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:
Calibration - what is that?
Functionality of semiconductor gas sensors, revision and deepening
Summary of the measuring process
Concentrations of gases
Experimental setup and implementation
Usage of the calibration software
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:
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!
Now use your calibrated thermometer to determine the temperatures of the three vessels.
confirm.
Follow the brief conversation between the teacher and a student discussing the essence
of calibrating a level thermometer.
assignment.
assignedto the rise.
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:
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.
confirm
again.
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?
confirmagain.
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.
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.
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...
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:
backagain.
continueto progress.
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.
The original measured concentrationc is converted into a time-dependent electrical variable R(t).
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 % |
| c = |
5
5 + 95
|
= |
5
100
|
= 5 % |
| c = |
2
2 + 92
|
= |
2
94
|
≈ 2 % |
| c = |
5
5 + 105
|
= |
5
110
|
≈ 4 % |
| c = |
9
9 + 91
|
= |
9
100
|
= 9 % |
| c = |
0
0 + 104
|
= |
0
104
|
= 0 % |
| c = |
5·100
5·100 + 95·100
|
= |
500
500 + 9500
|
= |
5
100
|
= 5 % |
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:
Exercise 2.
Convert the given concentration into smaller units!
Exercise 3.
Convert the specified concentration into larger units!
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:
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
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
Step 3: cGMC: Calculation of the concentration cGMC of vaporized ethanol in the gas mixing chamber in ‰ and ppm.
cGMC =
cGMC = = ‰ = ppm
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?
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 ·
Practice the conversion of the units again:
c1 ≈ 1,00·10-5 = ‰ = ppm
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 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 following methods, which are often used for calibration, are easy to understand:
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).
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:
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.
Apart from special cases, a system of equations with more equations than unknowns has no (exact) solution.
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.
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.
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:
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.
continueto get to the next slide.
continuebutton.