Introducing to Electronics
Martello I.I Pompe B. s093162 s097825
TU/e
Index Questions chapter 2 chapter 3 exercises 3 chapter 4 chapter 5 chapter 6 chapter 7 chapter 8 chapter 9 chapter 10
3 5 7 8 11 15 18 20 21 23
Practical Assignments 1 (chapter 3) 2 (chapter 4) 3 (chapter 6) 5 (chapter 8) 6 (chapter 9) 7 (chapter 10)
24 25 27 29 30 31
Building Blocks 1 (chapter 3) 4 (chapter 11)
32 33
Reflections Ivan Martello ` Brenda Pompe
35 36
P a g e | 3 Question 2.1 1.5V voltage supplied by 1 battery 300mA.Hr current over time supplied by 1 battery 50mA current used by MP3 player on average MP3 player uses two batteries, which can be connected in series or parallel. When two batteries are connected in series: I = I1 = I2 therefore Itotal = 300mA.Hr U = U1 + U2 therefore Utotal = 1.5V + 1.5V = 3V The MP3 player can play for 6 hours. When two batteries are connected in parallel: I = I1 + I2 therefore Itotal = 300mA.Hr + 300mA.Hr = 600 mA.Hr U = U1 = U2 therefore Utotal = 1.5V The MP3 player can play for 12 hours.
Question 2.2 1. Graph b depicts the ideal power source. The voltage is independent of the magnitude of the current. 2. Graph c depicts the non-ideal realistic power source. The voltage is dependent of the magnitude of the current.
Question 2.3 1. Graph c depicts the ideal power source. The voltage is independent of the duration of the current. 2. Graph b depicts the non-ideal realistic power source. The voltage is dependent of the duration of the current.
Question 2.4 50V 2A 50W 35%
voltage supplied by source maximum current supplied by source maximum power output amplifier efficiency amplifier
The amplifier requires 143W
The power supply is insufficient, because the amount of power required is larger than the maximum power that the source can supply.
Question 2.5 50Hz
electricity grid frequency
Tgrid = 1 / f = 1 / 50 The bulb goes on and off twice per Tgrid, therefore its frequency is twice that of the grid. f bulb = 2 · 50 = 100Hz
P a g e | 5 Question 3.1
Question 3.2 10V 1kΩ
voltage applied resistance
1. The current is 0.01A. 2. The power dissipation is 0.1W.
Question 3.3
Question 3.4 As a resistor dissipates power, it heats up. Larger surfaces allow for more heat dissipation, therefore large resistors can handle larger powers.
Question 3.5
0.225W < 15W It cannot dissipate more tan 0.225W, which is less than the maximum amount of power supplied by the source.
Question 3.6 To get a 500Ω, ±2% tolerance and .5W resistance, two 1000Ω, ±1% tolerance and .25W resistances connected in parallel are required.
Both resistors have a tolerance of ±1%, which is traduced as 1010Ω and into a total resistance of Rre =505Ω.
Therefore the tolerance of Rre is 1% that lies within 2%, requiered in the conditions of the system. And both resistors have a power dissipation of 1/4W, together they have a power dissipation of 1/2W.
Question 3.7
P a g e | 7 Exersise 3.1 For a voltage applied to resistors in parallel the following formula stands for: Substituting I for we get the following formula: By applying this in the formula
.
Exersise 3.2 When I > 0A, the output voltage is always lower than 4V because;
If I2 is bigger than 0: I1 = I2 + I3 I3 = I1 - I2
Vout will always be lower than 4V if I2 is different from 0.
Question 4.1 1. C = 100pF V = 12V 2. C = 100pF V = 12V 3.
Question 4.2
1. Considering the given formula, if f increases, the impendance will decrease. 2. Considering the given formula, if f decreases, the impendance will increase.
Question 4.3 Derive:
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P a g e | 9 Question 4.4
First calculate Cre1 which replaces C1 , C2 and C3 that are connected in series.
Second calculate Cre2 which replaces Cre1 and C4 that are connected in parallel.
Finally calculate Ctotal which replaces Cre2 and C5 that are connected in series.
