Do you remember the last module? Technical sensors have become indispensable in our modern world. Among other things, they are used for measuring and monitoring:

Radar control

For the safety of road users, but also to protect against noise speed controls are performed. This is done with optical sensors that can analyze the transit time of radar waves.

Air quality measurements

Dirty air makes you sick. To check the air quality, the concentration of various air pollutants is determined at many (but far too few) measuring stations.

In particular, when it comes to monitoring compliance with legal limit or benchmark values, it is not sufficient for sensors to only roughly record physical or chemical data of the environment.

At speed controls the exact speed of passing vehicles is needed.

In air quality monitoring, the concentration of a substance, e.g. of nitric oxide in the air.

What is wrong here?

Watch the video and tryto find out what is wrong! Type your explanation in the textfied below.

Explanation:

Your explanation is too short!
slow-motion normal time lapse
Theoretical fundamentals

Calibration - what is that?

Functionality of semiconductor gas sensors, revision and deepening

Summary of the measuring process

Concentrations of gases

Experimental basis

Experimental setup and implementation

Usage of the calibration software

Calibration on ethanol

Collection of training data.

A mathematical model of the least wrong solution.

Determination of an unknown concentration.

Arrange the temperatures ϑ1, ϑ2, ϑ3 according to their values in increasing orders. Write the indices 1, 2 and 3 in the lower inequality chain.

You can measure the temperature with the thermometer. Move it with the mouse.

Explain your answer:

The order is incorrect! Your explanation is too short! The order is correct!

You can read the temperature at the height of the liquid, as the liquid expands in the thermometer with increasing temperature. The higher the column of liquid, the higher the temperature.

We can not yet specify a value of the temperature. For this we would have to know which liquid level h at which temperature? established. Such mapping h ϑ is the basis for its calibration in the example of the liquid thermometer. This relationship is (approximately) linear.

According to the Celsius scale, ice water has a temperature of 0 ° C and boiling water a temperature of 100 ° C. The expansion of the liquid column is proportional to the temperature.

Label the thermometer!

Not all specified temperatures on the scale are correct! Right! Due to the linear relationship between rise height h and the temperature 𝜗 the distances of the heights of two temperatures are equidistant.

Now use your calibrated thermometer to determine the temperatures of the three vessels.

𝜗1 = 𝜗2 = 𝜗3 = Not all answers are correct. Correct them and press again on confirm. The answers are correct!

Follow the brief conversation between the teacher and a student discussing the essence of calibrating a level thermometer.

A level thermometer measures the temperature!
That is not quite right.
How so? The thermometer shows the temperature!
And where do you see the temperature?
The scale shows it. Depending on how high the liquid column is.
Right. Actually, we measure the height of rise.
Ah, exactly! Only the scale shows us which height corresponds to which temperature.
That's the way it is! The scale is the result of a calibration. It is based on the assignment of a rise height to a temperature.
Yes, I understood that. Mmmhh – But why actually this detour?
You mean: why do not we measure the temperature directly?
Yes, exactly.
At best, we can roughly estimate temperatures through our senses ...
... heights on the other hand can be measured accurately. Nice. That's a great idea.
This idea can also be described by the term assignment.
Yes, I know what you mean: The temperature is, so to speak, assigned to the rise.
And the calibration allows the way back from the height of rise to the temperature.
That's not hard at all.
Let's summarize the conversation and complete it with a diagram ...

The measuring principle with the steps assignment and calibration is quite analogous for technical sensors, but is usually more difficult to implement. Gas sensors (more precisely semiconductor gas sensors) are such technical sensors. Semiconductor gas sensors) are such technical sensors

Let's compare the principle for the liquid thermometer and the gas sensor:

The derived measured quantities in the semiconductor gas sensor are electrical (as with many technical sensors as well). This has the advantage that the raw data can be saved and processed with a computer. In fact, the calibration of gas sensors is possible only by computer-aided signal processing.

