EPO-1 Beukende Bass eindverslag final report

Booming Bass EPO-1 Project Group C1

Frida riday y 15th January January,, 2016

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Contents Foreword

1

Introduction

2

1 Power supply supply analysis analysis 1.1   Introduction   . . . . . . . . . . . . . . . . . . . 1.2 Calculating the the values of the power supply supply circuit. circuit. 1.2.1   Capacitors . . . . . . . . . . . . . . . . . 1.2.2   Resistors . . . . . . . . . . . . . . . . . . 1.3 Simulating the power supply supply using PSpice PSpice   . . . . 1.4 Building and testing the power power supply  supply    . . . . . . 2 Loudspeaker analysis 2.1   Introduction   . . . . . 2.2   Theory   Theory   . . . . . . . . 2.3 Loudspeaker Analysis . 2.4   Measurements   . . . . 2.5   Simulations . . . . . .

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3   Passive filter lter design design 3.1   Introduction   . . . . . . . . . 3.2   Design   . . . . . . . . . . . . 3.2.1 Impedance Correction . 3.2.2 Separation filters . . . . 3.2.3 Phase shift . . . . . . . 3.2.4 Volume adjustment adjustmen t   . . 3.2.5 Final schematics . . . . 3.3   Simulations . . . . . . . . . . 3.4   Measurements   . . . . . . . .

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4   Power Power ampli amplifier design design 4.1   Introduction   . . . . . . . . . . . . . . . . . . 4.1.1   Background   . . . . . . . . . . . . . . . 4.1.2 Objective and requirements requirements   . . . . . . . 4.1.3 Global description description of design steps . . . . 4.2 Amplifier Analysis. Analysis . . . . . . . . . . . . . . . . 4.2.1 Discussion schematic   . . . . . . . . . . 4.2.2 Derived transfer functions . . . . . . . . 4.2.3 Role of unknown unknown components components . . . . . . 4.2.4 Calculation of component component values values   . . . . 4.2.5 Recalculation Recalculation for available available components components . 4.3   Simulations . . . . . . . . . . . . . . . . . . . 4.3.1 Simulation in Matlab and PSpice PSpice.. . . . . 4.3.2 Final schematic   . . . . . . . . . . . . . 4.3.3   Results . . . . . . . . . . . . . . . . . . 4.4   Measurements   . . . . . . . . . . . . . . . . . 4.4.1 Measurement Measurement setup   . . . . . . . . . . . 4.4.2   Results . . . . . . . . . . . . . . . . . . 4.5 Volume Volume control control knob knob   . . . . . . . . . . . . . . 4.5.1   Introduction . . . . . . . . . . . . . . . 4.5.2   Schematic   . . . . . . . . . . . . . . . .

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Contents Foreword

1

Introduction

2

1 Power supply supply analysis analysis 1.1   Introduction   . . . . . . . . . . . . . . . . . . . 1.2 Calculating the the values of the power supply supply circuit. circuit. 1.2.1   Capacitors . . . . . . . . . . . . . . . . . 1.2.2   Resistors . . . . . . . . . . . . . . . . . . 1.3 Simulating the power supply supply using PSpice PSpice   . . . . 1.4 Building and testing the power power supply  supply    . . . . . . 2 Loudspeaker analysis 2.1   Introduction   . . . . . 2.2   Theory   Theory   . . . . . . . . 2.3 Loudspeaker Analysis . 2.4   Measurements   . . . . 2.5   Simulations . . . . . .

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3   Passive filter lter design design 3.1   Introduction   . . . . . . . . . 3.2   Design   . . . . . . . . . . . . 3.2.1 Impedance Correction . 3.2.2 Separation filters . . . . 3.2.3 Phase shift . . . . . . . 3.2.4 Volume adjustment adjustmen t   . . 3.2.5 Final schematics . . . . 3.3   Simulations . . . . . . . . . . 3.4   Measurements   . . . . . . . .

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4   Power Power ampli amplifier design design 4.1   Introduction   . . . . . . . . . . . . . . . . . . 4.1.1   Background   . . . . . . . . . . . . . . . 4.1.2 Objective and requirements requirements   . . . . . . . 4.1.3 Global description description of design steps . . . . 4.2 Amplifier Analysis. Analysis . . . . . . . . . . . . . . . . 4.2.1 Discussion schematic   . . . . . . . . . . 4.2.2 Derived transfer functions . . . . . . . . 4.2.3 Role of unknown unknown components components . . . . . . 4.2.4 Calculation of component component values values   . . . . 4.2.5 Recalculation Recalculation for available available components components . 4.3   Simulations . . . . . . . . . . . . . . . . . . . 4.3.1 Simulation in Matlab and PSpice PSpice.. . . . . 4.3.2 Final schematic   . . . . . . . . . . . . . 4.3.3   Results . . . . . . . . . . . . . . . . . . 4.4   Measurements   . . . . . . . . . . . . . . . . . 4.4.1 Measurement Measurement setup   . . . . . . . . . . . 4.4.2   Results . . . . . . . . . . . . . . . . . . 4.5 Volume Volume control control knob knob   . . . . . . . . . . . . . . 4.5.1   Introduction . . . . . . . . . . . . . . . 4.5.2   Schematic   . . . . . . . . . . . . . . . .

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Contents

5   Booming Booming bass bass design design 5.1   Introduction   . . . . . . 5.2   Design   . . . . . . . . . 5.2.1 Needed behavior. behavior . 5.2.2 The circuit . . . . 5.2.3 Final schematic   . 5.3   Simulations . . . . . . . 5.4   Measurements   . . . . . 5.5   Addition . . . . . . . .

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6 Acoust Acoustic ic chara character cteriza izatio tion n of the total total system system

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7   Conclusions Conclusions and recommendati recommendations ons 7.1 Power Power supply supply analysis analysis . . . . . . 7.2 Loudspeaker analysis . . . . . . 7.3 Passive filter design . . . . . . . 7.4 Power Power amplifier amplifier design design   . . . . . 7.5 Booming bass design . . . . . .

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Bibliography

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Appendix  A  Derivations of the equations   . . . . . . . . . . . . . . . . . . .  A.1 Loudspeaker analysis   . . . . . . . . . . . . . . . . . . .  A.2 Passive filter design   . . . . . . . . . . . . . . . . . . . . B Measurement results   . . . . . . . . . . . . . . . . . . . . . . . B.1 Loudspeaker analysis   . . . . . . . . . . . . . . . . . . . B.2 Passive filter design   . . . . . . . . . . . . . . . . . . . . B.3 Power Power amplifier design. design . . . . . . . . . . . . . . . . . . . C MATLAB MATLAB codes and simulations.   . . . . . . . . . . . . . . . . . C.1 Loudspeaker analysis   . . . . . . . . . . . . . . . . . . . C.2 Passive filter design   . . . . . . . . . . . . . . . . . . . . C.3 Power Power amplifier design. design . . . . . . . . . . . . . . . . . . . D   Components   . . . . . . . . . . . . . . . . . . . . . . . . . . . D.1 D.1 Used components for the the passive passive filters   . . . . . . . . . . E Assignments of power amplifier. amplifier . . . . . . . . . . . . . . . . . . E.1 Tutorial 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . E.2 Tutorial 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . E.3 Tutorial 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . E.4 Tutorial 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . E.5 Tutorial 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . F   Datasheets   . . . . . . . . . . . . . . . . . . . . . . . . . . . . F.1 LM3886 Overture™ Audio Power Power Amplifier. Amplifier . . . . . . . . . F.2 RØDE VideoMic   . . . . . . . . . . . . . . . . . . . . . .

37 37 37 38 46 46 49 50 51 51 53 56 59 59 61 61 63 65 67 69 71 71 75

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Foreword Over the course of 10 weeks a group of 12 students have designed, simulated and built the electrical components necessary to get a predetermined audio system working optimally. To accomplish this, the students have also build a booming bass extension to widen the ’dynamic range’ of the speaker. In this project, the students have tackled the basics of electrical engineering. The theoretical lecture courses have been used to understand the concepts of the loudspeaker. Furthermore, the system must comply to the following specifications: • The requirement ’best possible reproduction of an input signal’ may be interpreted as: the loudness of  a tone measured at a distance of one meter only depends on theamplitude of theelectrical input signal, and not (or as little as possible) on the frequency. In other words, within the displayed frequency range, the transfer of the system is as flat as possible. • The frequency of the audio system ranges from 20 Hz to 20 kHz (-3dB bandwidths). • When the signal source produces an electrical signal with an amplitude of maximum 0.4 V, the maximum amplitude of the signal presented to the speakers must be 10 V. • The system is built in the ’EPO-1 chassis’, which can be seen in Appendix B of the EPO book. [ 3]

1

Introduction In the EPO-1 project the main goal is to create an audio system which reproduces an input signal as best as possible. The system has to be built for a speaker which is provided by the TU-Delft. The audio system that must be built has to include the following parts:  [3] • A power supply  • A power amplifier • A three-way speaker filter • A ’booming bass’ extension The speaker has to be analysed as well which means that there were 5 main parts of the project. The project has been approached by dividing the group in 4 parts, with each group working on a different component of thesystem. As certain parts of the project got finished, the group members joined other groups or startedworking on thefinal part, the’booming bass’. When all 4 parts were finished everybody went to help  with the ’booming bass’ or making the finishing touches to the parts and the reports. In the further chapters of this report all the components of the system will be discussed and analysed. The complete system will also be discussed and analysed and conclusions will be drawn afterwards to see if the system satisfies all of the specifications.

2

 Power supply analysis 1.1. Introduction  Whether it is from a battery, or from the power outlet, every electronic amplifier needs power. For this amplifier, the obvious choice would be the power outlet. However, the power from the main outlet can’t be fed directly into the amplifier. So in order for us to supply the amplifier with the right power, a power supply  unit (PSU) will be needed that converts the mains voltage into a voltage that is usable by the amplifier. The schematic of our power supply unit is given in figure 1.1.

Figure 1.1: A schematic version of the power supply we built

There were a couple of requirements for the PSU given in the EPO book [3]: 1. Without any load, both the V +  and the V −  rails should be at least

±22 V .

2. When the amplifier would demand a current of 1 A , the ripple voltage should only be 5% of 19 V . 3. When the PSU is not loaded, the capacitors should be discharged within 2.5 minutes (5τ). 4. The PSU should be a double-sided rectifier with a positive and negative output voltage. Please note that these are minimum requirements. If in any way it is possible to lower the ripple voltage even below the earlier mentioned 5%, this should be aimed to do so. The main challenges of the PSU are finding the values of the components. The capacitors should be capable of keeping the ripple voltage below 5%, diodes with a reverse voltage higher than the current through our system. The bleeder resistors should be able to keep the discharge time below the 2.5 minutes.

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1.2. Calculating the values of the power supply circuit

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1.2. Calculating the values of the power supply circuit Before building the PSU, the components should be chosen and their values should be calculated.

1.2.1. Capacitors To obtain and reduce the ripple voltage above the 19  V , capacitors will be added at both rails of the PSU. To know what value to use for the capacitors, an equation is needed to calculate this value, starting off with equation 1.1.

Figure 1.2: V +  when adding a capacitor

q  C v 

 

=

(1.1)

 ≈ ∆∆q t  .

If both sides are divided by the time, the following equations are obtained using the fact that I  ∆q 

C ∆v  = ∆t  ∆t 

(1.2)

 = C ∆∆t v 

(1.3)

I 

Before going any further, the ripple voltage V r  should be defined.  V r  is the potential difference between the average voltage V R MS  and the maximum voltage. The difference between the maximum and the minimum voltage ∆v  is thus twice the V r .

 = 2C∆V t r 

I 

(1.4)

The ∆t  is half the period of the original power outlet, which equals T   f   −1 . Then it comes down to:

 =

1  = 2 f  

∆t 

(1.5)

Then the value of the capacitors can be calculated in the following way:

 = 4 fI V 

C 

(1.6)

r 

Using equation 1.6  the minimum value can be calculated to keep the ripple voltage below 5%. The ripple voltage V r  is 5% of the 19 V  and the frequency is f   50 H z .

 =

1.3. Simulating the power supply using PSpice

C mi n 

5

1 = 4 · 50 · 0.05 · 19 = 5300 µF 

 

(1.7)

This exact capacitor wasn’t available. However, a bigger capacitor should reduce the the ripple even more. That is why a 6800 µF  40 V  model capacitor is used in the PSU

1.2.2. Resistors  When switching off the amplifier, the capacitors theoretically will not discharge. Because charged capacitors can hold enough energy to damage components when handled incorrectly, bleeder resistors were added to discharge the capacitors in a safe manner to keep the discharge time  t ma x  below 2.5 minutes. Because this discharging process follows the exponential behavior of an RC-circuit, mathematically it is impossible to fully  discharge the capacitors. 5τ is considered empty enough and it should therefore equal the maximum time t ma x  allowed, which is the mentioned 2.5 minutes. 5τ  With τ

= t ma x 

 

(1.8)

= RC  in a parallel RC circuit, the maximum resistance can be calculated via the for mula:  = t m5C a x 

R ma x 

(1.9)

This means that a smaller resistance always means a shorter discharge time. However, having shorter discharge times is not always desirable. A shorter discharge time needs an smaller bleeder resistor resistor, and that resistor will also load the PSU down unnecessarily.  With the actually used capacitance being 6800 µF  and the maximum time being 2.5 minutes, the maximum resistance can be calculated using equation 1.9: 2.5 · 60  = 5 · 6800 · 10−6 = 4.4 k Ω

R ma x 

 

(1.10)

 A pair of 3.9 k Ω resistors were used, because they were the largest resistors available under 4.4  k Ω. They will keep the discharge time below 2.5 minutes and the load on the PSU the smallest.

