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I have been messing around with a signal generator, an RF current probe and an open ended coax cable.

From about 50 MHz onwards, I can measure a significant common mode current if I attach some wires (even short ones, about 2 cm in length) to the open ended coax cable, and it is my understanding that common mode currents are due to current waves traveling on the outside of the coax shield, as in this question. Attaching a 50 ohm termination, or any kind of small component at the end of the coax does not result in a common mode signal. Shorting the coax immediately at the end also does not give a common mode signal. The condition for common mode seems to be that the device at the end must be a radiating device.

Is it true that the common mode signal is caused by a radiating element attached to the coax cable? If so, why is that the case? And why would attaching a non-radiating element with the same resistance not cause these currents on the coax shield?

A picture for clarification. Depending on the termination, I measure a common mode current with the probe or not. If the termination is wires as in the picture, a current common mode current is measured. If the coax is instead terminated with a small component (be it capacitive or resistive) or immediately shorted, no current is measured.

Schematic of measurement

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3 Answers 3

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What causes common mode currents in a coaxial cable?

Common mode current in a transmission line fed by a differential voltage is the result of "mode conversion", which is the conversion of part of a differential signal into a common mode signal or vice-versa. Mode conversion occurs at either end of a (uniform) coaxial cable as a result of a difference between the "imbalance" of the transmission line and the imbalance of the components that connect to the end of the transmission line. This is described at LearnEMC.

The imbalance factor (also known as the current division factor), usually designated \$h\$, is the fraction of the total current that flows through a two conductor transmission line that flows through only one of the conductors, when the transmission line is driven by a pure common mode signal.

\$h\$ for an ideal coaxial cable is 1 (or 0). \$h\$ for an ideal twisted pair, or a balanced antenna, or indeed any balanced component is 0.5.

Mode conversion does not take place internally to a uniform transmission line, but always and only in places where the imbalance factor changes. In the example given in the OP, a coaxial cable, i.e. a transmission line with an imbalance factor of 1, is followed by a pair of wires, which for the sake of argument might be assumed to have an imbalance factor near 0.5. So, in that example, mode conversion will be present.

Attaching a 50 ohm termination, or any kind of small component at the end of the coax does not result in a common mode signal. Shorting the coax immediately at the end also does not give a common mode signal.

That is what we would expect.

The condition for common mode seems to be that the device at the end must be a radiating device.

Is it true that the common mode signal is caused by a radiating element attached to the coax cable. If so, why is that the case?

The answer to the first question is yes and no. Differential mode to common mode conversion results in common mode current at the junction between the coaxial cable, and the element to which it is connected. By continuity of current, there will be some common mode current in both the coax and the connected element in some neighborhood near their junction, and perhaps a long distance away as well. Any transmission line with unshielded common mode current will radiate. This includes the two wires connected to the coax, even if they are a twisted pair, but also, and this important, the coax as well. It is not so much the fact that the attached element is an antenna that makes common mode current present, but the fact that there is a change in imbalance that makes common mode present, which means that BOTH the two wires AND the coax will become radiators.

And why would attaching a non-radiating element with the same resistance not cause these currents on the coax shield?

Because (ideally) a "small" component simply terminates the transmission line, rather than serving itself as a transmission line with a different imbalance factor. (In fact, though, the leads of a resistor, say, will also be a transmission line, albeit a very short one.)

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  • \$\begingroup\$ I accepted this answer because it provides a useful explanation and source. The notion that the currents are caused by a transition of line modes with different imbalances is consistent with my measurements. \$\endgroup\$ Jan 26, 2022 at 7:44
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The condition for common mode seems to be that the device at the end must be a radiating device.

Correct. If the coax-inner current isn't all returned back along the coax-shield then, there must be some radiation (or extra loss or path) taking place that causes this: -

enter image description here

But, if the wires attached are close to that of a quarter wave monopole (length matches a quarter wavelength of the signal from the generator), then equal currents fill flow in both wires and there will be no common-mode current.

Is it true that the common mode signal is caused by a radiating element attached to the coax cable. If so, why is that the case?

Yes, it's true. Both limbs of the radiating element (what you describe as open-ended wires) do not necessarily radiate the same amount of energy. That means there is a difference in the current flow into both limbs of the radiating element (open-ended wires).

But, that "error" current (I've called it an error just for convenience) has to make a return path back to the source. It can make a return path via some awkward way or, it can make its return path back through the coax cable. The path of least impedance is via the cable and in particular, the shield. That gives rise to a common mode current flowing in the coax cable.

And why would attaching a non-radiating element with the same resistance not cause these currents on the coax shield?

Because there is no "error" current generated by a load like that.

