Electromagnetic Fields and Waves
Electromagnetic Fields and Waves ECEN 3400
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Date Created: 10/29/15
Design and Construction of a Magnetic Speaker with Audio Ampli er Adeel Baig Alex Starik and Sarah Makhija ECEN 3400 Electromagnetic Fields and Waves University of Colorado Boulder CO 80309 Email adeelbaigcoloradoedu starikalyahoocom sarahmakhijacoloradoedu Abstract This paper presents the design and fabrication of a magnetic speaker and audio ampli er using inexpensive and widely available materials I INTRODUCTION A speaker is a device which converts an electrical signal into vibrations we interpret as sound The diagram in Fig 1 shows the layout of the components that are essential to the construction of a functional magnetic speaker SIN1 39I Ham 1er nus I r M quot 1 tllfil lilill HASHJ l I r l l l 39xlf l 39 r li i Fig 1 Basic Speaker Diagram The voice coil functions as an electromagnet An electromagnet is characterized as a device that is wrapped with wire through which current ows to magnetize the core device Alternating current is passed through the wires and the eld around the coil either increases or decreases depending on the frequency of the current This varying eld then interacts with the constant eld of the permanent magnet which causes movement of the voice coil Movement of the voice coil results in movement of the cone vibrations in the air are produced which we interpret as sound When the electrical signal is applied and electrons ow through the coil and through the magnetic field of the permanent magnet they experience a Lorentz Force FqEvgtltB Equation 1 Lorentz force equation where q is the charge of the particle E is the electric eld v is the particle s velocity and B is the magnetic field Consequently an electron will be accelerated in the same linear direction as the electric eld but will spiral when traveling through the magnetic field This force is required to cause movement in the voice coil and cone thus creating sound waves Using trial and error we chose the components that resulted in best sound quality within our budget of under 10 The initial parts we tested were a Styrofoam cup cone ceramic magnets with a 12 inch diameter and 3 16 inch thickness permanent magnet 22 gauge enamel magnet wire and a 1 12 inch brass plated fastener voice coil Fig 2a illustrates how we constructed the rst speaker About 1 foot of wire was wrapped around the head of the fastener The coil was kept within 3 16 inches of the head in order to leave space for placement of the magnet The fastener was pushed through the center of the cup and the two ends of the fastener were bent as shown in Fig 2b Fig 2a Construction of speaker using 22 gauge wire with 1 12 inch fastener Fig 2b Fastener is bent on the inside of the cup to cause the bottom of the cup to vibrate The magnet was placed between the head of the fastener and the cone The two ends of the wire were connected to a headphone cable and connected to an audio source No sound was audible even though the volume was fully turned up Problems with some of our choices soon became obvious The wire chosen was too thick and heavy and there were few windings around the fastener The fastener itself was large and would require a greater force to move The placement of the magnet between the fastener and the cone was also a poor choice since the added mass limited the movement of the voice coil In order to fix the design we made the following adjustments 2 feet of 30 gauge wire was obtained this resulted in a great reduction of weight as well as increasing the number of windings around the fastener The l 12 inch fastener was replaced with a 3A inch fastener for further reduction of mass The solution for the reduction of the magnet s mass was to cut off the bottom end of another Styrofoam cup and place it over the end of the first cup a new and more powerful magnet taken from a hard disk drive was then placed as shown in Fig 3a Fig 3a Placement of larger magnet on the cut out bottom of a cup As the two ends of the wire were connected to an audio source faint but audible sound was heard as the cup was placed over an ear To further improve the design the length of the wire was increased to 3 feet to increase the number of windings The finished speaker is shown in Fig 3b As expected this resulted in an increased volume of the sound but it was still necessary to hold the cup over to our ears in order to hear the sound The next step was the design of an amplifier which would make the output sound louder Fig 3b Complete improved constmction with part from Fig 3a placed on the end of the Styrofoam cup with voice coil 11 AMPLIFER II Amplifier Circuit To amplify the sound in order to make it audible we added an amplifier circuit The schematic of such a circuit can be seen below R2 lOOkQ VCC15 v R 2 7 10139 6 3 i 4 LF356 1in T 55 Figure ampli er circuit As we can see from the diagram a signal sound such as music or a signal such as a sine wave is sent through Vin The inverting opamp acts as a main source of amplification After experimenting with several possible gains we determined to give the opamp a gain of around 7 this appeared to be the higher gain that the circuit could handle