1.3. Simulating the power supply using PSpice

Figure 1.3: The simulation model of the double sided rectifier

Orcad PSpice was used to simulate the schematic of the PSU that can be seen in figure  1.3.  To mimic the center-tapped transformer, two AC voltage sources in phase were used. Their amplitudes were 22 V  and the

1.4. Building and testing the power supply

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frequency was set to be 50 H z , to make them resemble the output voltages of the transformer box of the TU Delft. On the right side of the figure, there are two 20 Ω resistors to simulate a load of 1 A . The diodes used are not the ones that will be used to build the PSU, but are able to handle the voltage and the current delivered, so they are no problem in the simulation. This setup resulted in the graph that can be seen in figure  1.4.  The difference between the minimum and maximum voltage never exceeds 1.8 volts and therefore never exceeds the 10% of the 19 V . We can conclude from this simulation that the circuit meets all the requirements.

Figure 1.4: The voltage time diagram of the power supply 

1.4. Building and testing the power supply  The PSU was built on the PCB provided by the university and analysed to check if the simulations were correct. The measurements were done between the and instead of  and gr ound , because this way both the negative rail and positive rail were tested in one measurement. To measure at the 1  A  current at which there should be only 5% ripple voltage, a 40 Ω resistor was required, but such a resistance was not available. Thus, two measurements were executed, one with a 50 Ω and one with a 60 Ω. These were the results:

+

H resistance maximum voltage average voltage minimum voltage ripple voltage current (DC)

−

50 Ω 41.6 V    40.6 V    39.6 V    1.0 V  812 m A 

+

60 Ω 42.0 V  41.2 V  40.4 V  0.8 V  675 m A 

 As expected, the current is lower than the 1  A  at which the ripple is specified, because 50 tances were used instead of the calculated 40 Ω that wasn’t available.

Ω

and 60 Ω resis-

The ripple voltage as percentage of the total voltage is calculated using the following equation: V r i p p l e  (%)

= V maV x  − V av g  · 100%

(1.11)

av g 

Filling in the results of the first measurement gives us: V r i p p l e  (%)

− 40.6 · 100% = 2.46% = 41.640.6

(1.12)

− 41.2 · 100% = 1.94% = 4241.2

(1.13)

 And for the second measurement: V r i p p l e  (%)

1.4. Building and testing the power supply

7

Using PSpice, the first measurement was executed with a 50 Ω load connected. This simulation is shown in figure 1.5. In the simulation, the average current  i av g  810 m A  and the average voltage V av g  42 V . These results match with our real measurements within the tolerances of the used resistors.

 =

Figure 1.5: The simulated voltage and current with a 50

 =

Ω resistor, with a ripple voltage of 1

V 

The results of the simulation with a 60 Ω  load are show in figure 1.6. The average current i av g  680 m A , the average voltage V av g   41  V  and other results also match within tiny tolerances to our real measurements.

 =

 =

Figure 1.6: The simulated voltage and current with a 60

Ω resistor, with a ripple voltage of 1

V 

 Loudspeaker analysis 2.1. Introduction  A dynamic loudspeaker is a device that converts electric signals into sound. In order to produce sound, a certain pressure variation in the medium (air) is needed. For low frequencies, relatively large but slowly-vibrating  volumes of air are required and for high frequencies relatively small but quickly-vibrating air volumes are required. Separate types of speakers, optimized for certain frequency ranges, are needed to accurately produce the desired frequencies. A speaker element usually consists of a metal frame: the basket, in which a movable cone is attached (see figure 2.1 [4] ).

Figure 2.1: Loudspeaker intersection

Depending on the type of speaker, the cone can be made of paper, plastic, or other materials such as Kevlar, carbon, or other light metals. The rear of the cone ends in a circular bus around which a coil: the voice coil, is  wound. The voice coil is the electrical input of the speaker, with which it will be connected to the filter of the speaker. The voice coil is located in a strong permanent magnetic field and i s centered in here by the spring-mounted ’spider’, which allows the voice coil to be flexibly mounted to the speaker frame.  At the front, the cone is mounted by means of an edge made of rubber or foam in order to smoothly move up and down. With a tweeter, intended to sound the treble, the cone must be very light, and is therefore often made of silk, or a very thin metal foil that is glued directly onto the voice coil. For a loudspeaker it is important to analyze the behaviour of the different internal components at different frequencies, because sound consists of different frequencies. These components have their own resistances, inductances and capacitances. The loudspeakers’ electrical characteristics have to be analysed in order to properly understand 8

2.2. Theory

9

it’s behaviour when playing audio through it, which consists of signals with varying frequencies. These signals are essentially alternating current, which means the response of the circuitry inside the loudspeaker  will be very different because it contains components such as inductance’s and capacitance’s, which behave differently for different frequencies. In this report a basic explanation of impedance, an explanation of the electrical model of a loudspeaker and it’s behaviour under alternating current will be given. Measurements of the impedance of the loudspeakers  will be included, the corresponding models and the determined component values based on the measurements will also be included. To accomplish that, the impedance of the speakers will be measured for the frequency range of 0 Hz to 24 kHz. By using the impedance measurements a model can be made which includes: • Resistance of the voice coil • Self inductance of the voice coil • The values of the electrical equivalent of mass-spring system representing the cone Based on the measurements, the values of the model parameters can be determined in such a way that the calculated impedance corresponds as much as possible with the measured impedance. The simulation results will be compared with the measured impedance and the causes of the differences will be investigated. Furthermore, the behaviour of the loudspeaker at different frequencies, measured by the actual sound output of the loudspeaker in decibels will be measured. The purpose of these measurements is to investigate on which frequency a switch of output between the different speakers can best be made, to make sure the effectiveness of the entire system is maximized.

2.2. Theory   A circuit operating in steady-state with alternating current has impedance as an intrinsic property instead of  resistivity. Impedance can be seen as the equivalent of resistance for AC circuits. This means that impedance,  just like resistance, is the ratio of voltage over current (equation 1 in appendix  A.1). The impedance of certain components (such as inductors and capacitors) is dependent on the frequency of  the alternating current flowing through the circuit. Because the loudspeakers that are used are working under  AC, a proper understanding of impedance is needed. The impedance of a resistor is given in equation 3  in appendix  A.1. As is apparent from equation 3 the impedance of a resistance is independent of the frequency. The impedance of an inductance is given in equation 4 and the impedance of a capacitor is given in equation 5 in appendix  A.1. These equations show that the impedance is frequency dependant in these cases. The formulas for the resonant frequency for an RLC circuit and the formula for the bandwith of an RLC circuit are needed too. These are given in equations 6 & 7.

2.3. Loudspeaker Analysis To analyse the loudspeaker a simplified electric model of the speaker is needed. This model is shown in figure  2.2. In this model the resistance (R e ) and the inductance (L e ) in series represent the voice coil. The parallel RLC ( R p , L p ,C p ) circuit represents a cone. These two parts joined in series form the electric model of a loudspeaker. The simplified model comes from a more extensive model. The more extensive model includes two extra resistances in series with  R e . These resistances are R a , which represents the power losses to air, and R v , which represents the power lost to mechanical friction. Figure 2.2: Simplified electric model of a loudspeaker  [3] These losses however are so low that they can be ignored  which gives us the model in figure 2.2 to work with during the analysis the speaker.

2.4. Measurements

10

The value of the resistance (R e ) of the voice coil is equal to the DC resistance of the loudspeaker. This is because the frequency of DC is 0, which means Z L (0) 0, which means Z L e  (0) and Z L p  (0) 0. The impedance of the cone is also zero, so all the impedance from the loudspeaker comes from the resistance of the voice coil.

=

 =

=

 At high frequencies, the impedance of a capacitor is very low, to the point where it is negligible. So a RLC circuit at high frequencies will have an impedance which is practically equivalent to zero. As shown by equation 4 , the impedance of an inductance will be high at high frequencies. This means that at high frequencies, the impedance of the circuit of the loudspeaker is equal to R e  Z L e (ω).

+

The inductance of the voice coil can be calculated using this and equation  4, this is shown in equation 8 in appendix  A.1.  When a parallel LC circuit resonates, the inductance and capacitor resonate and ’cancel’ each other out, creating an open circuit. This is because together, their admittance(the reciprocal of the impedance) approaches 0. When the admittance is 0, the circuit becomes to a set of open terminals. This is shown in equations  6 through 18 in appendix  A.1. This means that at the resonant frequency, a parallel RLC circuit’s impedance is equal to the resistance of the resistor in the RLC. The resonant frequency is usually quite low, so that means that the added impedance from the inductance of the voice coil is low, to the point where it is negligible. This means that at the resonant frequency, the impedance of the loudspeaker is equal to R e  R p .

+

The resistance of the cone can be calculated with this, as is shown in equation  20 in appendix  A.1.  After calculating the resistance of the cone and measuring the bandwidth, the formula for the bandwidth (equation 7) will be used to calculate the value  C p , the inductance of the cone. This is shown in equation 21. Now that the value of  C p  had been calculated, the value of  L p  can been calculated by using the formula for the resonant frequency. This is shown in equation 22. Using all of the formulas the values of the components have been calculated for the low, mid and high speakers: Low speaker

Mid speaker

High speaker

• R e = 7.1 Ω

• R e = 4.13 Ω

• R e = 4 Ω

• L e = 0.38 mH

• L e = 0.17 mH

• L e = 53 µH

• R p = 9.9 Ω

• R p = 5.23 Ω

• R p = 0.91 Ω

• L p = 0.011 H

• L p = 1.9 mH

• L p = 96 µH

• C p = 0.61 mF

• C p = 1.8 mF

• C p = 0.16 mF

2.4. Measurements For measuring the complex frequency-dependent impedance of a speaker, the MATLAB program LS_Measure  was used, provided by the TU Delft. LS_Measure uses the sound card of the computer to generate a white noise signal containing different frequencies of a chosen range. It measures the response of the impedance on the frequencies and plots three graphs containing: • The impedance amplitude Z in ohms. • The phase shift of the returning signal in degrees. • The calculated electrical frequency response in decibels. The measurements are shown in Appendix  B.1. In figure 8 the measurement setup had been tested by measuring the impedance values for some components. The impedance values of these components can be calculated. The measurements can be compared to these calculations to determine if the setup is valid. They are the same so the setup is valid and can be used for further measurements. It can be seen that there is always a impedance peak at a certain frequency. In figure  5 the impedance amplitude and the and the impedance

2.5. Simulations

11

phase shift is shown. The peaks are caused by the frequency of the voice coil. As for figures  6  and  7  the same argumentation is applied. In 9, the amplitude of the frequency response of the loudspeakers is given in dB.

Figure 2.3: Measurement setup for measuring impedances  [3]

To measure the impedance of the speaker, the program LS_Measure, the circuit shown in figure  2.3 and a PC from theTellegen Hall in theEEMCS building of TU Delft have be used. The signal processing and-generation are being distributed among the PC and sound card. Two cables connected to the PC are needed to generate and process the signals. The cables are connected to the microphone- and the audio jack. The microphone  jack at the rear of the PC is used and the audio jack at the front due to disruptive environmental factors introduced by the other ports when used. The cables connected with the PC are routed to the Connection Box’s line in and line out. The V i n  and V ou t  ports are connected with the positive terminal of the high, mid or low speaker. Only one ground port of the V i n  and V ou t  is needed to connect the negative terminal of the speaker to ground it. The frequency input of the measuring system has a certain limit. That is determined by  the background noise of the frequency. The higher the frequency, the more background noise it produces. At 20 kHz the background noise is way to high to produce a normal sound. Furthermore, the soundcard of the computer used was not build to produce frequencies higher than 20 kHz.

2.5. Simulations In order to verify the measurements of the loudspeakers, a simulation of the loudspeaker has been created. The formulas for calculating the impedance of the components in the loudspeaker are already given. By  calculating the impedance of each component at a certain range of frequencies and putting them together, the total impedance of the loudspeaker can be calculated. There are two models of loudspeakers in the manual: Model 1, and model 2. Model 1 is the full representation of a the electrical circuit of a loudspeaker. Model 2 on the other hand, only represents the electric circuit of a loudspeaker for frequencies above those of the resonance peak. The values of the components of the models are already given and are given in list C.1 in appendix C.1. Using these values and the formulae given in the Introduction and ’Loudspeaker Analysis’ sections, the behaviour of this model in response to different frequencies can be simulated. First, the values are imported into MATLAB. Then, the impedance at a range of the frequencies has to be calculated. A range of frequencies that is logarithmically spaced will be chosen to get more samples in the lower frequency ranges, where the resonant frequency usually is. A frequency range of 20 kHz is chosen since that is aesthetically pleasing when plotted, for the simulations of the actual loudspeakers, different values will be used. The impedance of a loudspeaker is given in equation  23 in appendix  A.1.

