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HFSS - How can I determine dielectric between capacitor plates?

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drkirkby

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I've put some background information to my problem. If you don't want to read that, just skip the section marked "Background" and go directly to the section marked "HFSS problem".

Background.

I purchased cheaply an HP 16453A Dielectric Material Test Fixture. A picture of which is shown.
dielectric-between-plates.jpg

It is designed to allow one to determine the complex permittivity of a dielectric, by inserting the dielectric between two capacitor plate, and measuring the impedance.

An HP/Agilent/Keysight 16453A Dielectric Material Test Fixture, used in conjunction with a limited range HP/Agilent/Keysight instruments:

it is possible to determine the complex dielectric properties of the material

Er=Er' + j Er''

by converting measurements of impedance (R + j X) to permittivity. This requires 4 things:

  1. A flat material with a thickness in the range 0.3 to 3.0 mm and a diameter of at least 15 mm.
  2. Measurement of the thickness of the material with a micrometer.
  3. Measurement of R + j X on the impedance analyzer.
  4. Firmware support in the impedance analyzer to convert those R+j X values to Er' + j Er'' - that alone is over $5000 on the only current one of these instruments.

I don't have any of those instruments, but I do have other instruments able to measure impedance. But these instruments don't have the necessary firmware support to convert the R + j X values to Er' + Er''. I want to use one of my instruments, despite they don't have the right firmware, as they do have the basic measurement capability to measure R + j X.

HFSS problem
Can anyone help me set up a simulation in HFSS, where from these inputs

  • Diameter of the smaller capacitor plate is 7 mm
  • Diameter of larger capacitor plate is 10 mm
  • A known thickness of dielectric, which will be in the range 0.3 to 3.0 mm. The dielectric must be flat - unlike in my photo.
  • A known complex permittivity of the dielectric
  • A dielectric at least 15 mm in diameter - I assume this means fringing fields around the edges of the dielectric are negligible

HFSS is able to compute R + j X?

If I can compute R + j X with a number of different dielectric thicknesses, and complex permittivities, I'm sure I could create a lookup table or empirical equation to work the other way, and convert measured values of thickness, and complex impedance into complex permittivity.

I'm really now sure
  1. How I can compute R + j X of a capacitor.
  2. How to determine an appropriate criteria for convergence, since I suspect that the default will not be suitable

Can anyone help?

Dave
 

I tried to do something similar once, I did something along the following:

Once you have your geometry made in HFSS and material properties made for your dielectric (HFSS allows you to create your own material and input the dielectric properties), you can put two lumped ports on either side of the capacitor. Simulating driven modal analysis will result in the S-parameters. You can convert this into Z parameters from which you can extract the impedance of the lumped element capacitor.

Using the same simulation, you could setup a parametric sweep of the dielectric properties find the complex impedance as a function of only the complex permitivity.

Oh, except in HFSS I believe it asks for the loss tangent 'd' instead of the imaginary portion of permitivity where tan(d) = Er''/Er'

Convergence: I don't know any great insight for changing the convergence criteria, but from experience, I do usually change the default settings.
Maximum Number of Passes: 20
Maximum Delta S: 0.002 (increase this if your simulation is taking too long)
Minimum Number of converged Passes: 2 (makes false convergence more difficult)
 

I tried to do something similar once, I did something along the following:

Once you have your geometry made in HFSS and material properties made for your dielectric (HFSS allows you to create your own material and input the dielectric properties), you can put two lumped ports on either side of the capacitor. Simulating driven modal analysis will result in the S-parameters. You can convert this into Z parameters from which you can extract the impedance of the lumped element capacitor.
Why two lumped ports? Why is one not sufficient? I found this answer

https://www.edaboard.com/threads/227831/

to a somewhat similar problem, answered by tallface65, who seems an HFSS guru. He only used one lumped port. That was two square capacitor plate. I've changed this to a geometry which more closely resembles what I have - see picture. The dielectric is orange, and sandwiched between two cylinders which are the capacitor plates.

Screenshot-58.png


Using the same simulation, you could setup a parametric sweep of the dielectric properties find the complex impedance as a function of only the complex permitivity.

Oh, except in HFSS I believe it asks for the loss tangent 'd' instead of the imaginary portion of permitivity where tan(d) = Er''/Er'
Yes, that's not really a problem. In fact, my LCR meter can show capacitance and loss tangent directly.
Convergence: I don't know any great insight for changing the convergence criteria, but from experience, I do usually change the default settings.
Maximum Number of Passes: 20
Maximum Delta S: 0.002 (increase this if your simulation is taking too long)
Minimum Number of converged Passes: 2 (makes false convergence more difficult)

According to tallface65, converging on delta S is not the right thing to do. He wrote

"The port is always looking into an open so the S parameters do not change much at all from pass to pass causing premature convergence."

As far as I can see, and I'm not expert, he set up an expression for the capacitance, and converged on that. That's something I've never done myself, and don't really know what I'm doing.

