Report Introducing electronics

INTRODUCING ELECTRONICS DG220

STEPHANIE SARIS CHARLOTTE VAN DER SOMMEN JOERI HAHN CHAPTER 1 tm 10 ASSIGNMENTS 1 tm 7 BUILDING BLOCKS 1,3,4 REFLECTIONS INDUSTRIAL DESIGN

B1.1 and B1.2

05-04-2011

TABLE OF CONTENTS

CHAPTER 2 3 4 5 6 7 8 9 10 ASSIGNMENTS 1 2 3 4 5 6 7 BUILDING BLOCKS 1 3 4 REFLECTIONS Stephanie Saris Charlotte van der Sommen Joerie Hahn

Blz 3 5 7 9 11 14 15 16 18 21 24 28 30 31 33 34 37 39 41 45 46 47

CHAPTERS

2 tm 10

CHAPTER 2 Question 2.1 Given:

text battery 1.5 V / 300 mA.Hr IBattery = 50 mA

Solution: Parallel:

The voltage of the batteries remains the same: V= 1.5 V The current equals the sum of the currenst of the individual batteries: I1 + I2 = Itot 300 + 300 = 600 mA.Hr The current consumed by MP3 player: 50 mA mA.Hr / mA = Hr 600 / 50 = 12 Hr

Series:

The current remains the same I = 300 mA. Hr

mA.Hr / mA = Hr 300 / 50 = 6 Hr

The voltage equals the sum of the voltages of the individual batteries: V1 + V2 = Vtot 1,5 + 1,5 = 3 V

Conclusion: The MP3 player lasts 12 hours when the batteries are connected parallel and it lasts 6 hours when the batteries are connected in series.

Question 2.2 1.

Graph B depicts an ideal power source, because the voltage stays the same.

2.

Graph C depicts a non-ideal realistic power source, because the voltage drops after a certain amount of current is reached.

Question 2.3 1.

2.

Graph C depicts an ideal power source, because the voltage stays constant over time Graph B depicts a non-ideal realistic power source, because the voltage drops throughout time

Question 2.4 Given: for the DC-power supply Vout = 50 V

Imax = 2 A

For the audio-amplifier Vmax = 50 V

Pout = 50 W

Efficiency = 35% Solution: The power that the amplifier needs can be calculated:

Efficiency = Pout / Psource * 100% 35% = 50 / Psource * 100% Psource = 50 / (35% / 100 % ) = 142,86 W

The maximal power delivered by the DC-power supply can be calculated

Pmax = Vout * Imax = 50 * 2 = 100 W

Conclusion: The supply is not sufficient for the amplifier, because the amplifier needs a power supply that delivers 142,86 W. The DC-power supply delivers only 100 W.

Question 2.5 The frequency of the electricity grid in the Netherlands is 50 Hr. A light bulb will go on and of every cycle. So that means that it will flicker with a frequency of 100 Hz.

CHAPTER 3 Question 3.1

Question 3.2 1.

V = 10 V R = 1 k立 = 1000 立

I=V/R = 10 / 1000 = 0,01 A

2.

P=V*I = 10 * 0,01 = 0,1 W

Question 3.3 P=V∙I R = V/I V=R∙I P=R∙I∙I P = R ∙ I2

Question 3.4 The larger ones can handle more power. The larger the resistor the higher the power dissipation rating is. A bigger resistor can better get rid of heat.

Question 3.5 resistor = 1000 Ω Vsource = 15 V Imax = 1 A so P = 152/1000 = 0.225 W The source delivers 15 . 1 = 15 Watt 15 is bigger than 0,225

Question 3.6

You have to connect two resistors of 1 kΩ to get a resistor of 500 Ω.

Because parallel connected: Rre = 1/(1/R1+1/R2). resistance: 1/(1/1000+1/1000)= 500 Ω The tolerance will be 2 ∙1%= 2% and the maximum power dissipated will be 2.

Question 3.7 The two resistors that are closest to the the resistor needed for the voltage divider are: 4,7 kΩ for the 5 kΩ resistor and 3,9 kΩ for the 4 kΩ resistor.

CHAPTER 4 Question 4.1 1.

C = 100 pF = 100 * 10^(-12) V = 12 V

Q=C*V = 100 * 10^(-12) * 12 = 1,2 *10^(-9) = 1,2 nC

2.

E = ½CV² = ½ * 100*10^(-12) * 12 = 7,2*10^(-9) = 7,2 nJ

3.

