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ELECTRONIC MEASUREMENTS:
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b. Add squared and divide by n-1:
/"3-, w /"3./ ?v /"4+- w /"3./ ?v /"4,4 w /"3./ ?" "
v/"23- w /"3./ ?v /"22+ w /"3./ ?
0 w , y 13"04
c. 4.834V - 5.000V = -166 mV
d. No, the random error is 68.59 mV = 0.07 V, so two decimals is enough
Question 2.2 Error propagation and representation a. 10.1 Ω 0.5 Ω
b. 4.7 Ω 0.2 Ω
c. Absolute errors add up: 14.8 Ω 0.7 Ω
Question 3.1 Electric Currents and Potentials a. Resistance of a rod:
y * y *
)? y * ) -?
y ,3"+ 5 ,+HE ,"+ 5 ,+@ ) -"0 5 ,+- H??
y +".2&
b. Express the dissipated power in terms of the current through the cable:
c. y 5 y ? y 0++ ?+".2& 5 0+ y /"1-0
Question 3.2 AC Currents a. yRkonp
l8? y ,/",
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Question 3.3 Resistors in series and in parallel
a.
?@ y ?@ ?v @ >?@ y >v ?@
?v @
>?@A y >?@A
>?@v A z /./&
b. >@A y >v @v A >?@A y >@A?
>@Av ? y 20&
c. >?@ y >v ?@
?v @ y .++&
R R d R
d. Calculate power dissipated in R1, R2 and R3: Ok y Ok 5 Ok y JK >
>?@5 JK ,
>?@ { +"0
.
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Ol y Ol 5 Ol y JK ?@
>?@5 JK@#?@
>?@ { +"0 Om y Om 5 Om y JK ?@
>?@5 JK?#?@
>?@ { +"0
Look at similar terms and find that at constant UAB, the powers are: PR1 < PR2 <PR3. So R3 will dissipate most of the power:
p
Om y JK ? ?
>?@? y JK ? ,&
.++&?{ +"0 JK { +"0.++&?
,& z 1"2,
Question 3.4 The voltage divider a.
Figure 27: Voltage divider to make 5V out of 9V
Two boundary conditions:
y Ol yadc
? y , adc y ]` ?
>v ? So:
0
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Question 4.1 Kirchhoff’s laws
a. There are 6 nodes (points between two or more components) for the current law:
• One is 0V per definition,
• The two voltage sources eliminate one node each,
• The nodes between R1/Ra and R2/Rb do not have to be evaluated because series resistors can be taken together as R1+Ra and R2+Rb respectively (they have the same current).
This leaves only one node for the Kirchhoff’s current law, which is the node connecting R1, R2 and R3 when the node between Ua, R3 and Rb is taken as the grounded node.
There are two loops (mazes) for Kirchhoff’s Voltage Law.
b. Apply Kirchhoff’s Current Law applied to the node connecting R1, R2 and R3. Call this node “x” with potential Ux:
Okv Ol v Om y + Uw e
Uv >v+ w e
@ vwVw e Vv ? y +
Solve for Ux: U
Uv > w e
Uv >we
@w V
Vv ? w e
Vv ? y +
,
Uv >v ,
@v ,
Vv ? e y U
Uv >w V
Vv ? e y
U
Uv >w V Vv ? Uv , >v ,@v ,
Vv ?
z w-"+4;
l h h h R
Now we can calculate the current through R3: Om y Om
@ ye
@ z w,/"+
Where the minus sign means that the current is away from Ux (from the top to the bottom).
Question 4.2 Equivalent circuit of a power supply
a. A generic Thévenin equivalent circuit is given in Figure 28.
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Figure 28: Thévenin equivalent circuit
The Thévenin equivalent source can be found by determining the open loop voltage:
the ideal multimeter indicates 10.00V, so UTh = 10.00V.
