Example Sheet 3 With Solutions
1. Usingadiagramandequations,brieflydescribetheoperationandlimitationsofa Linear Variable Differential Transformer
Similar in concept to capacitive transducers instead of varying the dielectric permittivity use a device sensitive to the magnetic permeability
The simplest device to take advantage of inductive effects is a coil with a movable magnetic core
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The most common device is the Linear Variable Differential Transformer (LVDT) Amplitude of final signal proportional to displacement of core.
Direction of displacement obtained from the phase of the signal.
LVDTs usually run at 5V, in the low KHz frequency range
They can measure displacements from mm down to microns.
They have no moving parts so no mechanical wear, giving a long life. They are electrically isolated so can be used in hazardous environments.
Some limitations: Temperature affects the performance. Sensitive to stray magnetic field.
2. Discusswaysofsensingchemicaleffectswithelectronicsensors.
Chemical effects can be sensed through a twostep process, and recognition step where the chemical species is selected and interacts with the sensor and a transduction step where the interaction is transduced into some change on the sensor that can be measured.
Recognition – there are several ways for recognition processes to function, one example is through affinity interactions where an antigen is bound to an active site. An example of this is anti-body binding.
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Transduction – this can be physical or chemical and ideally provide an easy way to obtain an electronic signal from the sensor. It could be based on the resistance / capacitance of the sensor or optical effects, which can be converted via a photodetector to an electronic signal
3. Describe how a variable gap displacement sensor works, including a discussion of the response type.
The capacitance of a capacitor depends on the plate separation (d), the permittivity of the dielectric filling (𝜀) and the plate area (A).
Variable gap design: nonlinear in capacitance
Response is nonlinear it follows a 1/x form
4. A square variable gap capacitor has a relative permittivity of 2, a length of 1mm and an initial gap of 100 microns. If the object moves by 10 microns how much does the capacitance change?
We can work out C for both conditions and then subtract to see how much C has changed
We are told the device is square so l=w A=1*10-3*1*10-3=1*10-6, the value for 𝜀0𝜀𝑟 is 2*8.85×10-12, d1 = 100*10-6 , d2 110*10-6
Plugging in the values we get C1 = 0.177pF and C2 = 0.161pF giving the change in capacitance as -0.016pF
5. Describe how a resistive displacement sensor works.
Simple displacement sensor uses a resistive area and a contact on the moving object Source voltage (Vin) applied across full resistive bar.
Voltage at the object depends on the position on the bar.
Limited resolution depending on form, contacts wear out due to friction during use, and non linear response can result from loading.
Can be made linear or rotational.
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6. Fortheresistivedisplacementsensorsshowninfigurebelow.
(a) What is the output voltage if Vs is 10V,R = 60Komhs, dmin =0, dmax = 50cm and d = 25cm?
With dmin = 0 the voltage is just a simple voltage divider where all the output voltage is due to our displacement d.
This means we can get Vout = Vs (d/dmax) = 10*(25/50) = 5 Volts.
(b) If dmin = 5 cm what proportion of the output voltage is due to the displacement d?
Remember that d and dmax are references from dmin, so we now need to include this in the calculation.
dmin being nonzero means we will always have an output voltage even if d is zero.
We have Vout = Vs *(d+dmin) /(dmax+dmin) = 10* (30/55) = 5.45V
This value is different from above as the device is effectively 5cm longer due to dmin, so the resistance per unit length in this example is different to in (a)
The proportion from the displacement (using principle of superposition) is the voltage drop over the resistance due to the displacement d. = d/(d+dmin) = 25/30 *100 = 83% or 4.55V
(c) What is the value of R2 seen by the output in (b)? What power is dissipated by this resistance?
Here we use R of 60Kohms as total resistance and the proportion of the total length at the position of the wiper.
R2 = 60K * (d+dmin)/(dmax+dmin) = 60K*(30/55) = 32.7Kohms Voltage dropped across V2 is 5.45V
P2 = V2/R2 = 5.45^2/32.7K = 0.9mW
7. Whyisthemicrostructureofaerospacematerialsimportant?Brieflydescribea technique that can be used to measure the grain structure of materials.