Q=C*V = 100*10^(-12) * 90 = 9*10^(-9) C

dT = dQ/ I = 9*10^(-9) / 50*10^(-3) = 1,8*10^(-7) s = 18 µs

Question 4.2 1. If the frequency increases, the value below the bisector increases and the impedance will decrease.

This is because of the formula Zc = 1 / 2πfC 2. If the frequency decreases, the value below the bisector decreases and the impedance will increase.

This is because of the formula Zc = 1 / 2πfC

Question 4.3

Qre = Cre ∙ V Cre = Qre / V Qre = Q1 + Q2 + Q3 Cre = Q1 /V1 + Q2 / V2 + Q3 / V3 The voltage over all the resistors is the same. This means that Q = C1/V1 + C2/V2+C3/V3 Cre = C1 + C2 + C3 V is the same

Question 4.4 Cre1 = 1 / (1/C1 + 1/C2 + 1/C3 ) = 1 / (1/1*10^(-6) + 1/1*10^(-6) +1/1*10^(-6) ) = 3,33*10^(-7) F Cre2 = Cre1 + C4 = 3,33*10^(-7) + 1*10^(-6) = 1,33*10^(-6) F Cre3 = 1 / (1/Cre2 + 1/C5)) = 1 / (1/ 1,33*10^(-6) + 1/ 1*10^(-6)) = 5,7*10^(-7) = 0,57 µF

Question 4.5 Low-pass: Vout = Zc / Zre ∙ Vin(w) = (1/ wC) / √( R2 + (1/ w2C2)) High-pass: Vout = R / Zre ∙ Vin(w) = R / √( R2 + (1/ w2C2))

Question 4.6 Low-pass: Vout = Zc/Zre . Vin => Vout/Vin = Zc/ √( R^1 + Zc^2) = 1/wc/√( R^2 + (1/ w^2C^2)) = 1/√( 1^2 + 1^2) = 1√2 = 0,707 High-pass: Vout = R/Zre . Vin => Vout/Vin = R /√(R^1 + Z1^2) = R/√( R^2 + 1/w^2C^2) = R/√( R^2 + R^2) = 1/√1^2 + 1^2 = 1/√2 = 0,707

Question 4.7 1

Vmax = 10 Volt R = 15kΩ = 15 ∙ 103 Ω C = 100 nF = 100 ∙ 10-9 F Delay = 0,7 ∙ 15 ∙ 103 ∙ 100 ∙ 10-9 = 0,00105 s

2 The voltage over R is not constant, because the charging current for the capacitor is not constant. dV/dt is not constant and so V does not change in a nice linear way.

CHAPTER 5 Question 5.1 The inductor creates a high potential difference to keep the current flowing. Both the current and the flux depends on time. If there is a large flux in little time there will be a large voltage. A large voltage can damage the circuit because some components of the circuit are not made for a high voltage.

V = dflux/dt = L dI/dt

Question 5.2 E = 1/2L . I^2 E = 0,5 ∙ L ∙ I2 0,5 ∙ 1∙10-3 ∙ 0,12 = 5,0∙10-6 j

Question 5.3 formula for impedance: Zl=2 ∙ π ∙ f ∙ L If the frequency (f) increase, the impedance will also increase. When the frequency decrease, the impedance will also decrease.

Question 5.4 Lre < 68 µH is correct because by putting inductors in a parallel circuit, the replacement inductor will always be smaller than the lowest inductor 1/Lre = 1/L1 + 1/L2 + 1/L3, also if the inductance of an inductor is very low, the replacement inductor will be very small and it is not the only inductor.

Question 5.5 Lre = L1+1/((1/L3+L4)+(1/L2))+L5 Lre= 2∙10^-3 + 1/((1/(0,25∙10^-3 + 0,75∙10^-3)) + (1/3∙10-3)) + 1∙10^-3 Lre = 3,75∙10^-3

Question 5.6 The filter is a highpass filter, because when the frequency is high the impedance will also be high, so a high voltage will go to Vout. When the frequency is low , the impedance will also be low which means that the voltage will run through the inductor and their will be a small voltage over Vout .

Question 5.7 1.

2.

By interchanging the position of L and R in Figure 5.6 you will get a Low-Pass filter.

Vout = R/Zre ∙ Vin = R/(√(w2 ∙ L2 + R2) ∙ Vin

Question 5.8 I = Vin / Zre 1.

2.

3.

