Week 12 Tutorial

Question 1

The material PIC153 from PI Ceramic GmbH is being used in a force sensing application, as shown in Figure Q1. The material is poled along axis 3. The device is a 1 cm $\times$ 1 cm $\times$ 1 cm cube. Normal strains $T_{1}$ , $T_{2}$ and $T_{3}$ are applied as indicated. All shear strains are zero. Metal electrodes have been deposited onto the faces that lie in the 1-2 plane, allowing for voltage measurement $V_3$ along axis 3.

Figure Q1: The geometry of the device for Questions 1 and 2. Zoom:

(a) Calculate the voltage $V_{3}$ when $T_{1}=0$ , $T_{2}=0$ and $T_{3}=1000\ \text{N/m}^{2}$ .

(b) Calculate the voltage $V_{3}$ when $T_{1}=0$ , $T_{2}=1000\ \text{N/m}^{2}$ and $T_{3}=0$ .

(c) Calculate the voltage $V_{3}$ when $T_{1}=1000\ \text{N/m}^{2}$ , $T_{2}=0$ and $T_{3}=0$ .

(d) Determine the relationship between the stresses $T_{1}$ , $T_{2}$ and $T_{3}$ that will guarantee $V_{3}=0$ .

Answer

The electric displacement $\boldsymbol{D}$ is zero because the device is at open circuit conditions (no applied voltage). Hence the sensor equation is

\left[\begin{array}{c} D_{1}\\ D_{2}\\ D_{3} \end{array}\right]=\left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right]=\left[\begin{array}{cccccc} 0 & 0 & 0 & 0 & d_{15} & 0\\ 0 & 0 & 0 & d_{24} & 0 & 0\\ d_{31} & d_{32} & d_{33} & 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} T_{1}\\ T_{2}\\ T_{3}\\ 0\\ 0\\ 0 \end{array}\right]+\left[\begin{array}{ccc} \epsilon_{11} & 0 & 0\\ 0 & \epsilon_{22} & 0\\ 0 & 0 & \epsilon_{33} \end{array}\right]\left[\begin{array}{c} E_{1}\\ E_{2}\\ E_{3} \end{array}\right].

The measured voltage is determined by $E_{3}$ , so reading off the third row of this equation,

d_{31}T_{1}+d_{32}T_{2}+d_{33}T_{3}+\epsilon_{33}E_{3}=0.

Substituting $d_{31}=d_{32}$ and solving,

E_{3}=\frac{-d_{31}\left(T_{1}+T_{2}\right)-d_{33}T_{3}}{\epsilon_{33}}.

The device is 1 cm thick, so

V_{3}=0.01E_{3}=\frac{-d_{31}\left(T_{1}+T_{2}\right)-d_{33}T_{3}}{100\epsilon_{33}}.

Substituting the material properties,

V_{3}=\frac{295\times10^{-12}\left(T_{1}+T_{2}\right)-600\times10^{-12}T_{3}}{3.9825\times10^{-6}}.

Hence we can evaluate each scenario by substituting the given stresses.

(a) $V_{3}=-0.1507\ \text{V}$ .

(b) $V_{3}=0.0741\ \text{V}$ .

(c) $V_{3}=0.0741\ \text{V}$ .

(d) Substituting $V_{3}=0$ , we find

\begin{align*} 0 & =\frac{-d_{31}\left(T_{1}+T_{2}\right)-d_{33}T_{3}}{100\epsilon_{33}}\\ & =295\times10^{-12}\left(T_{1}+T_{2}\right)-600\times10^{-12}T_{3}\\ & =295\left(T_{1}+T_{2}\right)-600T_{3}\\ T_{3} & =\frac{295}{600}\left(T_{1}+T_{2}\right). \end{align*}

Hence any values of $T_{1}$ , $T_{2}$ and $T_{3}$ that satisfy this equation will ensure zero voltage is measured.

Question 2

Ceramic piezoelectric materials lose their inherent polarisation if they are heated above a temperature known as the Curie temperature. Suppose that the sensor in Question 1 needs to operate at 300 °C. Choose the material from the datasheet that will provide the highest sensitivity to strains along axis 3 (as per Figure Q1) but can tolerate these higher temperatures.

Question 3

A piezoelectric shear actuator is built using PIC151, as shown in Figure Q3. The height of an individual layer (when it is perfectly vertical) is 0.8 mm. The applied voltage is $V=100$ V.