Therefore, we supplement the scheme for the gas sensor with an important aspect:

thermometer
gas sensor
original measurement (not directly determinable)
temperature
conversion
concentration
In Module 1, you saw that when measuring gases, the electrical resistance of the sensor is detected. Compared to the liquid thermometer, the resistance assumes the analogous role of the rise height.
The gas-specific measured variable (for example, the concentration) was thus converted into the easily detectable measured variable of the electrical resistance.
In this form, the sensor corresponds approximately to the thermometer without a scale. As an application you have differentiated in module 1 different gases from each other.
An important application is also to record the concentration of a target gas.
However, this succeeds only after the sensor has been calibrated. For this we have to find a connection between the concentration and the sensor resistance.
derived measurement variable (raw data)
rising height
calibration
computer-aided signal processing
sensor resistance
original measurement variable (indirectly determined)
temperature
concentration

The derived measurement variable of semiconductor gas sensors is the sensor resistance R.

The sensor resistance depends primarily on the following primary variables:

  • Composition of the target gas (gas type)
  • Concentration of the target gas (gas concentration)
  • sensor temperature

The temperature of the sensor is regulated by a heating voltage.

The temperature dependence of the sensor resistance is very important for the desired calibration.

That is why in Module 1 you were able to distinguish different gases, as they have characteristic dependencies R(𝜗).

As an example, the graph below shows the dependence of the sensor resistance on the sensor temperature of Module 1 (for humid air over a water surface and for the effluents of apple juice).

By modeling the processes inside the sensor and on the sensor surface, you got to know other variables of how the sensor works (secondary variables).

The secondary variables are ...

  • ... the velocity () of the electrons that are responsible for charge transport in the sensor.
  • ... the coverage of the sensor surface with oxygen (). This oxygen binds free electrons and thus influences the number of free electrons.

Exercises

Decide in the context X → Y whether the influence of X on Y is in the same direction or in opposite directions. By clicking on the circle symbols you can switch between:

the larger X, the larger Y (concordant)

the larger X, the smaller Y (antithetical)

no general statement possible

I don't know

An example of the relationship between number of free electrons (X) and sensor resistance (Y) is already given:

The more electrons in the sensor material contribute to the charge transport, the greater the current flow I at a given voltage U which is applied to the sensor. If the number of free electrons increases, the sensor resistance R = U / I decreases.

gas type
gas concentration
sensor temperature
Coverage of the surface of the sensor with oxygen
heating voltage
number of free electrons
velocity of the electrons
sensor resistance

Some of your answers are incorrect. Correct them an click on confirm again.

Do not show mistakes Show mistakes
All answers are correct!
Remember the exercise from module 1, which investigated the temperature dependence of the sensor resistance (see right). After setting a new heating voltage (temperature change) you had to wait a few seconds before you have noted the value of the resistance. The reason for this is that the sensor resistance does not immediately assume a constant value after an (abrupt) temperature change, but gradually approaches it. Research has shown that this balancing process contains important information that can be used for the intended calibration. But let's start at the beginning and take a closer look at such a balancing process with the experimental setup of Module 1.

In the video, the heating voltage is suddenly increased from 1.0 V to 1.5 V (≙temperature jump from approx. 100 °C to 175 °C). The sensor resistance is measured over time.

Seht euch das Video an. Dabei könnt ihr zwischen normaler Abspiel­geschwindigkeit, Zeitlupe oder Zeitraffer wechseln.

Exercise: Which of the following five resistance-time curves fits the measurement in the video?

slow motion normal time lapse
The answer is wrong. Correct your answer and press confirm again. The answer is correct!

The next video builds on the end of the previous one. After 10 s, the heating voltage is reduced from 1.5 V back to 1.0 V (temperature jump from approx. 175 °C to 100 °C).

Watch the video and follow the progress in the diagram.

slow motion normal time lapse

As we mentioned earlier, the resistor-time curves can be used for calibration. But how does that work?

On the right, the idea is illustrated by two (fictitious) resistance-time curves.

In both measurements, the atmosphere around the gas sensor contained, besides air, a target gas with concentration c1 (green curve) and c2 (purple curve) respectively.

The resistance-time curves are different. In order to make this difference easier to understand, one does not consider the whole process.

Rather, one looks in the course for suitable characteristics, which one can express with individual numbers.

