Coaxial Cable vs Shunt, path of least impedance

Question

The website LearnEMC.com has a page called Identifying Current Paths which contains the following graphics.

The graphic on the left shows a coaxial cable arranged in a "U" shape, with a shunt connecting the shield at each end of the U. The graphic on the right shows that at low frequencies, current "prefers" to travel through the shunt, rather than in through the longer coaxial shield. However, at higher frequencies, the current "prefers" to travel through the coaxial shield, rather than through the shunt. That is, the current prefers to travel in oppositely travelling pairs, even though one of the pairs has a shorter alternate path.

The purpose of this graphic is to illustrate a special case of the principle that "current seeks the path of lowest impedance". (Obviously, current doesn't only travel through that path, but prefers it. However, this graphic seems relevant to the current discussion regarding Veritasium's video regarding the propagation of electricity. For example, this question and this question

How might one calculate the percentage of current flowing through the shield vs that through the shunt? Would the following procedure work?

1. Ignore the shunt, and calculate (or measure) the impedance of the coaxial cable by itself, then
2. Ignore the shield of the coaxial cable, and calculate (or measure) the (highly irregular) transmission line formed by the coaxial cable's inner conductor, together with the shunt.
3. Assume the ratio of current between the two paths is inversely proportional to the ratio of the two impedances found in 1 and 2?

I am inclined to believe that the shunt has little or no effect on the impedance of the coaxial cable. I have a doubt however, that the impedance found in step 2 is meaningful. After all, the inner conductor is shielded in reality, and that makes me question whether the impedance found in step 2 is at all meaningful.

Is there another method for calculating the current through the shunt as a function of frequency? Perhaps for special cases or special geometries?

Answer 1

How might one calculate the percentage of current flowing through the shield vs that through the shunt? Would the following procedure work?

You don't need to calculate anything to get close to the true answer.

Coax theory indicates that the preferred route for AC is through the coax shield. You don't need to measure anything either; you just need to recognize that when a load is attached to a length of ideal coax, the effective impedance of the shield is zero ohms.

Now let me repeat that because, it might appear problematic to some folk: -

$$\color{red}{\text{For an ideal coax, the impedance of the shield is zero ohms}}$$

For slightly non-ideal coax, the shield impedance is the DC resistance. For reasons of brevity, I'm not going to consider situations where the coax is significantly far from being ideal.

_{However, if you want to pursue this, there are several on-line articles explaining that the return-path of current through a ground-plane directly follows (as best it can) the forward-path current in order to minimize loop-inductance. Fundamentally, this is also why proximity-effect occurs in 2 wires.}

So, how can the shield of an ideal coax have zero impedance: -

• Ignoring the inner's forward-current, the shield return-current produces a magnetic field that exists only outside the shield; a magnetic field inside a "tube" is not made (proven theory).
• Ignoring the shield, the inner's forward-current produces a magnetic field that extends all the way from the inner (as it would for a normal wire that has no shield surrounding it).
• Combining the two; the magnetic field (outside the shield) produced by the shield's load return-current is exactly cancelled by the magnetic field from the inner's load forward-current.

_{We know this to be true because that is why we use coax cables for unbalanced signalling.}

• So, with no net field outside the shield, the effective inductance of the shield (when used properly with the inner) is zero.

There is of course a magnetic field between inner and shield but, that does not effect the impedance of the shield because magnetic fields inside a tube conductor have no impact on the inductance of the tube. Hence, the tube/shield has zero inductance.

In fact you could do an experiment to prove this (and it works): -

• Take 10m of coax and drive the shield (end to end) like a transformer primary.
• Let's say you used a moderate frequency like 10 kHz or 100 kHz
• Let's also say you applied a "primary" voltage of 10 volts RMS
• The end-to-end voltage on the inner would also be 10 volts RMS

_{You might say that if the shield produces no internal flux then how can anything be induced on the inner. You probably wouldn't say this if you know about induction of course.}

I'll leave that as a little thought but, back to this experiment.

• Now, instead of driving the shield you drove the inner (end-to-end) with 10 volts.
• If you then measured the end-to-end voltage on the shield, it would not have 10 volts RMS induced on it but, maybe only 9 volts.

This is because the flux that is produced inside the tube (by the inner) is not a valid flux for inducing a voltage in the shield/tube.

So, I hope I've demonstrated that the shield of a coax is ideally zero inductance and non-ideally it's zero inductance plus a bit of DC resistance.

This is why at low frequencies, the shorter "shunt" path is the preferred route because, it is likely to have a lower DC resistance compared to the longer shield of the coax. As frequency rises, due to the disposition of shunt and coax inner, the shunt will rapidly overtake the shield in terms of impedance (due to zero inductance in the shield) and be the far less preferable route for current.

$$\color{purple}{\text{This answer has got nothing to do with that stupid video}}$$

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2-dimensional QuickField simulations of flux and induced voltages in coax

The first scenario uses an excitation of 1,000 amps on the shield with the inner open circuit. Voltage measurements are "per metre" into the page (z-dimension): -

Ignoring slight discrepancies due to limited element structures in the student version of QuickField, we could make an argument that says the voltage on the inner is the same as the voltage on the shield. In other words, 95.906 volts is pretty damn close to 96.3 volts (within 0.5%). This is because both "wires" are subject to the same flux pattern (external to the shield).

Also note the lack of magnetic field between driven shield and inner core

The 2nd scenario puts 1,000 amps on the inner: -

Due to the inner carrying the excitation current and it producing quite a bit of flux inside the shield/tube, the voltage appearing across the inner is substantially more than the voltage across the shield. But, the important thing here is that the voltage across the shield is 96.382 volts i.e. pretty much the same voltage seen on the shield (per metre) when it was driven with 1,000 amps. 96.382 volts and 95.906 volts are within 0.5% of each other (despite the limited finite element sizes of the student model).

For both scenarios above, the magnetic flux density field pattern scaling was kept the same and clearly, both scenarios produce the same pattern and levels of flux density (the colours don't reflect the opposing polarities of course because they are RMS magnitudes).

And, just for completion, I dug up the old model for this rather simplistic and robust coax and plotted the fields for when the inner and shield were both set at 1,000 amps but with opposite polarities: -

Please take my word that the flux density outside the shield is miniscule in comparison with the two above scenarios. The voltage on the shield (despite 1,000 amps flowing) is only 1.08 volts and that, is due to the copper resistance. Maybe I'll simulate at 1 MHz next time.

Answer 2

I think AndyAka's answer is quite good, and I have accepted it. However, I think one point could be sharpened a little. Andy formulates that point as

for an ideal coax, the impedance of the shield is zero ohms

There is an article "Magnetic Coupling In Transmission Lines and Transformers" gives a formulation that I prefer.

Because of the absence of flux loops from the outer conductor in its interior, the self-inductance of the outer conductor exactly equals the mutual inductance between the two conductors.

Hence:

What this means is that no longitudinal voltage will exist on the outer conductor due to inductance, if the inner and outer conductor currents are equal and opposite!

One of the reasons why I prefer this formulation is that, rather than positing zero-inductive impedance, which seems to be a characteristic of the shield by itself, it gives a condition under which the voltage at the two ends of the shield will be equal. That is, the voltage at the two ends of the shield (in an ideal coax) will be equal if and only if the current through the shield is equal and opposite to that of the inner conductor. That is, apparent "zero inductive impedance" is not a feature of the shield alone, but a feature of the shield under certain circuit conditions.