EE3901/EE5901 Sensor Technologies
Week 9 Tutorial

Written by A/Prof Bronson Philippa
College of Science and Engineering, James Cook University
Last updated 2 March 2022

Question 1

An eddy current sensor is used to detect defects on a steel target. Steel has resistivity Ω.m and relative permeability . 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 H/m and the formula for the eddy current skin depth is given by

Answer

Calculate the skin depth:

Hence the target thickness should be at least (approximately), i.e. 7.2mm or more.

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 Ω.m and a relative permeability of . The tape has a thickness of 0.3 mm. What is the required frequency for the eddy current power source?

Answer

We require a skin depth of .

Hence we can calculate the required frequency:

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

where N/m and is the displacement from the zero point of the LVDT. Meanwhile the weight of the the mass in the direction of is

where . 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

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

Figure Q3
Figure Q3:

An angle sensor constructed from an LVDT, a mass, and a spring.

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Answer

The system is in static equilibrium so the net forces must sum to zero.

Converting the sensitivity to base SI units,

Hence the output voltage is

Question 4

Figure Q4 shows the equivalent circuit of an LVDT. The device produces a no-load output voltage of V (RMS) when measuring a deflection of 10 mm. The power supply is 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
Figure Q4:

Equivalent circuit of an LVDT.

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(a) You measure the DC resistance of the primary winding to be Ω. Next you measure its inductance using an LCR meter and find that the primary windings have an inductance of mH. Calculate the impedance of the primary winding at 2 kHz and hence find the magnitude of the excitation current .

(b) Calculate the net mutual inductance at 10mm deflection based upon the measured output voltage V.

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

(c) Assume that 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 because the impedance of the primary winding will change.

Answer

(a) The primary winding has an impedance . At 2 kHz, the impedance is

Hence the excitation current is

(b) Writing KCL around the output winding, you obtain . Therefore the mutual inductance is given by

(c) When the excitation frequency is changed, firstly the excitation current will decrease:

The output voltage becomes

Hence we have nearly doubled the sensitivity of the LVDT by the change in excitation frequency.

Question 5

You are measuring the strength of an electromagnet with the Hall effect. Your sensor element is a 10 mm 10 mm 0.1 mm wafer of p-type silicon with doping density . The geometry is shown in Figure Q5.

Figure Q5
Figure Q5:

Measuring the magnetic field strength using the Hall effect.

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(a) Is positive or negative as drawn?

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

(c) Returning to the p-type device, and assuming a geometry factor of , calculate for an injected current of 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 V/T.

(e) The sensor is now rotated about the axis as shown by the angle What is the minimum value of such that ?

(f) What is the minimum value of such that 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,

Since is positive, points along the axis. Using the right hand rule on the vector cross product , we find that points in the direction. Hence the carriers are deflected towards the axis, i.e. into the page. Therefore, there is a build-up of positive charge on the back surface, meaning that is negative as drawn.

(b) The conventional current flows along the axis, but in n-type silicon the majority carriers are negatively charged so they actually move in the opposite direction. Hence is along the axis. However, since is negative, still points along the axis. Hence, just like before, points along the axis.

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

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

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

μ

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

The required sensitivity is 1000 times larger, hence we require an amplification gain of . 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 is measured. Hence a degree rotation will ensure .

(f) The direction of the Lorentz force will be swapped if the sensor is rotated degrees. Hence the sign of will swap to positive under such a rotation.