Zre1=√(R^2 + w^2 + L^2) = √(1000^2 + (100000∙2∙π) + (1∙103)) = 1,181kΩ I = 10/1181 = 8,467 mA Zre2=√(R^2 + w^2 + L^2) = √(1000^2 + (0∙2∙π) + (1∙103)) = 1,0 kΩ I= 10/1000 = 10,0 mA Zre3=√(R^2 + w^2 + L^2) = √(1000^2 + (∞∙2∙π) + (1∙103)) = ∞ kΩ I = 10/∞ ≈ 0,0 mA

you forgot to take thesquare out of the L=1mH in the equation for subquestion 3. This means that thecurrent at 100kHz should be 8.467mA instead of your 7.8mA

Question 5.9 Ppri = Upri . Ipri = Psec = Usec . Isec R is 0 in the equation, so there is no heat at all and no energy transferred into heat.

Vprimary is not equal toVsecondary.

CHAPTER 6 Question 6.1 Va = 12 V Vc = 0 V I=V/R I1=I2

}

Question 6.2

(Va –Vb) / R1 = (Vb – Vc) / R2 (12 – Vb ) / 400 = (Vb – 0) / 290 290 (12 – Vb) = 400 (Vb – 0) 3480 – 290 Vb = 400 Vb 690 Vb = 3480

Question 6.3 Loop 1: a->b->c->a R1 * I1 + R2 * (11-I2) – Vb = 0 Loop 2: b->d->e->b R3 * I2 + R4 * I2 + R2 * (I1 – I2) =0 Loop 3: a->b->d->e->a R1 * I1 + R3 * I2 + R4 * I2 – Vb =0

Question 6.4 1.

Loop 1: abca R1 * I1 + R2 * (11-I2) – Vb= 0

Loop 2: bdeb R3 * I2 + R4 * I2 + R2 * (I1 – I2) = 0

Loop 3: abdea R1 * I1 + R3 * I2 + R4 * I2 – Vb = 0

2.

V1 + V2 = Vb 3,7 + V2 = 9 V2 = 5,3 V

3.

V3 + V4 – V2 = 0 V3 + 1,3 – 5,3 = 0 V3 = 4 V

4.

Loop 1: V1 + V2 – Vb = 0

3,7 + 5,3 – 9 = 0 } right

Loop 2: V3 + V4 – V2 = 0

4 + 1,3 – 5,3 = 0 } right

Loop 3: V1 + V3 + V4 - Vb= 0

3,7 + 4 + 1,3 – 9 = 0 } right

Question 6.5 1. Va = 10 V Vc = 5 V Vd = 0 Iab + Icb = Ibd (Va – Vb) / R1 + (Vc – Vb) / R3 = (Vb – Vd) / R2 (10 – Vb) / 2000 + (5 – Vb) / 2000 = (Vb – 0) / 500 10 – Vb + 5 – Vb = 4Vb 6 Vb = 15 Vb = 2,5 V 2. Loop 1: a->b->c->a

R1 * Iab + R2 * (Iab + Icb) – Vb1 = 0

2000 * Iab + 500 * (Iab + Icb) – 10 = 0

250 * Iab + 50 * Icb = 1

Loop 2: c->b->d->c R3 * Icb + R2 * (Iab + Icb) – Vb2 = 0 2000 * Icb + 500 * (Iab + Icb) – 5 = 0 2500 * Icb + 500 * Iab = 5 Loop 1: 250 * Iab + 50 * Icb = 1

250 * Iab = 1 – 50 * Icb

Iab = (1 – 50 * I cd) / 250 } substitute in Loop 2 Loop 2: 500 * Iab + 2500 * Icb = 5

500 * (1 – 50 * I cd) / 250 + 2500 * Icb = 5

2 * (1 – 50 * Icd) + 2500 * Icd = 5

2400 * Icd = 3

Icd = 0,00125

} use Icd to calculate Iab 250 * Iab + 50 * 0,00125 = 1 250 * Iab = 0,09375 Iab = 0,00375 V1 = Iab * R1 = 0,00375 * 2000 = 7,5 V V1 equals the voltage drop over resistor 1. This means that when Vbattery is 10 V 2,5 V remains on node b.

3.

The KCL solution and the KVL solutions are the same.

CHAPTER 7

Question 7.1

Voc = R2/(R1+R2) ∙ Vn Isc = Vin/R1 RT = RN = Voc/Isc = (R2/(R1+R2) ∙ Vin)/(Vin/R1) RT = (R1/Vin) ∙ (R2/R1+R2) ∙ Vin RT = (R1 ∙ R2)/(R1+R2)

Question 7.2 R1 = 1 kΩ R2 = 1 kΩ R3 = 1 kΩ R4 = 1 kΩ Voc = R3/ ((R1 + R4) R3) . Vin = 1/3 . 10 = 3.33V

R3 is parallel with R2 R23 = 1/R23 = 1/R2 + 1/R3 = 1/1000 + 1/1000 = 1/ (1/1000 + 1/1000) = 500 Ω I= I = Vin/Rre = Vin/ R23 + R1 + R4 = 10 / (500 + 1000 + 1000) = 0,004 A so Rt = Rt = Voc / I = 3.33 / 0.004 = 833.33 Ω

CHAPTER 8

Question 8.1

Output signal sine wave. The current can only flow through the diode in one way, so only the positive values of the sine wave.