2.5. Simulations

12

The equation is converted into MATLAB code and the impedance will be calculated. Now that the impedance is calculated, it has to be converted into, as complex numbers don’t have much meaning on their own. So the absolute value of the impedance is taken to get the amplitude of the impedance. The angle is also taken so that the phase shift caused by the loudspeaker can be obtained. Then the phase shift and impedance amplitude can be plotted (figure  15 in C.1) to see if it matches the simulation from the tutorial. Since this is identical to the results of the simulation on page 128 of the manual, it can be concluded that the code for the simulation works as intended. The code that’s already written to simulate the behaviour of the loudspeakers can be extended. By implementing the equations (2 through 7, A.1) that are used to calculate the values of the components in MATLAB, the process of creating a simulation of the speaker can be simplified. In the event of incorrect measurements, a newsimulation caneasily be created if this part will also be automated. First thevalues need to be added that can be (and need to be) measured. Then, equations 1 through 23 have to be implemented in MATLAB. Next, some of the code has to be rewritten so it works properly with vectors instead of scalars. This also means that what was previously a vector, is now a matrix, so the code has to be adjusted for that as well. The plotting functions should be updated too so that multiple vectors can be plotted.  While there is a fairly noticeable difference in the phase shift of the simulation (figure  16 ) compared to the phase shift of the actual model (figure  7), there is little reason to worry. This is because the unpredictable behaviour of the phase shift only starts to occur well after the frequencies where the signal transitions from one speaker to the other. This means that any interference caused by phase shift can be accounted for with less difficulty.  Aside from that, there is a minor difference in the minimum impedance amplitude. This could come from external sources such as the PC used to measure the loudspeaker’s impedance or background.

 Passive filter design 3.1. Introduction  Amplifiers take a signal and boost the amplitude. But in the case of an acoustic amplifier (a speaker) the electric signal gets amplified and converted into sound waves. To make sure that all frequencies produce as little distortion as possible, specialized speakers are used: two bass speakers for the low domain, one mid speaker for the middle domain and a tweeter for the highest domain. To use these, they must only receive the frequencies they were made to handle. The part of the frequency spectrum used for our system is equal to the HiFi range (this ranges from 20 H z  to 20 k H z ). [3] Not every speaker is able to give an accurate sound representation over a wide frequency range, due to physical limitations. To deliver the most accurate sound representation, specific frequency ranges have to be send to different speakers. In this section, a filter will be build that separates the frequency range, and sends the ranges to the best suited speaker. The best frequency ranges for our specific speaker have been analysed in the ’Speaker Analysis’ section, and will be used here. The speaker filter must satisfy the following requirements: [3] • • • •

Frequenciesthat are not in thefrequency range of a speaker must be attenuated by that specific speaker. The overall transfer function of the frequency and phase must be as flat as possible. All frequencies of the HiFi spectrum must be diverted to at least one speaker element. The frequency ranges for each speaker element must be based on the recommended values from the loudspeaker analysis section.

3.2. Design It is not easy to just start making the filters on the go. Therefore the components and schematics for every  filter are designed in this section. The measurement results of our loudspeakers are used as a baseline for the design procedure of our filters. The results of these measurements can be found in appendix  B.1.  The derivations of the equations used in this section can be found in appendix  A.2.

3.2.1. Impedance Correction The results of the impedance measurements show that the impedance of a speaker increases at high frequencies. This is caused by the inductor in a speaker, which impedance is linearly dependent of the frequency. The increase of the impedance means that the sound pressure increases with the frequency. [3] The compensation of this increase can be done with a Zobel network, which consists of a resistor and a capacitor in series. The Zobel network is placed in parallel with the speaker and since the frequency dependent part of the impedance is different for each speaker element, a different Zobel network has to be made for each speaker 13

3.2. Design

14

element. The result of a Zobel network is that the speaker element can be written as an impedance with only  a real part. The measured values of the resistor and inductor of the three different speakers are shown in table  3.1. The calculated values of the resistor and the capacitor of the three different Zobel networks are shown in table 3.2. The impedance of a speaker element ( Z l s ) will be, due to the Zobel network, equal to  R e . Table 3.1: The measured values of  R e  and L e 

Speaker Low Mid High

R e  7.1 Ω 4.13 Ω 4Ω

Table 3.2: The calculated values of  R z  and C z 

L e  380 µH  170 µH  53 µH 

Speaker Low Mid High

R z  7.1 Ω   4.13 Ω   4Ω  

C z  7.54 µF  9.97 µF  3.31 µF 

3.2.2. Separation filters  An easy way to keep the overall transfer function of the system as flat as possible, is to only use a speaker in the frequency range where its transfer function is as flat as possible. This can be achieved by separating the frequency spectrum into three regions, and passing each speaker with its matching region. In this section three filters will be designed, one for each speaker. The filters will allow certain regions of the frequency  spectrum to pass through, while attenuating the others. The frequency regions of each filter should match closely with each other, to avoid dead bands or volume peaks in the spectrum. The EPO book prescribes two different types of filters. [3] There is a choice between a first order and a second order low-pass filter. The differences between a first and a second order filter are: the maximum slope after the 3 d B   frequency and the availability of a quality factor. For a first order low-pass filter, the maximum slope is 20 d B   per decade. For a second order low-pass filter, the maximum slope is 40 d B   per decade, double as steep as the slope of the first order filter.  [3], [1] The advantage of a steeper slope is that less sound from outside the selected frequency region will ’leak’ to the ’wrong’ speaker. The quality factor gives control over extra attenuation or resonance at the 3 d B  frequency. Because of this, these systems only uses second order separation filters.

 −

 −

 −

−

The 3 d B  frequency of a filter is the frequency at which the power output is halved. It is important that the 3 d B   frequencies of the different filters match, because the two speaker will then both output half power, and make sure the total power remains 100%. On top of that it is important that the filter degree (first or second order) match, because the cut-off slope ( 20 d B  or 40 d B ) should match each other. The choices for the cut-off frequencies can be seen in table  3.3  and are based on the speaker analysis diagrams in appendix  B.1.

−

−

−

−

Table 3.3: The chosen cut-off frequencies

Switch point Low to mid Mid to high

Frequency  170 H z  950 H z 

 Woofer Results of the speaker analysis indicate that the woofer is able to deliver the flattest acoustic transfer in the lowest part of the frequency spectrum. For this reason only the lowest part of our frequency range should be sent to the woofer, while attenuating higher frequency signals. This can be achieved by using a low-pass filter (LPF). The schematic of a second order low-pass filter can be seen in figure 3.1. The following equations can be used to calculate the values of the capacitor and inductor of the second order lowpass filter:

Figure 3.1: A second order LPF

3.2. Design

15

 = Q Z ·lωs  0

L LP F 

 = Z  Q · ω

(3.1)

C LP F 

l s 

(3.2) 0

Filling in what we know gives us the values of the capacitor and inductor: L LP F  9.4 mH  and C LP F  93 µF 

 =

 =

Tweeter  As expected, speaker analysis shows that the tweeter excels at delivering high frequencies without too many variations. That is why a high-pass filter (HPF) is needed. This will filter out the low and mid frequencies that would come out distorted and lets through only the higher ones. The schematic of a second order high-pass filter can be seen in figure 3.2. The values of the components of the HPF can be calculated  with the same equations as used for the LPF. So the values of  the capacitor and inductor are as follows: C H P F  29.6 µF  and L H P F  948 µH 

 =

Figure 3.2: A second order HPF

 =

Midtoner Since the amplifier will feed three speakers, a circuit is needed to filter out the high and low frequencies, so the midtoner will only receive the middle ones. To achieve this kind of behavior, a band-pass filter (BPF) is being used. If the two cut-off frequencies are sufficiently far apart, the BPF can be treated as two separate filters in sequence: a high-pass and a low-pass filter. Then the same equations can be apFigure 3.3: A second order BPF plied for the inductances and capacitors as we used for the LPF and HPF. The frequencies differ enough  when the highest frequency is at least five times bigger than the lowest frequency, which is the case with 170 H z  and 950 H z . The values of the components of the BPF are as follows: C H P F  28.6 µF , L H P F  979 µH , L LP F  5.5 mH  and C LP F  160 µF 

 =

 =

 =

 =

3.2.3. Phase shift  When a current moves through an impedance with a complex part, it is delayed a little, which creates a current that is similar to the original, only shifted. The phase has been altered: phase shift. Normally this should not cause any trouble. The sound will only be delayed with such a tiny bit that it is humanly impossible to notice, but problems could arise when the delay gets too big. When using multiple speakers it is possible to create destructive interference. In this project two speakers will produce the frequencies near the -3dB frequency, but with a different phase shift. The two speakers have different filters and a different internal impedance,  which is likely to result in different delays. The closer the total phase differences are to 180 degrees, the more these will cancel one another out due to destructive interference.

Figure 3.4: An extreme example of two signals that will result in no sound at all

Re-syncing the phases is a difficult if not impossible task, as the phases are also dependent on the distance and angle from the speaker you measure it. So for now, as there aren’t any predetermined specifications of  this nature, this phenomenon was kept in mind while choosing the transfer frequencies.

3.2. Design

16

3.2.4. Volume adjustment The various speakers that are used, produce a different volumeat the same input voltage. [3] Fora plane acoustic frequency characteristic, these differences in volume should be corrected. The speakers that produce a sound which is too loud need an added network for damping. The equivalent impedance of the damping network has to be in combination  with Z l s  to be equal to  Z l s . This is to prevent a redimension of the filter, which is dimensioned for a constant output impedance, Z l s . The circuit for the damping network is shown in figure 3.5.

Figure 3.5: Volume adjustment

For a known value of the damping factor β (in decibel) and the speaker impedance  Z l s , the resistors R 1  and R 2  can be calculated with equations 3.3 and 3.4. R 1



= Z l s  1 − 10(β/20)



 

(3.3)

R 2

= Z l s 



1 10(−β/20)

−1



 

(3.4)

The damping factor β can be determined from the simulation of the filters, which can be seen in figure  18 in appendix C.2. A more detailed description of this simulation can be found in section 3.3. From this simulation the required damping factor equals 1.75 d B  for the mid speaker and 1.35 d B   for the high speaker. The impedance of the speakers, with the added Zobel networks, is also known. Z l s ,mi d  4.13 Ω and Z l s ,h i g h   4 Ω.  With this information the values of  R 1 and R 2  can be calculated, which are shown in table  3.4.

 −

 −

 =

=

Table 3.4: The calculated values of  R 1  and R 2  for the damping networks

Speaker Mid High

R 1 R 2 0.754 Ω   18.5 Ω 0.576 Ω   23.8 Ω

These damping factors were based on the electric measurements. After the acoustic measurements and human listening, all volume adjustments were removed. All members of the group found the mid- and hightones a bit too weak when listening different kinds of music. After removing all volume adjustments from the filters, the group concluded that the sound was more ’clear’ without the volume adjustments. On top of that, the acoustic measurements of the power transfer were flatter. These acoustic measurement results can be found in figure 6.3 in chapter 6.

3.2.5. Final schematics The final schematics turned out to the following figures 3.6, 3.7 and 3.8, where the values of the actually used components are shown. The internal resistances of the inductors are displayed as well. These schematics contain the filters themselves and the Zobel networks. At each circuit the input signal is presented to the terminals of V i n  and the speaker element is connected to the terminals of  V ou t . Low-pass filter Since the exact values for these components were not available, multiple capacitors, inductors and resistors are used together to create the correct values. These can be found in the appendix  D.1, next to the other components of this filter in the tables  1 and 2 in appendix  D.1. Band-pass filter

Figure 3.6: Low-pass filter schematic

3.3. Simulations

17

Figure 3.7: Band-pass filter schematic

 As with the components in the LPF, not all values for the BPF components were available either, so we combined components to get the correct values. These and all the other components can be found in appendix  D.1 in figures 3 and 4 High-pass filter

Figure 3.8: High-pass filter schematic

 As with the components in the LPF and BPF, not all values for the HPF components were available either, so we combined components to get the correct values. These components can be found in appendix  D.1 in figures 5 and 6.  Adjustments Considering the acoustic measurements a few changes were made to the values of the components. The volume adjustment circuits were changed as mentioned in the designated section  3.2.4.  Furthermore, the value of C H P F  used in the BPF was changed from 160 µF   to 190 µF  to get a flatter power transfer which can be seen in section 3.4.

3.3. Simulations The simulations are done using Matlab, because it hasmore possibilities to work with than PSpice. Thevalues of the capacitors and inductors were calculated first. To simulate the filters very accurately, the values of the actually used components were calculated with. The inductors have a small resistance too, so these values  were taken in account as well. Unfortunately, these resistances dropped the power transfer of the low-pass and high-pass filter, as can be seen in figure 3.9. Therefore the decision was made to use a volume adjustment network to get the band-pass and high-pass filters have the same maximum power transfer as the low-pass filter. Another simulation was made including the volume adjustment networks. The result of this simulation can be seen in figure 3.10. For both simulations, the speaker elements were simulated as a resistor with the Z l s   values. Those Z l s  values can be found in section  3.2.1. The Matlab codes used for this section can be found in appendix  C.2.

3.4. Measurements

18

Figure 3.9: Simulation of the filters

Figure 3.10: Simulation of the filters including volume adjustment

3.4. Measurements First, the electric power transfer response and phase of the filters, with the speakers attached, are measured. This is done using the Matlab program ls_measure . This resulted in figure 3.11.

3.4. Measurements

19

Figure 3.11: Measurements of the power transfer to the speakers using the filters

The little dimple in the sum where the midtoner takes on the signal from the woofer, could be easily fixed by reducing the cut-off frequency of the high-pass side of the BPF. This resulted in changing the value of the C H P F  in the BPF from 160 µF  to 190 µF  . The inductor didn’t have to be changed according to the Matlab simulation. This change resulted in figure 3.12.

Figure 3.12: Measurements of the power transfer to the speakers using the filters

 After this change, the acoustic measurements were done. The power amplifier, the filters and the speaker  were attached. The acoustic transfer of the total system is measured using a microphone on a distance of  approximately 1 meter. This measurement is also done using the Matlab program  ls_measure . The results can be seen in figure 3.13 and 3.14.

3.4. Measurements

20

Figure 3.13: The acoustic frequency response for the complete speaker system including the power amplifier and the three filters. Distance from the speaker to the microphone:

±1 meter.

