Multidielectric
coax
Updated May
19, 2012
Click
here to go to our main page on coax
Click
here to go to our main microwave transmission lines page
Click
here to try out our dualdielectric coax calculator
Click here to go to our page on dispersion (dispersion will exist in any multidielectric transmission line)
Thanks to contributor Alex R. from a major connector
manufacturer, we have now have a spreadsheet that can calculate
parameters for multidielectric coax, beyond just two dielectrics!
If you want to download Alex's spreadsheet, go to our download
area and look for a spreadsheet called Multidielectric_coax_Rev1.xls.
Alex solved the equations for two, three, four and five different
layers of dielectric. If you need more than five dielectric layers,
you are on your own, but if you modify the spreadsheet, send it
back and we might post your version too!
One of these days
we will fix up the graphics on this page, but don't hold your breath.
You can skip all the boring math and get your answer by using our
dualdielectric coax calculator!
When a coaxial cable is constructed
using two concentric dielectrics as shown above, the wave propagation
is no longer purely transverseelectromagnetic (TEM),
as in the singledielectric coax, but instead is quasiTEM, like
microstrip or CPW and can suffer from dispersion. But the "nonTEMness" is very slight
indeed. In the most general terms the characteristic impedance and
effective dielectric constant are very slight functions of frequency,
and if this is important to you, you should analyze your structure
using a 3D EM solver. However, the static Z0 and Keff
terms can be derived easily, and they are useful over a fairly wide
range of frequencies. The derivation of these terms is as follows...
Coax impedance is the squareroot
of the ratio of inductance/length divided by capacitance/length:
Note that because the ln(D/d)
term is a ratio, you can just as well use ln(R/r), and in either
case the dimensional units don't matter.
The "trick" to doing
the multidielectric calculation is to note that capacitance of
multiple concentric dielectrics can be easily be calculated by adding
their individual capacitances in series. The coax inductance calculation
does not change no matter what dielectric you put between the center
and outer conductors.
In the static (DC) case, all
of the electric flux lines are perpendicular to the center conductor
 dielectric 1, dielectric 1  dielectric 2, and dielectric 2 
outer conductor interfaces, as shown:
To calculate the static capacitance
between the inner and outer conductors, imagine a zerothickness
conducting cylinder placed inside the cable, at the interface between
the two dielectrics. Since all of the electric field components
are already perpendicular to this interface, this does not violate
the boundary conditions of the initial problem. With this conducting
cylinder in place, two capacitance values can now be calculated:
1) the capacitance (per meter length of cable) between the inner
conductor and the cylinder; and 2) the capacitance between the cylinder
and the outer conductor. The capacitance per meter of cable between
the inner and outer conductors is then the series combination of
the above two capacitance values.
The capacitance per length of
a singledielectric coaxial cable (dimensions shown above) is as
follows:
Therefore, the capacitance per
meter length between the inner conductor and the cylinder is:
The capacitance per meter length
between the cylinder and the outer conductor is:
The capacitance
per meter length of the cable itself is the series combination of
the above two capacitances:
Leaving out the algebraic steps,
this comes out to:
The effective dielectric constant
of the structure is the ratio of the capacitance of the structure
to the capacitance if the dielectrics were replaced with a vacuum
(or air), i.e.:
where:
This gives the following expression
for the effective dielectric constant:
With a bit of additional
effort, the characteristic impedance can also be found. For this
we need the inductance per meter length, which is unaffected by
the dielectric(s) in the structure. The expression for the inductance
is:
The characteristic impedance
of any transmission line is equal to the square root of the ratio
of the inductance per length to the capacitance per length. Again
leaving out the algebraic steps,
Dispersion in dualdielectric coax
I validated these expressions
using HFSS for the following geometry: r0 = 0.040", r1 = 0.100", r2 = 0.140", Er1 = 2.0, and Er2 = 5.0. For
those dimensions, the above expressions yield Z0 = 48.65,
and Keff = 2.384. HFSS results for the same configuration are as
follows:
Frequency,GHz 
Z0, Ohms (PV) 
Keff 
1.0 
48.62 
2.384 
2.0 
48.63 
2.385 
3.0 
48.64 
2.386 
4.0 
48.65 
2.387 
5.0 
48.67 
2.389 
6.0 
48.69 
2.391 
7.0 
48.71 
2.393 
8.0 
48.73 
2.396 
9.0 
48.76 
2.399 
10.0 
48.79 
2.403 
11.0 
48.82 
2.407 
12.0 
48.86 
2.412 
13.0 
48.89 
2.417 
14.0 
48.92 
2.423 
15.0 
48.96 
2.429 
As seen from the numbers, the
expressions for the static Z_{0} and K_{eff} are
valid at low frequencies, and for the dimensions used in this example,
the dispersion is mild below 15 GHz. Smaller dimensions and closer
values of Er1 and Er2 would reduce the dispersion further.
The derivation of the expressions
for Z_{0} and K_{eff} in the case in which the division
between Er1 and Er2 is angular, rather than radial, is probably
more complicated.
Good luck!
