EE3901/EE5901 Sensor Technologies
Chapter 4 Tutorial

Written by A/Prof Bronson Philippa and Laurance Papale

College of Science and Engineering, James Cook University

Last updated 24 February 2025

Question 1

Two rectangular metal plates of dimensions 10 cm x 5 cm are separated by 1 mm of air, which has . Calculate the capacitance of this structure.

Answer

Question 2

The device from Question 1 is being used as a displacement sensor. The distance between the plates is varied, and the capacitance is measured. Given a measured capacitance of 5.56 pF, what is the separation distance between the plates?

Answer

Rearrange the formula for capacitance to obtain

Question 3

Consider the capacitive strain gauge shown in Figure 1.

Figure 1:
A capacitive strain gauge formed by micropatterned metal foil electrodes on a flexible polymer substrate. **(a)** Top down view of part of the electrode structure. **(b)** The overall device consists of a long string of interdigitated electrodes. **(c)** Perspective view of the device.
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For a real example of such a device, see the scanning electron microscope image in Figure 2 of Kim et al, Science and Technology of Advanced Materials, 13, 035004 (2012).

The working principle of this strain gauge is that deformation of the sensor changes the distance between the electrodes. Each pair of electrodes forms a capacitor. The dimensions are as follows: overlap distance $μ$ , electrode width $μ$ , nominal separation distance $μ$ , total number of capacitors . Each capacitor is connected in parallel.

(a) Given a measured capacitance of 1.5 pF, calculate the thickness () of the metal electrodes above the polymer substrate. You may assume the relative permittivity is .

(b) Suppose that the gauge experiences a uniform strain of 0.02 in tension. Neglecting changes in electrode thickness, what will be the new capacitance?

(c) Suggest an interface circuit for this sensor using a single op-amp that will produce an output voltage whose RMS value is linearly proportional to the applied strain.

Answer

(a) Having 200 small capacitors in parallel is the same as having a single large capacitor with a surface area that is 200 times larger. Hence it is convenient to model this system as a single capacitor with an effective surface area . Therefore:

μ

(b) Strain is the relative change in length, i.e. . In other words, all lateral dimensions expand by a factor of 1.02. Therefore, the new separation distance is μm, and hence the new capacitance is

where is the capacitance at zero strain. Therefore the sensor’s impedance is linearly proportional to .

Hence you should place the sensor in the feedback path of the amplifier (Figure 2).

Figure 2:
The op-amp circuit that best suits this type of sensor.
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To check this, recall that the voltage output from an inverting amplifier is proportional to

Notice that this linear response is only achieved when the resistor can be neglected in the AC analysis, i.e.

which means

In practice you would need to choose and accordingly.

Question 4

A liquid level sensor is shown below (Figure 3).

Figure 3:
**(a)** A water level sensor based on plane-parallel electrodes. **(b)** The equivalent circuit.
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The electrodes are covered with a thin layer of insulating material to avoid shorting them when a conductive liquid is being stored in the container. The dimensions are as follows: the plate height is , the gap is , the width of the plates (in the direction into the page) is , and the liquid level is .

(a) Given that the total capacitance is measured to be 247.5 pF, find the relative permittivity of the liquid.

(b) Suggest an interface circuit for this sensor using a single op-amp that will produce an output voltage whose RMS value is linearly proportional to the level of the liquid.

Answer

(a)

Similarly

Given that and are connected in parallel,

(b) Notice that the total capacitance of this sensor is given by

Notice that is linearly proportional to . Hence the impedance is inversely proportional to , and therefore place the sensor in the input path of the amplifier:

Figure 4:
The op-amp circuit that best suits this type of sensor.
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Question 5

Capacitance is being measuring using the method of charge redistribution (Figure 5).

Recall that the charge stored in a capacitor is given by .

(a) For times , S1 is closed, S2 is open, and S3 is closed. Hence charges to the reference voltage , while completely discharges. Write down an expression for the stored charge in .

(b) At the moment , S1 is opened, S2 is closed, and S3 is opened. Hence and must come to the same voltage by exchanging charge. Using the principle of conservation of charge, calculate the voltage .

(c) Extension question, which is not examinable in this subject but is surprising if you haven’t seen it before: Choose , then calculate the energy in the system for and . Recall the energy in a capacitor is . Does this circuit violate conservation of energy??

Answer

(a) Only one capacitor is charged, so the stored charge is simply .

(b) The same charge is now distributed across the two capacitors:

The initial energy (with total capacitance ) is

However, the final energy (with total capacitance ) is

Notice that half of the energy has disappeared! We have .

There are no resistive elements in the circuit yet energy is being lost. Do you really believe in circuit theory if it contains inconsistencies like this?

This is called the “two capacitor paradox”.

One valid solution is to connect the capacitors with a resistance (with the view of taking the limit of later). You will find that no matter how small the resistance is, it will always dissipate the same amount of heat.

Question 6

Download this STL file, which is a classic parallel plate capacitor. When setting up a numerical method, it is essential to benchmark its performance on a problem for which you already know the answer. The geometry is a 2D slice through the plates. The plates are 20mm high and have 0.01mm separation.

Load the geometry into Matlab and examine its structure.

(a) Predict the capacitance using the equation for plane parallel capacitors. Note that you do not know the surface area (because the 2D model is only a slice through the real geometry). However, you can write the surface area as a height times an unknown width, and then divide by the width to get the capacitance in F/m. The meaning of this unit is the capacitance per metre of extrusion of this geometry.

(b) Numerically calculate the capacitance using the methods discussed in this week’s notes. Compare your answer to that in part (a).

Answer

(a) 17.71 nF/m

(b) You should obtain a value close to that in part (a), depending upon the size of your finite element mesh. Your relative error should be less than 0.5%. A more precise answer is obtained with a finer mesh.

Note that the answer from part (a) is an approximation that neglects edge effects, and therefore is subject to some uncertainty in its own right.

Question 7

Using the coplanar geometry considered in lectures, analyse the impact of the high permittivity region. Calculate the capacitance when the relative permittivity of this region is varied from 1 to 30. Produce a plot of this capacitance.

Answer

Figure 6:
The impact of the permittivity of the sensing region for the geometry used in lectures.
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Next steps

You are now ready to begin work on Assignment 2.