Sample smaller than parallel plates in capacitance setup

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SUMMARY

The discussion centers on measuring the capacitance of samples smaller than the parallel plates in a capacitance setup using an LCR meter. The primary challenge is accurately determining the dielectric constants of materials when the sample area is significantly smaller than the plate area, which introduces an air contribution to the capacitance readings. The user approximated the dielectric constants by treating the setup as two capacitors in parallel, but acknowledged the limitations of this method. Suggestions include using a substitution method to isolate the dielectric properties and considering alternative probe designs for better sample fitting.

PREREQUISITES
  • Understanding of parallel plate capacitance principles
  • Familiarity with LCR meters and their operation
  • Knowledge of dielectric materials and their properties
  • Basic circuit theory, specifically capacitor configurations
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  • Research the substitution method for capacitance measurements with varying sample sizes
  • Explore designs for custom probes to accommodate small samples in capacitance setups
  • Learn about advanced techniques for isolating air contributions in capacitance measurements
  • Investigate the effects of sample geometry on dielectric constant measurements
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Researchers, materials scientists, and electrical engineers involved in capacitance measurements and dielectric constant analysis, particularly those working with small sample sizes.

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Hi,

My setup for measuring parallel plate capacitance consists of an LCR meter and a parallel plate test fixture. The area of the parallel plates is much larger than the area of the samples I am measuring. From what I've seen in literature, and my physics classes, the sample area should be either equal or larger than that of the parallel plates. I haven't been able to find any information on capacitance measurements where the sample area is smaller than that of the plates. My goal is to obtain accurate values for dielectric constants of materials.

The problem: When measuring capacitance of a sample with smaller area than the area of the parallel plates, the obtained capacitance includes a contribution from both air and the sample.

The question: How does one "subtract" the air contribution? Is there a way to obtain accurate values for dielectric constant with the setup described above?

My attempt: The closest approximation of dielectric constants I was able to get was when I treat the entire setup (air + sample) as two capacitors connected in parallel; that is, C(total) = C(air) + C(sample). Using this approximation, most of the dielectric constants were within 10% error with a few being higher than 10%. However, I know that this is not correct because the area (A) in the parallel capacitance equation ( C = k e A / d) refers to the area of the parallel plates, not the sample. In my case, for C(sample), I am using the area of the sample itself.

Any thoughts would be helpful.
 
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Why can't you just make some new plates the right size to match the samples ?
 
I would do that but I wouldn't have a way to connect the leads to the newly made plates. Also, I have a variety of sample sizes, so one plate size/shape would not fit all. Keep in mind, the samples are tiny, areas of about 25 mm2 and the area of the plates on the test fixture is 3180 mm2.
 
(1) Something more like a probe than a plate as such . Copper rod . Machined end face is the plate . Reduce diameter of bar for a distance above plate then revert to full diameter for holding and connections .

(2) I have some memory of reading about a substitution method for the case of plates bigger in area than the dielectric separator . Can't give a reference but works something like testing with sample in place and then removing sample without altering plate separation and then inferring the dielectric properties from the difference of capacitance readings .
 
probe v2.png
 
Your approach of treating it as two capacitors in parallel should work. I don't follow why you think it doesn't. The area of C(sample) should be the area of the sample. The area of C(air) should be the area of the plates minus the sample. Is that what you were doing?
 

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