Calculating the capacitance of a capacitor

Summary
Two plates side by side, not parallel to each other.
Summary: Two plates side by side, not parallel to each other.

Hello everyone,

Purpose of this capacitor is to detect changes in water level. It is constructed of a single copper plated pcb on which middle I have made a 1 mm of space separating now two copper plates on a single pcb. So, plates are side by side with same width and height.

How would one calculate the capacitance of this type of a capacitor. I have only seen calculations for parallel plate capacitor.
 

Baluncore

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Two parallel tracks have inductance and capacitance per unit length, from which the characteristic impedance of a two track transmission line can be calculated. That is where you should go to find the capacitance between two parallel tracks.

I assume the tracks are on the surface of a single sided PCB, without a ground plane.

The case is actually more complex because there is capacitance through the PCB material and air on one side, with air only on the other. When immersed in water the capacitance on both sides of the PCB will increase. Since the back surface of the PCB does not follow an equipotential surface or field line, the math will be particularly difficult.

I would suggest you make one, then calibrate it.
 
The case is actually more complex because there is capacitance through the PCB material and air on one side, with air only on the other. When immersed in water the capacitance on both sides of the PCB will increase. Since the back surface of the PCB does not follow an equipotential surface or field line, the math will be particularly difficult.

I would suggest you make one, then calibrate it.

I have made the capacitor but I would like to know the calculation. I would like to be able to make a capacitor with specific capacitance based upon a calculation. Calculation does not need to be 100 percent precise.

I measured the change of capacitance when there is water on only one side. On the side with copper, the change is much greater.

Let us assume that only one side affects the change of capacitance. How would I then calculate the capacitance.
 

berkeman

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Summary: Two plates side by side, not parallel to each other.

How would one calculate the capacitance of this type of a capacitor. I have only seen calculations for parallel plate capacitor.
I did a Google search on PCB Trace Capacitance Calculator, and got lots of hits. Our PCB layout folks routinely do such calculations (and Zo impedance calculations) for traces on our PCBs that are sensitive to capacitance or characteristic impedance values. Have a look at the various calculators to see if you can enter in your plate dimensions and separation and the dielectric constant of your PCB material to get an estimate of the capacitance. :smile:

 

Baluncore

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I measured the change of capacitance when there is water on only one side. On the side with copper, the change is much greater.
Liquid water has a dielectric constant or about 80 while PCB material is closer to 4, so the PCB layer acts as a low capacitance in series with the higher water capacitance on the insulated side. There is no series capacitance on the copper side.
The capacitance you measure is always the sum of the two sides.
 

eq1

Gold Member
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How would one calculate the capacitance of this type of a capacitor.
The high-level strategy for non-symmetric shapes is to use Gauss's law to find the E-field, from the E-field calculate the change in voltage over the path of interest, then C=Q/V.

If you google "cylindrical capacitor calculation" (a.k.a. co-axial cable) I'm sure you can find an example of someone actually doing the calculation from Gauss's law.

P.S.
If it doesn't need to be precise, i.e. one can ignore the fringe fields, isn't your situation still basically parallel plates? You're just using the short side of the plate. Maybe the 1mm spacing is not well controlled?

P.P.S.
Actually, that's the strategy to calculate capacitance for symmetric shapes too. :) When the geometry is easy people tend to skip steps to avoid complicating explanations.
 
Last edited:

berkeman

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Here is a paper detailing exactly your question if I understood correctly.
Nice find. I'm not sure, but I think the OP's board does not have a ground plane to form a microstrip transmission line, so the equations in that paper may not be applicable.
 

Baluncore

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Unfortunately, the reference in post #7 refers to the situation with a ground plane. But the link in post #9 goes to an online calculator with the approximation equations for conductors on one side of a PCB without a ground plane. It references; “Analysis of Multiconductor Transmission Lines,” 2nd Ed., Clayton R. Paul, 2008. John Wiley & Sons, Inc. There is no derivation given, but the approximation for capacitance per unit length of coplanar lines, can be found on pages 146-8.

Unfortunately, there is no conductor length correction, so the approximation may be less accurate, and I expect overestimates the capacitance for shorter length lines.