He seems to have some other plots of convergence too, and also E-field. I need to work out what he was doing with those.

I've simulated it, and get the following graph

capacitance.jpg

based on
  • Diameter of the smaller plate is 7 mm
  • Diameter of the larger plate is 10 mm
  • Length of both plates are 5 mm
  • Spacing between the plates of 1.0 mm - set by the dielectric thickness.
  • Permittivity of dielectric is 2.1
  • Diameter of the dielectric is 15 mm
  • Material is gold - since the capacitor plates are all gold plated
  • Airbox is a cylinder, with a diameter of 40 mm and length of 40 mm, with the plates about central.

The result of the HFSS simulation is quite different from what I expect assuming a the usual formula:

C=Eo Er area/distance

where I assumed the area of the plates were 0.0000384845 m^2 (corresponding to a diameter of 7 mm), Er=2.1, with a spacing between the plates of 1.0 mm. That formula gives 0.71556 pF, whereas my graph shows a significantly higher 1.3285 pF at 1 MHz. I'm not sure whether to expect a value higher or lower than the usual equation, given the plates are of different diameters. If I assume 10 mm diameters for both plates, then the usual formula gives 1.46032 pF - all assuming I've done the maths right!!!

I think I'll have to repeat this with a much smaller spacing, so the fringe fields are negligible, as a spacing of 1 mm, with plates with diameters of 7 and 10 mm, the fringing fields are not going to be negligible.

I put a maximum mesh length of 2.0 mm , but perhaps I should reduce that too, but it took long enough at 2.0 mm !!! I suspect with plates that are only 7 and 10 mm in diameter, the mesh might be too course, but 2 mm is the maximum - HFSS should chose lower where it meshing algorithm determines it needs to be.

I'm just about to re-run with a maximum mesh of 0.2 mm, and see what happens. I expect it will take forever.

- - - Updated - - -

I just performed a rough measurement of the capacitance of 1 mm of PTFE, sandwiched between the two plates of the HP 16453A dielectric test fixture using an HP 4284A precision LCR meter. The problem is, this meter supports two types of calibration

  • Short and open
  • Short, open and load, where load is a known value of resistance, capacitance or inductance. The load should be close to what you are actually intending measuring. In other words, one should not use a 10 H inductor as a load, if trying to measure around 1 pF.
I could only do a short and open calibration of the LCR meter, which is not how this HP 16453A fixture is supposed to be used. It should have a short/open/load calibration. The LCR meter meter expect the load to be entered as a capacitance, not a thickness of PTFE. The instruments that support this dielectric test fixture allow direct entry of the thickness of PTFE.

Anyway, ignoring the errors the user calibration is probably introducing, the 4284A meter shows C=0.7811 pF and a dissipation factor of about 0.0005. That's much closer to the value (0.71556 pF) given by the usual formula, than it is to my initial HFSS simulation (1.3285 pF). I'm in a tricky position, as

  • I don't trust the user calibration of the meter is done well, as there's no way for me to do a load calibration with this fixture without having a known capacitor.
  • I don't trust my ability to set up the HFSS simulation properly

All I can say is that the LCR meter indicates a value of capacitance much closer to the usual formula than it does the HFSS simulation.

Hopefully, if I can get some convincing results from HFSS, I will be able to compute the capacitance capacitor, then enter that into the LCR meter as a known load. At that point the LCR meter should be able to measure capacitance properly with this fixture. The next step, of actually determining Er' and Er'' for an unknown material, seems a long way away at this point in time!

Essentially I need to do 3 things.
  1. Compute the capacitance of 1 mm of PTFE between the two capacitor plates using HFSS. That will give me a known capacitance that I can enter into the meter as a load.
  2. Calibrate the HP LCR meter with this fixture, using a short, open and load. That should get the meter reading properly.
  3. Find a way of converting the measured impedance into Er' and Er''.

I seem to have a long way to go.

I've just attached the .hfss file, if anyone has any comments. Changing to a smaller mesh has made virtually no difference in the value of C.
 

Attachments

  • 16453A.zip
    25.4 KB · Views: 58
Last edited:


"The port is always looking into an open so the S parameters do not change much at all from pass to pass causing premature convergence."


Hmm, interesting! I'm thinking there's another reason tallface65 chose to converge on capacitance. I find that to be a really clever idea, but if open circuit is the only reason, wouldn't that be easily fixed by either grounding the other side or doing as I have and shooting for two ports? I think I'll check out this simulation as well and take a look at the expression for capacitance used, I'm rather curious.

"I think I'll have to repeat this with a much smaller spacing, so the fringe fields are negligible, as a spacing of 1 mm, with plates with diameters of 7 and 10 mm, the fringing fields are not going to be negligible. "

Good idea, I'm curious how your simulation will work out after changing distance from 1mm to 0.1mm.
 

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