This question as originally written sounds a little bit insane: it was originally asked to me by a colleague as a joke. I am an experimental NMR physicist. I frequently want to perform physical experiments which ultimately boil down to measuring small AC voltages (~µV) at about 100-300 MHz, and draw the smallest current possible. We do this with resonant cavities and impedance-matched (50 Ω) coaxial conductors. Because we sometimes want to blast our samples with a kW of RF, these conductors are often quite "beefy" -- 10 mm diameter coax with high quality N-type connectors and a low low insertion loss at the frequency of interest.

However, I think this question is of interest, for the reasons I'll outline below. The DC resistance of modern coax conductor assemblies is frequently measured in ~1 Ω/km, and can be neglected for the 2 m of cable I typically use. At 300 MHz, however, the cable has a skin depth given by

$$ \delta=\sqrt{{2\rho }\over{\omega\mu}} $$

of about four microns. If one assumes that the centre of my coax a solid wire (and therefore neglects proximity effects), the total AC resistance is effectively

$$ R_\text{AC}\approx\frac{L\rho}{\pi D\delta}, $$

where D is the total diameter of the cable. For my system, this is about 0.2 Ω. However, holding everything else constant, this naïve approximation implies that your AC losses scale as 1/D, which would tend to imply that one would want conductors as large as possible.

However, the above discussion completely neglects noise. I understand that there are at least three main sources of noise I should consider: (1) thermal (Johnson-Nyquist) noise, induced in the conductor itself and in the matching capacitors in my network, (2) induced noise arising from RF radiation elsewhere in the universe, and (3) shot noise and 1/f noise arising from fundamental sources. I am not sure how the interaction of these three sources (and any I may have missed!) will change the conclusion reached above.

In particular, the expression for the expected Johnson noise voltage,

$$ v_n=\sqrt{4 k_B T R \Delta f}, $$

is essentially independent of the mass of the conductor, which I naïvely find rather odd -- one may expect that the larger thermal mass of a real material would provide more opportunity for (at least transiently) induced noise currents. Additionally, everything I work with is RF shielded, but I can't help but think that the shielding (and the rest of the room) will radiate as a black body at 300 K...and therefore emit some RF that it is otherwise designed to stop.

At some point, my gut feeling is that these noise processes would conspire to make any increase in the diameter of the conductor used pointless, or down right deleterious. Naïvely, I think that this has clearly got to be true, or labs would be filled with absolutely huge cables to be used with sensitive experiments. Am I right?

What is the optimum coaxial conductor diameter to use when carrying information consisting of a potential difference of some small magnitude v at an AC frequency f? Is everything so dominated by the limitations of the (GaAs FET) preamplifier that this question is entirely pointless?

  • 2
    \$\begingroup\$ The emission coefficient for blank metals in the IR region is very low (you can use it as mirror and measure -40°C with an IR thermometer by pointing the metal at the sky), so maybe that helps with respect to the black body radiation (and it's around 30 THz). I'm also wondering if the thermal mass is effectively being taken care of as the mass will have an influence on the resistance, an increase in mass would lead to a smaller resistance, I've never tried to calculate that... Tough question (maybe better for physics.SE?) \$\endgroup\$
    – Arsenal
    Nov 20, 2015 at 14:49
  • \$\begingroup\$ About the LNA/pre-amp, yes, I let a good low noise amplifier do the heavy lifting and compensate for the losses and hence, the additional noise is very minimal and designed to be of no consequence. Interesting question \$\endgroup\$
    – johnnymopo
    Nov 20, 2015 at 17:41
  • \$\begingroup\$ Interesting also to consider impedance as the circumference of the wire approaches a resonant size - BIG at 300 MHz but following the spirit of the question \$\endgroup\$
    – johnnymopo
    Nov 20, 2015 at 17:47
  • \$\begingroup\$ As to black body radiation, the isolation of the cable probably (didn't calculate) leaks vastly more power at kW powers (60+ dBm). Cheaper cable is maybe 30 dB and really good maybe 90 dB isolation. \$\endgroup\$
    – johnnymopo
    Nov 20, 2015 at 19:10

3 Answers 3


You're substantially correct on everything you've mentioned. Bigger cable has lower losses.

Low loss is important in two areas

1) Noise

The attenuation of a feeder is what adds Johnson noise corresponding to its temperature onto the signal. A feeder of near zero length has near zero attenuation and so near zero noise figure.

Up to a meter or several (depending on frequency), the noise figure of a typical cable tends to be dominated by the noise figure of the input amplifier you are using, even cables of pencil diameter (you can get really thin cables, sub-mm even, and in these you do have to worry about meter lengths).

To get signals down off your roof into the lab, any feasible cable will be so lossy, even unusually thick ones, that the solution is almost always an LNA on the roof, straight after the antenna.

That's why do tend not to see really fat cables in labs, they're not needed for short hops, they're not sufficient for long drags.

b) High power handling

In a transmitter station, you tend to have the amplifier in the building, and the antenna 'out there' somewhere. Putting the amplifier 'out there' as well is usually not an option, so here you do have fat cables, as fat as possible given that they have to remain TEM, without moding. That means <3.5mm for 26GHz, <350mm for 260MHz etc.

The impedance of the cable also matters, as well as the size. Have a look at this cable manufacturer's tutorial on why we have different cable impedances, so 75\$\Omega\$ for lowest loss, and 50\$\Omega\$ as a compromise that has settled itself in as a standard.


For most folks posting answers on this particular stack, the answer to optimum cable size generally has a lot to do with economics, service life, ease of use and such. Each individual problem has its own set of defining parameters, which in turn will be used to create a specification that will be met or exceeded.

This is an important step to take, because premature optimization is a real problem. I can absolutely guarantee several things about electronic design that are always true. Larger diameter cables experience less heat waste due to improved conductivity, higher voltages allow more power to be transmitted per unit current, and larger batteries have more capacity. But the solution must actually fit the problem, so frequently you'll find yourself using the specification to choose just exactly what is acceptable for the particular problem you are having at the moment.

You have demonstrated a more than adequate understanding of the issues at hand, and I humbly submit that you are likely better suited to the details than I am at the moment. You also seem to be engaged in research, rather than design. That being the case, I would offer this advice - having a firm understanding of the noise terms and how they are affected by increasing temperatures over time, decide on a firm, non-zero value of Johnson noise that is currently acceptable for your work, and design around that as a specification. Set conductor sizes and types, and if necessary consider active cooling (provided, of course, that it doesn't interfere with or invalidate your research).


While you're correct in your details, I think you've missed the forest for the trees. With 50 ohm loads, you don't need to worry about losses in the cable due to resistive effects. at least not for RF measurements.

Consider your N connector example. An effective conductor resistance will give a voltage drop of approximately $$\Delta v = \frac{0.2}{50} = 0.4\text{%}$$which as about 48 dB down. To put it another way, a 10 uV signal would give a nominal -100 dBV, but a 0.2 ohm conductor will produce a signal at the load of 9.96 uV, or -100.035 dBV, and somehow I have a hard time believing that will be a problem.


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