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I found two calculators on internet for calculating the self inductance of a wire and an other calculator for calculating the inductance of twisted pair.

Here are the results for the following parameters :

  • Diameter of wire: 0.2 cm
  • Length of wire: 100  cm
  • Distance between wires: 0.3 cm

Here are the results :

  • Wire self inductance = 1.37  μH
  • Twisted pair inductance = 441 nH

What I am not sure about is what is really the twisted pair inductance given by the calculator? I said that the length of the cable was 1 meter but in a twisted pair cable there are two conductors, so the inductance of the twisted pair represent the "return" path and also the "going" path? So what I really calculate is the inductance for a cable of 2 meters and not for only one meter? I mean to make the comparison between the two kinds of cable I should take 2 meters of the wire self inductance and 1 meter for the twisted pair inductance. And so we got the following results :

  • Wire self inductance 200 cm = 2,74 μH
  • Twisted pair inductance 100 cm (+100 cm) = 441 nH

And so twisting pair reduce the inductance by a factor of 6!
It appears to me a lot and I am not sure that it is correct.

Did I make a mistake?

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2 Answers 2

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Twisted pair is not meant to transport a current and the return current to convey information - the information travels as the energy between the two conductors, and always in one direction.

So, the value you're being presented with us the inductance per unit length of the twisted pair cable understood as transmission line. It answers the question of "if I suddenly changed the differential current I use to excite this transmission line, what happens?".

So, you wouldn't compare these numbers altogether. They are not describing the same thing. You can feed a common-mode current into both conductors of a twisted pair cable and calculate something that is comparable to what you've got for the single conductor. But I wouldn't know what that would be useful for.

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  • \$\begingroup\$ Thank you for your answer. If you send a data across a twisted pair, there will be a current flowing into the "going" path and a current flowing into the "return" path. I do not see it as a common mode current. I do not really understand your answer. Do you know how can I modelize the inductance of a twisted pair for "going" and "return" current ? \$\endgroup\$
    – Jess
    Commented Dec 5, 2023 at 7:54
  • \$\begingroup\$ Yes, as a transmission line. No, there will not be a global current flowing back (as long as bits are balanced, which usually is taken care of). That's not how that works. The information is carried as guided radio wave between the two conductors. You will need to learn about transmission lines. \$\endgroup\$ Commented Dec 5, 2023 at 7:57
  • \$\begingroup\$ You say "if bits are balanced", so you mean there are more "going" path than "return" path ? \$\endgroup\$
    – Jess
    Commented Dec 5, 2023 at 8:04
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    \$\begingroup\$ No, I don't. I mean that there is such a shaping of data that you modulate into the analog waveform that there's net balance between positive and negative voltages or currents. That's because you don't want to transfer a net average current from source to sink. Again, the transport of information here happens in the electromagnetic wave in the space between the two conductors, not as current that reaches the receiver. \$\endgroup\$ Commented Dec 5, 2023 at 8:13
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    \$\begingroup\$ again, that is not what's being modeled here. Sigh. \$\endgroup\$ Commented Dec 5, 2023 at 8:31
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The inductance of a lone wire in free space is something of a misnomer.

There are useful measures of free-segment inductance, but it must be applied carefully, as there is no physical meaning until a complete closed loop has been formed. We can calculate an inductance this way, and even hand-wave a close-to-meaningful result, but we must always keep in mind what we're really doing, how we're doing it, and where that loop finally goes.

When we close a loop, we generally end up with a lower measure of inductance, essentially because the magnetic field from different regions of the loop partially cancel out. In particular, if you have two wires close together making a loop, the field at a distance of several multiples of the spacing will be very small -- it drops off as roughly the inverse square of distance (as opposed to the inversely-proportional lone-wire case).

When we send a signal down a twisted pair, we specifically send a differential signal, complementary (equal and opposite) between the two wires; the loop is closed at both ends, by whatever circuits we have attached there, and so the parallel-wire case applies.

If we sent a signal common-mode (both wires acting together), we have to close the loop somehow, even if that means closing it through capacitance to free space, or radiation resistance: in general, we won't have a nice neat pure-reactive inductance, but some amount of coupling (mutual inductance) to whatever other conductors happen to be nearby, and resistance due to, essentially, some of the AC magnetic field being lost as it radiates off into the aether (or absorbed in nearby lossy elements, whatever). In practical cases, the common-mode loop might be closed on nearby metallic structures -- ground wires, nearby equipment, metal beams, the mildly conductive surface of the Earth [moist soil], etc., and so we might have a situation more akin to a much wider parallel-wire transmission line (where the separation distance is the cable's height above nearby conductor(s)), or microstrip (cable above ground plane). The larger distances mean the inductance per length is relatively high (near \$\mu_0\$).

The common-mode case is similar to your one-wire calculation, but again -- at least to the extent we can conceive of a closed loop -- it's actually a loop, not an imaginary lone wire, and perhaps if we're lucky, can be reduced to segments of transmission lines, without having to do a full-field analysis (but that is also very likely given the complexity of a real environment i.e. everything in the room matters).

Because there are so many more sources of noise, and such poorly controlled impedance, in the common mode, we prefer to transmit signals differentially; whether that's between the core and shield of a coax cable (the shield isolates the interior currents from external common-mode fields, and the common-mode geometry is explicitly given by the outer dimensions of the shield), or symmetrically as with twisted pair.

Note, because the system is linear (we're only considering fields around ohmic wires, no funny business here), it obeys superposition, and so we can represent the voltages on both wires (with respect to some imagined distant reference[1]) as not just two voltages \$a\$ and \$b\$, but their sums and differences, \$V_{CM} = (a + b) / 2\$ and \$V_{DM} = a - b\$. (Well, customarily it's the average not the sum, but the average uses a sum, alright.)

[1] And along some given path, because in general, AC voltage is path-dependent -- again, loops matter a great deal to general fields.

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  • \$\begingroup\$ Thank you very much for your answer. I better understand the concept :) \$\endgroup\$
    – Jess
    Commented Dec 5, 2023 at 12:24

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