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I was looking for information on coplanar waveguides (curiosity prompted by another question here) and found this calculator.

In the description of use, it mentions that the units for the dimensions do not matter as long as the same units are used for all dimensions.

Note: Units do not matter for this calculation as long as they are consistent.

That seems to imply to me that scaling all the dimensions by the same factor would still leave you with the same impedance.

So, if I enter dimensions and get an impedance of 50 Ohms, then I could multiply all dimensions by, say, 137.52 and still have a waveguide with an impedance of 50 Ohms.

That also implies that I could make a waveguide out of meter thick copper bars, as long as everything is proportional to the same wave guide using .1mm copper traces (as long as I also have a substrate of the same proportions with the same permitivity.)

Is waveguide impedance really governed by the proportions rather than the absolute size?

This bugs me because it seems to me that the absolute size determines which frequencies the waveguide can work at. It doesn't seem reasonable to me that a wave guide 5mm across would have the same impedance as one that is 5M across.

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  • \$\begingroup\$ The description says "Units do not matter for this calculation as long as they are consistent". That doesn't necessarily imply that "scaling all the dimensions by the same factor would still leave you with the same impedance" (this depends on the mathematical expression of Z), but simply that you cannot insert S in cm, W in inches and H in mm. This happens in this case, but it's not a general implication of that description of use. Moreover, if I remember correctly, absolute dimensions are related to the frequency that you can transmit, so such a huge guide wouldn't be much useful. \$\endgroup\$
    – DavideM
    Commented Oct 21, 2016 at 13:51
  • \$\begingroup\$ @DavideM That's what I would have thought. But, putting in values of 5 meters and so on then entering values of 5mm would still result in the same impedance for both cases. \$\endgroup\$
    – JRE
    Commented Oct 21, 2016 at 13:54
  • \$\begingroup\$ There is a better calculator that lits the formula (for CPW with ground - without ground is a bit more involved): chemandy.com/calculators/… Doubling track and gap yields a large change in impedance as you suspect. \$\endgroup\$
    – Peter Smith
    Commented Oct 21, 2016 at 14:09
  • \$\begingroup\$ @PeterSmith: Thanks. I'll have to tackle those equations and see if the units really all cancel out. From the looks of them, I won't be doing it while waiting for the compiler to run like I do most things here. :) \$\endgroup\$
    – JRE
    Commented Oct 21, 2016 at 14:11
  • \$\begingroup\$ @PeterSmith I've just tried the chemandy.com calcuator, doubling all dimensions including the substrate yields the same impedance to all displayed decimal places. \$\endgroup\$
    – Neil_UK
    Commented Oct 21, 2016 at 14:27

2 Answers 2

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Characteristic impedance is \$\sqrt{L/C}\$ where L and C are the "per distance" values. Think about the plates of a capacitor: -

enter image description here

If you make "d" twice as deep and you double the plate area (because the plate width has doubled) then capacitance remains constant.

A wire of a certain length has a certain inductance that is somewhat affected by conductor width but not greatly so, I would say that as a first approximation, the impedance largely remains constant as dimensions scale.

However, due to the inductance not remaining perfectly constant with scaling I would say that their formulas are somewhat non-representative of reality.

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Scaling all dimensions by the same factor leaves you with the same impedance. However, it does not leave you with the same frequency range. These two are completely independent.

Beware that you have to scale all dimensions. So comparing two coplanar waveguide strips that are (say) 5mm wide and 10mm wide, with proportional gaps, the second one will only have the same impedance as the first if the substrate and the foil thickness double, and the distance to any screen doubles as well. If you keep the same substrate thickess, or foil thickness, then the impedance will shift. If the foil thickness isn't specified, then it's 'thin', and can be neglected in simple models.

In doubling the size, the frequency range over which it works will halve.

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