Lecture Examples Week 9 Resistive displacement sensors
If a resistance displacment sensor has a total resistance of 3400 , the minimum distance measurable corresponds to zero displacment and the maximum range of motion that can be sensed is 100mm. The supply voltage to the sensor is 5V. What is the displacement if 3.6V is measured at the output terminals of the sensor?
Vin = 5; %V
R = 3400; %ohms
Copyright By PowCoder代写 加微信 powcoder
dmax = 100 ;%mm
Vout = 3.6 ;%V
d = (Vout/Vin)*dmax % in mm d = 72
What is the resistance seen by the output
R2 = (Vout/Vin)*R % in ohms R2 = 2448
Capacitive displacement sensors
A Square variable gap capacitance sensor has side length of 4mm, a permativity of 2, and a plate separation of 50 microns.
What is the change in capacitance if the gap increases to 100 microns?
Eo = 8.85e-12
Capacitance is :
here d is initially 50 microns and then increases to 100 microns. We need to work out both capacitances and find the difference.
Area = 4e-3*4e-3; % square device so 4mmx4mm
Ep_0 = 8.85e-12; % permativity of pfree space
Ep_R = 2; % filling permaitivity
d = 50e-6 ; % zero displacement separation
x = 50e-6 ; % amount moved (100 is the final separation)
C0 = Ep_0*Ep_R*Area/d %F
C0 = 5.6640e-12
C1 = Ep_0*Ep_R*Area/(d+x) %F C1 = 2.8320e-12
DeltaC = C0-C1 %F
DeltaC = 2.8320e-12
How could we measure this change in capacitance?
We could montior the RC time constant of a circuit.
If we use a 10M resistor we can monitor the voltage as the capacitor discharges. Using a large resistor allows the discharge to be slow and in a range that a microcontroller can montior.
R = 10e6; %Ohms
Vin = 5; %V
t = 0:1e-5:0.5e-3; % some time values to calculate for
Vo = Vin.*exp(-t/(R*C0));
V2 = Vin.*exp(-t/(R*C1));
plot(t*1000,Vo,t*1000,V2);
xlabel(‘time (ms)’)
ylabel(‘Output Voltage (V)’)
legend(‘5.6pF’,’2.83pF’)
We can relate the voltage across the sensor to the capacitance and therefore the dispalcement by montioring the time to discharge to a set level or measure the voltage after a certain time.
Inductive displacement sensors
A LDVT is used to measure dispalcement. The device is driven from a supply operating at 100Hz with 50V amplitude.
Plot the output voltage (rms) versus displacement for this device.
We assume the same number of turns for each secondary winding, and that the total number of turns from primary to secondary is the same so that the coupling ratio is 1.
t = 0:1e-3:0.1; %time vector
Vs = 50; %ampltiude of driving voltage
k1=(0:0.01:1).’; %coupling depends on position
k2 = k1(end:-1:1); %assume k2 is the opposite of k1 (total coupling is always 1 in this case)
V1 = k1*Vs*sin(200*pi.*t+0); %sinusoidal voltage is 50*sin(200*pi*t), 100Hz, initial phase her
V2 = k2*Vs*sin(200*pi.*t+0);
Vo = V1-V2; %output Voltage is v1-v2
Vrms = sqrt(mean(Vo.*Vo,2)); %work out Vrms to see how varies with position
plot(k1-0.5,Vrms)
xlabel(‘displacement from centre (Arb. Units)’)
ylabel(‘RMS of output voltage’)
title(‘RMS voltage at output for different positions’)
imagesc(t,k1-0.5,Vo)
xlabel(‘time (s)’)
ylabel(‘displacement from centre (Arb. Units)’)
title(‘output voltage versus time for each displacement position’)
We can see from looking at the output voltage waveform that the RMS voltage is mirrored from the zero position line. However if you look at the waveforms the phase of each side is different (peaks are opposite) so the phase can be used to determine the direction.
An absolute position encoder needs to cover the range of 300mm linear travel.
If we require a positional resolution of 10 microns. How many bits does each position need to encode? How might this be achieved?