Question 8.2 R = V / I = 7/0,015 = 467 立 ( so 470 立 is the resistor)

CHAPTER 9

Question 9.1

The NPN-transistor is replaced by a PNP-transistor. For a PNP-transistor the potential difference between the emitter and the base must be negative.

If the subcircuit would deliver a voltage of 5 V the potential difference would be 5 -5 = 0V For a PNP-transistor the potential difference must be negative, so this option is excluded.

If the subcircuit would deliver a voltage of 0 V the potential difference would be 0 -5 = -5 V. The potential difference is negative so the output voltage of the subcircuit should be 0 V

Question 9.2 1) When the power is disconnected, the current will decrease very fast. In the following formula V = L * dI/ dt, dI will become real large. dt is a small number, because the time in which the current decreases is very short. This means that V will be really large, because a large number divided by a number close to zero results in a large number.

2) V= 1∙10-3 ∙ 100∙10-3/1∙10-6 = 100 V Only the 2N3439- transistor has a Vmax value that is large enough to handle the 100 V.

Question 9.3 The potential difference over the base and the emitter must be 0,6 V. In a darlington pair transistor there are two transistors. This means the potential difference will be the equivalent of the two. This is 0,6 * 2 = 1,2 V

Question 9.3 The maximum of Ic is 200 mA. The BC550 transistor can only handle 0.1 A, therefore this transistor is not usable.

The maximum of Ib is 0,4 mA. With the current ß can be calculated. ß = Icmax / Ib, ß = 200 * 10 ^ (-3) / 0,410 ^ (-3), ß = 500 The only transistor that is applicable is the BC618, because it is the only one that has a ßmax larger than 500.

The resistance value of Rb can be calculated: Ib = Ic / β = 200 ∙ 10 ^ -3 / 50 000 = 4,0 ∙ 10 ^ -6 Is ≈ 10 ∙ Ib = 10 ∙ 4,0 ∙ 10 ^ -6 = 0,0 ∙ 10 ^-3 Vbe = 1,6 V => (Vx - 1,6) / Rb = Ib Rb = (5,0 - 1,6) / 0,04 ∙ 10 ^ -3 = 85 kΩ

CHAPTER 10

Question 10.1

I = Vin/R1 Vout = I(R2+R1) Vout = Vin/R1 ∙ (R2+R1) Vout = R2 ∙ (Vin/R1) + R1 ∙ (Vin/R1) Vout = R2/R1 ∙ Vin + Vin Vout = (1 + R2/R1) ∙ Vin

Question 10.2 The formula for power is: P = U . I = u^2/R So if the resistance (R) is higher, the current (I) will be lower. They use large resistors so the current will be small through the circuit and that is better then a high current through the circuit.

Question 10.3 It will start clipping to + or - of the power supply if the input voltage Vin is higher than the supply voltage, but the transistor will not break immediately.

ASSIGNMENTS 1 tm 7

ASSIGNMENT 1 the circuit:

Calculate Rretotal, I1, I2, I3, I4, and Vout.

R=V.I Vout = I1 . R2 I1 = Vin/R1 + R2 Rre = 1/ (1/R1 + 1/R2 + 1/R3) R1 = 2200 Ω R2345 = 1/ (1/10000 + 1/4700 + 1/5500) = 1/0,000492 = 2032 Ω R6 = 3300 Ω Rre = 2200 Ω + 2032 Ω + 3300 Ω = 7532 Ω = 7,5 k Ω

Confirm your calculation by measuring Rretotal, I1, I2, I3, I4, and Vout.

Make sure that you remove the source from the circuit before you measure Rre. I1 = 10V / 7,52 k 立 = 1,33 mA I2 = 2,7 / 10 = 0,27 mA I3 = 2,7 / 4,7 = 0,57 mA I4 = 2,7 / 5,5 = 0,5 mA Vout = I1 . R6 + I2 . R5 = 1,33 . 3,3 + 0,5 . 3,3 = 6,039 Volt We measured values and they were the same as the calculated values.