Question 4.5
Question 4.6
R=Zc
Question 4.7
R = 15Ω C = 100nF Vin = 10V
The potential difference across a charging capacitor is defined by the exponential
, which makes a linear outcome impossible.Question 5.1
P a g e | 11 Question 5.1 I=0 When the current is cut off, it drops in a small amount of time. The current stays the same but the time is very small and this generates a big potential difference that could damage the circuit.
Question 5.2
Question 5.3 1. -When the frequency increases also the inductance impedence (Z) increases within the same factor. Because it is a linear function.
2. –When the frequency decreases also the impedence (Z) decreases within the same factor, because it is a linear function.
Question 5.4
This relation shows that the replacement inductance should be smaller than the smallest inductor. Therefore answer 2, < 68µH, is the correct answer.
Question 5.5
Question 5.6 Is a high-pass filter, because when ω tends to Vout equals Vin . When ω goes to 0 Vout becomes very small or near to 0. Therefore when ω is high no voltage is blocked characteristic of a high-pass filter.
P a g e | 13
Question 5.7 1.– By changing the position we obtain a Low pass filter because when ω increases Vout decreases. 2.-
Question 5.8
1. R= 1kΩ L=1mH =100KHz
2. R= 1kΩ L=1mH =0Hz
3.
R= 1kΩ L=1mH = Hz
Question 5.9 Pprimary=Psecondary
P a g e | 15 Question 6.1 I1 =I2
Question 6.2
Question 6.3
Question 6.4 1.-
2.-
3.-
4.-
Question 6.5
P a g e | 17 Loop1
Loop2
Both results are equal
Question 7.1
1.-
2.-
3.-
4.-
Thevenin and Norton resistances are equal. Thevenin voltage is equal Norton current times Norton resistance( ). The Norton current is equal to Thevenin voltage divide by Thevenin resistance (
 Â
).
P a g e | 19 Question 7.2
Question 8.1 Therefore the voltage will create a current after the diode starts working as a semiconductor equal to the knee voltage (0.6V for silicon diodes).
The negative part of the sine wave is filtered
Question 8.2
A good R for the circuit would be 466.6Ω and based on the E12 series the nearest value is 470Ω.
P a g e | 21 Question 9.1
The Vout from the subcirciut should be 0V, because the NPN type of transistor should have a negative potential difference in order to work. Therefore in order to switch on the lamp the Voltage output should be 0V.
Question 9.2
The potential difference is 100V and in case no bleeding diodes where available, the only transistor from the above mentioned that can stand this voltage, without getting damaged, is the 2N3439.
Question 9.3 (Base emitter potential difference) For Darlington transistors are used NPN type, this means that between the base and the emitter is a positive value for the potential difference. For one single transistor the potential difference is 0.6V. In the Darlington transistor we multiply 0.6V per 0.6V (the potential difference of both transistor separately) and we obtain the potential difference between the base and the emitter, equal to 0.36V.
Question 9.4
1. -
The only transistor from the above mentioned that fulfills the requirements when switched on saturation is the BC618.
2. -
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P a g e | 23 Question 10.1
Question 10.2 When using large resistors the opamp works better, this happens because operational amplifiers work better with low currents. The gain of 11 is equal for low and high resistors.
Question 10.3 The input voltage (Vin) is limited by the delivered Voltage of the Vsupply. Therefore the input voltage cannot be higher.
Practical Assignment 1 (Chapter 3) 2.
Measuring voltage in the circuit built for practical assignment 1.
3. Rretotal I1 I2 I3 I4 Vout
Calculated 7.52kΩ 1.32mA 0.26mA 0.53mA 0.48mA 5.93V
Measured 7.51kΩ 1.32mA 0.26mA 0.56mA 0.48mA 6.01V
4. A-C : 0kΩ-10.83kΩ B-C : 10.83-kΩ-0kΩ 6. A-C : 7.14kΩ B-C : 3.95kΩ 7. New resistances chosen based on the potmeter values.: R4 = 6.8kΩ & R5 = 3.9kΩ New Vout = 5.27V
P a g e | 25 Practical Assignment 2 (Chapter 4)
2. f 100Hz 1kHz 100kHz
Vout 1.1V 1.1V 1.1V
3.
For 100Hz:
For 1kHz:
For 100Hz:
100Hz 1kHz 100kHz
4. For 100Hz:
For 1kHz:
Calculated 0.13V 1.06V 2V
Measured 0.10V 1.10V 2V
For 100Hz:
100Hz 1kHz 100kHz
Calculated 2V 2V 25mV
4. 5. 6. A low-pass filter: resistance first, capacitor second.