Several important material properties are structure sensitive:
• yield strength,
• fracture toughness,
• thermal conductivity
These are all sensitive to microstructure parameters, such as:
• Mean grain size
• Degree of randomness, (both size and orientation)
• Clusters of grains all oriented in the same direction
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SRAS can be used to measure the grain structure because the speed of sound of surface wave depends on the crystallographic orientation of the material. By measureing the speed of sound across a sample the grains can be image.
SRAs does this by measuring the frequency of the surface wave that is generated with a fixed wavelength. The speed of sound (c) is then determined by using c=fL with a known wavelength(L) and the measured frequency (f)
8. Whyisultrasoundusefulforimaginginthebody?Howisanimagereconstructedina simple linear array ultrasound machine?
Ultrasound will pass through the body and allow us to sense in places where we cannot see directly with optical techniques. Differences in the mechanical properties cause differences in reflection and absorption of the sound leading to contrast in the image.
Ultrasound is non-ionising – the energy the sound carries is sufficiently low that it does not cause damage to the tissue it is propagating in (there is an exception to this for HIFU where it is used to destroy cancerous tissue).
Ultrasound scanners provide real time, low cost, with reasonable resolution and contrast and are widely used in hospitals around the world for a wide range of applications.
You could scan the transducer around (in a line) to produce a cross sectional image or B scan.
However, transducer arrays are usually used so that the scanning can be done electronically.
The transducer array emits ultrasound pulse on each element in turn. It records the trace and then moves onto the next element. Each signal is processed and forms an image of x vs z (space vs depth)
This gives an image of a slice through the object.
9. A quarter bridge, with a supply voltage of 5V, is used with a platinum strain gauge (100% Pt) to measure the strain on an object.
(a) if the applied strain is 100 microstrain, what is the output voltage? You may
assume the circuit was balanced before the strain was applied.
(b) If this is to be interfaced to an ADC with a +/-1V input range, how much voltage
gain do you need? (expressed in dB)
(c) Using a differential amplifier configuration suggest values for the resistors
R1,R2,R3 and R4 to achieve this. Why are these values suitable is there a better way to do this?
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(a) G for the 100% Pt device is 6.1 (see table in notes), initially balanced means we can use Vo = 0.25*Vs*G* εl = 0.25*5*6.1*100*10-6=0.7625mV
(b) Voltage gain we need to consider the range of ADC which is +/-1V. assume we want access to both positive and negative strains so will use half the full range = 1V;
Gain is 1V/0.7625mV = 1311, this is 20 log(1311) = 62.3dB of gain.
(c) Assume R1=R2 and R3 = R4 in the diagram below.
𝑉𝑜 = 𝑅3 (𝑉2 − 𝑉1) 𝑅1
If we use R3 = 1.3MΩ and R1 = 1kΩ we will get a gain of 1300. Which is good enough.
We need to ensure that the load impedance seen by the bridge is high, this was one of our assumptions when we derived the operation of the circuit. The choice of 1kΩ for R1 may be sufficient, here the load will be the differential input impedance of the circuit above which is R1+R2 = 2*R1. It would be better to use an instrumentation amplifier so that the circuit uses the op-amp impedance as the load (which is very high) rather than the differential amplifier. The high values of the resistance will likely mean the circuit has a slow response time (in combination with any capacitance found in the real circuit made with non-ideal components) this could also be an issue depending on the application.
10. Show via the use of partial derivative that the expression for the combined output for a full bridge is:
We start with:
𝑉=𝑉[ 𝑅3 − 𝑅4 ] 𝑜 𝑠 𝑅3 + 𝑅2 𝑅1 + 𝑅4
We will use principle of superposition to find the output for each R varying, this is only ok if the change in the resistances with strain are small. Let’s start with R3 as we did this in class using the other method.
