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