EE3901/EE5901 Sensor Technologies
Week 7 Tutorial

Last updated 7 April 2022

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 Q3.

Figure Q3
Figure Q3:

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

(c) Given a strain , the impact on 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 A3).

Figure A3
Figure A3:

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 Q4
Figure Q4:

(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 Q4
Figure Q4:

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 Q5).

Figure Q5
Figure Q5:

The circuit for Question 5. The unknown capacitor is , while is a fixed capacitor whose value is already known.

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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:

(c) Substituting into the result from part (b), the output voltage is

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.