without breaking down burning BJT s resistors or opamps A gain that was lower while safer would not amplify the sound to a desired level A BIT buffer was added to further improve on the performance of the amplifier this stage further improved the gain and made the signal less distorted We decided to use BJ T transistors over MOSFETs due to the fact that they produced a slightly less distorted output as seen on the oscilloscope hence the sound produced should be closer to the original This was especially true for higher frequencies We made improvements to this circuit by adding decoupling opamps which grounded the power supplies pin 4 amp pin 7 of the opamp Decoupling capacitors can greatly reduce distortions Fig4a Output of a BJT amplifier Eig 5b output of a OFET amplifier For the same signal input MOSFET has noticeable distortions clearly visible around the Xaxis The first problem we encountered was large power dissipation across the output Since all our parts are designed to handle power of no more than 25W in the initial stages of testing several BJTs were destroyed Since we were able to hear the sound for a few seconds before the circuit broke down we concluded that the problem lay in excessive power supplied to the output To prevent this problem we added a 1W resistor all other resistors used are 25W between the output of the amplifier located at the emitter of both BJT s to divert excessive power from the output the resistor in a sense acted like a current divider We tested several possible resistor values with resistors over 100 2 the circuit was safe from breakdown yet the sound was too low When the resistor was reduced to around 159 the sound was very clear yet after some time passed 25W resistors would burn We finally settled for a resistor value of 302 the sound level was high enough to be easily audible and during testing the speaker operated without breakdown we ran it for over 5 minutes III ANALYSIS Without calculation we were able to conclude that the speaker operated successfully Simply listening to music was proof However calculating the force on the Styrofoam cup proved to be an interesting and useful method in further understanding our theory and design By finding the self inductance of the voice coil we were able to calculate the energy and therefore the force using the following values Radius of coil 25 cm Length of wire 9144 cm Number of windings 58 Length of solenoid 316 inch 0047625 In Input current 130 mA L uNZSb L u5827z39002520047625 174e5 H Equation 2 Inductance of a solenoid where u is permeability N is the number of windings S is cross sectional area and b is length w 5LI 2 w 5174e513 2 147e7 Equation 3 Energy where L is inductance and I is current intensity a 4 Equation 4 Force and energy with displacement d very small WFd We were unable to measure the displacement of the cone as it was extremely small However for curiosity s sake we can make a rough estimate of d 1e6 m in order to get an idea of what the value of force may be In this case F Wd 147e71e6 147 N From this we can further realize that the electromagnetic system results in a force on the Styrofoam cone producing movement In addition measurements of frequency and output voltage of the complete system were taken and produced the Bode diagram in figure 5 1 m mu mun mnnn mnnnn Output voltage V Frequency Hz dB 9 6 1 80000 1964542 85 1 50000 1858838 65 1 30000 1625827 53 1 20000 1448552 49 1 15000 1380392 46 1 12000 1325516 43 1 10000 1266937 4 1 8000 120412 36 1 5000 1112605 28 1 3000 8943161 23 1 1000 7234557 22 1 700 6848454 21 1 500 6444386 21 1 300 6444386 21 1 100 6444386 21 1 80 6444386 21 1 50 6444386 21 1 10 6444386 Fig 5 Bode plot and values of frequency and output voltage A trend is seen in the system s response to varying frequencies With rising frequencies output voltage increases with leveling occurring at low frequencies 10 to 500 Hz Values above 80000 Hz were not emphasized since audible sound waves are noticed only up to about 20000 Hz IV DISCUSSION This paper demonstrates the ability to construct a functioning electromagnetic speaker and amplifier with inexpensive components We can conclude that following our design for a speaker and building it using common materials is an effective way to create an operational sound system ACKNOWLEDGEMENTS We would like to thank Karl Erickson for suggesting this project to us and for his guidance along the way We would also like to thank Dr Richard Mihran for discussing aspects of speaker design theory with us We would like to thank Professor Dragan Maksimovic for his interest in our project and for helping us obtain the necessary parts in order to build a functional amplifier REFERENCES 1 D Weems Designing Building and Testing Your Own Speaker System New York NY McGraw Hill 1997 2 H Davidson SMD Electronics Projects Indianapolis IN PROPMT Publications 2000 3 M Ryerson Acoustic Troubleshooting of Audio Systems Reston VA Reston Publishing Company Inc 1979 4 S Matt Electricity and Basic Electronics Tinley Park IL The GoodheartWillcox Company Inc 1998 5 Z Popovi and B Popovi Introductory Electromagnetics Upper Saddle River NJ PrenticeHall 2000 6 A Sedra and K Smi Microelectronic Circuits New York NY Oxford University Press 2004 