Note: The subject of calibration of a semiconductor gas sensor is to trace characteristics of resistance-time curves to the underlying concentration.

For instance ...

  • The (constant) resistance before temperature jump is greater for green than for purple.
  • The resistance at green reaches a higher value due to the temperature jump.
  • In addition, the sudden increase in resistance is greater for green than for purple.
  • Immediately after the temperature rise, the green curve drops more slowly than the purple curve.
  • Toward the end, however, the green curve drops faster than the purple curve.

One could still find many more, if not quite so obvious features. To keep things simple, this module will focus on exactly four features.

We want to take a closer look at the example from the video shown above. Of the total resistance-time curve, we only consider a short excerpt of 10 s duration (red area between 160 s and 170 s) following the temperature jump from 175 ° C to 100 ° C.

The four features inside the red area are...

... the maximum MAX
= R1 = 156 kΩ
... the minimum MIN
= R2 = 128 kΩ
... the mean value MV
= (R1 + R2)/2 = 142 kΩ
... the slope SL
= (R2 - R1)/10 s = -2,8 kΩ/s
more

In the period of t1 = 160 s to t2 = 170 s, within 10 s, the resistance decreases around 28 kΩ, within a second on average only 2.8 kΩ. Note: The mean slope of the resistance curve in the marked red area is neagtive. Therefore SL = -2,8 kΩ.

Determine the four characteristics of the resistance-time curve in the red marked interval using the cursor in the diagram:

Maximum MAX
=
Minimum MIN
=
Mean value MV
=
Slope SL
= kΩ/s
Tipp Check your information! For the calculation of the mean and the slope, see the previous slide with back again. Nice! Your information is correct. Click on continue to progress.
Maximum MAX
= R1
Minimum MIN
= R2
Mean value MV
= (R1 + R2)/2
Slope SL
= (R2 - R1)/30 s

With the knowledge of the last slides we supplement or refine the so far developed conversion process of the semiconductor gas sensor in four places.

Instead of a constant temperature, the sensor is operated with variable temperatures.

The velocity of electrons is affected by a immediate temperature change.

The coverage of the sensor surface with oxygen and consequently the number of free charge carriers is influenced by a temperature change only gradual.

For a given gas type and concentration in the studied atmosphere, a temperature change leads to a characteristic course of the sensor resistance.

gas type
gas concentration
course of the sensor temperature
coverage the sensor surface with oxygen
number of free electrons
velocity of the electrons
course of the sensor resistance
gradually immediate
gas underground
concentration of a target gas
variable temperature of the gas sensor
course of the sensor resistance
characteristics of the course
concentration of the target gas
Finally, we summarize the entire measurement process, including the calibration. Our goal is to measure the concentration of a target gas. In general, the target gas is part of a gas mixture, also called underground. Frequently, ordinary air is the background of a target gas. The gas sensor is operated with a variable heating voltage U or temperature 𝜗. By changing the heating voltage one can extract resistance-time-curves:
The original measured concentrationc is converted into a time-dependent electrical variable R(t).
The resistance-time curves determine characteristics (e.g., maximum, minimum, average and slope). The features allow conclusions about the concentration c ( calibration ). We've used the term concentration more often.
What does
mean? Concentration?

We want to calibrate the sensor to the measured quantity concentration. In doing so, we have assumed that you have an intuitive idea of ​​the concept of concentration of a gas. It is important to specify the term.

Concentration as mass fraction is very often found in the specification of limit values.

For example, the WHO's recommended annual mean limit value of nitrogen dioxide (NO 2) in the outside air is 40 μg / m³ . This information means that the total mass of all NO 2 molecules in 1 m³ (= 1000 l) of outdoor air may not exceed 40 μg per year.

In this module, however, we will report the concentrations asparticles per volume.

We consider e.g. the concentrations (as the proportion of particles) of the gases that make up dry air. Dry air consists almost exclusively of nitrogen (78%) and sorbent (21%).
Only about 1%, ie 1 of 100 particles, belongs to another gas (eg argon, helium, hydrogen or CO 2).

