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Hey all. in a typical design of a capacitive sensor for level measurements, compensation capacitors are added as a reference. I struggle to understand why we need a reference capacitance to measure the level of the liquid in a tank, shouldn't only the main capacitor be enough for this job? like we calculate the equivalent parallel capacitance and then we figure out the level of the water.

and here, can we do our measurements using DC voltage ? or we need an AC voltage for this application to keep our circuit flowing? Thanks ahead.

  • 1
    \$\begingroup\$ The capacitance of \$C_2\$ is known, except for the uncertainty of \$\epsilon_r\$. So you want to develop an experimental setup and consequent derived equation that cancels out this uncertainty. \$\endgroup\$
    – jonk
    Commented May 8, 2021 at 6:53
  • \$\begingroup\$ DC won't work to measure C . there must be a dV/dt value , ramp pulse or sine \$\endgroup\$ Commented May 8, 2021 at 7:32
  • \$\begingroup\$ Thanks a lot. it's clear now. \$\endgroup\$
    – Zero304
    Commented May 8, 2021 at 7:37

1 Answer 1


One of the key parameters that will vary a lot with respect to some liquid being used is the value of \$\epsilon_r\$. Who knows? One thing you do know is that it is the same for both capacitors: reference and measuring.

If only you could work out a way where a lack of knowledge of that factor could be taken out of the equations.

Well, that's not too difficult to do. If you construct two capacitors where both of them have the same width and both of them have the same distance between the plates (this isn't all that hard to achieve), then perhaps something could be done about the problem.

Let's construct two capacitors: a reference capacitor \$C_2\$ and a measuring capacitor \$C_1\$:

$$\begin{align*} C_1 &= \epsilon_0\cdot \epsilon_r\cdot\frac{h_1\cdot w}{d}\\\\ C_2 &= \epsilon_0\cdot \epsilon_r\cdot\frac{h_{ref}\cdot w}{d}\\\\ &\therefore\\\\ \frac{C_1}{C_2}&=\frac{\epsilon_0\cdot \epsilon_r\cdot\frac{h_1\cdot w}{d}}{\epsilon_0\cdot \epsilon_r\cdot\frac{h_{ref}\cdot w}{d}}\\\\ &=\frac{h_1}{h_{ref}} \end{align*}$$

Please note that \$\epsilon_r\$ has dropped out of the equation, now. So different liquids, even minor differences, are eliminated. At least, to the degree that \$w\$ and \$d\$ can be carefully managed to be equal in both cases. Luckily, that kind of manufacturing isn't difficult to achieve.

So now the level measurement isn't confounded by the details of the liquid. As long as it is the same liquid through-out, the concept remains valid and:

$$h_1 = h_{ref}\cdot\frac{C_1}{C_2}$$

Since \$h_{ref}\$ is known by the design itself and since \$C_1\$ and \$C_2\$ are in a continuous process of measurement, the device works well.

It is a requirement, though, that \$C_2\$ be under continual submersion. And this also discounts the idea that the portion of \$C_1\$ that uses air as its dielectric contributes significantly to \$C_1\$. That can be included in the above analysis, but then the result becomes a little more nuanced.

  • \$\begingroup\$ Thanks, that was helpful :) \$\endgroup\$
    – Zero304
    Commented May 8, 2021 at 7:31
  • \$\begingroup\$ @Zero304 If you feel this is what you wanted and you don't feel a need to wait for more (and better) answers, then selecting my answer will help others avoid wasting their time thinking you need something more. But if you want to wait for better, then don't select this and wait for a while. Either way, I'm glad I was of some help and thanks for the kind words! \$\endgroup\$
    – jonk
    Commented May 8, 2021 at 18:09
  • \$\begingroup\$ oh, I'm new to SE, I didn't figure this out by myself. done. \$\endgroup\$
    – Zero304
    Commented May 9, 2021 at 9:45

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