𝜕𝑉 𝜕 𝑅 𝜕 𝑥 𝑜=𝑉[3]=[]
𝜕𝑅3 𝜕𝑅3 𝑠 𝑅3+𝑅2 𝜕𝑥𝑥+𝑎
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Use the quotient rule to work out the differential
𝑓(𝑥) = 𝑔(𝑥) = 𝑔′(𝑥)h(𝑥) − 𝑔(𝑥)h′(𝑥) h(𝑥) [h(𝑥)]2
𝑑[𝑥](𝑥+𝑎)−𝑥𝑑[𝑥+𝑎] 1(𝑥+𝑎)−𝑥(𝑑 [𝑥]+ 𝑑 [𝑎])
𝑥 𝑥= 𝑑𝑥𝑑𝑥=
𝑥+𝑎−𝑥(1+0) (𝑥 + 𝑎)2
The add back in our variables
𝜕𝑥 𝑥+𝑎 (𝑥+𝑎)2
𝜕𝑉 𝑅 𝑜=𝑉[2]
𝜕𝑅3 𝑠 (𝑅3+𝑅2)2
Assume we start with the initial conditions that it is balanced so: 𝑅2=𝑅0 and𝑅3 = 𝑅0 +∆𝑅
𝜕𝑉 𝑅 𝑅 𝑜=𝑉[ 0 ]≈𝑉[0]
𝜕𝑅3 𝑠 (𝑅0 +𝑅0 +∆𝑅 )2 𝑠 (2𝑅0)2
Assume ∆𝑅 is small
So the change in our output due the chance in R3 is:
𝑉𝜕𝑅 𝑉𝜕𝑅 𝜕𝑉 ≈ 𝑠 0 ≈ 𝑠 3
This process will be the same for device in R1 position. For R2 and R4 the initial part is different but follows a similar approach. Here we have:
𝜕𝑥 𝑥+𝑎 (𝑥+𝑎)2
And so we can see we have the same overall form of the solution as above but we have the minus sign to include.
This gives us:
When we include all four terms and keep the Vs/4 as a common factor at the front.
11. A bridge circuit is to be used with two identical strain gauges (platinum tungsten). The devices replace R2 and R3 in the bridge as numbered in the lecture notes. What is the output voltage if the sensors are used in the below configurations?
(a) In this configuration the two sensors see the same strain but the sensors have opposite signs in their response so the change due to the strain will cancel out giving and output of zero.
Vo = 0.25 * Vs *[ G1εl – G2ε2] as εl = ε2 and G1 = G2, then Vo = 0 V.
(b) Here the response is still opposite but one sensor sees compression and the other tension and so the reponses will add.
Vo = 0.25 * Vs *[ G1εl – G2ε2] as εl = -ε2 and G1 = G2
This will give Vo = 2*0.25* Vs Gεl as G=4.0 for this device type the final expression is: Vo = 2Vs εl
12. A MEMS accelerometer has the following characteristics: Active Mass = 5 μg
Spring constant (k) = 50×10-3 Nm-1
What was the acceleration applied to the device if the extension of the spring was 1
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Acceleration = Force / Mass
The force comes from the spring in the device to restore the mass to its central position.
Force = spring constant x extension.
A = (50/1000) * 1μ / (5 μ/1000) = 10 m.s-2
Convert grams to Kg.
This is ~ 1G
13. A manometer is used to measure the pressure of a gas line. The height of the fluid (water ρ=1000kg.m-3) in each side of the loop are h1 = 51cm and h2 = 52 cm.
(a) what is the difference in pressure between the gas line (PA) and the reference (PB)?
(b) If the reference is atmospheric pressure (101kPa) what is the absolute pressure of the gas line?
(a) PA- PB =ρg(h1-h2) = 1000kg.m-3*9.8ms-2*0.01m=98Pa
(b) PB =101kPa then PA =101kPa+98Pa=101.098kPa
14. The sensors shown in the figure below are used to measure the fluid flow of a lubricant oil (ρ=900kg.m-3). If the pressure measured for sensor 1 is 250kPa and for sensor 2 is 251kPa, what is the velocity of the lubricant flowing past the sensors?
Sensor 1 measures static pressure Po, sensor 2 measures the total pressure Ps.
2(𝑃 − 𝑃 ) 𝑉 = √ 𝑠 𝑜
2(251k − 250k) 900
2(1000) 900
= √2.22 = 1.5𝑚/𝑠
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15. How does a Bourdon tube sense pressure? How could the output of this sensor be modified to produce and electrical signal?
This is a curved tube (typically 250 degrees) of oval cross section. One end is sealed and is free to move.
The open end is connected to the pressure to measure.
As the pressure increases the cross sectional shape of the tube changes which causes the tip to move.
The tip is often connected to a link and pinion to a needle to indicate pressure on a dial.
Other forms exist such as the spiral or helical bourdon tube which are more sensitive.
This could be used with a resistive displacement sensor in a circular form where the needle provides contact so that a change in voltage will be obtained with changes in pressure
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