I conducted a complex electrical impedance measurement for a water partially saturated rock.My complex electrical impedance data (Z' and Z'') between 500 ~10000 Hz can be well fitted by a Debye circuit. Through the fitting, I obtained the resistance R1 ,R2 and capacitance C.
For example, R1= 112 ohm, R2=24 ohm, and C = 3.2e-6 F.
The capacitance is proportional to total permittivity (E) or dielectric constant (Er) according to this relation C = E* k = E0*Er*k
Here k is the geometrical factor equaling 0.03 m, E0 is the vacuum permeability (8.854e-12) and Er is the dielectric constant. Thus the dielectric constant D should be calculated as Er = C/E0/k. So I calculated the Er as 12047285. This value is very huge compared to the common literature values.
If I am wrong, could anyone help me to point out where did I go wrong?
Thank you very much.
The only part, after the measurement and curve fitting, that looks the reason you got such dielectric constant is geometrical factor. In electric capcaitors it is in the range of 1e3 to about 1e5.
I agree with Mr. Etemadrezaei, the A/d factor is usually a few orders of magnitude higher. It could also be the units you are using for the variables. Simple, yes but easy to make mistakes. Be sure the units of vacuum permittivity, geometric (A/d) and capacitance all share common units. For example, if you put capacitance units in terms of pF, A/d units in terms of mm then E0=0.00885418782 pF/mm.
Mohammad and David, thank you very much for your answers.
I used the Zview software did the fit. It doesn't show the unit, but I think the it is in SI unit. Because I can use the fitted parameter to calculate the complex impedance rightly.
The geometrical factor was calculated by area/length (A/d).
The area is 3.14*(0.035/2) ^2 m^2 and the length is 0.032 m; so I get the geometrical factor as 0.03 m.
I also calculated the geometrical factor using FEM modeling by COMSOL software. It gave a similar value.
Could it be then that your value of capacitance is off? 3.2E-6 Farad is a very large capacitance. With your given geometrical values of the capacitor, I have calculated an empty cell capacitance (capacitance with only an air-gap as dielectric) of 0.26621022pF or 2.66E-13 Farad. It might be a good idea to recheck your C-value.
Thank you David.
I found an explanation. The frequency dispersion in my data is not related to dielectric nature, but may related to the EDL polarization.
There is a basic relation between capacitance and the peak frequency of dispersion.
The larger the capacitance, the smaller the peak frequency of dispersion is.
The dielectric relaxation should correspond to a small capacitance and show a dispersion peak at very high frequency.
However, the EDL relaxation leads to a dispersion peak appearing at low frequency, which corresponds to a large capacitance.
I have a rectangular sample for which i have measured the capacitance. I want to calculate the dielectric constant. I have some doubts regarding calculation of geometrical factor k.
As k = A/d.where A is the area and d is the length of sample.
1) Is area A is here the area of cross section?
2) For a rectangular sample, A = length * breadth. So k = length * breadth / length = breadth. Is this correct?
Thanks.
Dinesh,
The answer to these questions depends on the geometry of your capacitor used to measure the capacitance of your sample. Could you explain the measurement?
If you have a parallel plate capacitor with your sample "sandwiched" between them acting as a dielectric then:
1) The variable "A" is the area of the capacitor plates. If your sample's cross-sectional area is less than that of the capacitor plates, then the dielectric constant becomes much harder to calculate, so hopefully your sample's cross-sectional area is equal to or greater than the area of the plates.
2)See the answer for 1). The variable "d" is the gap between the capacitor plates, so it would be the height of your sample. A is the area of the capacitor plates. See the attached image for reference.
Hello David,
Thanks for your response.
I have a very thin (few mm thick) rectangular pellet sample. I am measuring the capacitance using NF ZM2376 LCR meter with four probe measurement. How do I calculate correct A and d?
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