The Thévenin internal resistor can be found by interpreting the set-up as a voltage divider with a resistor of 100Ω. We find:
adc y ]` ? >v ? 4"32 y ,+ ,++&
>v ,++& : > y ,++&,+ w 4"32
4"32 z ,".-&
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b. First of all: RTh = RN = 1.32Ω. IN = UTh/RTh | 7.89A c. P = U·I = U2/R = (9.87V)2/100Ω | 0.974W
Question 4.3 Norton and Thévenin source transformations
a. Drawing: same as Figure 28. The Thévenin equivalent source can be found by calculating the open loop voltage:
Q\ y JK y U @
>v @ y -"0
and the Thévenin internal resistor is the resistance between node A and B when the internal source Ua is set to 0V:
Q\ y ?v >@
>v @ y -0+&
b. M yQ\
Q\ y ,+
M y Q\ y -0+&
c. Drawing: same as Figure 28. The Thévenin equivalent source can be found by calculating the open loop voltage:
and the Thévenin internal resistor is the resistance between node A and B when the internal source Ua is set to 0V:
0V:
Q\ y ?>v @
?v >v @ z -,"4&
d. The circuit is loaded with a 1kΩ resistor:
JK$R`_aUXYX y Q\ y ,",-0
JK$LaUXYX y Q\ LaUX
Q\v LaUX z ,",+,
The reduction of the output voltage of 24mV is 2.4%
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Question 4.4 Equivalent circuit of a battery
a. First of all, the UTh is equal to the open loop voltage 8.5V. Next, RTh can be found by observing the circuit as a voltage divider as shown in Figure 29.
Figure 29: Lightbulb connected to the Thévenin equivalent of a battery
3"-0 y Q\ L][\cVd_V
L][\cVd_V v Q\ y 3"0 ,++&
,++& v Q\
Q\ y ,++& , w 3"-03"0 3"-0
3"0
z /,"4&
Question 4.5 Superposition
a. Evaluate two circuits: one with Ua = 0V and one with Ub = 0V.
Figure 30: Partial circuits with Ua = 0V (left-hand) and Ub = 0V (right hand)
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With a voltage source, “0V” means a short circuit:
T$RqI=S y V
>@ >v @ ?v >@
>v @
z ,"/-4
T$RrI=S y U
?@ ?v @ >v ?@
?v @
z -".3, T y T$RqI=Sv T$RrI=S z ."3,+
b.
Ok v Ol v Om y + Uw e
> vVw e
? v+ w e @ y + e y
U
>v V? ,>v ,?v ,@
z ."3,+
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Question 4.6 The Wheatstone bridge
a. The left branch with R1 and R2 is 5V (half of the power supply because R1 = R2). The right branch is a voltage divider with R3 and RT and is
O][\cy U @
Qv @ z /"4/.
So the difference is
LYZcw O][\c z 0 w /"4/. z 01"3 b. In balance when RT = 1kΩ:
Q y /"4. 5 ,+D5 H Q?D"=y ,&
y w-2"+ 5 ,&
/"4. 5 ,+D z -4,"20 z ,3"16
c. Half of the supply voltage: 5V, or (5.000V+4.943V)/2 = 4.9715V exactly d. By re-using the results of question a) and b):
yfRfQ y BC"EjiH=i
?F>">BhH?F>"DBh z84%6'
Question 4.7 Time dependent circuits a. The final potential is Ua = 5V b. W = R·C = 1kΩ·10µF = 10ms
c. Maximum potential difference over R, so I = Ua/R1 = 5mA d. dU/dt = 0, so I becomes 0mA
e. No change: Ua = UC = 5V, IC = 0mA
Question 5.1 Terminology a. Sensitivity
b. Saturation
c. Offset or systematic error
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g. One-point calibration h. Differential measurement
i. Common-mode rejection ratio (CMRR) j. Bandwidth
Question 5.2 The transfer function of a Pt100
a. Draw a line in Figure 11 or use Excel to fit the curve over the range of 0°C-100°C. With ΔR = 230Ω and ΔT = 600°C, we find:
y %
% y +".3	
< y100&
Note that the accuracy based on this graphical method of determining a and b will be poor. Based on the given raw data, a better fit could be made.
b. The sensitivity is a = 0.38Ω/°C
c. The offset is b = 100Ω for x in °C, which means the sensor is 100Ω for T = 0°C d.
Figure 31: Fit the line over the range 0-100°C, read the slope out over a larger range
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Question 5.3 A resistive sensor
a. The sensitivity can be read from the fitted line: dy/dx = -0.0656 MΩ/mm
= -6.56·10-7 Ω/m.
b. Over the range from 0 to 5.5 mm deflection, the non-linearity can be found from the maximum deviation which is about 0.3 of a vertical unit (MΩ). SO the non-linearity is 0.3 MΩ/5.5 mm · 100% =
Question 5.4 The candle clock a.