I would expect a full analytic capacitance solution for coplanar rectangular conductors, including a length variable, but without the PCB, to have been solved between 1850 and 1930, by an applied mathematician such as Maxwell or one of his followers. Unfortunately that solution will not help here as the presence of one-sided PCB material is a recent complexity.
 

sophiecentaur

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I have a feeling that the thread has got a bit out of hand and could be addressing a problem that's far harder than the OP suggests. There will be the capacitance between the two plates and then two series capacitances between plate 1 and water and water and plate 2. There is bound to be information about a simple pair of co-planar plates and the capacitance between a plate and water is easy.
If the parasitic capacitance is of the same order as the plate / water capacitance then there would be a measurable change with separation. Wouldn't it be a bit af a waste of effort to calculate things when it could all be measured. In any case, no one would risk trying to do the exercise without calibrating.

The weak link in this method is surely the use of a simple DMM for the C measurement. Capcaitances of only a few tens of pF are not really on the menu for regular electrical measurements. Look at the available, run of the mill DMMs for sale and 50pF is down amongst the last couple of sig figs on the display, at best. This Capacitance meter
"B&K Precision 830C Dual Display Handheld Capacitance Meter",
Is more likely to do the job for you. (0.1pF claimed resolution)
 

Baluncore

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There is bound to be information about a simple pair of co-planar plates and the capacitance between a plate and water is easy.
The problem is that there are two dielectrics. The boundary between the two dielectrics does not follow an equipotential, so the assumed simple series capacitance is not exactly there.

The boundary can be forced to become an equipotential by placing virtual “shadow” plates on the back of the PCB behind the sensor plates.

That will require a way of approximating the one-sided capacitance between coplanar plates in air or water. That is half of the two sided capacitance.
https://www.emisoftware.com/calculator/coplanar-capacitance/
And the mythical fixed series capacitance of each plate with it's shadow, through the PCB dielectric. https://www.emisoftware.com/calculator/rectangular-plates/

Computation procedure.
Solve for Coplanar capacitance with water or air as dielectric. Halve it to make one sided = C1.
Solve for Cparallel capacitance between a plate and it's shadow = C2.
The approximate approximation of Ctotal = C1 + Series( C2, C1, C2 )
 
23
7
Unfortunately, the reference in post #7 refers to the situation with a ground plane. But the link in post #9 goes to an online calculator with the approximation equations for conductors on one side of a PCB without a ground plane. It references; “Analysis of Multiconductor Transmission Lines,” 2nd Ed., Clayton R. Paul, 2008. John Wiley & Sons, Inc. There is no derivation given, but the approximation for capacitance per unit length of coplanar lines, can be found on pages 146-8.

Unfortunately, there is no conductor length correction, so the approximation may be less accurate, and I expect overestimates the capacitance for shorter length lines.

I would expect a full analytic capacitance solution for coplanar rectangular conductors, including a length variable, but without the PCB, to have been solved between 1850 and 1930, by an applied mathematician such as Maxwell or one of his followers. Unfortunately that solution will not help here as the presence of one-sided PCB material is a recent complexity.
You may be able to neglect the effect of the ground plane if you simply pick the distance to be extremely large. This would yield a small capacitance according to the parralel plate model. In calculus, you would take the limit as the distance to the ground plane approaches infiinity. Naturally, you would want to verify these results.
 

Baluncore

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Naturally, you would want to verify these results.
The derivation of equations for coplanar capacitance of two rectangular patches comes through Schwarz-Christoffel transformation from the parallel plate equivalent.
It seems like factors of two are being lost in some solutions. Maybe the parallel plate solution ignores the fringing field, having only one sided capacitance, while the SC transform to coplanar arrangement assumes a two sided capacitance.
I have tried to solve this problem before for another application, and now I don't trust any of the numerical solutions I have seen.

Physical calibration of the capacitive element is still the only way to be sure where reality lies.
 

tech99

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Assuming conductive water (ie not distilled water), I think all we have is two capacitors each consisting of one plate and a ground plane. This is equivalent to a single parallel plate capacitor with twice the separation.
I suspect the PCB thickness is small compared to the spacing above the water. Otherwise Baluncore has suggested the required correction.
Using dimensions in metres,
C=8.85 Area/(2 * Height above water ) (pF)
 

Baluncore

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Assuming conductive water (ie not distilled water), I think all we have is two capacitors each consisting of one plate and a ground plane.
I think you have assumed or invoked a ground plane where there is none. There is no water on either side of the PCB when the sensor is dry. How can you solve for the dry capacitance?

Also, the real resistive current and the reactive capacitive current are orthogonal and independent. The concept of conductive water as part of a capacitive circuit is therefore misleading.
 

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