I'm very new to EE, so please excuse me if this question is bad or has an obvious answer. After reading an overview of differential signalling, it left me wondering:

Why does differential signalling use "complementary" signals for D+ and D- instead of simply making D+ the input voltage and D- the "ground" voltage (or whatever the reference voltage is)? In other words, instead of:

D+ = +½ V_SIGNAL
D- = -½ V_SIGNAL

(as this site states is the case for differential signalling), why not simply make the signal pair as follows:


External interference would still affect both wires the same, and the receiver could still do V_SIGNAL = (D+) - (D-) to remove common-mode interference and recover the original signal.

Update: This graphic I made should help clarify what I'm asking - Why is the DS sent as "complement" signals (like on the left) instead of sending the orignal signal "as is" and ground reference (like on the right)? Would external interference not be "induced" into the ground wire in the same way?

enter image description here

  • \$\begingroup\$ Two behavioral reasons: driving both gives you twice the differential voltage swing of only driving one, which makes life easier for the receiver. Next, if and only if there is an external electrical connection (for example through a power supply) or greater capacitance to an external object from one half of the pair than the other, and this leads to unbalanced currents, then the radiated fields can be greater. \$\endgroup\$ Jul 21, 2014 at 18:34
  • \$\begingroup\$ @ChrisStratton: If the wires are connected to a low-impedance signal source and low-impedance ground, each via 75 ohm resistor, and the receiver impedances likewise match, would the pair of wires radiate or accept any common-mode noise? I find myself thinking the combination should radiate, but I don't see how the received differential signal wouldn't be immune to induced common-mode noise. Would this be an exception to the rule that radiation and noise-susceptibility go hand in hand, or am I missing something? \$\endgroup\$
    – supercat
    Jul 27, 2014 at 16:34
  • \$\begingroup\$ Sorry for bumping an old post, but I found this document fantastic for learning about Differential pair routing, specifically Appendix A - Transmission Line Theory and Channel Information oiforum.com/wp-content/uploads/2019/01/OIF_CEI_03.1.pdf \$\endgroup\$
    – Andrew
    May 22, 2019 at 16:59

5 Answers 5


"Ground" as a concept needs some clarification. If you have just one signal line and one ground, then yes, it's hard to tell the difference. But if there's anything else going on, it matters.

All AC signalling involves current flow, even if it's being measured as a voltage on the receiver. At a minimum, you have to charge/discharge the parasitic capacitance of the receiver, and the signal wire will have a capacitance to ground as well. Note that the common mode behavior only holds if the two wires have the same length and are a constant distance from each other.

So if you have a signal wire and a ground wire, then the current in the signal wire must be matched by a corresponding current in the other direction in the ground wire. If you have lots of signals, the ground wire will contain a mashed-together copy of all the signals. Therefore it's advantageous for each signal to have its own ground. If you look at VGA, you'll notice that each signal gets its own ground because of this. If you look at 80 pin IDE, each pair of signals has a ground wire between it in the ribbon cable. Those are to prevent the signal wires inducing currents in each other ("crosstalk").

Once you've accepted that each signal must have its own matched ground, it's more natural to embrace the two as a matched pair, disconnect one from ground and connect the two of them together via a termination resistor network, and drive / read them as a differential signal.

  • \$\begingroup\$ This image (which I just made in photoshop) may help you understand my question: postimg.org/image/h63pwhzuv - why is the DS sent as "complement" signals (like on the left) instead of sending the orignal signal "as is" and ground reference (like on the right)? Would external interference not be "induced" into the ground wire in the same way? \$\endgroup\$
    – etherice
    Jul 21, 2014 at 14:01
  • 4
    \$\begingroup\$ IF the "ground" wire runs parallel to the signal wire, and there's no other signals between the two ends, and the ground wire is not also part of a power/ground system, and there is no ground loop, then yes you would expect the same interference. My point is that that's a long list of conditions that may be hard to meet. \$\endgroup\$
    – pjc50
    Jul 21, 2014 at 14:53

The difference is that the differential lines are matched in common mode impedance (wrt ground), so that noise induced or capacitively coupled to the lines tends to be common mode noise and thus rejected.


A differentially driven cable creates no effective far field interference - what is created by one wire is cancelled by the other. Everyone so far is talking about susceptibility but the big deal to me is that two antiphase signals being carried along two wires (such as twisted pair) may generate local E and H fields but these fields effectively cancel out at some short distance away.

If you transmitted a single ended signal like this the far field interference would be significantly greater.

  • \$\begingroup\$ Are you sure? You seem to be assuming the existence of "absolute voltage" or, more practically, substantial capacitance or current path to something other than the opposite half of the pair. \$\endgroup\$ Jul 21, 2014 at 16:40
  • \$\begingroup\$ @ChrisStratton I'm certainly assuming the existence of magnetic fields emanating from both wires hence at some rather small distance the magnetic fields are practically opposite each other and cancel out. When it comes to the single-ended transmission with 0V on one wire and a signal on the other there is no absolute voltage presumed and as a practicality this type of transmission would tend to be grounded on the 0V wire. \$\endgroup\$
    – Andy aka
    Jul 21, 2014 at 18:02
  • \$\begingroup\$ Why do you insist on thinking there is a difference? What do you mean by "grounded"? Unless you have some other conductor, the currents in the conductors of the pair will be complementary, regardless of how many of them are driven. \$\endgroup\$ Jul 21, 2014 at 18:04
  • \$\begingroup\$ @ChrisStratton I'm not insisting on anything. "Grounded" meaning is the usual one as in earth plane etc.. \$\endgroup\$
    – Andy aka
    Jul 21, 2014 at 18:07
  • \$\begingroup\$ So your system would behave differently in space? \$\endgroup\$ Jul 21, 2014 at 18:07

why not simply make the signal pair as follows:



I assume you mean that the V_GROUND signal would be connected to the circuit ground at one end or the other of the link. Maybe at both ends.