Figure 32: Burning of the candle with a fitted line
b. The value for t = 0 minutes is equal to b, this is where the line crosses the vertical axes. So b = 39.65 cm. The slope of the fitted line is parameter a and is -0.1972 cm/min: almost 2 mm per minute. In SI units this is a = -32.9 µm/sec.
c. The standard deviation is a good measure for the error. In this case it is the standard deviation of the measured instantaneous candle lengths with respect to the estimated length at that time using the fitted line. Take the sum of these error squared, divide by the number of points minus one, and take the square root.
d. Equal to the absolute value of the slope “a”: 32.9 µm/sec.
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Question 6.1 From sensor to knowledge
a. When signals x and y have the same common error signal, this error will be removed when the difference between x and y is taken. If x = x’+e and y = y’+e, then x-y is equal to x’-y’ and will not be sensitive to e. A practical example is the use of an inaccurate spirit level to measure whether a construction project is absolutely horizontal. If we use the spirit level twice, where the second time it is rotated 180 degrees, a systematic error in the spirit level is removed.
Other examples are in sensor systems where we want to cancel out temperature interference: take a second indifferent sensor that is not measuring the quantity of purpose, but only the cross interference of temperature.
b. In a stimulus-response measurement we measure how a system responds to a known actuator input. For example when tapping on a melon to hear whether there is a high water pressure inside, an acid-base titration, etc.
Question 6.2 Differential measurements a.
}!> y > v >v '
!? y ? v ?v '
!? w !> = y ? v ?v ' w > =v >v '
!? w !> = y ? v ?v ' w > =w >w '
!? w !> = y ? w > =v ?w >
Assuming that a1 ≈ a2, this is an output signal where is completely eliminated. The output y2-y1 is a signal proportional to x-x0.
b.
}!> y > v ' v >
!? y ?v ' v ?
!? w !> = y ~?v ' v ? w ~>v ' =v >
!? w !> y ? v ' v ?w > =w ' = w >
!? w !> y ? w > =v ? w > v ' w =
( ) d f h d ff l l
The last term (x-x0) is not removed from the differential sensor signal.
c. The left and right branch of a Wheatstone bridge are in fact the two inputs of a differential set-up. As an amplifier to measure a difference, the differential OpAmp amplifier of Appendix B of the book can be used, or a dedicated
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Question 6.3 A resistive measurement bridge for a train gauge
a. There are several options, but the resistors on one side (“View A” or “View B”) must be diagonally opposite. For bending downwards, R1 and R3 will become larger and R2 and R4 smaller. Therefore, in all valid solutions, there is a loop R1 Æ R2 Æ R3 Æ R4 Æ R1. Only then the differential signal is the result of bending in the positive or negative x direction.
Figure 33: Correct positioning of the strain-gauge resistors in the Wheatstone bridge
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b. The response will be an s-shaped curve. Around x = 0, the second derivative is zero, meaning this is a deflection point where the curve is linear locally. In practice, for small changes of the resistances of the gauges, a very high linearity is seen.
Figure 34: An s-curve is linear around x=0
c. With the full bridge of Figure 33, the temperature effect is minimize. If R1, R2, R3 and R4 are designed with the same nominal resistance (resistance at the same temperature), the temperature effect is common and equal. The bridge cancels these effects out.
d. The differential signal is -5mV..+5mV and the common signal is half the power supply Us/2 = 5V. This means that the required Common-Mode Rejection Ratio must be better than 10mV/5V = 0.002. The amplifier should amplify 1000x to map the differential signal onto the AD converter range.
e. Not needed: the strain gauges are very close to each other, and the wires to the voltmeter are not ideally not carrying a current.
Question 7.1 The Wheatstone bridge
a. The LDR must be 10kΩ to make the left branch equal to the right branch b. The theoretical linearized sensitivity of a single bridge is
%P
% z w S
/= y ,+
/ 5 ,+& z +"-0#&
assuming all resistors are 10kΩ. In this case the sensitivity is positive because the output increases with increasing resistance of the LDR. Since not all resistors are 10kΩ, we can also calculate the output for 10kΩ and for 10kΩ+1Ω and determine the difference:
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P$OI>=^g y +
P$OI>=^gG>g y ,+ -+&
,+& v -+& w ,+
-+&
,+& v ,& v -+& z +"--#&
The differential voltage between the right and left branch is about 0.22mV/Ω, which is slightly less sensitive than the first estimation.
