means that there will be less cross-talk.

That is an possible advantage of twisting a single-ended signal wire with a ground wire in a cable with many signal wires. However, at the same time, that ground wire may not have a current that matches the current in the signal wire. In that case, inductive coupling will tend to add noise/crosstalk into the signal wire.

Whether twisting single-ended signal wires with ground wires in a multi-signal-wire cable is a net win, or a net loss will depend upon the specifics. What are the voltages in the wires? What are the currents? What are the insulation thickness and dielectric constant? The details of the cable and how it is used will determine whether or not the practice described in the original question makes sense, or is misguided.

  • 1
    \$\begingroup\$ Telephone cables have been twisted pair for more than a century for crosstalk reduction. \$\endgroup\$
    – grahamj42
    Jun 24, 2022 at 15:55
  • \$\begingroup\$ @grahamj42 This is true. However it doesn't quite match the scenario of the question. Neither the tip, nor ring wire of a plain old telephone system (POTS) is permanently connected to ground. So there is not a multitude of signal wires each one twisted with ground wire where are the grounds are common. \$\endgroup\$ Jun 24, 2022 at 19:11
  • \$\begingroup\$ It's still fairly effective against inductive crosstalk even if the line is unbalanced. \$\endgroup\$
    – John Doty
    Jun 25, 2022 at 12:21

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