We define the concentration of ethanol in a ethanol-air mixture:

The concentration of ethanol in the ethanol-air mixture (gas mixture for short) is 1%, if in 100 particles of the gas mixture there are on average 99 air particles and 1 ethanol particle .

The concentration c of ethanol (or a target gas) of a gas mixture in the volume V is thus defined by the ratio of the ethanol particle number (or the particle number of the target gas) to the total particle number in the gas mixture:

c =
Particle number of the target gas in volume V
Total particle number in volume V

If, on average in 100 particles of the gas mixture, one ethanol particle is present, the concentration of the ethanol in the gas mixture is:

c =
1
100
= 0,01 = 1 % = 1 %

This volume contains 5 ethanol particles out of 100 particles in the gas mixture (including 95 air particles ). Therefore, the gas mixture has the concentration
c =
5
5 + 95
=
5
100
= 5 %
In experiments, we are dealing with much larger volumes and thus with much larger numbers of particles. Does the concentration c change when we refer to other or larger volumes of the gas mixture? The volume considered so far is only one of many sub-volumes. In this sub-volume, 2 ethanol particles and 92 air particles are included. That means the concentration is:
c =
2
2 + 92
=
2
94
≈ 2 %
Although this volume contains only 5 ethanol particles , , there are 105 air particles . That means the concentration is
c =
5
5 + 105
=
5
110
≈ 4 %
Here you can see 9 ethanol particles and 91 air particles . Therefore the concentration is
c =
9
9 + 91
=
9
100
= 9 %
This sub-volume, contains 0 ethanol particles and 104 air particles. The concentration is
c =
0
0 + 104
=
0
104
= 0 %
The concentration may differ for different partial volumes. In our example, it varies by 5%, sometimes more and sometimes less. In larger volumes, these deviations compensate on average. The large volume (100 times larger) contains about 500 ethanol particles and 9500 air particles:
c =
5·100
5·100 + 95·100
=
500
500 + 9500
=
5
100
= 5 %
Therefore the concentration c of a gas mixture is independent of the considered volume.

Considering air pollutants we will usually find much smaller concentrations than 1%. In these cases smaller units than percent are used. You surely know the next smaller unit of percent: per thousand (short ‰).

Example: A drunk driver contains 1 ‰ alcohol. This means, on average, there is 1 alcohol particle (ethanol) and 999 other particles in the blood. That means that the concentration of ethanol in the blood is:

c =
1
1 + 999
=
1
1000
= 0,001 = 1 ‰

It gets even smaller: After 1000 times dilution of a drunk driver's blood sample (eg with water), 1,000,000 (one million) particles contain on average 1 alcohol particle and 999,999 blood and water particles :

c =
1
1 + 999.999
=
1
1.000.000
= 10-6 = 1 ppm

The unit ppm stands for parts per million.

The already 1000-fold diluted blood is diluted once more 1000 times. The concentration of alcohol drops from 1 ppm to 0.001 ppm.

That means 1.000.000.000 (one billion) particles of the gas mixture contain on average 1 ethanol particle and 999.999.999 blood and water particles.. Die alcohol concentration isc = 1 ppb.

The abbreviation ppb stands for parts per billion

Exercise 1.
In one milliliter of gas mixture are 2.55·1019 air particles and 1.02·1013 of ethanol particles. Calculate the concentration of ethanol in ppb with an accuracy of two decimal places: Tip

c = ppb Check your solution again! Very good! Your solution is correct!

Exercise 2.
Convert the given concentration into smaller units!Tip

c = 0,0012 % = ‰ = ppm = ppb Check your solution again! Very good! Your solution is correct!