Question 7.2 An LDR in a voltage divider
a. The transitions are 100 and 1000 lux, that means from 10 to 10000 lux is more than sufficient to see al events.
b. The range from 1.0 to 10000 lux is equivalent to 100MΩ down to 0.1kΩ, from 10 to 10000 lux is equivalent to 20MΩ down to 0.1kΩ.
c. We want to have hysteresis to avoid a transition state where the screen goes up and down within seconds. When the two decision levels are switched, the screen will be stuck to one of the sides.
d. The middle between 100 and 1000 lux is roughly at 2kΩ: that is the correct resistor.
e. The light level of 100 lux gives about 3kΩ, the level of 1000 lux will be 0.7kΩ.
With the series resistor of 2kΩ this gives transition levels of 3/5 = 0.4 times the power supply and 0.7/2.7 = 0.3 times the power supply. With a Schmitt trigger type of comparator, these two levels can be realized.
Question 7.3 Sensor characteristics of an NTC
a. The voltage divider is probably more non-linear than the sensor over that small range.
b. To sense Utemp without loading the voltage divider resistively.
c. Two degrees is 0.03·0.03 = 0.0009, so the resistance becomes 970Ω·(1-0.0009)
= 969.1Ω. It gets lower with increasing T because it has a negative temperature coefficient. Utemp (and so Uout) will go up from 5V·1000Ω /1970Ω ≈ 2.5381V to 5V·1000Ω /1969.1Ω ≈ 2.5392V. This is a difference of only 1.13mV on top of an offset of about 2.5V.
d. Make a Wheatstone bridge with more resistors of 1kΩ to reduce the
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e. As long as the temperature dependency of R2 is not equal to that of the sensor R1, the voltage divider ill still give a voltage change with changing temperature, only smaller. The lowered change of voltage can be registered with calibration, and if desired removed by increasing the amplification factor of a differential amplifier.
Question 7.4 Schmitt trigger
a. The OpAmp is a comparator where the decision level Uc is dependent on the history. For a signal going from low to high, Uc is higher than a signal going from high to low.
b. If Uout is initially low (0V), the circuit is effectively as in Figure 35a, when Uout is high (5V), the circuit is effectively as in Figure 35b.
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Figure 35: In a Schmitt trigger, one of the resistors determining the decision level of the comparator is effectively moved in place
For Uout = 0V:
W$RsutI=S y
?@ ?v @ >v ?@
?v @
WW y +"4;
For Uout = 5V:
W$RsutIBS y ? >@
> v @v ?WW y ,"2
h h k d
These are two equations with three unknown resistors R1, R2 and R3. Take for example R1
= 10kΩ, and we find:
?@
?v @ ,+& v ?@
?v @
0 y +"4;
?
,+& 0 y ,"2
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Question 8.1 AD Conversion a. The resolution is defined as:
y
-Mw , z
-M So:
| ?
y
>=
>=- y 4"42
Which means that N = 10 to make the outcome larger than 9.97. And indeed, with 10 bits over a range of 5V, we find a resolution of 4.89mV which is better than 5mV.
b. The Nyquist theorem says a sampling rate is needed which is at least higher than two times the highest frequency in the signal. So, this is 2kHz. Only then, the original analog signal can be resolved from the digital samples.
Question 8.2 Measurement of light intensity a. 5 lux Æ 74kΩ, 50 lux Æ 15kΩ,
b. No, when the transfer curve looks linear on a log-log scale, it must be strongly exponential on a linear scale. There is no single slope for the given range.
c. A voltage divider, for example as given in Figure 36.
Figure 36: A voltage divider can be used to convert the value of the resistive temperature sensor into a voltage
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d. The value for R1 can be chosen in the middle of the operational range of the sensor: 47kΩ for example. With Ua = 5V, the output voltage will be between 1.92V and 3.75V for the range from 5 to 50 lux.
e. An AD converter with a range of 5V and 12 bits has a resolution of 5V/(212-1)
| 1.22mV, so this is not enough for detecting 1mV variations.
f. The day-night variation of the light intensity is probably not a sine wave. We are interested in the exact transition from dark to light or the other way around.
That signal if faster than a period of 24 hours. Normally, there is no big issue with sampling every minute.
Question 8.3 A temperature measurement
a. For measurement around room temperature the NTC has the largest variation.
In numbers: from 10 to 25°C the Pt100 changes from 107Ω to 119Ω (11%) while the NTC changes from 590Ω to 350Ω (28%). This is much easier to work with in a voltage divider. However, when a high linearity is needed, the Pt100 is better.