Take a potmeter with a resistance value of 10 k立. Put it somewhere on your breadboard and measure the resistance value between pins a and b, and between pins b and c wile you turn the knob. a-b

b-c

Replace R4 and R5 by the potmeter. Notice that the resistance value between pins a and c replaces R4 and the resistance value between b and c replaces R5. Adjust the knob on the potmeter until you get an output voltage Vout = 5 V.

Take the potmeter away from the circuit and measure the resistance value between pins b and c and between pins a and c. b-c a-c

Verify your results by replacing the resistance value between pins a and c by R4 and the resistance value between pins b and c by R5 and calculate Vout.

R2345 = 2431 Ω (calculate again) R45 = 10140 Ω R1 = 2200 Ω R6 = 3300 Ω Rre = 7931 Ω V = 10 Volt I1 = V/R = 10 V / 7931 Ω = 1,26 mA V1 = 1,26 mA . 2200 Ω = 2,77 V V6 = 1,26 mA . 3300 Ω = 4,16 V V2345 = 10 -2,77-4,16 = 3,07 V V45 is also 3,07 Volt I5 = 3,07/10140 = 0,303 mA V5 = 0,000303 . 1670 = 0,51 V Vout = 0,51 + 4,16 = 4,66 V I measured 4,4 V but it is almost the same

ASSIGNMENT 2

Measure the amplitude of Vout at 100 Hz, 1 kHz and 100 Khz with an oscilloscope.

At 100 Hz, 1 kHz and 100 kHz the amplitude was everywhere 0,6 Volt. Because of resistors.

The resistors are not affected by frequency, so there was no change in voltage.

Build circuit (b) by replacing R1 with a capacitor of 100 nF. First: calculate the amplitude of Vout for 100 Hz, 1 kHz and 100 kHz. Then: verify your result by measuring Vout at the three frequencies with the oscilloscope.

The circuit is a high-pass filter: Vout = Rz/√(Rz^2+ 1/(w^2∙C^2)) ∙ Vin Vin = 2 V R=1kΩ C = 100 nF w = 2 πf

Hz 100 Hz 1 kHz 100 kHz

100Hz: calculated: Vout = 1∙10^3/√((1∙10^3)^2+ 1/((2∙π∙100)^2∙(100∙10-9)^2)) ∙ 2 = 0,125 V measured: 0,1 V 1 kHz calculated: Vout = 1∙10^3/√((1∙10^3)^2+ 1/((2∙π∙1∙10^3)^2∙(100∙109)^2)) ∙ 2 = 1,06 V measured: 1 V 100kHz calculated: Vout = 1∙10^3/√((1∙10^3)^2+ 1/((2∙π∙100∙10^3)^2∙(100∙109)^2)) ∙ 2 = 2,0 V measured: 2 V

calculated 0,125 V 1,06 V 2,0 V

measured 0,1V 1V 2V

Build circuit (c) by interchanging the resistor and the capacitor.

First: calculate the amplitude for Vout for 100 Hz, 1 kHz and 100 kHz. Then: verify your result by measuring Vout at the three frequencies with the oscilloscope.

The circuit is a low-pass filter: Vout = 1/(w∙C)/√(1/(w^2∙C^2) + Rz^2) ∙ Vin Vin = 2 V R=1kΩ C = 100 nF w = 2 πf

Hz 100 Hz 1 kHz 100 kHz

calculated 1,996 V 1,693V 0,03V

100Hz: calculated: Vout = (1/(2 ∙ π ∙ 100 ∙ 100∙10^-9)) /√(1((2∙π∙100)^2∙(100∙10^-9)^2) + (1∙10^3)^2) ∙ 2 = 1,996 V measured: 2V 1 kHz calculated: Vout = (1/(2 ∙ π ∙ 1∙10^3 ∙ 100∙10^-9)) /√(1((2∙π∙1∙10^3)^2∙(100∙10^-9)^2)+(1∙10^3)^2) ∙ 2 = 1,693 V measured: 1,7 V 100kHz calculated: Vout = (1/(2 ∙ π ∙100∙10^3 ∙ 100∙10^-9)) /√(1/((2∙π∙100∙10^3)^2∙(100∙10^-9)^2)+(1∙10^3)^2) ∙ 2 = 0,03 V measured:

measured 2V 1 ,7V V

Calculate the cut-off frequency fc of the last filter. Verify your result by measuring Vout at this frequency with the oscilloscope. wcut-off = 1/(2 . π . R . C) R = 1.10^3 c = 100.10^-9 1/(2 . π . 1.10^3 ∙ 100.10^-9)= 1591 Hz What type of filter is circuit (c) (high- or low-pass)?