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  • \$\begingroup\$ To be clear, it is the difference between the currents flowing into the radiating element at the feed points that defines the error, right? That is, if I connect the two wires at the ends I still measure a common mode current, because the length of this loop is significant in terms of the wavelength and therefore the current flow in the limbs at the feed point is different. \$\endgroup\$ Jan 25, 2022 at 13:12
  • \$\begingroup\$ You said this in your question: Attaching a 50 ohm termination, or any kind of small component at the end of the coax does not result in a common mode signal. Shorting the coax immediately at the end also does not give a common mode signal. The condition for common mode seems to be that the device at the end must be a radiating device. - are you now saying something different i.e. are you moving the goalposts of the question? I mention this because you are a new user of this site and this site is not a progressive learning resource i.e. it isn't a forum or talking shop. \$\endgroup\$
    – Andy aka
    Jan 25, 2022 at 13:15
  • \$\begingroup\$ So, on this occasion I will entertain your comment (now that I've got site rules/recommendations off my chest). If the loop radiates due to it having a significant length compared to the generator frequency then it will produce an "error" current (I'm using the term "error" just to make it clear what I'm talking about). \$\endgroup\$
    – Andy aka
    Jan 25, 2022 at 13:19
  • \$\begingroup\$ The results are still consistent. If I short the coax immediately at the end of the cable I measure no common mode current. When I connect the two (electrically fairly short) wires, I do measure a common mode current. In light of this answer, I'm trying to make sense of my measurements because it seems to me that the currents going in and out must be fairly similar if the wires are connected. \$\endgroup\$ Jan 25, 2022 at 13:20
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    \$\begingroup\$ Or maybe try this \$\endgroup\$
    – Andy aka
    Jan 25, 2022 at 13:30
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EDIT: something to see ?

from "Balanis_Antenna_Theory_ Analysis and Design"

9.8.3 Baluns and Transformers A twin-lead transmission line (two parallel-conductor line) is a symmetrical line whereas a coaxial cable is inherently unbalanced. Because the inner and outer (inside and outside parts of it) conductors of the coax are not coupled to the antenna in the same way, they provide the unbalance. The result is a net current flow to ground on the outside part of the outer conductor. This is shown in Figure 9.28(a) where an electrical equivalent is also indicated. The amount of current flow I3 on the outside surface of the outer conductor is determined by the impedance Zg from the outer shield to ground. If Zg can be made very large, I3 can be reduced significantly. Devices that can be used to balance inherently unbalanced systems, by canceling or choking the outside current, are known as baluns (balance to unbalance).

As stated in the post shown, the equivalent schematic is this (from reference), because of the radiating "part", the currents are not "equal" as in the case of 2 leads loads. A EM simulator would be necessary help to prove.

enter image description here

How to suppress this "current" : use of "bazooka", from reference.

enter image description here

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  • \$\begingroup\$ Thank you for the source but it does not answer my question. If I terminate the coax with a small capacitor instead of a dipole, there is still a path from the inside of the outer conductor to the outside surface of the outer conductor, yet I don't measure a common mode current when I do this. I believe there is something fundamentally different in the element being radiating, and it is this distinction that I'm interested in. Is I3 only there in the case of radiation, and why is it only there when the element is radiating? \$\endgroup\$ Jan 25, 2022 at 9:55
  • \$\begingroup\$ If the load is "really" a 2 pins (R, L or C ... "well" wired), the currents are the same ("inside out conductor" or "outside internal conductor", "closed" system). If "radiation" occurs ("open" system), then it can be a current on "outside out conductor" (induced by an "external" current, same as receiving signal on "receiver antenna", external side of wire). It is why "baluns" exist and are used. I think that only a EM simulation would justify that clearly. NB: the two wires external to the coax do the "same thing", but not at the same distance, so not the same contrary "value". \$\endgroup\$
    – Antonio51
    Jan 25, 2022 at 10:08
  • \$\begingroup\$ NB: ZL is the radiating "resistor" as usual ... but Zg is also a "radiating" resistor ... on a "certain length" of the "external side" of the outside conductor, but it is "perpendicular" to the other two wires. So, if one goes far away from the dipole, this effect will disappear? \$\endgroup\$
    – Antonio51
    Jan 25, 2022 at 10:14
  • \$\begingroup\$ This is done with a "bazooka" : i.stack.imgur.com/iobTV.png \$\endgroup\$
    – Antonio51
    Jan 25, 2022 at 10:25
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    \$\begingroup\$ My coax cable is 1m long. I have tried measuring the common mode current when the cable is quarter wavelength (at 75 MHz, since propagation medium is air in this case), thinking Zg must be close to infinite since the cable is grounded at the generator. The common mode was indeed very low at that frequency, so no confusions there. My questions are more related to the origin of the common mode current. \$\endgroup\$ Jan 25, 2022 at 10:49

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