If you do this, you are no longer doing differential signalling, you are doing plain old single-ended or unbalanced signalling.

So the question you're really asking is, When do we use differential signals and when do we use single-ended signals?

Single-ended signals are used in lots of situations. Most traces on a PCB are typically unbalanced signals. For short connections with slowly-changing signals over a ribbon cable between boards, we often use single-ended signals. Even for long connections between boxes, unbalanced signals can still be used.

But to achieve good noise immunity and low radiated emissions, for reasonably fast signals over "long" distances, unbalanced signals typically require a shielded cable, such as coax.

But coax is more expensive than the now-ubiquitous unshielded twisted pair (UTP) cable.

So we'd rather, if we can get away with it, use UTP.

Now, if we drive a signal on one line of a UTP connection, we can, like you say, eliminate common-mode interference by receiving the signal at the other end with a differential receiver (V_SIGNAL = V+ - V-). But look at the signal we're sending out on the cable. The signal we're generating has a substantial common mode component.

V_CM = 0.5 * (D+ + D-)

Because of this common mode signal, our system is likely to radiate strongly, which will make it difficult to sell in most jurisdictions.

So really, we are using differential signalling to allow us to use low-cost UTP wiring with reasonable (hopefully conforming to standards) radiated emissions. In some cases we might still prefer unbalanced signals on shielded cables when the system requirements (very low emissions or very stringent immunity requirements) justify the cost.

Even over short distances within a box, we might use differential signalling on untwisted cable (such as common ribbon cable) to reduce emissions at a lower cost than using shielded cables.


What is the typical return path for current driven on the differential pair?

The currents on the two conductors of a differential pair (for a differential signal) are equal and opposite. You could say that each one of the conductors is the return path for the other one, or you could say that the net current is zero, so no return path is necessary.

Of course if you launch a common-mode signal on the differential pair, its current will need to return somehow. If there's no designed return path for the common-mode signal, it could travel over a large loop and cause significant EMC issues.

What, exactly, about the differential pair signalling causes the EMR cancellation?

  1. Because the two wires are closely coupled to each other, there is a small loop area between them and so the possibility to generate magnetic fields is small. But you could achieve the same thing with a well-designed single-ended transmission line.

  2. In twisted pair, you have alternating loops where the magnetic field will be in opposite directions. In the far field, the contributions from the alternating loops will tend to cancel each other out, resulting in very little radiation. There will also be a similar effect for susceptabilty.

  • \$\begingroup\$ Thanks. +1. I have 2 questions: (1) What is the typical return path for current driven on the differential pair? (2) What, exactly, about the differential pair signalling causes the EMR cancellation? Is it because one wire has a decrease in current that is equal (in magnitude) to the other wire's increase in current, and these equal-but-opposite changes in current result in equal-but-opposite magnetic fields? \$\endgroup\$
    – etherice
    Jul 26, 2014 at 23:04
  • \$\begingroup\$ Would it be practical to use wires as quartets rather than pairs, such that at any given time each quartet would have two signals high and two low? If at any given time one always switched two wires of opposite polarity, one could deliver two bits per cycle synchronously, with embedded clocking on every cycle. \$\endgroup\$
    – supercat
    Jul 27, 2014 at 15:12

What you're missing is the noise immunity.

Say you have 2 lines, one at +5V and one at -5V. The difference between them is 10V.

Now say 1V of noise gets induced into each line equally. The +5V line becomes +6V, and the -5V line becomes -4V (5 + 1 = 6, -5 + 1 = -4).

The difference between the two lines is still 10V. The noise has been cancelled out and the original signal is still intact.

That is the main point of differential signalling.

  • \$\begingroup\$ Here is a picture illustrating this: upload.wikimedia.org/wikipedia/commons/thumb/e/e7/… \$\endgroup\$
    – ACD
    Jul 21, 2014 at 12:47
  • 2
    \$\begingroup\$ I understand the concept of DS (interference affects both wires the same, and it cancels out at the receiver). But the same thing should happen if you're using the original input voltage (e.g., +10v) for D+ and the ground potential (0v) for D-. For example: Let's say the input voltage (at the transmitter) is 3 volts, so it sends +3v (relative to its ground) on D+, and ground on D-. If 1-volt of noise is induced into both wires, then the receiver gets +4v and +1v, the difference is still 3 volts (noise cancelled, signal in tact). \$\endgroup\$
    – etherice
    Jul 21, 2014 at 12:54
  • \$\begingroup\$ Except, ground is ground. It's 0V, not 1V. So, the difference from 0V to 3V is 3V - the difference from 0V to 4V is 4V. The signal is not the same, and if 4V is interpreted as a different value to 3V, then your data is corrupt. \$\endgroup\$
    – Majenko
    Jul 21, 2014 at 13:35
  • \$\begingroup\$ @Majenko: So, to be clear, you're saying that a wire connected to "ground" will always have the same electric potential, regardless of what external interference that wire is exposed to? \$\endgroup\$
    – etherice
    Jul 21, 2014 at 13:48
  • 1
    \$\begingroup\$ @etherice not exactly, no (see pjc50's comment), but the difference in induced noise between a signal wire and a ground wire will be different compared to the noise induced into a pair of signal wires. \$\endgroup\$
    – Majenko
    Jul 21, 2014 at 14:53

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