Exercise 3.
Convert the specified concentration into larger units!

c = 3.120 ppb = ppm = ‰ = % Check your solution again! Very good! Your solution is correct!
The experimental setup for the following experiment consists of four components: The gas storage chamber is there for several working groups. The gas storage chamber contains a heating plate. If the heating plate is drizzled with a liquid (here ethanol) and the heating plate is in operation, the liquid is vaporized and evenly distributed by a fan in the gas storage chamber. The hot plate has already been drizzled with ethanol. The ethanol is completely vaporized and distributed evenly throughout the chamber by the fan. With the help of a syringe, a small amount of the ethanol-air mixture can be extracted through a septum on the housing of the gas storage chamber. After extraction, the septum closes automatically. The measuring chamber is only filled with air. Via the septum an ethanol-air mixture can be transferred into the measuring chamber by means of a syringe. The measuring chamber houses a gas sensor (BME680). The gas sensor is supplied with a variable heating voltage U and controlled with a microcontroller. The PC reads out the changing heating voltages U, as well as the sensor resistance R via the microcontroller. A software can record the resistance-time curves and determine their features. The software uses Machine Learning to create a mathematical model to associate the features with the corresponding concentration.
The experimental procedure consists of the following steps: First, a resistance-time curve for air without ethanol is recorded in the measuring chamber. The initial concentration of ethanol is c0 = 0 and the resistance-time curve is recorded for temperature jumps. A syringe is used to remove some of the ethanol-gas mixture from the gas storage chamber through the septum. The ethanol-gas mixture of the syringe is transferred into the measuring chamber by means of the other septum. A concentration of c1 > c0 of ethanol is then present in the measuring chamber. A new resistance-time curve for the ethanol concentration in the measuring chamber c1 (with the same temperature jumps) is recorded. With the same step sequence the ethanol concentration in the measuring chamber is increased to c2... Some ethanol-air mixture is extracted from the gas storage chamber. The ethanol-air mixture is transferred into the measuring chamber. Again, a resistance-time-curve of the ethanol concentration c2 > c1 is recorded.

For our calibration, it is important to calculate the concentration in the measuring chamber after the etanol gas mixture has been transferred from the gas storage chamber to the measuring chamber with the syringe. We want to go through the calculations step by step.

A drop of liquid ethanol (Vethanol = 2 ml) was dripped onto the heating plate, causing it to vaporize and evenly distribute in the gas storage chamber (VGMC = 62,500 ml, concentration cGMC).

A volume VSyringe = 1 ml of the ethanol-air mixture is then removed with a syringe and transferred to the measuring chamber (VMK = 1,250 ml).

Further mixing with air dilutes the ethanol-air mixture in the measuring chamber to the concentration c1.

Also use the following data:

nAir = 2,55·1019 particles/ml
nEthanol = 1,03·1022 particles/ml
NAir,GMC = 1,59·1024 particles
NEthanol = 2,06·1022 particles
cGMC = 12,7 ppm

Step 1: NAir,GMC: Calculation of the number of air particles NAir,GMC in the gas mixing chamber:

NAir,GMC = nAir,GMC · VGMC

NAir,GMC = ·10 · ml = particles

Check your solution! Your solution is correct!

Step 2: NEthanol: Calculation of the number of ethanol particles in 2 ml of liquid ethanol NEthanol in the gas mixing chamber.

NEthanol = nEthanol · Vdrop

NEthanol = ·10 · ml = particles

Check your solution! Your solution is correct!

Step 3: cGMC: Calculation of the concentration cGMC of vaporized ethanol in the gas mixing chamber in ‰ and ppm.

cGMC = NEthanol NAir,GMC + NEthanol = particles ( + ) particles

cGMC = = ‰ = ppm

Check your solution! Your solution is correct!

step 4: Equation of proportion: Calculation of the concentration c1 of vaporized ethanol in the measuring chamber.

To calculate c1 it is not necessary to calculate the particles in the syringe and in the measuring chamber. The ethanol-air mixture from the small syringe can now be distributed into the much larger volume of the measuring chamber. This results in a simple ratio equation. What does the ratio look like?

c1 cGMC = VSyringe VMK c1 cGMC = VMC VSyringe c1 cGMC = VSyringe VMC + VSyringe c1 cGMC = VMC + VSyringe VSyringe Check your solution! Your solution is correct!

Schritt 5: c1: Calculation of the concentration c1 of vaporized ethanol in the measuring chamber in ‰ and ppm.