Magnetics Q1 Z 1 A X If I does not change in time and the charge is stationary the magnetic eld at the charge is A B0 B BBuX C BBuz By the BiotSavart B 11 law and the right 2 hand rule E B B uy Magnetics QZ Z 1 A X Q If I does not change in time and the charge is stationary the force on the charge is By the BiotSavart e F 0 law force is proportional to B F 11 Charge velocity X C F F uz D F F uZ E F F uy Magnetics QB 1 I If the positive charge is moving as shown what is the force on the charge Force is Q V crossed F 11X into B B FFux C FZFuZ D FFuz E FZIFuy Magnetics Q4 A X b V Q If the current is magically restricted to a length 11 and is r away from the charge as shown what is the magnitude of the force A Not enough information 1339 Z 0 By BiotSavart F Q V itO4 7t1dla2 D F 7ta21Q E F Q V itO4 n 1 d1 212 b2 Magnetics Q5 r O 8 ltlgodlzll IOIenc o 8 L o N turns per length L 8 O O O gt a If we agree by symmetry that B must be along the axis of the in nite solenoid and use Ampere s law to evaluate B outside the coils on the path as shown what is the value of B A Not enough information B 0 By Ampere s Law total enolosed C B IaoNIL current 1s zero D B ZuONIL E B Z oNIZ r Magnetics QG 4B1 ll lOIeIZC N turns per length L r OOOOOOOOOOOO a If we agree by symmetry that B must be along the axis of the in nite solenoid and use Ampere s law to evaluate B inside the coils on the path as shown what is the value of B A Not enough information B B 0 B uONI L By Ampere s law D B ZuONIL E B ZuONIZ r Magnetics Q7 N turns per length L OOOOOOOOOOOO If the coil is now led with iron of permeability Hr what is B inside A Not enough information BBO B 23231 to last D B IUONILur E B uONIL Farfield radiation What directions do you expect E and H to be in far from the antenna A E2 H B EZ H Co E29 1 Mustbe in spherical d39 t f f E9 HQ 236212 E E H give power outwards 0 39 Plane waves and dielectrics Material 1 Material 2 A plane wave is incident from Material 1 onto Material 2 If we know that Material 1 and 2 are identical polar materials except that the separation between charges may be different in the two materials we can conclude Separation 2 lt Separation 1 B Separation 2 Separation 1 C Separation 2 gt Separation 1 D Not enough info Refraction angle and wavelength greater in 2 therefore index is lower Index is sqrt of dielectric constant which is proportional to dipole moment which is charge separation times charge thus separation 2 lower Ante n n a i n k No radiation along I I 3 axis b JL Transmit Cross polarized Which receiver gets a signal A a B bampc C aampb b Ec Vector calc You re going on a hike in the Swiss alps Given that the local terrain is at an altitude AltXy What quantity is of most interest to you Alt A VAlt VAlt C VgtltAlt D V2Alt You want the gradient slope and direction Which surface has negative flux of E How much energy is required to move one Coulomb of charge between point A and point B on the path shown FT lV G gt Q 600 Free charge is placed uniformly throughout an insulating body We then magically let the body become a perfect conductor How does the B eld in the second case compare to the rst Field zero inside metal far away only total charge matters Via Gauss law Inside Near A O 0 e o lt Smne D Same Same Far 0 Same 0 Same Charge accumulation and thus eld greater at small radius zero inside Free charge is on an oddly shaped conductive body We measure the force on test charges placed at the labeled locations Which is true A FAFBFCFDO B FDgtFAFBFC FAR FB gtFCgt FD FCgtFAgtFBgtFD gt X X By image method Plane must be moved in with insulating gloves so it doesn t pull charge from say ground The B eld in the left hand diagram is measured at point X Then a conductive plane is moved into the space as shown on the right The B eld is now measured to be A The same B O 12 it s former value Twice it s former value it Iosinat Area A l l l D1electr1c er Dt The displacement vector eld across the capacitor plates is A Dt 10 A sinoo t Dt 10 a A cosoo t Dt 10 a A cosoo t Dt 10 er A sinoo t By charge continuity and the sign shown for I charge on the upper plate is the time integral of the current Surface charge density D Q over A A1 erl A2 gr2 A39r3 Q d1 d2 d V II II 0V Three capacitors are attached in series as shown Given a known charge Q on the first capacitor plate What is the total voltage across the stack of thee capacitors VdeiQeoer iAi V2808riiAliQdi C V 2 di Q Ai D V2diQ808ri Charge on each plate must be either or Q D in each cap is thus 6 QA E in each cap is D 8 and V across each cap is d E Sum them up to get the total V V 0V A tub of salt water a moderately good conductor is placed in an electric eld creating a force on the free ions in the salt water Given Newton s law F m A the ions accelerate What limits their speed A Nothing Relativity Collisions Molecular forces between dipoles This is the basis of Ohm s law and Joule heating Q Closed Wire loop in the Bl xy plane 3Z0 A Wire loop hanging in zero gravity is normal to an increasing B eld What happens to the loop A It moves in the Z direction B It spins clockwise in the plane of the page It spins around the B eld t contracts slightly Nothing I threw this one out due to my sign error Find emf and current in loop going in