Range is 300mm, we need a resolution of 10 microns, this is 300mm/10um unique positions. These will each need their own unique code and so we need at least this many codes. Assuming we use binary pattern we can take the log base 2 of this number and get the number of bits.
unique_positions = 3.0000e+04
bits = log(unique_positions)/log(2)
bits = 14.8727
needed_bits = ceil(log(unique_positions)/log(2))
needed_bits = 15
We need at least 15 bits encoded at each position to cover the full range with 10 micron resolution. This could be achived by using 15 bit code of reflective / absorbing material and a photodiode head to read each bit. Having many bits like this is expensive to manufacture and more complex to read.
final_resolution = range / 2^needed_bits
final_resolution = 9.1553e-06
Our final resolution was 9.12 microns using 15 bits for the 300mm travel range.
Lecture Examples Week 10
An object is square with side lengths of 10cm and is 10mm thick. The Youngs modulus of the material the object is made from is 69GPa.
What is the stress applied to an object if a force of 10kN is applied to the top (normal to the surface)? What is the longitudinal strain in the object? Express your answer in micro-strain.
How much does the material compress to generate this strain?
Stress = Force / Area:
Stress = Strain * Young’s modulus
range = 300e-3; % 300 mm travel
res = 10e-6; % 10 micron resolution
unique_positions = range./res
Force = 10e3;%N
Area = 0.1*0.1 ;% m2
Stress = Force / Area %N/m2
Stress = 1.0000e+06
Strain = 14.4928
Longitudinal strain = dx/x
dx = 0.1449
A water tank contains a column of liquid, if the base of the tank has a diameter of 1m, and the tank contains 20000 litres of water, what is the stress in the base of the tank?
You can assume that the liquid has a mass of 1kg per litre and that acceleration due to gravity is .
We need to use F = ma , where a = acceleration due to gravity (g) in this case. Force = mg
Then we use
Stress = Force / Area
Force = 20000*10 ; %1 litre has a mass of 1 kilo for water, acceleration due to gravity is 10m
r = 1/2; % 1m diameter
Area = r*r*pi ;
Stress = 2.5465e+05
What proportion of the response of a platinum strain gauge comes from the change in resisitivty with strain term?
You may assume the Gauge factor for the 100% pt strain gauge is 6.1, and the Poisson’s ratio is 0.385
Strain = Stress/E *1e6 % in microstrain
x = 10e-3;
%Strain = dx / x
dx = Strain*1e-6 * x *1e6 % in microns
Stress = Force / Area %Pa N/m2 or kg/(m.s2)
Recall the form of the Gauge factor G
The first term is not related to the resitivity and we have the possions ratio we can work out that part, and subtract it from the known gauge factor. What is left is due to the change in resistivity with strain term.
G = 6.1000
Nu = 0.385
Nu = 0.3850
ans = 1.7700
dp_p = G-(1+2*Nu)
dp_p = 4.3300
prop = 100* dp_p / G
prop = 70.9836
71% of the response comes from the change is resistivity.
If this device is used in a quarter bridge configuration, what is the strain felt by the sensor if the supply voltage is 5V and the measured output voltage from the bridge is 0.244mV.
What assumptions must you make in order to answer this question?
Recall the quarter bridge equation is:
We assume the bridge was intiially balanced, with no strain applied Vo = 0. We also assume the strain is small as we are using the small strain approximation form of the equation. We also assume that the strain gauge is in a location in the bridge where an increase in strain leads to an increase in Vo.
Strain = 32
If we were to add another strain gauge and use a half bridge, how should we arrange the devices? How will the responsivity change in this case?
This depends on the shape of the object and whether compression or tension is applied to the strain gauge. If both devices see the same type of strain then they both need to be in the part of the bridge with the same sign of response (diagonally opposite).
If they experience opposite strains (for example on top and below a beam) then they should be in the same half of the bridge vertically.
If we get the positions in the bridge correct then the change in Vo with strain will double
If we get the positions wrong, then the output will not change with strain, as the changes in each device cancel out.
Vo = 0.244e-3;
Strain = Vo / (0.25*Vs*G) *1e6 %micro strain
What is the resistance of a platinum strain gauge with an effective length of 300mm (15mm*20 lines), Where the foil lines have a width of 50 microns and the foil thickness is 5 microns? The resistivity of platinum is
. Solution
Rho = 10.6e-8
Rho = 1.0600e-07
l = 20*15e-3
l = 0.3000
area = 50e-6*5e-6 %m2 area = 2.5000e-10
R = Rho * l /area %Ohms R = 127.2000
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com