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b. See Figure 36 for the circuit. When the NTC is used, R1 becomes 400 Ω because that is the sensor value at the room temperature of 20°C.
c. If the non-linearity over the needed range is more than accepted, a
microcontroller can be used to calculate the temperature from the instantaneous sensor signal using a calibration curve.
Question 9.1 Bus interfaces
a. (1) Digital signals are less susceptible to noise, (2) Standardization, (3) all sensors can be placed on the same bus, (4) meta data for identification and calibration can be sent over the same bus, (5) reduction of the number of wires.
b. Standardization, low noise, simplified maintenance, error checking, internet over the same bus.
Assignment 1: The position sensor
Question 10.1 First measurement – Does it work?
a. The experience is that with the given materials and sizes, the resistances are in the tens of MΩ’s.
Question 10.2 Calibration curve a. An example is given in Table 3.
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d [cm] R [MΩ]
10 12,07
9 11,53
8 11,15
7 10,52
6 8,92
5 7,24
4 6,02
3 0,524
2 0,274
1 0,277
Table 3: Resistance measurements using a Voltcraft VC170 multimeter in auto
b. The values of Table 3 are plotted in Figure 37. The linear fit and the error bars will be explained later.
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Question 10.3 Analyzing
a. The dependency is clearly linear from d = 4cm.
b. Yes, we are measuring a homogeneous strip (U is constant) with a fixed with and a changing length L.
c. The resistance is somewhat high. We can make it lower by adding more carbon or by using a wider strip.
d. The values below d = 4cm cannot be trusted: we expect a linear behavior and these are obviously not on the curve.
e. The sensitivity is determined in Figure 37, but only starting from the fourth point. The first three points, for distances smaller than 4cm, are seen as outliers.
The fitted line gives a slope of 1.0343MΩ/cm. So the sensitivity is 1.0MΩ/cm, and there is an offset of 2.4MΩ.
f. The slope “a” is the linearized sensitivity of the sensor. It is the first
derivative of the resistance R on the distance d (or length L of the resistor).
The resistor equation is y *
so
y7 7 y
*
g. The offset is the resistance for L = 0cm. This can be the resistance of the cables plus the contact resistance of the paperclip to the paper.
Question 10.4 Error estimation
a. To obtain better insights, the experiment is repeated many more times. This is summarized in Table 4.
b. Also in Table 4.
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Second measurement: error analysis
d = 2cm d = 6cm d = 10cm d = 2cm d = 6cm d = 10cm
R [MΩ] R [MΩ] R [MΩ] R [%] R [%] R [%]
0,274 8,92 12,07
0,159 18,91 37,82 100,0% 100,0% 100,0%
0,188 17,99 37,91 118,2% 95,1% 100,2%
0,235 18,75 38,91 147,8% 99,2% 102,9%
0,204 24,45 40,23 128,3% 129,3% 106,4%
0,287 24,24 40,01 180,5% 128,2% 105,8%
0,410 19,38 34,47
Mean [MΩ] Mean [MΩ] Mean [MΩ]
0,215 20,868 38,976
St.Dev [MΩ] St.Dev [MΩ] St.Dev [MΩ]
0,049 3,194 1,131
45,6% 30,6% 5,8%
Table 4: The position experiment repeated
c. The data of Table 4 is plotted in Figure 37. With this data we can determine standard deviations in order to plot error bars in Figure 37. Now we are allowed to say that the fitted linear approximation fits definitely within the error range of the measurements, again, only for distances above 4cm.
The repeated experiments are also plotted in Figure 38. We can see that over time, the sensor value is drifting.
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Figure 38: The sensor value is drifting when the experiments are repeated
Question 10.5 Repeat calibration experiment
a. The characterization (calibration) of the sensor as was done in Figure 37, is repeated after all experiments of Table 4 are done. The result is the calibration curve Figure 39.
b. See Figure 39.
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Question 10.6 Analysis
a. There are obviously two types of drift. First of all, the sensitivity (slope “a”) was 1 MΩ/cm, and after some use it became 3.7 MΩ/cm. An explanation can be that the carbon becomes crunched and mixed. If that is the case, the drift will stop because this process will stabilize. Besides drift in the sensitivity, there is also drift in the offset: this was 0.7MΩ, and became 3.7MΩ. An explanation of a change in the offset can be a change in the contact resistance.
b. Answered under a): it is probably the grain size of the carbon that is changing.
c. In fact, the measurement error of this experiment is much smaller than the drift.
The drift is that large that we must conclude the sensor cannot be calibrated in a useful number of x MΩ/cm.