I have a piece of double sided PCB that I am plan to use a capacitor to measure soil moisture (somewhere between 133-170MHz).

Two prototype boards

I used the "two-pin method" of measuring small capacitance on the Arduino (calibrated against several values of ceramic capacitors between 2pF and 1nF).

Using the insulated board, I got the following results:

  • Air: 62pF
  • Hand grip: 26-27pF
  • Immersed 1/4 in water: 48pF
  • Immersed 1/2 in water: 60pF
  • Immersed 3/4 in water: 69pF
  • Immersed fully in water: 97pF


Firstly, why would the capacitance value initially decrease as the dielectric constant went from 1 (air) to >1 (my hand and partially submerged in water)?

Secondly, why would I only see such a small change when the board went from air (dielectric constant of 1) to fully immersed in water (dielectric constant of 80)? I accept that the leads from the boards will have some effect on the capacitance, but I doubt that they will have enough effect so as to swamp out the effect of water.

  • \$\begingroup\$ I suspect your measurement method is flawed. \$\endgroup\$ – Andy aka May 20 '18 at 13:46
  • \$\begingroup\$ The dielectric constant you mention for water is for water with no salts (distilled in other words). Ordinary tap water is definitely not without salts. See this question: chemistry.stackexchange.com/questions/16434/… \$\endgroup\$ – Peter Smith May 20 '18 at 13:49
  • \$\begingroup\$ Okay, from my old chemistry days sea water is approx 0.6 molar which from the graph in that answer would approximate to a constant of 70. We definitely don't have a problem with hard water where we live. \$\endgroup\$ – talikarng May 20 '18 at 14:01
  • \$\begingroup\$ Your hands are fairly lossy (resistive) which will have some effect on the measurement (depending on the measurement technique). Same would be true of salt water. \$\endgroup\$ – Brian Drummond May 20 '18 at 14:16
  • \$\begingroup\$ @BrianDrummond How could the resistivity of my hands affect the insulated board? \$\endgroup\$ – talikarng May 20 '18 at 14:19

Your dielectric constant is roughly a geometric average of the PCB and solder mask constants and the outer medium's. To see the 80 fold increase, you need two parallel plates, not just parallel traces.

  • \$\begingroup\$ I don't think that's how this product work. It has two parallel traces, the way you increase the capacitance is by placing a third plate (or something else that is conductive) that bridges the two parallel traces, forming two capacitors back to back. \$\endgroup\$ – Harry Svensson May 20 '18 at 14:26
  • \$\begingroup\$ @HarrySvensson Unfortunately parallel plates are not an option because this probe will be going into soil. I believe that I am relying on fringing capacitance. \$\endgroup\$ – talikarng May 20 '18 at 14:40
  • \$\begingroup\$ @talikarng The two parallel traces on the PCB is already there, that is what you got. That is not a part of the option. A third plate is the water, or something else, like moisture. I am not sure if it is "fringing capacitance". - If it looks like I'm saying something that doesn't add up, chances are I'm being misunderstood, which I think I've just been. \$\endgroup\$ – Harry Svensson May 20 '18 at 15:01
  • \$\begingroup\$ @HarrySvensson Now I understand what you mean. Thankyou. \$\endgroup\$ – talikarng May 20 '18 at 22:28

I used the "two-pin method" of measuring small capacitance on the Arduino (calibrated against several values of ceramic capacitors between 2pF and 1nF).

This measurement technique may work well for measuring components that are designed to be capacitors but for measuring capacitance where the "plates" are surrounded by lossy materials with various resistive conduction your measurement method becomes flawed.

The reason is because the arduino two-pin method doesn't actually measure capacitance; it measures a signal being attenuated and possibly time delayed by capacitance and that of course would also measure resistive losses in the material. So your method doesn't implicitly measure capacitance; it measures losses in the dielectric of the capacitor.

The better way is to put the "capacitor" into a tuned circuit with an inductor in typically a colpitts oscillator like this: -

enter image description here

The change in oscillation frequency is a much better way of measuring the capacitance - see where the picture shows "fine adjustment" - that is where you would add your capacitor plates. It is fairly resiliant to conduction losses compared to the simplistic and flawed measurement principle of the 2-pin arduino method commonly spouted to be a "capacitance meter".

This method is used in several industrial capacitance probes where resistive/conduction/dielectric losses are significant.

See also this related question and answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.