Figure 3.14: The acoustic phase response for the complete speaker system including the power amplifier and the three filters. Distance from the speaker to the microphone: 1 meter.

±

The result in figure 3.14 is not reliable, because the microphone-to-speaker distance was 1 meter. At this distance, some frequencies have made more periods then others, which changes the phase. Therefore the phase response at a distance of 1 meter is not equal to the phase response at a close distance to the speaker.

±

 Power amplifier design 4.1. Introduction 4.1.1. Background Now that the power supply is finished, it is time to build the power amplifier. The power amplifier has three purposes: the amplification of the input signal, blocking a possible DC-signal, and blocking a range of frequencies; the amplifier acts as an active bandpass filter. This is perhaps one of the most complex units of the system, requiring advanced knowledge of linear circuits. But luckily a lot of the hard work is already given in theEPO student manual[3], so it’s more a matter of understanding than designing a whole new circuit.

4.1.2. Objective and requirements The end goals of the power amplifier are as following  [3]: • The amplifier should have a non-inverting configuration • The passband of the amplifier is 20Hz-40kHz (which is the -3dB bandwith) • The voltage gain in the passband equals 25 • A possible DC input signal should be blocked and may not be present at the output • The DC offset of the op-amp may have a maximal amplification of 1.  Also, the power amp should be experimentally tested with either MATLAB or PSpice. Futhermore, the restrictions of the LM3886TF should be taken into account. The datasheet of the LM2886TF can be found in appendix  F.1

4.1.3. Global description of design steps The students were first required to follow tutorial 5 of the EPO Student Manual  [3] to get acquainted with all the possible active filter configurations, before designing the power amplifier. After the assignments we are required to come up with a design. However, it should be specified that the design of the power amplifier isn’t fully done by the students. A PCB and a few possible designs without component values are already  given. It is the student’s task to choose the design, calculate the correct component values to match the given specifications, and solder the calculated components to the PCB. The power amplifier itself needs its own supply circuit; this should be soldered to the PCB as well.

21

4.2. Amplifier Analysis

22

4.2. Amplifier Analysis

Figure 4.1: The schematic of the final power amp circuit.

4.2.1. Discussion schematic This schematic is chosen instead of the other possible schematics, because all the specifications can be met using this circuit. For a clear explanation of this schematic, it can be can divided in two parts. Part one consists of the components R 2 , R a , C a  and C  1 ; part two consists of the operational amplifier and the components R b , R c  and C  3 . The components R 1  and R 2  were already given from the EPO student manual [3].

∗

∗

Part one is actually a low-pass and a high-pass filter connected in series, while part two is the actual amplifier of the circuit, consisting of the operational amplifier and the components used to specify the gain. So when a signal enters the circuit it will: 1. First come across the capacitor C*1, filtering out a possible DC signal. If the signal is AC however, the signal will go through the low-pass filter and the high-pass filter respectively. 2. After the filters a voltage potential is present at the positive pin of the op amp, causing the op amp to amplify the signal on such a way specified with the components Rb, Rc and C*3. 3. Finally; if a DC offset is present in the output of the circuit, because of the capacitor C*3, it’s maximum amplification will be 1.

4.2. Amplifier Analysis

23

4.2.2. Derived transfer functions There are four transfer functions to be derived. The transfer function of the high-pass filter, thelow-pass filter, the amplifier and of the whole system. The transfer functions are as follows: V ou t  V i n   lowpass  V ou t  V i n   highpass  V ou t  V i n  ampli f ier  V ou t  V i n   whole 

= V V ou t 

=

R 2 1  j ωC 1

(4.1)

+ R 2

1 = 1 + RC   j ω

 

(4.2)

a 

+ Rc ) j f   2πC 3 = 1 + 1(Rb  + j f   2πRbC 

(4.3)

3

· V V ou t 

i n   lowpass 

i n   highpass 

· V V ou t 

(4.4)

i n  ampli f ier 

4.2.3. Role of unknown components The role of the possible components varies. R 1  functions as a pull down resistor. C  1 and R 2  are there to both provide a high-pass filter and to block a possible DC input. R a  and C a  are combined a low-pass filter. R b  and R c  are components that determine the amplification by the opamp. C  3  is necessary to make sure a DC component is not amplified by the opamp, because DC can’t flow through the capacitor. And last, C b  and C c  can be used to counter the effects of  C  3  or to implement another filter. However, this isn’t done in our final circuit, because of all the hard calculations needed for the values, and the advantages probably won’t up way  the disadvantages.

∗

∗

∗

4.2.4. Calculation of component values The calculation of the components was done using Matlab. We wrote a program that could calculate certain components if values of other components where already given. This is done using the transfer functions derived in section 4.2.2 and choosing a cut-off frequency. The Matlab program can be found in appendix  C.3. It is impossible to calculate all the values at once using a system of equations, because only the ratio of the values counts, not an absolute number. So there are an infinite possibilities of values. The calculated results are therefore only a possible configuration. A configuration with other values, doesn’t mean immediately that the eventual system will comply less to the specifications.

4.2.5. Recalculation for available components The Matlab program calculated the values based on values we thought were decent to use. However the component values we initially came up with weren’t available. So a recalculation was needed to come up  with values that were actually available. This was done by calculating these new values based on the values that are available. Eventually the component values listed in table  4.1 were calculated.

4.3. Simulations

24

Table 4.1: The initially calculated component values and the values we’ve finally chosen

Specified component Initial calculated value R a  100Ω R b    12.5k Ω R c  300k Ω C  2   6.77µF  C a    15.9nF  C  3 4.7µF 

∗ ∗

Best available component value 1k Ω 20k Ω 470k Ω 4.7µF  0.88µF  1.5nF  4.7µF 

+

4.3. Simulations 4.3.1. Simulation in Matlab and PSpice The circuit is simulated twice, namely with the initially calculated values, and with the values calculated for the available components. The schematics used for the PSpice simulation are shown in figure 4.2 and 4.3. The code used for the simulation in Matlab is found in Appendix  C.3.

Figure 4.2: The schematic of the power amplifier circuit with the initially calculated values in PSpice.

Figure 4.3: The schematic of the final power amplifier circuit with the calculated available component values in PSpice.

4.3. Simulations

25

4.3.2. Final schematic Thecircuit that was eventually made is presented in figure 4.4 with all the component values clearly marked.

Figure 4.4: The schematic of the final power amp circuit with the used values clearly marked.

4.3.3. Results The results of simulation of the final schematic is shown in figure  4.5 and 4.6. The other results of the schematics with the initial values are found in appendix  C.3.

Figure 4.5: The result of the PSpice simulation of the final schematic

Figure 4.6: The result of the Matlab simulation of the final schematic

4.4. Measurements

26

The calculated results agree with the simulated results, since the figures look practically the same. No real differences can be pointed out. As would be expected; in both cases the opamp is ideal and the values of the components are the same.

4.4. Measurements The final circuit is hooked up to a function generator and a oscilloscope to test the basic functions of the system. The results of the measurements are also found in appendix  B.3.

4.4.1. Measurement setup The measurements were performed using an oscilloscope and a function generator. The signal from the function generator is split and one part is put into the oscilloscope, while the other part is connected to the input of thepower amplifier. Finally theoutput of thepower amplifieris connectedto theoscilloscope as well. Then any frequency can be send out from the function generator through the power amplifier and compared to the input on the oscilloscope.

4.4.2. Results

(a) Result with the frequency in the (b) Result with the frequency below  (c) Result with the frequency above passband the passband the passband Figure 4.7: Results of the measurements with CH1 being the output from the power amplifier and CH2 being the reference signal from the signal generator

Note that the circuit indeed amplifies by 25 if the frequency is in the passband; while a phase shift is present and the gain is less if the frequency is out of the passband. With this information the following conclusions can be made: Table 4.2: Conclusion of the results of the final circuit of each of the requirements

Measurable specifications  A non-inverting configuration The passband is 20H z  - 40k H z  Voltage gain in the passband is 25 DC input should be blocked

Measurements Satisfied f   20H   z  and  f   40k H z   are suppressed The voltage gain equals 25 0 V is measured with DC input

 <

 >

4.5. Volume control knob 4.5.1. Introduction  Although the design specification state that the amplifier can only amplify the 25 times, we also liked to have the ability to adjust the volume on the amplifier itself, and not rely completely on the audio source. This has several advantages; for example we can easily mute the amplifier when switching between audio sources to avoid the 50Hz mains hum.

4.5. Volume control knob

27

4.5.2. Schematic  We could have changed one of the 2 resistors near the power op-amp, but to keep things simple, we added an potentiometer right after the input jack. The potentiometer we ended up adding to the circuit was an logarithmic 10k potmeter. The resistor value of the potmeter is not that important, as everything before the power amp is ’voltage-driven’. We did use a logarithmic potentiometer because sound intensity is measured in decibels, which is a logarithmic scale. The schematic can be seen in figure  4.8.

Figure 4.8: Schematic of the volume control knob

 Booming bass design 5.1. Introduction Overall the sound from the system was quite nice at this stage, but the spectrum was not equally present. Especially the 20Hz to 80Hz range was lacking volume with respect to the other tones.

5.2. Design To achieve the better bass sound, a few solutions could be applied. The higher frequencies could be damped,  which would get all tones to sound as good as the bass does. A better solution is to boost the bass tones to the level of the mid and high tones. This implies that an extra op-amp is needed to boost a certain range of frequencies before amplifying with the Power Amplifier.

5.2.1. Needed behavior To keep the setup easy, it should boost the low frequencies and send the higher frequencies on to the power amplifier without changing their amplitude. Thus the wanted dB output response should look like the one shown in figure 5.1.

Figure 5.1: The wanted dB output for the Booming Bass Extension

5.2.2. The circuit This behavior can be achieved using a non-inverting active shelving low-pass filter, which can be seen in figure 5.2a. This filter is not invented by us, but found on a website [ 2]. 28

5.2. Design

29

(a) A non-inverting shelving low-pass filter

(b) The corresponding graph

Figure 5.2: The circuit and its behavior

This filter consists of an op-amp and a passive shelving low pass filter underneath. T his whole circuit follows the next few equations. d  and ω 0  are the same variable as can be seen in figure  5.2a.  This site gives us the circuit including the following two equations to calculate with. d  20 l og 10 (1

= ·

B  + R  ) R 

(5.1)

 A 

= C ·1R 

ω0

(5.2)

B 

Using these two equations the C  value can be chosen and the  R  A  and R B  are dependent on that capacitance according to the following equations. R  A 

= ω · C · (101 d /20 − 1) 0

(5.3)

 = ω 1· C 

(5.4)

R B 

0

5.2.3. Final schematic In our case the bass tones should be boosted by 12  d B  up until 60 H z . This will cause a large section of the mid to be boosted a little as well, but we agree with that. Thus  d  12 d B  and ω 0 2π 60 H z . So we can choose the value of the capacitance, therefore we choose the available value of  C  680 nF , the resistances have to be the following values.

 =

R  A 

 = ·  =

= 60 · 2π · 680 · 10−19 · (10−12/20 − 1) = 1.2k  Ω 1  = 60 · 2π · 680 · 10−9 = 3.9 k Ω

R B 

 

 

(5.5)

(5.6)

 With these values the full schematic can be drawn, it can be seen in figure  5.3. There are two voltage dividers in there as well to get the both the positive and negative voltages  V  20 V  of the power supply down to the maximum allowed voltage V ma x  15 V  of the op-amp.

 =

 =

5.3. Simulations

30

Figure 5.3: The full booming bass extension circuit

5.3. Simulations  When simulating the expected acoustic measurement, the following graph is acquired. The power dB output of the booming bass is summed up with the original acoustic measurements of the speaker  without the booming bass. This results in the graph in figure  5.4.

Figure 5.4: The simulated acoustic measurements including the booming bass extension

5.4. Measurements The low frequencies are boosted enough and the mid tones look still good as well, therefore we agree to build this booming bass. With this booming bass added, another two acoustic measurements are done, they can be seen in chapter 6. These acoustic measurements look very good in our opinion. But the high tones will need some tweaking according to this graph.

5.5. Addition The ’booming bass’ extension works great, but while listening to music the bass can sometimes overpower the other tones. Although we found 12 d B   as an ideal number during the measurements and simulations, we wanted the option to reduce the bass. So we added an option to reduce the boost of  the booming bass by 4 d B . This was done by adding a switch and a 3.9  k Ω  in parallel to the 3.9  k Ω resistor that exists in the circuit already. Thus there are two switches, one boosts the bass by 8 d B  and then another switch can be pulled to boost it with another 4  d B .

  Acoustic characterization of the total system In Figure 6.1 the entire circuit is shown, with the ’booming bass’ extension, power amplifier and filter system from left to right.

Figure 6.1: The whole circuit

By simulating this circuit we get the results in figure 6.2, with the simulation of the system without the ’booming bass’ on the top and the simulation with ’booming bass’ on the bottom. These simulations are very  good as the transfer is as flat as possible. According to these simulations, the system should perform very   well.

31

32

Figure 6.2: The Bode-plot of the whole circuit

The acoustic measurements were done and the results are shown in figure  6.3, with the first measurement being the one without booming bass, the second one with the damped-booming bass, and the last one being  the one with the booming bass. These measurements were done at 5 cm distance for the high, mid, and low speakers individually and a measurement at 1 meter distance for all speakers at once. At first we used a microphone provided by the TU Delft but for these measurements we used a better microphone. The datasheet for this microphone is shown in appendix   F.2.  The graph in figure 6.3  is summation of the measured acoustic responses of all speakers separately at a a distance of 5 cm. The measurements done at a distance of 1 meter were affected by sound of  other sources than our speaker. That is why the other measurement results were used.

