# Does a twisted pair cable radiate more EM radiation than non-twisted pair?

I am looking at many electronic forums and people always come up with different answers, so I would like to settle this question once and for all.

Does a twisted pair cable radiate more EM radiation than non-twisted pair? In the sense of generating EMI interference for nearby equipment and/or radiating the signal from inside the cable to a larger distance.

Let's just ignore shielding for a moment and focus on the twisted pair/non-twisted pair types, which setup radiates less EM radiation?

We know that the twisted pair cable shields the signal from outside interference, but does it shield other cables from its own generated interference? Is it vice versa?

• This question, and the accompanying answers, assume the the two wires are carrying perfectly balanced differential signals. Apr 6 at 17:08

Yes, it is vice-versa, through the principle of reciprocity.

Any "antenna" is just as (in)effective at transmitting as it is at receiving, so the advantages of twisting the conductors together apply in both cases.

• Could you add more details? This is more like a comment Oct 1, 2016 at 1:08
• @jbord39: It's a direct answer to the question. I could excerpt some of the Wikipedia page, but it's just as easy to follow the link. Oct 1, 2016 at 1:46

Does a twisted pair cable radiate more EM radiation than non-twisted pair?

No

In the sense of generating EMI interference for nearby equipment and/or radiating the signal from inside the cable to a larger distance.

In the special case of two twisted-pair cables laying side-by-side, if the twistings are "exactly wrong", they can have EMI between them. (This is why modern ethernet cable has twisted-pairs that have different twisting rates).

With twisted pair, the electromagnetic forces cancel out over the length of the pair as each 360 degree twist completes. This ends up cancelling the inductance -- there literally is no transfer of energy from electrical to magnetic, or the other way either, from magnetic to electrical, when the twisting averages them out. This property keeps the twisted-pair from sending out EMI, and also keeps the twisted-pair from receiving EMI. Except for the special case. Or the case where something is really close to the twisted-pair, such that the averaging out, or cancelling-out of electromagnetic forces, does not happen.

An example of an instance of bypassing the twisting electromagnetic effect, would be using a current clamp-meter around only one conductor of the pair to measure a 0.3A 60 Hz AC current going through the twisted-pair. The magnetic field created around the single conductor gets measured. Similarly, any wire or inductor or transformer wrapped around just one of the conductors can both send and receive EMI with that conductor. But where the inductance of the twisted-pair has been effectively canceled by sufficient distance, there is no EMI to or from the pair. No inductance, no EMI.

Let's just ignore shielding for a moment and focus on the twisted pair/non-twisted pair types,

Okay...

The twisted-pair.

We know that the twisted pair cable shields the signal from outside interference, but does it shield other cables from its own generated interference?

Yes. Except for the very special exception already told above.

Is it vice versa?

HTH.

Does a twisted pair cable radiate more EM radiation than non-twisted pair?

"It depends."

If the driver sends balanced signals using a balanced drive impedance, then each wire will have equal and opposite current and voltage, so the far-field radiation of both wires will cancel out. If the wires are twisted, then each twist will radiate EM waves of opposite polarity, and they will cancel much better in the far field. In the near field (say, 1cm away from the cable) they don't, unless there are many twists per cm. The twisting effect is only effective at a distance which is relatively large to the length of a twist.

However if the driver is unbalanced (in impedance or voltage) then some common mode component will be present on the wires, which will then act as a single wire carrying this common mode component.

This occurs with most drivers, as the two resistors, and the components inside the driver, can't match perfectly. If there is a few % mismatch then the same amount of differential signal will be converted into common mode. Thus e common mode choke can be used on the differential pair to add common mode impedance without affecting the differential signal. Adding impedance reduces common mode current, and thus reduces emissions.

For example, Ethernet uses a common mode choke and a transformer. The transformer provides isolation, but also decreases the amount of common mode current sent into the pair, which is essential for long cable runs, which make very good antennas.

Common mode emissions are the same whether the wires are twisted or not.

Any "antenna" is just as (in)effective at transmitting as it is at receiving, so the advantages of twisting the conductors together apply in both cases.

Yes, but in this case the driver and the receiver also matter: a twisted pair driven by a single ended signal will radiate due to the common mode component, but if the impedances are balanced and the receiver is differential and has good CMRR, then it can have good immunity.

So, the "low emissions" advantages of a twisted pair only occur when properly driven. If the driving signal is single-ended, then a coax would be better. Twisted pair requires a balanced driver, or a balun transformer (ie, common mode choke).

• I only just saw this (someone came along and upvoted my answer). If you wanted me to see your "reply", you should have left me a comment. Most of this answer has nothing to do with whether the pair is twisted -- the difference between balanced and common-mode signals applies regardless. And twisting is just as important in the "near field" as in the far -- it's what allows me to run a long microphone cable alongside a power cable and still experience essentially zero interference. Apr 7, 2020 at 11:48

Twisted pair has lower differential impedance and lower common mode noise.

Impedance can be regulated by gap and twists per unit length.

CM chokes are needed for inputs when GBW product degrades CMRR and emissions rise from imbalanced inputs or outputs. Hence the term BALUN, which is a bi-directional two-port. A BALUN raises the CM impedance to attenuate stray EMI and a CM choke or BALUN is used in all line filters, and ethernet interfaces, industrial stepper motor cables.

• This doesn't seem to answer the question. It just seems like an explanation of twisted pair. Sep 30, 2016 at 23:57