EE3901/EE5901 Sensor Technologies

Week 7 Tutorial

Question 1

Two rectangular metal plates of dimensions 10 cm x 5 cm are separated by 1 mm of air, which has $\epsilon_{r}=1.0006$. Calculate the capacitance of this structure.

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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?

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Question 3

Consider the capacitive strain gauge shown in Figure Q3.

For a real example of such a device, see the scanning electron microscope image in Figure 2 of this paper by Jiseok Kim et al.

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 $v=400\ \text{μm}$, electrode width $w=40\ \text{μm}$, nominal separation distance $d=20\ \text{μm}$, total number of capacitors $n=200$. Each capacitor is connected in parallel.

(a) Given a measured capacitance of 1.5 pF, calculate the thickness ($h$) of the metal electrodes above the polymer substrate. You may assume the relative permittivity is $\epsilon_{r}=1$.

(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.

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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 $A = 200vh$. Therefore:

$$\begin{align*} C & =\frac{\epsilon_{r}\epsilon_{0}\times200vh}{d}\\ 1.5\times10^{-12} & =\frac{8.85\times10^{-12}\times200\times400\times10^{-6}\times h}{20\times10^{-6}}\\ h & =42.37\ \text{μm.} \end{align*}$$

(b) Strain is the relative change in length, i.e. $\frac{\Delta L}{L_{0}}=0.02$. In other words, all lateral dimensions expand by a factor of 1.02. Therefore, the new separation distance is $d=20\times10^{-6}\times1.02=20.4$ μm, and hence the new capacitance is

$$\begin{align*} C & =\frac{\epsilon_{r}\epsilon_{0}\times200vh}{d}\\ & =\frac{8.85\times10^{-12}\times200\times400\times10^{-6}\times42.37\times10^{-6}}{20.4\times10^{-6}}\\ & =1.4705\ \text{pF}. \end{align*}$$

(c) Given a strain $\epsilon$, the impact on capacitance is

$$C_{x}=\frac{C_{0}}{1+\epsilon},$$

where $C_0$ is the capacitance at zero strain. Therefore the sensor’s impedance is linearly proportional to $\epsilon$.

Hence you should place the sensor in the feedback path of the amplifier (Figure A3).

To check this, recall that the voltage output from an inverting amplifier is proportional to

$$\frac{Z_{feedback}}{Z_{input}}\approx\frac{1/j\omega C_{x}}{1/j\omega C}=\frac{C}{C_{x}}=\frac{C}{C_{0}}(1+\epsilon).$$

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

$$R\ ||\ \frac{1}{j\omega C_{x}}\approx\frac{1}{j\omega C_{x}},$$

which means

$$R\gg\frac{1}{\omega C_{x}}.$$

In practice you would need to choose $\omega$ and $R$ accordingly.

Question 4

A liquid level sensor is shown below.

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 $h_{max}=100\ \text{mm}$, the gap is $d=1\ \text{mm}$, the width of the plates (in the direction into the page) is $w=15\ \text{mm}$, and the liquid level is $h=65\ \text{mm}$.

(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.

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Answer

(a)

$$C_{1}=\frac{\epsilon_{r}\epsilon_{0}A}{d}=\frac{8.85\times10^{-12}\times0.015\times(0.1-0.065)}{0.001}=4.6462\ \text{pF}.$$

Similarly

$$C_{2}=\frac{\epsilon_{r}\epsilon_{0}A}{d}=\frac{\epsilon_{r}\times8.85\times10^{-12}\times0.015\times0.065}{0.001}=\epsilon_{r}\times8.6288\times10^{-12}.$$

Given that $C_{1}$ and $C_{2}$ are connected in parallel,

$$\begin{align*} C_{1}+C_{2} & =247.5\times10^{-12}\\ 4.6462\times10^{-12}+\epsilon_{r}8.6288\times10^{-12} & =247.5\times10^{-12}\\ 8.6288\epsilon_{r} & =242.854\\ \epsilon_{r} & =28.14. \end{align*}$$

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

$$\begin{align*} C & =C_{1}+C_{2}\\ & =\frac{\epsilon_{air}\epsilon_{0}\left(h_{max}-h\right)w}{d}+\frac{\epsilon_{liquid}\epsilon_{0}hw}{d}\\ & =\frac{\epsilon_{0}w}{d}\left(\epsilon_{air}h_{max}-\epsilon_{air}h+\epsilon_{liquid}h\right)\\ & =\frac{\epsilon_{0}w}{d}\left(\epsilon_{air}h_{max}+(\epsilon_{liquid}-\epsilon_{air})h\right). \end{align*}$$

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

Question 5

Capacitance is being measuring using the method of charge redistribution (Figure Q5).

Recall that the charge stored in a capacitor is given by $Q=CV$.

(a) For times $t<0$, S1 is closed, S2 is open, and S3 is closed. Hence $C_{x}$ charges to the reference voltage $V_{r}$, while $C_{1}$ completely discharges. Write down an expression for the stored charge in $C_{x}$.

(b) At the moment $t=0$, S1 is opened, S2 is closed, and S3 is opened. Hence $C_{x}$ and $C_{1}$ must come to the same voltage by exchanging charge. Using the principle of conservation of charge, calculate the voltage $V_{0}$.

(c) Extension question, which is not examinable in this subject but is surprising if you haven’t seen it before: Choose $C_{x}=C_{1}=C$, then calculate the energy in the system for $t<0$ and $t>0$. Recall the energy in a capacitor is $W=\frac{1}{2}CV^{2}$. Does this circuit violate conservation of energy??

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