33

Figure 6.3: Acoustic measurement without booming bass

 Conclusions and recommendations 7.1. Power supply analysis Because a 40 Ω resistor was not available, it was not possible to load the PSU with specified 1 A , but it was possible to check the validity of the pSpice simulations with the measurements from the 50 Ω and 40 Ω load. Both simulations are within the 5% tolerances of our load resistors to the measurements, thus we can conclude that thesimulations with the 1 A -load are valid andtherefore our PSU will meet all the conditions in real measurements. To optimize the PSU even more, we could have used bigger capacitors to reduce the ripple voltage even more. However, these were not available. To conclude, the requirements enumerated in the introduction were reached. The ripple voltage stays under the 5% required, the discharge time after shutting off the circuit stays under the 2.5 minutes and the power supply is two-sided.

7.2. Loudspeaker analysis The final purpose of analyzing the loudspeakers, is to give recommendations on how certain filters should be made and placed. At first, the transition from the LOW-speaker to the MID-speaker will be discussed. It is desirable for the quality of the sound that the used range of the speaker does not lie in the range of the impedance peak. This is, because the at this frequency, the whole speaker resonates, generating a variety of  undesirable reflections and other resonances. Regarding the LOW, speaker, it is unavoidable that the used range of the LOW-speaker lies in the range of the impedance peak. This has to be solved with a filter. Regarding the MID-speaker, the peak of the impedance has more or less no effect at approximately 120 Hertz. Therefore, when the MID-speaker has an acceptable frequency response a switch can be made. This is at approximately 170 Hertz. Second, the transition from the MID-speaker to the HIGH-speaker will be discussed. The impedance peak of  the MID-speaker has no effect around the range of the switch. The impedance peak of the HIGH-speaker lies around the 1200 Hz. After approximately 1000 Hertz the frequency response of the MID-speaker fluctuates a lot, resulting in disruptions in the eventual sound quality. Therefore, the choice has been made to solve the effects of the HIGH-speaker impedance peak using a filter, anduse a transition frequency of 950 Hertz.  Also, because the frequency response after the 1000 Hertz is increasing significantly, this needs to be elevated, so that this is on the same level. Furthermore, after the 9000 Hertz the HIGH speaker decreases significantly, so this needs to be fixed as well.

7.3. Passive filter design The goal of the passive filter design was to make the overall transfer function as flat as possible. The used speaker consists of three different speaker elements, for the low, mid and high frequencies. Therefore three 34

7.4. Power amplifier design

35

different passive filters are designed and built. Parallel with each different speaker element, a Zobel network  is placed, which makes the impedance of the speaker element frequency independent (for frequencies higher than its resonance frequency).  After designing the Zobel networks, the low-pass, high-pass and band-pass filters are designed. For these filters, the cut-off frequencies are based on the measurement results of the loudspeaker.  According to the simulations of the filters, the high-pass and band-pass filters needed an added damping net work to get a simulated plane electric frequency characteristic. These desired damping networks are achieved  with an volume adjustment network. During the simulations of the filters, first it was impossible to create a flat line when adding the speakers up. This was due to a fault in the simulation, not the design. To simulate a speaker, a parallel RLC circuit was drawn, but because of the Zobel network that was added, the speaker should be simulated as a resistance.  When this was altered, the line became flat like it was supposed to.  While measuring the filters, the filters were connected to the speakers, so that the Zobel network would be combined with the speaker impedance. The electronic signal coming off the output of the filters was measured. The alternative, measuring with a microphone, would be more inaccurate, since the microphone isn’t as accurate as the electronic measurements. The measurements did not correspond completely with the simulations, mainly because of physical properties of the coils. The inductors started to interfere with each other, as these were rather large and were not placed perpendicular to each other and relatively close to each other. This problem was solved by placing the inductance’s away from each other, but the measurements were still a bit off. Further tweaking and testing   with slightly different inductance’s and capacitors ultimately lead the results as seen in section  3.4.

7.4. Power amplifier design The goal of the power amplifier is to amplify the input audio signal with a gain of 25 while suppressing the signal out of the bandwidth. To achieve this goal, first we got acquainted with active filters by doing the assignments in the student manual [3]. Then a schematic was chosen and the component values were calculated to comply with the specifications given also in the student manual. Finally simulations were made, before building the actual circuit.  When the circuit was build, measurements were made to make sure the circuit complied to all the given specifications. The results of these measurements agreed with the results of the simulations and the calculations.  All the requirements are met and when the power amplifier was connected to the rest of the system finally  sound came out of the speakers, which means the power amplifier works successfully. It became clear during the design process that a lot of time was wasted in the project hall figuring out what to do. So it is recommended to read up on the subject before the project time and to spend in the overall process less time on the assignments and begin sooner with the design of the actual power amplifier.

7.5. Booming bass design The aim of the booming bass extension is to boost the lower range of frequencies, because the signals with these frequencies are acoustically not as powerful as the others. There is more than just one possibility to achieve this. In our case, an active shelving low-pass filter is used, which boosts the low frequencies by 12 d B  or 8 d B  and lets the higher frequencies through without boosting. This is a big advantage, because it means that the extension can easily be added to the system, without any summing amplifiers for example. Our cut-off frequencies are quite low, which has as result that the woofer does not carry a big range of frequencies. The booming bass extension does therefore boost the mid-toner a little as well, which could be considered a disadvantage. However, the difference between the sound with and without the extension is big. The extension gives the low frequencies a nice saturated sound and hence it can be considered a successful circuit.

Bibliography  [1] C. K. Alexander and M. N. O. Sadiku. Fundamentals of Electric Circuits, international edition . McGraw-Hill Education, New York, 5 edition, 2013. [2] ElectroSmash. (2015). Boss CE-2 Analysis. ElectroSmash. [Online]. Available: http://www.electrosmash.com/boss-ce-2-analysis. Consulted: 2015-12-16. [3] G. J. M. Janssen, J. F. Creemer, D. Djairam, M. Gibescu, I. E. Lager, N. P. van der Meijs, S. Vollebregt, B. Roodenburg, J. Hoekstra, and X. van Rijnsoever. Lab Courses EE Semester 1 – Student Manual. Course  Labs of EE1C11, EE1P11, and EE1M11. TU Delft, 2015-2016. [4] Rohitbd. (2014). Cross-section of a subwoofer drive unit. Wikipedia. [Online]. Available: https://upload.wikimedia.org/wikipedia/commons/c/cd/Speaker-cross-section.svg. Consulted: 2015-11-20.

36

 Appendix   A. Derivations of the equations  A.1. Loudspeaker analysis Following is a summary of the most important formulas used in this report. Z ( f   )

ω

 ) = U ( f   I ( f   )

(1)

= 2π f  

Z R (ω)

(2)

= R 

(3)

The formula for the impedance of a resistance  Z C (ω)   ωL 

=

(4)

1 =  ωC 

(5)

The formula for the impedance of an inductance  Z C (ω) The formula for the impedance of a capacitance  ω0

=  1

(6)

C L 

The formula for the resonant frequency of an RLC circuit  B ω

1 = RC 

(7)

The formula for the bandwidth of an parallel RLC circuit  Following is the derivation of the formulas and answers used during the section of Loudspeaker Analysis.  For  high values of  ω:  Z (ω) R e  Z (ω) R e  Z L e  (ω) Z DC   ωL e  L e  (8)  ω

= +

=

+

 →  =|

=   L  C L    1 C L  Z C (ω0 ) =   C  =  C  Z L (ω0 )

 

C L 

37

− |

(9) (10)

 A. Derivations of the equations

38

1 =  ωL  + ωC  2 2 =   ω LC +

Y RC (ω)

 

(11) (12)

 ωL 

2

LC + 1 = −ω ωL  − LC   + LC  1 Y RC (ω0 ) =   L 

(13) (14)

 

LC 

= −1  L + 1

(15)

 

LC 

0

=

(16)

  L  LC 

 

=0

(17)

 = 1 1  + R  0 = R 

R eq 

(18)  

Z ω0

(19)

 = R e  + R p = R p =   R DC  + R p →     Z ω − R DC 

 

0

B ω

ω0

(20)

1 = R  1C  → C p =   B  R 

=

p  p 

1

 

C p L p 

Z l oudspeaker 

(21)

ω p 

1 → L p =   ω 2C  0

 = Z R  + Z L  + e 

e 

(22)

p 

1 1 Z R p 

= R e  + ωL e  +

(23)

+ Z 1 + Z 1 L p 

C p 

1 1 R p 

+  ω1L 

(24)

1

1 p   ω C p 

 A.2. Passive filter design Speaker impedance To obtain the equation of the speaker impedance, three general equations are written down. The impedance model for frequencies above the resonant frequency of the speaker consists of an resistor and an inductor in series. The total impedance for the speaker equals the sum of the impedance of this resistor and inductor. This can be seen in equation  25.  Furthermore the impedance of the resistor and the inductor are known. [1] Z l s 

= Z R  + Z L 

 

(25)

Z R 

 = R 

 

(26)

Z L   j ωL 

=

 

(27)

The substitution of equation 26 and 27 in equation 25 gives us the equation of the speaker impedance. Z l s 

= R + j ωL 

 

(28)

 A. Derivations of the equations

39

Zobel network 

Figure 1: Speaker element with Zobel network 

The values of the resistor and the capacitor of a Zobel network can be determined as following. The Zobel network is in parallel with the speaker element. The total impedance of the Zobel network can be written as 1 Z z obel  R z obel  . The equivalent impedance of the total circuit will be Z eq  Z z obel  Z l s . In the next  j ωC z obel  calculations R z obel  and C z obel  will be written as R z  and C z .

 =

+

 =

Z eq 

 =

Z z obel  Z l s  Z z obel  Z l s 

· +

R e R z  Z eq 

 =

R e   j ωL e 

+

R z 

+

1  j ωC z  1  j ωC z 

+

R e   j ωL e  R z 

+ +

z 



(29)

z 

 +  +  +

R e  R z   j  ωL e 

+

Z eq 

·

R e   j ωL e  +  j ωC  + j ωL e R z  +  j ωC 

R e R z 

 =

=



−

1 ωC z 

L e   j  ωL e R z  C z 

+

R e  R z   j  ωL e 

+

−

−



R e  ωC z 

1 ωC z 







 

(30)

 

(31)

The equivalent impedance has only a real part. So the imaginary part can be set to zero. Furthermore the equivalent impedance is equal to R e , because it is desirable that the equivalent impedance to equals the real part of the speaker impedance just after its resonance frequency. The real part of the speaker impedance  just after its resonance frequency can be approximated by the value of  R e . This can be done, because the influence of the inductor on the speaker impedance is relatively low at that frequency. This leads to equation 32.

+ C L e  z  = R e  Z eq  = R e  + R z  R e R z 

R z 

+ R L Ce   = R e  + R z 

 

 

(32)

(33)

e  z 

 = R L e2 

C z 

(34)

e 

Equation 34  gives us the equation for the value of  C zobel , which is dependent on the values of the speaker element. For the value of  R z obel  following theory has been made. When the input frequency is very low, the branch  with the Zobel network will behave like an open circuit and the branch with the speaker will behave like it

 A. Derivations of the equations

40

has only a resistor  R e . That is because the inductor approximates an open circuit at these circumstances. So the equivalent impedance at very low frequencies equals R e . Furthermore, when the input frequency is very high, the branch with the speaker element will behave like an open circuit. The branch with the Zobel network will behave like is has only a resistor, so the equivalent impedance at very high frequencies equals R z . The equivalent impedance has to be independent of frequency. In other words, this impedance has to have at very low frequencies the same value as at very high frequencies. Because of that reason, R z  is equal to R e . To conclude, the values of the components for the Zobel network can be calculated with equation  35 and 36.  Where R e  and L e  are the measured values of the resistor and inductor of the speakers. R z obel 

 = R e  = Z l 

 

 = R L e2 

(35)

C z obel 

(36)

e 

Low-pass filter To find the output voltage in the second order low-pass filter, which is shown in figure  2, the inductor and 1 capacitor are rewritten to impedances. The impedance of a capacitor can be written as Z C  and the  j ωC LP F  impedance of an inductor can be written as  Z L   j ωL LP F . [1] Nodal analysis in node 1 leads to equation  37. Further calculations on this equation gives us equation 40.

 =

 =

Figure 2: A second order LPF

i L 

= i C  + i l s 

U i n  U ou t  Z L 

−

U i n  Z L 

= U ou t 

Z L 

 = j ωL LP F  and Z C  =  j ωC 1

Substitution of  Z L 

(37)

= U Z ou t  + U Z ou t  C 

·

1 Z L 

+

1 Z L  Z C 

U ou t 

 = Z L 

 

(38)

l s 

+

1 Z C 

+

Z L  Z l s 

+

1 Z l s 



· U i n 

 

(39)

 

(40)

in equation 40  leads to equation  41.   By simplifying this

LP F 

equation, equation 42 is made.

1

U ou t 

 =

1

+



 j ωL LP F  1  j ωC LP F 

+

 j ωL LP F  Z l s 

· U i n 

 

(41)

 A. Derivations of the equations

41

1

U ou t 

 =

1

−

ω2 L LP F C LP F 

+

 j ωL LP F  Z l s 

· U i n 

 

(42)

Equation 43 is provided in the EPO book [3]. 1

U ou t 

 =

1

ω2

− ω2 + 0

ω  j  Q ω0

· U i n 

 

Equation 43 together with equation 42 leads to the equations for ω0  and Q .  ω 0

=  L 

(43)

1

LP F C LP F 

 

, which is both

C LP F  , which is the quality factor L LP F  of the filter. With these two equations, the equations for L LP F  and C LP F  can be derived. the resonance and 3 d B  frequency (in radians per second), and Q 

−

= Z l s 

Calculations on the equation for ω0  leads to equation 45.