The type of the filter in the circuit is a low-pass, see the measurements and calculations.

ASSIGNMENT 3 Calculate the voltage drops across the resistors and the currents through them. Use KVL and indicate the loops you’ve chosen. Take node ’e’ as reference.

LOOP 1 LOOP 2

V=R.I

VR1 + VR2 - Vbattery1 = 0 R1 = 2,2 k Ω R2 = 470 Ω Vbat = 10 V 2,2∙10^3 ∙ I1 + 0,47∙10^3 ∙ (I1 + I2) -10 = 0 2,2∙10^3 ∙ I1 + 0,47∙10^3 ∙ I1 + 0,47∙103 ∙ I2 -10 = 0 2,67∙10^3 ∙ I1 + 0,47∙10^3 ∙ I2 -10 = 0 2,67∙10^3 ∙ I1 = -0,47∙10^3 ∙ I2 + 10 I1 = -0,18 ∙ I2 + 3,7∙10^-3 VR3 + VR2 + VR4 - Vbattery2 = 0 R3 = 2,2 k Ω R2 = 470 Ω R4 = 5,6 k Ω Vbat2 = 5 V 2,2∙10^3 ∙ I2 + 0,47∙10^3 ∙ (I2 + I1) + 5,6∙10^3 ∙ I2 - 5 = 0 2,2∙10^3 ∙ I2 + 0,47∙10^3 ∙ I2 + 0,47∙10^3 ∙ I1 + 5,6∙10^3 ∙ I2 - 5 = 0 8,27∙10^3 ∙ I2 + 0,47∙10^3 ∙ I1 - 5 = 0 8,27∙10^3 ∙ I2 + 0,47∙10^3 ∙ I1 = 5 8,27∙10^3 ∙ I2 + 0,47∙10^3 ∙ (-0,18 ∙ I2 + 3,7∙10^-3) = 5 8,27∙10^3 ∙ I2 + -84,6 ∙ I2 + 1,74 = 5 8,19∙10^3 ∙ I2 = 3,26 I2 = 3,26 / 8,19∙10^3 = 0,4 mA I1 = -0,18 ∙ I2 + 3,7∙10^-3 = -0,18 ∙ 0,398 ∙10^-3 + 3,7∙10^-3 = 3,6 mA

V (loop 1): VR1= 2,2∙10^3 ∙ I1 = 2,2∙10^3 ∙ 3,6∙10^-3 =7,92 V VR2= 0,47∙10^3 ∙ (I1 + I2) = 0,47∙10^3 ∙(3,6∙10^-3 + 0,4∙10^-3) = 1,88 V V(loop2): VR2= 0,47∙10^3 ∙ (I1 + I2) = 0,47∙10^3 ∙(3,6∙10^-3 + 0,4∙10^-3) = 1,88 V VR3= 2,2∙10^3 ∙ I2 = 2,2∙10^3 ∙ 0,4∙10^-3 = 0,88 V VR4= 5,6∙10^3 ∙ I2 = 5,6∙10^3 ∙ 0,4∙10^-3 = 2,24 V

Build the circuit and verify the calculated values. resistor 1 2 3 4

calculated 7,92V 1,88V 0,88V 2,24V

measured 8,07V 1,94V 0,88V 2,25V

Draw your conclusions The Voltage we got by the calculation are almost the same as the voltage we got by the measuring. The difference can caused by the not that exact measurement and because of the resistors and wires.

ASSIGNMENT 4

Your function generator has an equivalent Thevenin/Norton resistor (Zout). Determine its resistance value using your multimeter, a resistor of 100 Ω (1/3 W ± 1%) and not more than two voltage measurements.

Draw some schematics in which you indicate how you’ve measured the voltages.

Show how you’ve calculated Zout.

measuring without resistor: 2,0 Volt = Vin

measuring with resistor: 1,3 Volt

Vout = R2/(R1/R2) ∙ Vin 1,3 = 100/(R1 + 100) ∙ 2 0,65 = 100/(R1 + 100) R1 + 100 = 100/0,65 R1 = 100/0,65 - 100 = 53,84

ASSIGNMENT 5 The circuit we built is the following:

From the following measurements we can conclude that the knee voltage is at 2,5 V. We can see this in the graph as the point that the current increases rapidly.

ASSIGNMENT 6 Saturation is reached when Rb is larger than 4,7 k立. In the table we can see that this is when Ic is no longer independent of Vce.

