I am interested in measuring a small \$d\$ between two objects shown below by measuring the capacitance between the Terminals denoted as T0 and T1:

Sketch of Geometry

The distance \$d\$ is roughly 1mm +/- 0.5mm large. In this case, the capacitance should be given by (\$r\approx 6.5\mathrm{mm}\$ is the radius of the base of the cone).

$$C=\frac{\epsilon_{0} \pi r^2}{d} \approx 1.2\mathrm{pF}$$

My idea is to measure the distance with the following setup:


simulate this circuit – Schematic created using CircuitLab

If the current and voltage are 45° out of phase at a frequency \$f\$ we have that:

$$\frac{1}{\omega C_d}=\frac{d}{2 \pi f \epsilon_{0}r^2\pi}=R_1$$

From which \$d\$ can be determined. I have the following questions regarding this setup:

  • Is it ok if I solder two small "pins" on the two objects (cylinder and cone) and simply connect the wires from the 30MHz signal generator via alligator clips? (I'm using a "BNC-To-Banana" adapter like this one here: link adapter example and the generator has a 50Ohm output impedance).

My main question here is mostly whether I need to pay attention to any parasitic quantities (capacitances/inductances) or other things which might affect the precision of the distance measurements?


I know did this measurement using the setup shown in the following picture (I used a 1MOhm resistor and the gap is around 2mm):

Measurement Setup

I'm using the following circuit:


simulate this circuit

The issue is that if I set the frequency of the function generator in the range of 200kHz and 10MHz \$f \in [0.4,10]\mathrm{MHz} \$ I am only measuring voltages of around 100mVpp over the resistor (with an input voltage of around 3Vpp [peak-to-peak]). At 10MHz I have that:

$$Z_{Cap}=\frac{1}{2*\pi*10^7\mathrm{Hz}*2*10^{-12}\mathrm{F}}\approx 8\mathrm{kOhm}$$

So basically at 10MHz, the voltage over the resistor should be approximately equal to the input voltage but I'm measuring only a very small voltage over the resistor, see the plots below (note that on the bottom left plot, I use another voltage scale to show that there seems to be a slight phase shift between input voltage and voltage over the resistor). The colors in the oscilloscope plots are the same as the colors on the photo of the measurement setup.

enter image description here

  • \$\begingroup\$ what amount of separation-error will you tolerate? Efield fringing will be massive. \$\endgroup\$ – analogsystemsrf Mar 22 '19 at 17:23
  • \$\begingroup\$ Your picture makes no sense to me. You say d is about 1 mm then the picture shows it at over 70 mm. \$\endgroup\$ – Andy aka Mar 22 '19 at 18:10
  • \$\begingroup\$ What is in the gap? Air? something else? \$\endgroup\$ – uglyoldbob Mar 22 '19 at 20:48
  • \$\begingroup\$ Hello, the gap consists of air. By "gap", I mean the tiny space between the cylinder and the cone. I think I should have an accuracy of around 0.2mm. \$\endgroup\$ – Mantabit Mar 23 '19 at 8:47
  • \$\begingroup\$ Okay, I did this experiment now but I have the issue that I don't measure a lot of voltage over the resistor at very high frequencies (10MHz). At 10MHz, the capacitor should essentially be "shorted" and all the voltage should appear over the resistor but I can't measure a lot of voltage over the resistor at those high frequencies. \$\endgroup\$ – Mantabit Mar 23 '19 at 18:39

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