# Measuring a small distance via capacitance

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:

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?

## Update

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

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.

• what amount of separation-error will you tolerate? Efield fringing will be massive. – analogsystemsrf Mar 22 '19 at 17:23
• Your picture makes no sense to me. You say d is about 1 mm then the picture shows it at over 70 mm. – Andy aka Mar 22 '19 at 18:10
• What is in the gap? Air? something else? – uglyoldbob Mar 22 '19 at 20:48
• 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. – Mantabit Mar 23 '19 at 8:47
• 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. – Mantabit Mar 23 '19 at 18:39