ASSIGNMENT 7 Given: R1 = 10 kΩ R2 = 10 kΩ R3 = 100 kΩ

Vout,high = V+ = 10 V Vout,low = V- = 0 V

Solution:

case 1: Vout = 10 V Vin+ = R2 / (R2 + (R1 ∙ R3 / (R1 + R3 ))) ∙ Vout

= ( 10 / (10 + 10 * 100 / (10 + 100)) * 10

= 5,238 V

Case 2: Vout = 0 V Vin+ = (R2 ∙ R3 / (R2 + R3 )) / (R1 + R2 ∙ R3 / (R2 + R3 )) ∙ Vout

= (10 * 100 / (10 + 100)) / (10 + 10 * 100 / ( 10 + 100 )) * 10

Building Bloks 1, 3, 4

Building Block 1

Vin = 12 V Vout = 0 - 10 V I = 0,5 - 5 mA We need a potentiometer te have an adjustable resistor to get an adjustable Vout.

Because the voltage have to be at max 10V we need to put a resistor between Vin and the potential meter. Than the circuit will look like this:

Rtotal = Umax / Imax Rtotal = 12 / (5 ∙ 10 ^ -3) Rtotal = 2400 Ω = 2,4 kΩ The resistance of the potentiometer: Rpotmeter = Umax Vout / I Rpotmeter = 10 / (5 ∙ 10 ^ -3) Rpotmeter = 2000 Ω = 2 kΩ We used a 10 kΩ potmeter, this one will work because we took Imax, and when I decrease the Rpotmeter increase.

To calculate Rtotal we need to calculate I first: Rpotmeter = Umax vout / I = 10 000 Ω I = Umax vout / Rpotmeter I = 10 / 10 000 I = 0,001A = 1 mA Rtotal = Utotal / Imax Rtotal = 12 / 0,001 Rtotal = 12 000 Ω = 12k Ω R1 = Rtotal - Rpotmeter R1 = 12 000 - 10 000 R1 = 2 000 Ω = 2 kΩ

Building Block 3

This is our PWM circuit:

We used a comparator, a potmeter and a function generator.

The circuit compares the signal from the voltage divider with the signal from the function generator. When the signal from the function generator is higher than the one from the voltage divider, Vout will also be high. When the signal from the voltage divider is higher, Vout will be 0.

Building Block 4

We used a NTC resistor, a comparator, a potmeter (10 kΩ) and a LED. We measured the Rvalue at 2 different temperatures, at first the temperature in the room and secondly the temperature while heating up.

Our measurements: R at 23,4 ˚C = 12,38 kΩ R at 33,2 ˚C = 3,25 kΩ so by increasing the temperature the resistance decrease.

Because the resistance decrease with increasing the temperature, there will be send a smaller part of the voltage towards the comparator. And the other way around. If the voltage is higher that the reference point, the comparator will let a signal true.

This signal will light the LED and will heat up the power resistor.The power resistor will heat up the box around the circuit so the NTC will become warmer.The resistance of the NTC will derease and more current is sent towards the comparator.

When the signal has become smaller than the reference point, the power resistor gets lower in temperature and the led will be turned off.

Reflections

Reflection: Stephanie Saris In my last project, Vogue, the high-tech fashion issue I didn’t do much with electronics and integrating technology. This competency area, integrating technology, I really had to work on that this semester. I chose the assignment Introducing Electronics because I thought it would be important to first get the basics of electronics and learn all the different components. I know you can find a lot on the internet about how to build a circuit for your project, but I think I will not really understand exactly what I am doing and just recreate it.

I liked the construction of the assignment’s booklet because I didn’t know anymore all the formula from secondary school, so the first chapter was about what I already learned last year.

By making the questions I began to remember all the formula. For me the most difficult part of remembering it was that this time everything was in English, so some of the letters in the formula were different and sometimes I didn’t understand completely the explanation. Before every class I read the text and made a little summary about the most important things of that chapter so I would understand the explanation during the class better. I think it really helped because I already knew what is was about and it was not all new for me during that class.

Last semester I went once to the E-lab to make a circuit with LEDs. This was not the first time I build a circuit. At secondary school I made a doll with Lights, buttons and resistors, but that was different. This time it was more about how different components of a circuit for example a resistor works, what their role is in a circuit. I think for me is the best way of learning by building the circuit and not only calculating. By building it you can see how it will look like in real and how to measure. I also didn’t like to go to the E-lab because I don’t like to ask things and I mostly want to figure it out on my own, but I learned that if I just ask I will understand it better than if I just try things out till it will work.