 = ω1

 

L LP F C LP F 

C LP F 

(44)

0

1

 = L 

(45)

2 LP F ω0

Calculations on the equation of Q  leads to equation 48.

 

C LP F  L LP F 

C LP F  L LP F 

= Z Q 

(46)

l s 

Q 2

(47)

= Z 2

l s  2

 = L LPZ F 2Q 

(48)

C LP F 

l s 

The combination of equation 45 and equation 48 leads to equation 51. 2

1 L LP F 

L LP F Q  = 2 ω2

(49)

Z l s 

0

L 2LP F Q 2 ω20

= Z l2s 

 = Q Z ·lωs 

L LP F 

 

(50)

(51) 0

Substitution of equation 51 in equation 45 leads to equation 53. C LP F 

 =

1

1

Z l s  ω2 Q  ω0 0

Z l s  ω0 Q 

  = ·

 = Z  Q · ω

C LP F 

l s 

0

·



 

(52)

(53)

 A. Derivations of the equations

42

1 Here the Q   equals   , because that gives a critically damped response. Filling in what we know gives us the 2 values of the capacitor and inductor:

L LP F 

 =  1

7.1

· 170 · 2π 2

= 9.4 mH 

 

(54)

 1

 = 7.1 · 1702 · 2π = 93 µF 

C LP F 

 

(55)

High-pass filter

Figure 3: A second order HPF

To find the output voltage in the second order high-pass filter, which is shown in figure  3, the inductor and 1 capacitor are rewritten to impedances. The impedance of a capacitor can be written as  Z C  and  j ωC HP F  the impedance of an inductor can be written as  Z L   j ωL HP F . [1] Nodal analysis in node 1 leads to equation 56. Further calculations on this equation gives us equation 59.

 =

=

i C 

 = i L  + i l s 

U i n  U ou t  Z C 

−

U i n  Z C 

= U ou t 

Z C 

 =  j ωC 1

Substitution of  Z L   j ωL HP F  and Z C 

=



L 

1 Z C 

+

+

1 Z C  Z L 

1 Z L 

+

(57)

l s 

+

Z C  Z l s 



 

(58)

· U i n 

 

(59)

1 Z l s 

in equation 59 gives equation 60.

HP F 

1

U ou t 

 =

(56)

= U Z ou t  + U Z ou t 

U ou t 

 = Z C 

 

1

− ω2L 

1

H P F C HP F 

+  j ωC  1

· U i n 

 

(60)

HP F  Z l s 

 −ω2 L HP F C H P F  leads to equation 61. −ω2 L HP F C H P F  −ω2 L HP F C H P F  · U i n  U ou t  =  j ωL HP F  2 1 − ω L HP F C HP F  + Z 

Multiplication of equation 60 with

l s 

Equation 62 is provided in the EPO book [3].

 

(61)

 A. Derivations of the equations

43

2

− ωω2 0

U ou t 

 =

1

ω

2

− ω2 + 0

ω  j  Q ω0

· U i n 

 

Equation 61 together with equation 62 leads to the equations for ω0 and Q .  ω 0 the resonance and 3 d B  frequency (in radians per second), and Q 

−

1

=  L 

HP F C H P F 

 

= Z l s 

(62)

, which is both

C H P F  , which is the quality factor L H P F 

of the filter. These two equations are the same for the LPF and the HPF, so the equations for L HP F  and C H P F  are the same as well. The derivation of  L LP F  and C LP F  can be found in section Low pass filter of this appendix.

 = Q Z ·lωs 

L HP F 

 = Z  Q · ω

(63)

C H P F 

0

l s 

(64) 0

Filling in what we know gives us the values of the capacitor and inductor:

 1

2

 = 4 · 950 · 2π = 29.6 µF 

C HP F 

 

(65)

L HP F 

4

 =  1

· 950 · 2π 2

= 948 µH 

 

(66)

Band-pass filter P F  The bandpass filter consists of a second order low-pass and high-pass filter. If the high-pass frequency  f   H −3dB  P F  and the low-pass frequency  f   L −3dB  are sufficiently far away from one another, the values of their components P F   H P F  can be calculated as if thefilters were separate. Thefrequencies differ enough when f   L −3dB  5 f  −3dB  according  P F   H P F  to the product requirements. In this case, f   L −3dB  950 H z  and  f  −3dB  170 H z . So the values of the components can be calculated as if the filters were separate. And like in the second order low-pass and high-pass filters the following equations hold for both low-pass and high-pass part of the band-pass filter:

 =

L 

 With Q 

=  1

2

and ω0

= Q Z ·lωs 

 >

 =

(67)

 = Z  Q · ω

(68)

 =   Z l Hs  P F 

(70)

  1

(72)

C 

0

l s 

0

= 2π f  −3d B , the next four equations have to hold:

C HP F 

 =

  1

Z l s 

(69)

P F  8π f   H3dB 

−

 =   Z l Ls  P F 

L LP F 

L HP F 

(71)

2π f  −3d B 

C LP F 

 =

2π f  −3dB 

P F  Z l s  8π f   L −3dB 

The values of the components of the BPF can be calculated as follows:

 

 = 4.1 · 170 · 2π = 28.6 µF 

C HP F 

L LP F 

 =  1

2

4.1

· 950 · 2π

= 5.5 mH 

 

 

(73)

(75)

4.1

= 979 µH 

 

(74)

 = 4.1 · 9502 · 2π = 160 µF 

 

(76)

L HP F 

1 2

 =  1

C LP F 

· 170 · 2π 2  1

 A. Derivations of the equations

44

 Volume adjustment

Figure 4: Volume adjustment

To calculate the output voltage of the circuit in figure  4  as a function of the input voltage, the equivalent Z l s R 2 impedance has to be determined. This equivalent impedance is  Z eq  R 1 (R 2 Z l s ) R 1 . The Z l s  R 2 voltage over the speaker element  Z l s  is calculated as following.

 = +

U l s 

Multiplication of equation 77 with

= R 2Z  Z l s  · U i n  = eq 



= +

Z l s R 2 Z l s  R 2 U i n  Z l s R 2 R 1 Z l s  R 2

+

+

 

·

+

(77)

+

Z l s  R 2 gives equation 78. Z l s  R 2

+ +

U l s 

= R  (Z  +Z lR s R )2+ Z  R  · U i n  1 l s  2 l s  2

 

(78)

 As a result of this voltage divider, the power has decreased by a factor of  β.

β

U l2s 

2

l s R 2 ) = U 2 = (R  (Z  (Z  + R  ) + Z  1

i n 

l s 

2

 

2 l s R 2 )

(79)

The factor β expressed in dB is as following.

β

U l2s 

U l s  = 10 · l og 10 U 2 = 20 · l og 10 U  = 20 · l og 10 R  (Z  +Z lR s R )2+ Z  R  1 l s  2 i n  l s  2 i n 

(80)

To make sure that the input impedance of this circuit is equal to the speaker impedance an expression can be derived for R 1  and R 2 . R 1

Z l2s 

2

R 2

(81)

= R  + Z 

l s 

= Z l s (Z R l s  − R 1 )

(82)

1

To get the equations of R 1 and R 2 as functions of β and Z l s , the equation 80 has to be solved for the part inside the l og 10 . Z l s R 2 R 1 (Z l s  R 2 ) Z l s R 2

+

+

= 10(β/20)

Next equation 82 is substituted into equation 83 to get equation 84.

(83)

 A. Derivations of the equations

45

10(β/20)

10(β/20)

=

R 1

R 1 Z l s 

+ Z l s (Z l s  − R 1 ) R 1

−

+  

−



=

Z l2s 

−

Z l s (Z l s  R 1 ) R 1 Z l2s 

+ Z l s  +−R 1 + R  + Z l s 

R 1



Z l s (Z l s  R 1 ) Z l s  R 1 Z l s (Z l s  R 1 ) Z l s (Z l s  R 1 ) Z l s  R 1 R 1





=



−



 



= Z l s Z − R 1

(84)

(85)

l s 

R 1

1

−

= Z l s  − Z l s  · 10(β/20) = Z l s  1

10(β/20)



 

(86)

Substituting equation 86 in equation 82 leads to the equation for R 2 .

R 2

=

Z l s (1 10(β/20)

 − −

Z l s  Z l s 

−

Z l s  1 10(β/20)



R 2



=

= Z l s 



Z l s 

−



Z l s (1 10(β/20) )

−

1 10(β/20)

−

1 10(−β/20)

−1





l s  = 10(−βZ /20) −1

 

(87) (88)

B. Measurement results

46

B. Measurement results B.1. Loudspeaker analysis Following are the results and graphs of the section measurements

Figure 5: Measurement results for the low-speaker

B. Measurement results

47

Figure 6: Measurement results for the mid-speaker

Figure 7: Measurement results for the high-speaker

B. Measurement results

Figure 8: Impedances for a 1uF capacitor, a 1mH inductor and 12, 56 and 100 ohm resistors

48

B. Measurement results

49

Figure 9: Frequency response graphs for LOW, MID and HIGH speakers

B.2. Passive filter design

Figure 10: The electric frequency response for all separate filters, with the speakers attached

Figure 11: The electric phase response for all separate filters, with the speakers attached

B. Measurement results

50

B.3. Power amplifier design The measurement results of the power amplifier are shown in figures  12, 13 and 14

Figure 12: Result with the frequency in the passband

Figure 13: Result with the frequency below the passband

Figure 14: Result with the frequency above the passband

C. MATLAB codes and simulations

51

C. MATLAB codes and simulations C.1. Loudspeaker analysis Following are the results and MATLAB codes obtained in the section Simulations. R e 

 = 6Ω L e  = 4mH  R p =   30Ω L p =   20mH  C p =   1mF 

Figure 15: The result of simulating models 1 and 2

R_e L_e R_p C_p L_p

= = = = =

[ 6 ]; [ 0 .0 0 4 ]; [ 3 0] ; [ 0 .0 0 1 ]; [ 0 .0 2 0 ];

f _ ve c =   logspace(0 ,   log10 ( 2 0 00 0 ) , 1 0 0 0 0 ); % F o r c o n v en i e n ce s s a ke : s = 1 i . * 2 .* pi   . * f _ v ec ; Z _v ec = R _e + s . * L _ e + 1 ./ (1 ./ R _p + 1 ./ ( s . * L _p ) + ( s . * C _p ) );  '

Z _ a m p l i tu d e _ v e c = ab s ( Z _ v e c ) ; Z _ p h a s e _v e c = r a d 2 de g (angle( Z _ v e c ) ) ; subplot (2,1,1); semilogx( f _ v ec , Z _ a m p l i t ud e _ v e c ) ; axis ( [0 , 2 00 0 0 , 0 , 3 0 ]) ; ylabel( I m p e d an c e a m p l i t ud e [ \ O m e g a ] ); xlabel( F r e q u en c y [ H z ] ); ' '

'

'

subplot (2,1,2); semilogx( f _ v ec , Z _ p h a s e _v e c ) ; axis ( 0 , 2 0 0 00 , - 90 , 9 0 ); ylabel( I m p e d an c e p h a s e [ \ c i r c ] ); '

'

C. MATLAB codes and simulations

52

xlabel( F r e q u en c y [ H z ] ); '

'

Simulating model 1 g r a ph _ na m e = [ Woofer ; Midrange ; Tweeter ]; f _ r es o na n t = [ 6 2. 3 , 8 6 .6 1 , 1 2 72 ] ; f _ ma x = [ 1 22 0 0 , 2 0 00 0 , 2 0 00 0 ]; Z _ f _r e s on a n t = [ 17 . 0 , 9 .3 6 , 4 . 90 8 ]; Z _ f_ m ax = [ 3 0. 0 , 2 2. 0 , 7 . 76 5 ]; Z _ dc = [ 7. 1 , 4 .1 3 , 4 . 0] ; g r a ph _ c ol o u r = [ r , b , g ]; f _ B _o m eg a = [ 4 9. 2 0 , 7 7 . 66 , 8 9 2. 3 ; 7 5 .4 4 , 9 4 . 40 , 1 9 66 ] ; '

'

'

'

'

'

'

'

'

'

'

'

B _ om e ga = ( f _ B _o m e ga ( 2 , :) - f _ B _o m eg a ( 1 , :) ) . * 2 . * pi ; R _e = Z _d c ; L _e = ab s ( (( ( ( Z _ f _m a x ) .^ 2 - ( R _ e ) . ^ 2) ) . ^( 1 / 2) ) . /( f _ m ax . * 2 * pi )) ; R _ p = Z _ f _r e s on a n t - Z _ dc ; C _ p = 1 . /( B _ o m eg a . * R _p ) ; L _ p = 1 . /( ( ( f _ re s on a n t . * 2 . * pi ) . ^2 ) . * C _p ) ; f _ ma t =   zeros ( length ( R _ e ) , 1 0 0 0 0 ); fo r   n = 1:( length (R_e)) f _ m a t ( n , :) =   logspace (0 ,   log10 ( 2 4 00 0 ) , 1 0 0 0 0) ; en d s = 1 i . * 2 .* pi   . * f _ ma t ; Z _ ma t =   zeros ( size( f _ m a t ) ) ; fo r   n = 1: length( R _ e ) Z _ m a t ( n , :) = R _ e ( n ) + s ( n , : ) .* L _ e ( n ) + 1 . / ( ( 1 ./ R _ p ( n ) ) + ( 1 . / ( s ( n , : ) .* L _ p ( n ) ) ) + s ( n , : ) .* C _ p ( n ) ) ; en d Z _ a m p l _m a t = ab s ( Z _ m a t ) ; Z _ p ha _ ma t = r a d2 d eg (angle( Z _ m a t ) ) ; subplot( 2 , 1 , 1 ) fo r   n = 1:( length ( R \ _ e ) ) semilogx( f \ _ m a t ( n , :) , Z \ _ a m p l \ _ m a t ( n ,: ) , g r a p h \ _ c o lo u r ( n ) , hold on en d hold of f legend( g r a p h \ _ n a m e , Location , Best ); se t ( gc a , Fontsize , 1 3 ); grid on grid   minor xlabel( F r e q u en c y ( H z ) ); ylabel( I m p e d an c e ( \ t e x t b a c k sl a s h \ { \ } O m e g a ) ); title( I m pe d a nc e a s f u nc t io n o f f r eq u e nc y ); axis ( [0 , 2 40 0 0 , 0 , 3 0 ]) ; '