For the dilution from c0 to c1 we will do the calculation for you:

c1 = c0 · VSyringe VMC + VSyringe = 1,27·10-2 · 1 ml 1.250 ml + 1 ml

Practice the conversion of the units again:

c1 ≈ 1,00·10-5 = ‰ = ppm

Check your solution! Your solution is correct!

If 1 ml of the ethanol-air mixture is injected from the gas storage chamber into the measuring chamber, the ethanol concentration in the measuring chamber is c1 = 10 ppm.

What about the ethanol concentration if another ml of the ethanol-air mixture is injected from the gas storage chamber into the measuring chamber?

The ethanol concentration remains almost the same. The ethanol concentration increases by a further 10 ppm. The ethanol concentration increases by more than another 10 ppm. Your solution is correct! Check your solution!

The injected quantities of ethanol-air mixture and the ethanol concentration in the measuring chamber are approximately proportional to each other. Proportionality is shown in the diagram.

Now start the software for the first measurements!

Ask your supervisor for further information!

You have chosen the features and . The two diagrams show the values of the two characteristics as a function of the concentrations.

All data points are so-called training data. We use them to train a mathematical model that correlates characteristic values with concentrations. The better the training data you select, the better your calibration will be.

The question now is: How can concentrations and characteristic values be assigned?

The following methods, which are often used for calibration, are easy to understand:

Search for a calibration curve that fits optimally to the training data (also called Best Fit).

In the pictures below this is indicated for two fictitious connections. Note that optimal fit does not mean that the training data must lie exactly on the curve. It is an approximate solution.

For an unknown concentration, we could now determine the characteristics of the characteristics and assign the calibration curves to a concentration.

However, the application of the best-fit method is subject to a condition: At least in principle, we need to know the functional relationship between the variables.

For example, with liquid thermometers, the relationship between temperature and head is linear (left picture). The other curves are based on a exponential (middle picture) and a square (right picture) relationship.

However, characteristic expressions and concentrations can be related in very different ways and model equations generally cannot be given so easily.

For the calibration we therefore choose another mathematical method.

How can concentrations and characteristic values be assigned?

In order to understand this mathematical method, we briefly leave the concrete application case of the gas sensor and instead look at an illustrative example:

Imagine a scientist trying to deduce the mass (M) of an adult human from the two characteristics body size (S) and waist circumference (W).

body size
waist circumference
features
minimum
slope
mass
target
concentration

The mass as a target corresponds to the concentration.

Body size and waist circumference are to be regarded as the characteristics of the resistance-time curves.

Do you realize that the prediction of mass from only one of the two characteristics body size or waist circumference causes problems? So there are thick small or lean tall people.

Both types will probably weigh similarly much. But if only the waist circumference was chosen as a characteristic, in such a prediction the tall man would weigh less than the small one, because he is leaner.

This also applies to the gas sensor: if we consider two characteristics of the resistance-time curve, we can improve the calibration.

For the prediction, the scientist is looking at a total of 100 people, 20 of whom weigh exactly 60 kg, 80 kg, 90 kg and 100 kg each, and also determines their body size and waist circumference. The average values of each weight class are compiled by the scientist in a table:

mass [kg] height [cm] waist [cm]
60 159 66
70 168 70
80 178 77
90 184 87
100 187 104

Graphically displayed, it looks like this:

Instead of determining two calibration curves, the scientist uses an approach with the following model equation:

M = g1·W + g2·S

The mass M is determined here as weighted sum of body size S and waist circumference W. A piece of cake, provided we know the weights g1 and g2. However, these are initially unknown and must be determined using the training data in an equation system.

mass [kg] body size [cm] waist circumference [cm]
60 159 66
70 168 70
80 178 77
90 184 87
100 187 104

Insert data in model equation

Solve the equation system

A piece of cake! Or is it not?

Follow the discussion of the three students.

