I know several applications where twisted pairs are used and why they are used in those applications, but I would like to know more about the advantages and disadvantages of using twisted pair wires so I can choose appropriately based on what I am designing.

As an example, I understand the use of a twisted pair in differential signaling. This helps reduce noise because both wires receive the same interference which can then be removed by a differential amplifier.

I also remember reading quite a while ago about an induction forge where the power output was significantly lower than anticipated. The guy figured out that the separation of the power leads to the electronics enclosed a large area, creating extra inductance (which acted as a low-pass filter decreasing the power factor). By taping the power wires in a bundle, the stray inductance was removed.

It seems that twisted pairs offer benefits when inductance or induction are concerns, but this is where my understanding terminates.

What are the advantages and disadvantages of twisted pairs, and for what applications does this make them beneficial?

EDIT: The issue of inductance does not appear to be addressed elsewhere, indicating one application for which twisted pairs are helpful, which isn't addressed outside this question. Similarly, a comment raised the issue of added capacitance, which is not addressed in the linked question.

  • \$\begingroup\$ see my answer here electronics.stackexchange.com/questions/290310/… \$\endgroup\$
    – Trevor_G
    Mar 15, 2017 at 20:44
  • \$\begingroup\$ Thanks for posting the link. I missed it when scanning if my question is a duplicate. That said, it doesn't add any details beyond what I describe in my question, so I believe my question is still unique. \$\endgroup\$
    – Hari
    Mar 15, 2017 at 20:49
  • \$\begingroup\$ Its not that's your question is not unique, I just wanted to bring your attention to my answer. BTW the forge thing is probably because induction forges use HUGE currents. Any wire loop in the vicinity will pick up a LOT of energy from those... Even the metal band on your wrist watch if you are wearing one.... \$\endgroup\$
    – Trevor_G
    Mar 15, 2017 at 20:56
  • \$\begingroup\$ ^ My point exactly. That's why I am hoping for more general design considerations. \$\endgroup\$
    – Hari
    Mar 15, 2017 at 20:58
  • 2
    \$\begingroup\$ Possible duplicate of Is using twisted pair wiring makes sense only for differential signalling? \$\endgroup\$
    – Trevor_G
    Mar 15, 2017 at 21:18

2 Answers 2


There are two key advantages to twisted pairs

  • Reduced inductance \$ L = N^2 \cdot \frac{\mu \cdot Ae}{l}\$ by twisting the wires together you are reducing \$ Ae \$ the enclosed area and so the inductance.

  • Twisting the wires together means they are close together and so any noise picked up in one conductor should also be picked up in the other. A differential measurement should not see it.

The obvious down-side is cost. You will pay a little more if you buy a twisted pair or you can twist them your self but if you consider your time is not free this is an extra cost too.

  • 1
    \$\begingroup\$ Both the advantages you cite would also apply to 2-wire ribbon cable. The benefit of twisting is it reverses the direction of the loop every few mm, so that an interfering field that intercepts two or more twists partly counter-acts itself. \$\endgroup\$
    – The Photon
    Mar 15, 2017 at 21:57
  • \$\begingroup\$ @ThePhoton I was comparing a twisted pair with two separate wires which is what you usually see instead. Can you expand on this comment as a full answer? I don't instantly see why twisting the wires reverses the effect of the field. But I am always keen to learn more than I already know. \$\endgroup\$ Mar 15, 2017 at 22:07
  • \$\begingroup\$ Aside from cost, are there other disadvantages? I'm a little confused because I don't see them more often in noise-sensitive equipment when non-differential signaling is used. If there aren't any other disadvantages, then it seems the simple answer is used twisted pairs wherever the added cost isn't an issue... \$\endgroup\$
    – Hari
    Mar 15, 2017 at 22:18
  • \$\begingroup\$ Another obvious downside is higher capacitance, as a twisted pair is longer than flat strands. \$\endgroup\$
    – Janka
    Mar 15, 2017 at 22:35

Elaborating on @ThePhoton's comment by @WarrenHill's request.

enter image description here

The black X's are an increasing magnetic field pointing into the page. The integral form of Faraday's law says $$\oint \textbf{E} \cdot dl = -\int_S \frac{\partial{\textbf{B}}}{\partial{t}} \cdot d \textbf{s}$$ In words this means that there is voltage induced around loops of the wire. As illustrated below, the voltage induced on one wire in one loop is cancelled by the voltage induced in the same wire on the next loop.

This works best if the magnetic field is changing at a constant rate along the length of the twisted pair. If the change is not constant along the length of the twisted pair, then the induced voltage in one loop will imperfectly cancel the induced voltage in the next loop. The twist rate must be smaller than the wavelength of the EMI you are trying to block. You can imagine if you were dealing with an electromagnetic wave with a wavelength equal to the twist rate, you would get a lot of coupling as the sign of the magnetic field would change loop to loop, causing the induced voltages to add up rather than cancel.

  • \$\begingroup\$ To better tolerate uneven magnetic fields, use commercially-twisted high-density-twist pair; avoid hand-twisted with each twist being 1 inch long. \$\endgroup\$ Mar 16, 2017 at 2:57
  • \$\begingroup\$ The arrows represent... what? The two blue arrows are in opposite directions, as are the two red ones, so it's not current. \$\endgroup\$
    – Whit3rd
    Mar 16, 2017 at 6:39
  • \$\begingroup\$ @Whit3rd why can't the two blue arrows be currents? The whole point is that they are equal and opposite so the net induced current is 0. \$\endgroup\$
    – DavidG25
    Mar 16, 2017 at 15:57
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    \$\begingroup\$ A varying magnetic flow induces voltage, not current. \$\endgroup\$
    – Oskar Skog
    Mar 16, 2017 at 15:59

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