'

'

'

Linewidth , 2 ); '

'

'

'

'

'

'

'

'

subplot( 2 , 1 , 2 ) fo r   n = 1:( length ( R \ _ e ) ) semilogx( f \ _ m a t ( n , :) , Z \ _ p h a \ _ m a t ( n ,: ) , g r a p h \ _ c o lo u r ( n ) , hold on en d hold of f legend( g r a p h \ _ n a m e , Location , Best ); ; se t ( gc a , Fontsize , 1 3 ); grid on grid   minor xlabel( F r e q u en c y ( H z ) ); ylabel( I m p e d an c e p h a s e ( \ t e x t b a c ks l a s h \ { \ } c i r c ) ); title( I m pe d a nc e p h as e a s f u nc t io n o f f r eq u e nc y ); axis ( [0 , 2 40 0 0 , - 90 , 9 0 ]) ; '

'

'

'

'

'

Linewidth , 2 );

'

'

'

'

'

'

'

'

Calculating the component values of and simulating the speakers 

'

C. MATLAB codes and simulations

Figure 16: The simulation of the actual loudspeaker

C.2. Passive filter design The Matlab code used for the simulations of the filters can be seen below: clear; j

= 1i;

f w

= 20 : 40 0 0 0 ; = f .* 2 .* pi ;

f _ l = 1 7 0; f _ h = 9 5 0; b _ b = - 1. 7 5; b _ h = - 1. 3 5; Q

= 1 ./   sqrt (2);

% I m p e d a nc e s s p e a k er s Z _ sl = 7 . 1; Z _ sm = 4 . 13 ; Z _s h = 4 ;

53

C. MATLAB codes and simulations

54

% L PF C _l = 9 2. 7e -6 ; L _l = 9 50 0e -6 ; R _ Ll = 1 . 1; Z _C l = 1 . / ( j . * w .* C _l ); Z _L l = j . * w . * L _l + R _L l; V _l = Z _C l . * Z _s l . / ( Z _C l . * Z _s l + ( Z_ Cl + Z _s l) . * Z _L l ); d V_ l = 1 0 . *   log10 (( ab s (V_l)).^2); p h i_ l =   angle ( V _l ) . / pi   . * 1 8 0; % B PF C _b h = 19 0 e -6 ; L _b h = 5 50 0 e -6 ; R _L bh = 1 ; Z _C bh = 1 . / ( j . * w . * C _ bh ); Z _L bh = j . * w . * L _ bh + R _L bh ; C _b l = 29 e -6 ; L _b l = 1e -3 ; R _ Lb l = 0 . 6; Z _C bl = 1 . / ( j . * w . * C _ bl ); Z _L bl = j . * w . * L _ bl + R _L bl ; V_m

= 1 . / ( Z _ C bh . * ( ( Z _ L b l . / Z _ C b l + Z _ Lb l . / Z _ s m + 1 ) . * ( 1. / Z _ C b h + 1 . / Z _ L b h + 1 ./ Z _ L b l ) - 1 . / Z _ L b l ) ) ;

% V AB R 1_ b = Z _s m * ( 1 - 1 0^ ( b_ b /2 0) ); R 2 _b = Z _ sm / ( 10 ^ ( - b _b / 2 0) - 1 ) ; V _m = V _m . * ( Z _s m * R 2_ b / ( R 1_ b * ( Z _s m + R 2_ b ) + Z _s m * R 2_ b )) ; dV _m = 10 . *   log10 (( ab s ( V _ m ) ) . ^ 2 ) ; p h i_ m =   angle ( V _m ) . / pi   . * 1 8 0; % H PF C _h = 30 e -6 ; L _h = 1e -3; R _ Lh = 0 . 6; Z _C h = 1 . / ( j . * w .* C _h ); Z _L h = j . * w . * L _h + R _L h; V _h

= Z _L h . * Z _s h . / ( Z _L h . * Z _s h + ( Z_ Lh + Z _s h) . * Z _C h );

% V AH R 1_ h = Z _s h * ( 1 - 1 0^ ( b_ h /2 0) ); R 2 _h = Z _ sh / ( 10 ^ ( - b _h / 2 0) - 1 ) ; V _h = V _h . * ( Z _s h * R 2_ h / ( R 1_ h * ( Z _s h + R 2_ h ) + Z _s h * R 2_ h )) ; d V_ h = 1 0 . *   log10 (( ab s (V_h)).^2); p h i_ h =   angle ( V _h ) . / pi   . * 1 8 0; % P l ot a m pl i t ud e V _s = (ab s ( V _l ) ) . ^2 + (ab s ( V _m ) ). ^2 + (ab s (V_h)).^2; d V_ s = 1 0 . *   log10( V _ s ) ; subplot (2,1,1); semilogx( f , d V_ l , semilogx( f , d V_ m , semilogx( f , d V_ h , semilogx( f , d V_ s , legend( LP F , '

'

'

' ' ' '

k: k: k: k-

BP F , '

'

, , , ,

'

HP F ,

' ' '

' ' ' '

LineWidth LineWidth LineWidth LineWidth '

'

Su m  , '

' ' ' '

, , , , '

2 );   hold on ; 2 ); 2 ); 2 ); Location , '

'

southeast ); '

grid on ;   grid   minor; x l i m ( [ 20 4 0 0 0 0 ] ); y l i m ( [ - 30 5 ] ) ; xlabel( frequencies ); ylabel( s i mu l at e d d B o u tp u t b y t h e t h re e d i ff e r en t f i lt e rs ); '

'

'

'

% P l ot p h as e subplot (2,1,2); semilogx( f , p hi _ l , semilogx( f , p hi _ m , semilogx( f , p hi _ h , legend( LP F , '

'

'

' ' '

BP F , '

grid on ;   grid   minor;

k: , LineWidth , 2 );   hold on ; k - , LineWidth , 2 ); k - - , LineWidth , 2 ); '

'

'

'

'

'

' '

'

HP F , '

'

'

Location , '

'

northeast ); '

C. MATLAB codes and simulations

55

x l i m ( [ 20 4 0 0 0 0 ] ); y l i m ( [ - 1 80 1 8 0 ] ) ; xlabel( frequencies ); ylabel( s i mu l at e d p h as e o u tp u t b y t h e t h re e d i ff e r en t f i lt e rs ); '

'

'

'

This Matlab code result in figure 17.

Figure 17: Matlab output

The electric power transfer graph for the filters, without volume adjustment is shown in figure  18.

Figure 18: Simulation of the filters, without volume adjustment

C. MATLAB codes and simulations

C.3. Power amplifier design Simulation results  All the simulation schematicsand results from the power amplifier are given in figures 19, 20, 21 and 23.

Figure 19: The schematic of the PSpice simulation of the circuit with the initial values.

Figure 20: The result of the PSpice simulation of the circuit with the initial values.

Figure 21: The schematic of the PSpice simulation after the calculation of the available component values.

56

C. MATLAB codes and simulations

Figure 22: The result of the PSpice simulation after the calculation of the available component values.

Figure 23: The result of the PSpice simulation after the calculation of the available component values.

Matlab code close all clear % I n it i al v a lu e s f _ hp _ 0 = 0 . 5; f _ hp _ 1 = 0 . 00 1 ; f _ hp _ 2 = 0 . 02 5 ; f _ lp _ 0 = 1 0 00 0 0; k = 1; % RA = 1 00 0; C A = 1 .5 e - 9 ; R c = 4 7 00 0 0; C 1 = 8 80 e - 9 ; % R c = 1 : 5 00 0 00 ; ZC = Rc; f = 0 : 10 0 0 00 ; % f = 2 0; f _ ma x = ma x (f);

57

C. MATLAB codes and simulations

58

% C a l c u l at e d v a l u es % C 1 = 1 / ( f _ h p _ 0 * p i * 9 4 00 0 ) ; C 3 = 4 .7 e - 6 ; % C A = 1 / ( 2 * p i * RA * f _ l p _ 0 ) ; R A = 1 /( 2* pi * C A * f _ l p _ 0 ) ; % RB = 2 4* R C; R b = ( ( 1 . / 2 4) . * Z C ) - ( 1 . / ( 1 i . * 2 .*pi .*f.*C3)); Rb =   real ( R b ( 2 0 0 0 ) ) ; % R b = r e al ( R b ); , m  

% P o si t iv e p in V _ t r a n s_ l p = ( ( 4 7 0 0 0 . / (( 1 . / ( 2 . *pi . * f . * 1 i . * C 1 ) ) + 4 7 0 0 0 ) ) . * ( 1 . / ( 1 . + ( 1 i . * 2 . *pi .*f.*CA.*RA)))); V _ t r a n s_ h p = ( 1 + ( 1 i . *( f . / ( 1 . / ( ( R b + R c ) . * 2. *pi . * C 3 ) ) ) ) ) . / ( 1 + ( 1 i . * ( f . / ( 1 . / ( R b . * 2 . *pi .*C3))))); % V _ t r a n s = V _ t r a n s_ l p . * ( ( 1 + (1 i . * ( f . / f _ h p _ 1 ) ) ) . / ( 1+ ( 1 i . * ( f . / f _ h p _ 2 ) ) )) ; % V _ t r a n s = V _ t r a n s_ l p . * ( ( 1 + (1 i . * ( f . / ( 1 . / ( ( R b + R c ) . * 2. * p i . * C 3 ) ) ) ) ) . / ( 1+ ( 1 i . * ( f . / ( 1 . / ( R b . * 2. * p i . * C 3 ) ) ) ) ) ); V _ tr a ns = V _ t ra n s_ l p . * V _ tr a n s_ h p ; % P o we r t r an s fe r f u nc t io n % G f _ l p = 1 . / ( 1 +( f . / f _ l p _ 0 ) . ^ 2 ) ; % W e h eb be n f _1 e n f _ 2 e n a . D us n u d e v er g el i jk i ng e n o pl os se n o m d e % w a ar d en t e k r ij g en . % P l o t t i n g m e c h a n ic s % f = l o g1 0 ( f ); % G f _ l p = 1 0 . * ( l o g 10 ( G f _ l p ) ) ; loglog( f , V _ t r a ns ) ; %plot(f, abs(Rb)) % p l o t ( R c , V _ t r a ns ) ; x l i m ( [ 0 1 0 0 0 0 0] ) ; title( V o l t a ge t r a n s f er ); ylabel( Uout/Uin ) xlabel( Frequency ) '

'

' '

'

'

D. Components

59

D. Components D.1. Used components for the passive filters Since the exact values for these components were not available, multiple capacitors, inductors and resistors are used together to create the correct values. The total value of a component is the measured value of the combination of the first, and if added second component. This combination is in parallel for capacitors and in series for inductors, therefore their values can be summed. Low-pass filter

Figure 24: Low-pass filter schematic

Table 1: The filter components in the LPF the first two components in figure 24 from left to right

L LP F  C LP F 

First component 4.5 mH  47 µF 

Second component 3.3 mH  33 µF 

Third component 1.5 mH  4.7 µF   

Total value 9.5 mH  92.7 µF 

Table 2: The Zobel components in the LPF the last two components in figure 24 from left to right

C Z  R Z 

First component 6.8 µF    6.8 Ω

Second component 0.68 µF 

Total value 7.5 µF  7.1 Ω

Band-pass filter

Figure 25: Band-pass filter schematic

Table 3: The filter components in the BPF the first four components in figure 25 from left to right

C HP F  L H P F  L LP F  C LP F 

First component 150 µF  3.3 mH  1.0 mH  22 µF 

Second component 4.7 µF  2.2 mH  5.6 µF 

 

Total value 160 µF  5.5 mH  1.0 mH  28.7 µF 

D. Components

60 Table 4: The Zobel components in the BPF the last two components in figure 25 from left to right

C Z  R Z 

First component 10 µF  2.2 Ω

Second component

Configuration

1.8 Ω

in series

Total value 10 µF  4.1 Ω

High-pass filter

Figure 26: High-pass filter schematic

Table 5: The filter components used in the HPF the first two components in figure 26 from left to right

C HP F  L H P F 

First component 22 µF  1 mH 

Second component 8.2 µF   

Total value 29.6 µF  1 mH 

Table 6: The Zobel components in the HPF the last two components in figure 26 from left to right

C Z  R Z 

First component 3.3 µF  1.8 Ω

Second component

Configuration

2.2 Ω

in series

Total value 3.3 µF  4Ω

E. Assignments of power amplifier

61

E. Assignments of power amplifier E.1. Tutorial 5.1

Figure 27: An inverting amplifier with a first-order lowpass filter.