(a) Consider initially a device made from a single layer (of height 0.8 mm). Use the piezoelectric actuator equation

\boldsymbol{S}=\left[d\right]^{T}\boldsymbol{E}+s\boldsymbol{T}

to calculate the strain $\boldsymbol{S}$ on a single layer of the actuator. Assume no mechanical load (i.e. the stress is given by $\boldsymbol{T}=0$ ).

(b) Now consider the multi-layer structure shown in Figure Q3 (b). Let $n$ be the number of layers. Calculate $\Delta L$ , which is the distance travelled by the tip of the actuator when the voltage is varied from $V=0$ to $V=100$ V.

Hint: Engineering shear strain has units of radians, shown in the figure by the angle $S_{5}$ . You may approximate the geometry as that of a right-angled triangle.

(c) How many stacked actuator layers would be required to achieve a working range of 2 μm given an applied voltage range of 0 to 100 V?

Figure Q3: A piezoelectric shear actuator. (a) A single element with the shear greatly exaggerated for clarity. Applying a voltage in direction 1 causes a bending of the tip. (b) A complete actuator has multiple layers with each subsequent layer having an opposite polarisation direction and opposite voltage polarity. The overall movement distance is

n\,\Delta L

where

n

is the number of layers. Zoom:

Answer

(a) The electric field is

\boldsymbol{E}=\left[\begin{array}{c} 100/\left(0.8\times10^{-3}\right)\\ 0\\ 0 \end{array}\right]=\left[\begin{array}{c} 125000\\ 0\\ 0 \end{array}\right]\ \text{V/m}.

Using zero stress ( $\boldsymbol{T}=0$ ), the actuator equation is

\begin{align*} \left[\begin{array}{c} S_{1}\\ S_{2}\\ S_{3}\\ S_{4}\\ S_{5}\\ S_{6} \end{array}\right] & =\left[\begin{array}{ccc} 0 & 0 & d_{31}\\ 0 & 0 & d_{32}\\ 0 & 0 & d_{33}\\ 0 & d_{24} & 0\\ d_{15} & 0 & 0\\ 0 & 0 & 0 \end{array}\right]\left[\begin{array}{c} 125000\\ 0\\ 0 \end{array}\right]=\left[\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 125000d_{15}\\ 0 \end{array}\right]. \end{align*}

The only nonzero strain is the shear strain $S_{5}$ . Shear strain $S_{5}$ is a bending moment around axis 2. From the material datasheet, $d_{15}=610\times10^{-12}$ C/N. Note that the units C/N are equivalent to m/V, so $S_{5}=d_{15}E_{1}$ has no units (i.e. is measured in radians).

Substituting the actual numbers, $S_{5}=7.6250\times10^{-5}$ radians.

(b) Since the shear strain is small, we can approximate the geometry to be a triangle. Hence

\Delta L=0.8\ \text{mm}\times\sin S_{5}=61\ \text{nm}.

Note that shear strains are so small that $\sin S_{5}\approx S_{5}$ , so some references state the equation for travel distance as $\Delta L=LS_{5}=d_{15}V$ .

(c) Each actuator provides 61 nm of travel distance, so to achieve 2 μm we need

n=\frac{2\times10^{-6}}{61\times10^{-9}}=33\ \text{layers}.

Question 4

A sample of quartz has a piezoelectric coefficient matrix of the form

\left[\begin{array}{cccccc} d_{11} & d_{12} & d_{13} & d_{14} & d_{15} & d_{16}\\ d_{21} & d_{22} & d_{23} & d_{24} & d_{25} & d_{26}\\ d_{31} & d_{32} & d_{33} & d_{34} & d_{35} & d_{36} \end{array}\right]=\left[\begin{array}{cccccc} -2.3 & 2.3 & 0 & -0.67 & 0 & 0\\ 0 & 0 & 0 & 0 & 0.67 & 4.6\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]\times10^{-12}\ \text{C/N}.

Also, the material has a relative permittivity of 4.5.

(a) Sketch a 3D cube and draw the axes 1, 2, and 3 arbitrarily on your figure. Make sure that you use a right-handed coordinate system (i.e. use your right hand to determine the polarity of each axis).

(b) You are designing a force sensor and want to maximise sensitivity to normal stresses. Choose two opposite faces of the cube to deposit metal electrodes. Indicate how you would connect a voltage meter in order to create a force sensor.

(c) Suppose your cube has dimensions 1 cm $\times$ 1 cm $\times$ 1 cm. Based on the geometry of part (b), you measure a voltage of 0.05 V. You know that the only stress acting on the quartz crystal is a normal stress along axis 1. Calculate the value of this stress.