Oscilloscope reading during practical assignment 2.
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Measured 2V 1.90V 25mV
P a g e | 27 Practical Assignment 3 (Chapter 6)
1.- Loop1
Loop2
2.- Voltage drops
Circuit built for practical assignment 3.
V VR1 VR2 VR3 VR4
Calculation 7.92V 1.88V 0.88V 2.24V
Measurement 8.04V 1.91V 0.89V 2.28V
Conclusion: In small voltages there was not so much difference but in higher voltages the difference between the values calculated and the measurements changes. We think that partly the voltage is lost in the connections and wires and also inside the circuit in the node b at R2 where the two currents get together. We think that at this point current increases and some voltage is lost because if this increment.
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P a g e | 29 Practical Assignment 5 (Chapter 8) V 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4
mA 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.2 1.4 4.9 10 14.9 17.3 21.3
The current was 0 until the diode reached the knee voltage, then while the voltage was increased by 0.1 steps the current increased more or less quadratic as can be seen in the graphic. The knee voltage was detected around 1.7 and 1.8 volts. So the relation between voltage and current should when voltage increases current increases quadratic.
Practical Assignment 6 (Chapter 9)
Resistances 100kΩ 10kΩ 4.7kΩ 2.2kΩ 1kΩ 470Ω
Ic 0.4mA 4.5mA 9.1mA 16.4mA 28.8mA 44.5mA
Vce 5.03V 4.98V 4.89V 4.41V 3.3V 1.24V
Vbe 0.59V 0.68V 0.71V 0.74V 0.71V 0.81V
BC550 transistor (data)
Circuit built for practical assignment 6. Observations: Large resistances create a large voltage drop that seems not to permit the transistor reached the saturation level when the saturation level was reached, the light turned on and the value of Vce decreased. The resistance that would permit the lamplight was the 470Ω and also the 1kΩ but the light emitted was very low. So we figured out that with large resistances the saturation level is not reached therefore the lamp won’t be switched on until saturation is reached.
P a g e | 31 Practical Assignment 7 (Chapter 10)
Circuit built for practical assignment 7. 1. -LM324
2.
After verifying the values calculated was observed that during the measurements when Vin decreased to 4.7V, V0ut would change to 5.12V. This values were almost the same as the one’s calculated.
Building Block 1 (Chapter 4)
Vin=12V 0V ≤ Vout ≥ (Vin-2V) 0.5 mA ≤ I ≥ 5 mA Rre(max)= Vin / Imin Rre(max)= 12/0.5*10-3 Rre(max)=24kaΩ Rre(min)=Vin / Imax Rre(min)=12/5*10-3 Rre(min)= 2.4kΩ 2.4kΩ ≤ Rre1 +R3 ≥ 24kΩ R3=(pot)0kΩ -10.64kΩ Rre1=24kΩ - 10.64kΩ Rre1=13.36kΩ The nearest values in the E 12 series are 8.2kΩ and 5.6kΩ connected in series in order to have: Rre1= R1+R2=8.2kΩ+5.6kΩ Rre1= 13.8kΩ The building block is formed by two resistors, 8.2kΩ and 5.6kΩ connected in series, and a 10kΩ potmeter. The potmeter permites to adjust the Vout smoothly. This way the minimun I is between the limits Imin= Vin/Rre1 Imin=12V/13.8kΩ Imin=0.86 mA
P a g e | 33 Building Block 4 (Chapter 11)
One DC source supplies 15V to the system. R1 Matches the value of R2 when the switch-off temperature is reached. R1 = R2(40°C) = 7.5kΩ R2 NTC, when the temperature rises, the resistance drops causing the voltage at the side of the comparator to drop. Needs to be far away from R4 , otherwise it will measure the heating elements temperature instead of the ‘room’ temperature. R2max = 10kΩ R2(40°C) = 7.5kΩ R3 Potmeter, functions as reference setting. Should be the value of R1 + R2 when the NTC measures 40 degrees Celcius. R3 = 2R2 = 15kΩ R4 Power resistor that heats up the ‘room’. SBCHE4 P = 4W R4 = 100Ω P = UI = 4 = 15V• I I = 4/15 = 270mA R5 Makes sure the right amount of current flows into the transistor. R5 = U/I = 15V/0.36mA = 42kΩ R6 Functions as hysteresis. R2 /R6 defines how much the comparators sensitivity is reduced. This needs to be a strong resistor, or the margin will be too big. 7.5k/300k = 0.025 It expands the margin for switch-off with +0.025V and -0.025V. Comparator LM324 Transistor BD679 hFE = 750 270mA / 750 = 0.36mA
Heating the system up quickly is possible, because R2 will not slow down the process until the ‘room’ temperature is 40°C. Cooling down cannot be done quickly, because the heating element does not immediately lose all its heat, nor does the ‘room’.