1. Derive the transfer function H ( f   ) =

v o  v s  .

v o  v s 

= − Z Z 2

Z 1

Z 2

= R 2||C 2 =

v o  v s 

=−

R 1

= R 1

 

1  j ωC  R 2 1  j ωC  R 2

R 2 = R   j ωC  +1

 +

R 2 R 2 j ωC  1

+

(89)

1

(90)

(91)

2

R 2 = − R  (R   j ωC  + 1) 1

(92)

2

2. Show that the transfer function is of the following form: H ( f   )

=

α

(93)

f  

1  j  f  0

+

and express α and f  0 in the component values R 1 , R 2  and C 2 . From assignment 5.1.1, we can conclude that α

= − R R 

2 1

and f  0

= 2πC 1 R  . 2 2

3. Show that the power transfer equals G ( f   ) = H ( f   ) 2 = 0.5 α2 for f   = f  0 .

|

|

·

E. Assignments of power amplifier

G ( f   )

62

2

 = +

α f  

1  j  f   0

 | |  =    =     = +  + α2

1

α2

f   2  j  f  0

1

f   (  f  0 )2

2

α2 f  

1 (  f   )2

+

(94)

0

4 & 5. Sketch the power transfer function G ( f   )

Power transfer as function of the frequency

4

10

1

10

0

Power transfer in logarithmic decibel scale

3.5

3    ) -1    )    f    ( 10    G    (   g   o    l    *    0    1   r -2   e10    f   s   n   a   r    t   r   e   w   o -3    P10

   ) 2.5    f    (    G   r   e    f   s   n 2   a   r    t   r   e   w   o    P1.5

1 10

-4

10

-5

0.5

0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Frequency f

(a) G ( f   ) on a linear scale.

1.6

1.8

2 10

×

10

0

10

1

Frequency log(f)

5

(b) G ( f   ) on a logarithmic decibel scale

Figure 28: The power transfer as a function of the frequency 

6. Show that the value of 10 log 1 0G ( f   ) for the frequency  f   = f  0  is equal to -3dB relative to the value at  f   = 0. 7. Show that for frequencies greater than f  0  = f  −3dB  the value of G ( f  ) decreases by a factor of about 4 (or 6dB) for each doubling of frequency.

·

Both can be read from the sketches.

E. Assignments of power amplifier

63

E.2. Tutorial 5.2

Figure 29: An inverting amplifier with a first order highpass filter.

1. Derive the transfer function H ( f   ) =

v o  v s  .

v o  v s 

= I I 2 Z Z 2

I 2

= −I 1

v o  v s 

= − Z Z 2

Z 2

Z 1

(95)

1 1

 

(97)

1

= R 2

 

1 = R 1 + C 1 = R 1 +  j ωC  v o  v s 

(96)

= − C  R +2R  1

(98)

(99)

(100)

1

2. Show that the transfer function is of the following form: f  

H ( f   )

= α·

 j  f  

0

f  

1  j  f  0

+

and express α and f  0 in the component values R 1 , R 2 , and C 1 . From 5.2.1 we can conclude that α

= − R R 

2 1

and f  0

= 2πC 1 R  . 1 1

3. Show that the power transfer G ( f   ) = H ( f   ) 2 = 0.5 α2 for f   = f  0 .

|

|

·

(101)

E. Assignments of power amplifier

G ( f   )

 = ·  α

64

2

f    j  f  0 f  

1  j  f  

+

0

2

2

   | |    =  | + |  =    +  = f  

α j  f  0

f  

α  f  0

f  

1  j  f  

12

0

f   (  f  0 )2

f  

α2 (  f  0 )2

(102)

f  

1 (  f   )2

+

0

4 & 5. Sketches of G ( f   )

(b) G ( f   ) on a logarithmic decibel scale

(a) G ( f   ) on a linear scale

Figure 30: The power transfer as a function of the frequency 

6. Show that the value of 10 log 10 G ( f   ) for the frequency  f   = f  0  is equal to -3dB relative to the value for  f   . 7. Show that for frequencies less than  f  0 = f  −3d B  the value of G ( f   ) decreases by a factor of about 4 (or 6dB) for each halving of the frequency.

·

 → ∞

f  

10log 10

   + =

10log 10

α2 (  f   )2 0

1

α2 2

f   (  f  0 )2

f  

  +   ·∞  = α2 (  f   )2 0

10log 10

f  

1 (  f  0 )2

 

(103)

2

α = x lim 10log 10 →∞ 1 +∞

0

(104)

E. Assignments of power amplifier

65

E.3. Tutorial 5.3

Figure 31: A non-inverting amplifier with a first-order low-pass filter.

1. Derive the transfer function H ( f   ) =

v o  v u  .

= v 

The goal is to find the function  H ( f   ) v os   for the first order non-inverting low-pass filter seen in figure  31. This function can be derived with the knowledge that v n  v p . Since v n  and v p  can be expressed in terms of  v s  and v o  the following equations can be derived:

 =

= R  R +1R  v o 

 

(105)

 = R  Z +4Z  v s 

 

(106)

v n 

1

v p 

2

3

4

 With Z 4  being the impedance of the capacitor C 4 So if  v n 

= v p  is true, the following is also true: R 1 v o  R 1 R 2

+

= R  Z +4Z  v s  3

 When we reorganize equation 107 and bring along the fact that Z 4 v o  v s 

3 = R  (1R +1 R + R  j ωC  ) 1

 

(107)

4

3

=  j ω1C  , the equation looks like this: 4

(108)

4

2. Show that the transfer function is of the following form: H ( f   )

= α·

1 f  

1  j  f  0

+

and express α and f  0 in the component values R 1 , R 2 , R 3  and C 4 . From 5.3.1 we can conclude that α

= R  R +R  1

3

1

and f  0

= 2πR 1 C  . 3 4

(109)

E. Assignments of power amplifier

66

3. Show that the power transfer G ( f   ) = H ( f   ) 2 = 0.5 α2 for f   = f  0 .

|

G ( f   )

|

·

2

 = +

α f  

1  j  f  0

 | |  =  +   =  +   = α2

α2

f   2

1  j  f  0

1

f   (  f   )2 0

2

α2

(110)

f  

1 (  f  0 )2

+

4 & 5. Sketching G ( f   )

Power transfer as function of the frequency

4

10

1

10

0

Power transfer in logarithmic decibel scale

3.5

3

   )    f    (

   )    )    f 10 -1    (    G    (   g   o    l    *    0    1   r -2   e10    f   s   n   a   r    t   r   e   w   o -3    P10

2.5

   G   r   e    f   s   n 2   a   r    t   r   e   w   o    P1.5

1 10

-4

10

-5

0.5

0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Frequency f

10

×

10

0

10

1

Frequency log(f)

5

(a) G ( f   ) on a linear scale

(b) G ( f   ) on a logarithmic decibel scale

Figure 32: The power transfer as a function of the frequency 

6 & 7. Prove that the -3dB frequency is  f   = f  0  and that for f    f  −3dB  power decreases 6dB.

 >

±

In case the frequency  f   is equal to the cutoff frequency  f  0  the power transfer will become like this: G ( f  0 )

=

α2

1

   = +  f  0 2  f  0

α2 1 1

= 0.5 · α2 +

(111)

E. Assignments of power amplifier

67

E.4. Tutorial 5.4

Figure 33: A non-inverting first-order high-pass poweramplifier

1. Derivation of transfer function

= u 

The goal is to find the function  H ( f   ) u os   for a first order active high-pass filter for tutorial 4 of the EPO1 tutorials. To achieve this function, first the standard transfer function is derived in section 3.1 and then applied to the circuit. First, u p  and u n  need to be calculated. It can be clearly seen that this is a voltage divider, so  u p  equals u s  minus the voltage drop over R 3 , so in other words, u p  equals the voltage over R 4 . This becomes

 = R  R +4R  · u s 

u p 

3

 

(112)

4

For u n  it’s essentially the same concept, namely a voltage divider. In this case we consider the voltage  i t  flowing back to the opamp as negative feedback. u n  equals u 0  minus the voltage drop over R 2 , so  u n  is the voltage over the impedance Z 1 . We define Z 1  = R 1  +  j ω1C 1 . u n 

= R  Z +1Z  · u 0 2

 

 And because we’re dealing with an opamp we regard as ideal, we can say that  u n  Z 1

R 4 · · u s  u 0 = R 2 + Z 1 R 3 + R 4 The ratio

u 0 u s 

(113)

1

= u p . This becomes  

(114)

is needed so we rearrange the terms and this becomes u 0 u s 

= R  R +4R  · Z 1Z + R 2 3

4

(115)

1

R 

Then the term R 3 +4R 4 is separated from the original fraction and the term  j ωC 1  is factored out from both the numerator and denominator. This leaves u 0 u s 

1 (R 1 R 4 + R 2 R 4 ) = R  R +4R  ·  j ωC   j ωC  (R  R  + R  R  ) 3

4

1

1 3

1 4

(116)

E. Assignments of power amplifier

68

 Applying some more calculation the formula can be written as u 0 u s  f  

f  

1

2

1 + R 2 ) + 1 = R  R +4R  ·  j ωC (R   j ωC R  + 1 3

4

(117)

1

2. Defining α,  f   and  f  

It can be seen that equation 117 is of the same type as the wanted formula

+ f   H ( f   ) = α · f   1 +  j  f   1  j  f   1

(118)

2

α is defined as

R 4 R 3 R 4 .

+

f  1  and f  2  are defined as

1 1 2πC (R 1 R 2 )  and 2πC R 1

+

respectively.

3. Defining the transfer function Itis very easy to showthat f  1 < f  2 becausethe term (R 1 R 2 ) in the denominator of  f  1 is larger than the term R 1 in the denominator of  f  2 . Then it needs to be proven that the transfer function G ( f   ) H ( f   ) 0, 5G ( f   ) if  f    f  2 if  f  1  f  2 .

+

 =

<<

=|

+ f  1 2 2  f   f  21 2 1 f  2 2 | = α · 2 = 2 α ·  f  1 and because 1+ j 

1  j   f  2

= |α · f   2 the other terms, the function G ( f   → ∞) = α2 ·  f   . So, if  f    f  2  the function will be G ( f   )

 =

1 2  is

|=

 → ∞

negligible relative to

2 1

4. Defining critical point of G ( f   )

 <<  f  1 , then the term j  f  f    will be very low so we can neglect in in the equation. Because f  1 <  f  2 , the term

If  f   f    j  f   2

1

becomes negligible as well. So the equation becomes G ( f   )

= |α|2 ·| 11 |2 = α2

(119)

 And if  f    f  2  then both imaginary terms become very large, so that the 1 is negligible in the nominator and the denominator. Then the term becomes

 >>

G ( f   )

2

 f   2 2

→ α ·  f   2

(120)

1

It should be mentioned that there shouldn’t be an equal sign because it is a limit, because for f   ).

 >>  f  2 , ( f   →

∞

5. Plotting the transfer function

Figure 34: The graph of the transfer funcion

E. Assignments of power amplifier

69

 Above the transfer function is shown. It is plotted in MATLAB.

E.5. Tutorial 5.5 The circuit

Figure 35: The first order high-pass active filter the transfer function is derived for

1. Derivation of transfer function

= u 

The goal is to find the function  H ( f   ) u os   for a first order active high-pass filter for tutorial 5 of the EPO-1 tutorials. To achieve this function, first the standard transfer function is derived in the section below and then applied to the circuit. The standard transfer function is in the form: v o  v s 

4 (R 1 + R 2 ) = R  R  (Z  + R  ) 1

3

(121)

3

2. Applied transfer function Since the standard transfer function for an active high-pass filter is derived, we can now apply our custom values to this equation. To do this we must find the impedance  Z 3  for the circuit displayed above. The impedance of the capacitor is  Z 3

=  j ω1C  . ω

= 2π f  

3. Power transfer Derivation of the power transfer function Equation 121 is of the form: f  

H ( f   )  With α

= − R  R +R  1

2

1

and f  0

= 2πC 1 R 

3 4

The equation to calculate the power is transfer is

=α

 j  f  0

f  

1  j  f  0

+

(122)

E. Assignments of power amplifier

70

G ( f   )

= |H ( f   )|2

(123)

To get the power transfer for this circuit equation 122 equation  122 is  is substituted in equation 123 equation  123 as  as follows:

G ( f   )

 = 

α

2

f    j  f  0 f  

1  j  f  0

+

 |  = | | | + α

f   2

 j  f  0

2

f  

|

f   2

1  j  f  0

|

=α

f  

(  f  0 )2

2

 +

(

1

f   2  f  0 )

=α

(  f  0 )2

2

(124)

f  

1 (  f  0 )2

+

4 & 5.Sketches of the power transfer function G ( f   ) as a function of the frequency  The power transfer function if sketched both with a normal linear scale and a logarithmic decibel scale in figure 36a figure 36a and  and figure 36b. figure  36b.

(b) On a logarithmic decibel scale

(a) On a linear scale

Figure 36: The power transfer as a function of the frequency 

6. The -3dB frequency To show that the value of the 10 l og ( og (G ( f   )) for the frequency  f    f    f  0  relative to f   to  f    we need to use the formula:

·

f  

10log 10

   + =

10log 10

α2 (  f   )2 0

1

α2 2

f   (  f  0 )2

 =  =

 →  → ∞

f  

  +   ∞ =

10log 10

α2 (  f   )2 0

f  

1 (  f  0 )2

 

(125)

2

 α = x lim 10log 10 →∞ →∞ 1+∞

0

(126)

F. Datasheets

F. Datasheets F.1. LM3886 LM3886 Overtur Overture™ e™ Audio Audio Power Power Amp Amplifie lifierr Due to the size of the report, only the first four pages of the datasheet is included. These were also the only pages used during the project.

71

F. Datasheets

72

F. Datasheets

73

F. Datasheets

74