The workshop at the beginning of the assignment was very useful. I learned to work with a function generator and an oscilloscope. This was completely new for me. I learned at secondary school what you can do with it but never used it. After the workshop we had to do some assignments ourselves. At the beginning I found it difficult or it really took a long time to build the circuit and to do the measuring right. The first day we only made the first chapter, we didn’t have time anymore to do assignment two. But the second day we worked on the assignment we did a lot more. If I look back at assignment one I understand it directly and it will not cost me a whole day to make the circuit and to do the measuring.

The building blocks were the most difficult part of the whole assignment, so we asked sometimes for help from people who already did this assignment to help us. Also at the end Joeri joined our group because his partner quit. I think this also helped because now we were with 3 people who tried to build the circuit. The teamwork also went well. We did all the assignments and building blocks together and most of the chapters too. I made the lay-out of our booklet and worked out some chapters and 4 of the assignments. I put a lot of work in this assignment because for me it is very important to do a great job for integrating technology and I think this was also a very helpful and important assignment to understand the basics of electronics. In my project I also want to focus on the integrating technology part and I think this assignment helped me understanding how to build a circuit. I think people will see me now more sitting in the E-lab building circuits for prototypes and for my project.

Reflection: Charlotte van der Sommen My name is Charlotte van der Sommen and I am a B1.2 student at the moment. During my B1.1 semester I already started to do a little with electronics, but I just did something and didn’t really understood what I was doing. So I decided to develop myself this semester in the competency area Integrating Technology by doing the assignment Introducing Electronics.

In the beginning I thought this was very hard assignment because I had an hard time understanding what it was going about but when we started to make the assignments and the building blocks I started to understand everything a lot better. I learned a lot new things about the devices in the E-lab, like how the function generator works, and how to work with a breadboard and a multi-meter. Together with Stephanie I figured out a lot of problems we bombed into. Most of the time when we didn’t understand something we both looked at the problem and most of the time together we knew enough to understand the problem or by discussing we figured out the solution. Sometimes we couldn’t figure out the problem together, than we went to ask other groups and tried to figure it out together. This way we eventually understood it all.

By working together in a group I learned that it is not wrong to ask other students how they figured out a problem, because together you will get the best results. In the future I am going to ask more around when I don’t understand something.

I learned a lot things about circuits, all kind of objects in a circuit and about how to work with the devices in the E-lab. Now I know a lot more about integrating technology and am I able to make my own circuits. This knowledge will be very handy in my future projects and assignments for instance for making prototypes.

Now I have done this assignment I have made a few important steps forward in becoming very good in the competency area integrating technology.

Reflection: Joerie Hahn I followed the introducing Electronics assignment this semester. In this assignment we learned some of the basics of electronic circuits. In the lessons we first got a presentation and explanations about the chapter, with a powerpoint presentation. After some time we had to do practical assignments and so called building blocks.

The practical assignments really helped to understand the theory better. At a certain point I noticed that I was having a hard time understanding what the teacher was talking about.

When doing the practical assignments, I got to understand the subject matter a lot better, because this way we got to use the theory in practice. Also when a circuit wasn’t right, we could see it immediately, because then the measurements would be far off. On the one hand this was sometimes really frustrating. I remember that I got really frustrated during the first practical assignment, because we just couldn’t get it right. Luckily the teachers in the e-lab are very helpful and they showed us what we were doing wrong. This also helped to understand the theory better and we could built later assignments more easily.

I was really happy with the teamwork. Me and Iris were a good couple. We didn’t always know everything from the beginning, but a lot of the problems we solved ourselves. We were helped by fellow students with some other problems and even sometimes tried to tackle a problem with another group. This really helped increase the teamwork and communication. I found out that we know more when were with four than we do with two.

Everyone has something that they are good at and this way we can help each other learn.

For instance, I am really good at algebra. So some of the calculations we had to do were easy for me, but harder for Iris.

Unfortunally Iris has quit the education. The last bits of the assignments and chapters I finished with Charlotte and Stephanie. The teamwork with Charlotte and Stephanie went really good as well.

What I was really proud of is that Iris and I were really good planners. Every Wednesday we spend time on doing the assignments and we stayed on schedule the whole time we worked together. For me this is really good, because I have a hard time planning sometimes.

I am not sure yet how to use the knowledge in the future. My opinion is that this assignment has a lot of basic knowledge, but we haven’t really learned what to do with it. For instance, I know now how to build a circuit with resistors and how the calculate the voltage or the current, but I do not know when I would need to build such a circuit. I think it would be a good thing to do the following assignment in electronics. This can help me to go deeper into the subject and to learn to put knowledge into practice better.

Also I think that when I am at the point of building a circuit, I will notice some problems and thanks to this assignment I will be able to solve them much quicker.