lefthand sense then use BiotSavart to nd force on moving charges The self inductance of a solenoid is measured to be L Without changing anything else the number of turns on the coil are doubled using more Wire of course The inductance A Doubles Stays the same Quadruples Halves One factor of N comes from ux generated by N loops of the same current A second comes from the need to integrate N times around the coil to nd the EMF So L goes like N squared I Measurement Va point after Sohorot re ection Cll Clllt Short reverses voltage but time must have same sense t VAV A voltage pulse shown is introduced onto a lossless transmission line terminated in a short The voltage pulse travels along the line bounces off of the short and is then measured vs time at the point shown The measured voltage trace of this re ection is A We measure the VSWR on a transmission line to be 2 How much power is owing backwards relative to the amount of power owing forwards A 12 B 14 U3 U9 Solve VSWR expression for p VlV1 13 This gives ratio of backward voltage amplitude to forwards voltage amplitude Power is voltage squared over impedance so backwards power to forwards power is like p2 The characteristic impedance of a transmission line is The ratio of the forward voltage to the forward current B The ratio of the total voltage to the total current C The ratio of the backwards voltage to the backwards current D A and C but not B This is a bit of a trick question but since the sign de nition of voltage and current is the same for a forwards and backwards wave either the voltage or current must have the opposite sign for a backwards wave and thus the characteristic impedance is minus V39 I39 Air Em Plastic A gtEout B ltEout Em D Not enough information Boundary condition is that E is conserved tan Air Air DOW Plastic lt Dout C Dout D Not enough information Boundary condition is that Etan is conserved amp D e E a will be greater in plastic than in air thus Din gt Dout Air Air Plastic Boundary condition is that D D Not enough information norm IS conserved Air Air Eout Plastic A gt Eout E 2 g m C Eout D Not enough information Boundary condition is that Dnorm is conserved amp D e E a will be greater in plastic than in air thus Ein lt Eout Air Midterm review Q1 A charged body is in the middle of nowhere infinitely far from anything else The electrostatic force on the body A Depends on the charge distribution on the body B Can not be determined IS zero Coulomb s law always produces equal and opposite forces between charges so charges on body produce no net force D Depends on the reference point for potential Midterm review QZ A magician designs a trick in which a charged insulating wand will be used to 1 levitate a rabbit via electrostatic attraction 93 d Assuming the rabbit is small what is the e W 5 force A Fz L c F 9192 4780 d dL 47reoltdu2gt2 Treat small rabbit as point integrate Coulomb s law over wand with linear charge density Q2L F Q1Q2 L l DF Q2 2 4 39 0L d dL 4780dL2 Midterm review QB 57 Will the trick work A Sure and it will be C Nope rabbit will cool die due to high Electrostatic forces are not large enough to move a rabbit without exceeding breakdown voltage of air I I B Need values for Q Nope alr WI d L to decide break down first Didn t do this one as a Clicker Q end of Class Midterm reVIew Q4 Write the expression for the total charge given a distribution of charge p p0 ra cose lt mum Wit it llIl39 a l I quotWNW u Iquot I I n quotIil 39c a in it39s 139 Radius a A Q pickfifdg zjfd p0 10036 0 0 0 a 7 27 B Q Jrsin dg jrdapoicosa a 0 0 a 7 27 Q Jdrjrd l rsin6d poicos l 0 0 0 2 a D QJdrjasin6d Jad polcos l a 0 0 0 Standard integral over spherical volume of p Midterm review Q5 A current density J is flowing through a S J block of conductive material with area S G 8 and length L If the conductivity is o 39 and its dielectric constant is a what is 39 L 39 the resistance of the block A RzaL C RzSLO39 Use unit analysis and a little logic OR find voltage and current and plug into Ohm s macroscopic law RLSO D R808L0 a Midterm reVIew Q6 A block of cherry jeIIo is inserted between two metal spheres with charges 0 and Q Sketch the total charge distribution Induced surface charge 0 density will vary in both bodies a B Midterm review Q7 A nite volume of space is lled with a charge density p Cm3 that is finite everywhere The potential relative to in nity is A Undetermined from C Finite outside the this information charged region Only get in nite potential for in nite charge density eg ideal point charge Finite everywhere D Zero Midterm review Q8 The charged sphere from the last problem is in empty space The equallyspaced equipotential lines lookae V is like 1r so circles with increasing spacing going outwards review QQ A charged metal sphere is placed inside a dielectric sphere The E field beyond the dielectric is A E 211 C E Q ll 47rgogrr 4350 Use Gauss law on a sphere outside the dielectric yielding Coulomb s law Q E u Q B 47229an r E r 4723980r2 Midterm review 010 Lightning is about to strike a tree You should stand To reduce the voltageditterence between yourteet given a radial current out from the strike increase the t conductivity C Doesn t matter
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