Week 9 Tutorial

Question 1

An eddy current sensor is used to detect defects on a steel target. Steel has resistivity $\rho=1.18\times10^{-7}$ Ω.m and relative permeability $\mu_{r}=100$ . If the driving frequency is 50 Hz, what is the minimum thickness of the steel target to ensure proper sensing with the eddy current sensor?

Hint: the magnetic permeability of free space is $\mu_{0}=4\pi\times10^{-7}$ H/m and the formula for the eddy current skin depth is given by

\begin{equation} \delta=\sqrt{\frac{2\rho}{\omega\mu}}. \end{equation}

Question 2

A thin metal tape is being affixed to a non-conductive target to facilitate distance measurements using an eddy current sensor. The tape is made of aluminium, which has a resistivity of $\rho=2.65\times10^{-8}$ Ω.m and a relative permeability of $\mu_{r}\approx1$ . The tape has a thickness of 0.3 mm. What is the required frequency for the eddy current power source?

Question 3

The inclination of a plane is measured with an LVDT that has a 2 kg mass attached to its rod (Figure Q3). The mass is supported by a spring which exerts a force

F_{s}=-kx,

where $k=2000$ N/m and $x$ is the displacement from the zero point of the LVDT. Meanwhile the weight of the the mass in the direction of $x$ is

F_{g}=mg\sin\theta,

where $g=9.81\ \text{m}/\text{s}^{2}$ . Assume that the friction between the mass and the plane is negligible. The LVDT has a sensitivity of 150 mV/cm/V when powered by a 2.5 kHz, 3 V RMS sine wave.

Derive the transfer function between output voltage and angle $\theta.$

Hint: assume static equilibrium to find the relationship between $\theta$ and $x$ , then use the sensitivity of the LVDT to find the relationship between $x$ and the output voltage.

Figure Q3: An angle sensor constructed from an LVDT, a mass, and a spring. Zoom:

Question 4

Figure Q4 shows the equivalent circuit of an LVDT. The device produces a no-load output voltage of $V_{0}=0.5$ V (RMS) when measuring a deflection of 10 mm. The power supply is $V_{i}=5$ V (RMS) at 2 kHz. You would like to increase the output voltage (i.e. increase the sensitivity) by raising the excitation frequency.

Figure Q4: Equivalent circuit of an LVDT. Zoom:

(a) You measure the DC resistance of the primary winding to be $R_{1}=200$ Ω. Next you measure its inductance using an LCR meter and find that the primary windings have an inductance of $L_{1}=10$ mH. Calculate the impedance of the primary winding at 2 kHz and hence find the magnitude of the excitation current $i$ .

(b) Calculate the net mutual inductance $M=M_{1}-M_{2}$ at 10mm deflection based upon the measured output voltage $V_0 = 0.5$ V.

Hint: write KCL around the output winding to obtain an expression for $V_0$ as a function of $M$ .

(c) Assume that $M$ is constant with respect to frequency (i.e. the magnetic media in the core is not saturated). Calculate the new output voltage when the excitation frequency is raised to 20 kHz.

Hint: you will need to recalculate the excitation current $i$ because the impedance of the primary winding will change.

Question 5

You are measuring the strength of an electromagnet with the Hall effect. Your sensor element is a 10 mm $\times$ 10 mm $\times$ 0.1 mm wafer of p-type silicon with doping density $N_{c}=10^{18}\ \text{cm}^{-3}$ . The geometry is shown in Figure Q5.

Figure Q5: Measuring the magnetic field strength using the Hall effect. Zoom:

(a) Is $V_{H}$ positive or negative as drawn?

(b) Suppose instead that the sensor material were n-type silicon. What would be the polarity of $V_{H}$ ?

(c) Returning to the p-type device, and assuming a geometry factor of $G_{H}=1$ , calculate $V_{H}$ for an injected current of $i=50$ mA and a magnetic field strength of 100 mT.

(d) Suggest a type of amplification circuit that could be connected to this sensor to obtain a sensitivity of $-3.1211$ V/T.

(e) The sensor is now rotated about the $y$ axis as shown by the angle $\theta.$ What is the minimum value of $\theta$ such that $V_{H}=0$ ?

(f) What is the minimum value of $\theta$ such that $V_{H}$ has the opposite sign but the same magnitude?

Answer

(a) In p-type silicon, the charge carriers are positively charged. The polarity is determined by the direction of the Lorentz force,

\boldsymbol{F}=q\boldsymbol{v}\times\boldsymbol{B}.

Since $q$ is positive, $q\boldsymbol{v}$ points along the $+y$ axis. Using the right hand rule on the vector cross product $q\boldsymbol{v}\times\boldsymbol{B}$ , we find that $\boldsymbol{F}$ points in the $-z$ direction. Hence the carriers are deflected towards the $-z$ axis, i.e. into the page. Therefore, there is a build-up of positive charge on the back surface, meaning that $V_{H}$ is negative as drawn.

(b) The conventional current flows along the $+y$ axis, but in n-type silicon the majority carriers are negatively charged so they actually move in the opposite direction. Hence $\boldsymbol{v}$ is along the $-y$ axis. However, since $q$ is negative, $q\boldsymbol{v}$ still points along the $+y$ axis. Hence, just like before, $\boldsymbol{F}$ points along the $-z$ axis.

However, what is different is that the carriers accumulating on the back surface are negatively charged. This matches the polarity of $V_{H}$ , hence the Hall voltage is positive as drawn.

(c) Note that we must convert the doping density to SI units.

N_{c}=10^{18}\ \text{cm}^{-3}=10^{18}\ \frac{1}{\cancel{\text{cm}^{3}}}\times\left(\frac{100\ \cancel{\text{cm}}}{1\ \text{m}}\right)^{3}=10^{24}\ \text{m}^{-3}.

Using the negative sign from part (a), the Hall voltage is then

V_{H}=-G_{H}\frac{iB}{qN_{c}L}=-1\times\frac{0.05\times0.1}{1.602\times10^{-19}\times10^{24}\times0.1\times10^{-3}}=-312.11\ \text{μV}.

(d) The sensitivity of the semiconductor element alone is given by

S=\frac{\partial V_{H}}{\partial B}=-\frac{i}{qN_{c}L}=-3.1211\times10^{-3}\ \text{V/T}.

The required sensitivity is 1000 times larger, hence we require an amplification gain of $G=1000$ . A suitable circuit would be an instrumentation amplifier.

(e) A 90 degree rotation will place the Lorentz force completely perpendicular to the direction along which $V_{H}$ is measured. Hence a $\pm90$ degree rotation will ensure $V_{H}=0$ .

(f) The direction of the Lorentz force will be swapped if the sensor is rotated $\pm180$ degrees. Hence the sign of $V_{H}$ will swap to positive under such a rotation.