Actual resistance values used (if different from the calculation): R1 4.7kΩ + 2.7 kΩ = 7.4kΩ R5 3842kΩ + 2.742kΩ = 40.7kΩ
P a g e | 35 Ivan Martello s093162
Reflections on the assignment DG220 The word electronics before I started this assignment on meant wires, electricity and power. Just by getting familiar with the terms and components, one can start “playing” with the electronics. Before I started practicing the first assignments I had some difficulties to understand completely the theory in the reader of the course. But after the first time I started practicing in the electronics atelier you realize the simplicity of electronics. I see now the world of electronics as a game where some rules have to be taken in consideration to make things work, but also you should be curious and explore. By practicing the theory and physically test it most of the information is processed and the after succeeding on making work something I always start thinking where in my daily life could find that application and how could I use it for a different purpose. One thing I did not achieve entirely during this assignment was the possibility of analyzing electro domestics or other objects with electronics inside to see how it was applied. The length of the course is very short but it is a hard working assignment. I think that with a little more time would be possible to gain the ability of analyzing complex electronics in daily use objects but I guess that is going to be my next step during the weeks to come. I am very pleased about the content of the assignment and also motivated to go further with my studies on electronics. I would only suggest some kind of blog or online library was created where the students that accomplish this assignment could join and then use this platform to share new discoveries or inventions that students after or during the course would like to show. I think this way students could get more motivated and inspired to create new applications or experimentations. On this assignment I had the opportunity to know a little bit of the electronics’ world inside the daily use objects. This was my first formal approach to the electronics and the structure of the classes where motivating. The amount of work for me is difficult to compare it with other assignments, since I am an exchange student, I do not have any references to judge the amount of work, but certainly is a hard working assignment. I expected from this assignment an overview of the basic electronics behind simple circuits and I gained more than this. I have learned to analyze simple and more complex circuits but most of all I have realized that great amount of the content on electronic circuits is acquirable on the Internet. By applying the basic skills developed in these weeks one can investigate, learn and apply more knowledge on electronics. From the reader most of the information could be understood but the possibility of practicing the technical part ensured the acquisition of knowledge.
Brenda Pompe S097825
DG220 Introducing Electronics reflection When I signed up for this assignment, I thought I’d be taken by the hand, shown how to build simple circuits, get to build those simple circuits myself and learn practical skills like proper soldering. Although it was quite shocking to find that it was more like ‘really advanced physics’, I am happy I got to do this assignment. I learned to calculate values in more than just the most basic circuits you get to see at secondary school. This will come in handy if I need to build a circuit for another assignment or project. It was interesting to see what different tools can be used in building and testing a circuit. Although I’m still no expert and I still can’t remember every function of every component we discussed, I do know where to look if I want to know what I need and how I should use it. I had great difficulty comprehending the theory within the specified time, so I was very grateful this was a team assignment. This encouraged sharing knowledge and made it easier for me to ask for help. Before this assignment I might have felt intimidated at the prospect of having to build a difficult circuit, now I know I can ask for help and figure it out together. Before I never thought about the insides of the electronic devices I see every day much. Now, when I look at my tv, I can’t help but wonder what’s inside of it. With all the functions it can perform, how complex must the wiring be? How much thought went into building and calculating the circuit? What steps were taken? Even though I probably won’t understand everything, I do want to see this process from beginning to end. This assignment was a bit overwhelming. All the new information still hasn’t set in my mind quite yet, so it’s a bit hard for me to write a coherent reflection at the time. I’m very pleased with what I’ve learned and experienced through this assignment, it would have been nice to have more time to let all the theory sink in a bit before moving on to the next subject.