*doing math* ...
*doing math* ...
Why are they doing the math?
g1 = 1 und g2 = -1,5 solves equation I!
Huh? I have a completely different solution for equation I, namely g1 = 0 and g2 = 10/11
You're both right. But use your solutions in equation II. Both do not fit!
*doing math* ...
*doing math* ...
Right. 168·1-70·1,5 equals 63 but not 70.
And my result doesn't match either. But wait - I got it...
Well, I'm looking forward to that!
So with g1 = -10 and g2 = 25 both equations are true.
(Checks with a calculator) ...
I'm a genius!
They just don't get it!
So, am I right?
Yes, it works! Calculator test passed!
And what about the other three equations?
Oh! That has to be true for all equations?!
Okay. That's challenging. But I'm almost there...
Well, have fun!
There is always a contradiction...
Huh? You can't do that!
This can't even work! The system of equations is overdetermined!
Oh, yeah! More equations than variables!
What now? There is no solution?

Apart from special cases, a system of equations with more equations than unknowns has no (exact) solution.

But: One can show that there is a least wrong solution. In other words, there is a g1* and g2 < with which the predictions are still incorrect but the mean error for all equations is minimized.
What now? There is no solution?!.
Well, yeah. There is no right solution, but a least wrong one.
Cool!

In the diagram the calibrated mass is plotted against the true mass.

The calibrated mass M is calculated with our model approach by

M = g1·S + g2·W

We have chosen the weights g1 = 0.8 and g2 = -1.4.

Unfortunately the weights are still very bad. E.g. the model predicts mass of 4 kg instead of 100 kg, when given a body size of S = 184 cm and a waist circumference of W = 104 cm. Do the math yourself!

The calibration is perfect if the training data are on the bisector of the angle. Here the calibrated mass is equal to the true mass.

But as we have just learned: There is no perfect solution. We allow an error tolerance of ±10 kg, which should be sufficient for our purpose.

Exercise: Find weights g1 and g2, so that all training data are within error tolerance. With the controls you can vary g1 and g2.

M = (0,8) · S + (-1,4) · W

Good job! By the way, in the lower right corner you can see in which area (green) the error is small. The optimal (but not perfect) weights are g1* = 0,03 and g2* = 0,92.

Exercise: Determine your mass from your body size S (in cm) and waist circumference W (in cm) using the optimized model equation

M = 0,03· S + 0,92·W.

Use tape measure and ruler. Enter your results into the table. If you do not know your mass, estimate it.

Group member
Body size [cm]
Waist circumference [cm]
Calibrated mass [kg]
True mass [kg]
1
0,0
2
0,0
3
0,0

Are you satisfied with the prediction or was the deviation from your true mass too high? Name at least one idea how the mass calibration could be improved.

Please enter an explanation!

Thank you for your explanation!

You can compare your ideas with our suggestions for better calibration:

  • More training data (e.g. not only data of 100, but 1000 people)
  • More representative training data (including people weighing more than 100 kg)
  • Integration of more or other characteristics (e.g. consider body fat percentage)

The calibration software works in principle similar to our example. We will explain roughly how the calibration software works:

The software takes an approach to a model equation that is much more complex and contains many weight factors. The recorded training data is then inserted into the model equation. This results in an overdetermined system of equations with unknown weight factors. The software then uses several steps to search for optimal weight factors that lead to an optimal model equation. Finally, the features of unknown concentrations can be used in the optimized model equation to predict a concentration.

You already know this diagram from a moment ago. This time the calibrated concentration is plotted against the true concentration. The predictions are still quite wrong. A click on Start training will execute an algorithm that automatically searches for optimal models and the least wrong solution. The result is getting closer and closer to this solution.

Step: /10000

Best error tolerance: %

Congratulations! You calibrated the sensor. In conclusion two more tasks:

How good is your model compared to other groups? Compare the number of steps it took to go below the error tolerance. What could be the reasons for possible differences?

Check the calibration at an unknown concentration:

  • Open your measuring chamber to remove the ethanol-air mixture you used for calibration.
  • Close the measuring chamber again and insert the contents of a syringe (you get it from the supervisor) into it.
  • Records the resistance-time curve with the calibration software and determines the features. The software automatically displays the ethanol concentration. Now ask the supervisor what was the real concentration.
  • Enter the features in the two input fields on this page and click predict. The program uses your model for the prediction.
concentration in ppm
Click on continue to get to the next slide. ↓ This is the continue button. ← Click here to open or close the menu.