I.3.5 Ringing. Ringing=Unwanted oscillations of voltage and/or current

Transcription

1 I.3.5 Ringing Ringing=Unwanted oscillations of voltage and/or current Ringing is caused b multiple reflections. The original wave is reflected at the load, this reflection then gets reflected back at the generator, etc, etc. We will illustrate this b looking at the step change in voltage when a device is switched on. 7 Generator P oad I F F G G 1.At switch-on, a pulse F is generated and travels towards the load At the generator: and GDEPARTMENT F OF ENGINEERING I F G G 73

2 I Thus: F F G But, for an unidirectional (un-reflected) wave: F F G F G I F F. Part of the pulse is then reflected at the load as Where 3. is reflected at the generator as Where G DEPARTMENT G OF ENGINEERING G B F 3 G G F The amplitude at the oad asmptoticall approaches F G G F attice Diagram or Bounce Diagram 75

3 1 P F What is the asmptotic value? (1 ) P F F F (1 ) 3 P F F G F F G 4 P F F G F G F F(1 G G )... (1...) n 3 P F G G G G 76 n P can be rewritten as: n P F[1 G ( G)...] F[1 G ( G)...] Hence, the asmptotic value, for n is: [1 ] ( ) n P F G Since, b definition 1 and G 1 and for 1 n P F[1 ] 1 G 77

4 Substituting the definitions of G and : P G Thus, if we wait long enough, an "transmission line" effects should go awa, and we converge to what we would have if the line was just some wire connecting the source to the load. In this case, the load resistor and the source resistor would form a voltage divider the voltage across the load is determined b the voltage divider equation. If, open circuit, then: P 78 Ringing 79

5 I.3.6 ¼ Wave Matching The impedance of a line is onl in the absence of reflections. With reflections the impedance at point B is a function of: Intrinsic impedance Impedance of the load Distance from the load Wavelength 8 The general epression of impedance at is: Remembering that: ( ) e e F j j B I( ) I e I e F j j B ( ) ( ) I( ) I F F I B B Then: ( ) ( ) I( ) e F F e e e j j B j B j e e j B j F j B j F e e 81

6 Since, from (3.1a), (3.1b): B F We get: e e j j ( ) e j e j ( ) e ( ) e j j ( j j ) e ( ) e Remembering that: j e cos( ) j sin( ) j e cos( ) j sin( ) and cos(-)=cos() sin(-)=-sin() 8 We can replace the eponential with sin and cos and substitute =-b sin sin tan tan cos b j b b ( b) cos b j b j b b ( b ) j b (3.3) A quarter of a wavelength back from the load we have b 4 Remembering that: b 4 j We get tan / b j tan / 83

7 Since tan / The impedance at point b = /4 is: b 4 b (3.4) This epression is important when we are tring to connect two lines with different impedances, and we do not want to have an reflections This leads to the concept of quarter wave transformer 84 I.3.7 Quarter wave transformer Two lines are to be linked. The first has an impedance ine1 = 5, while the second has an impedance ine = 75 What should the impedance of a quarter wavelength section of line be, in order to eliminate reflections? Second line appears as = ine =75 to the ¼ wave link 85

8 Remembering that B F To have no reflections we need We want at b to equal 5, the of line 1, so there is no reflection back along the line. Hence ine1 b 5 75 = The graph below shows how varies along the ¼ wavelength section. Note: this solution is onl valid for one frequenc b = /4 b= 75 7 b [] b b 4 b 87

9 II.1 Electromagnetic Fields Aims Define E, B, D, H, Recall basic maths and Mawell s equations Objectives At the end of this section ou should be able to describe the relation between E, B, D, H, understand the meaning of displacement current densit and its role in Mawell s equations 88 II.1.1 Definitions q - E + A charge q placed in an electric field eperiences a force F which is dependent on the charge itself and on the electric field strength E We define Electric Field, E, such that F qe For an electron: q= C Units E N C m 89

10 Magnetic Flu B q A charge moving with velocit v in a magnetic field also eperiences a force We define Magnetic Flu Densit, B, such that F qvb orentz Force For an electron: q= C N s Cm N ma Units B T ( Tesla) 9 Electric and magnetic fields are closel related. One can give rise to the other and vice versa Electric fields are not onl created b charges (such as the charge on the plates of a capacitor) but also b a changing magnetic field Magnetic fields are created not onl b moving charges, i.e. current in a coil or aligned spins in an atom (as in a permanent magnet) but also b changing electric fields (Mawell s Displacement current, as discussed later) In addition to the above, we have to allow for charges and currents in materials. We thus define two new quantities: 91

11 Electric Flu Densit: D [D]=C/m Magnetic Field Intensit: H [H]=A/m In linear materials, D and E; B and H, are directl related b the permittivit and permeabilit of such materials D=E B=H Permittivities and permeabilities are often epressed relative to those of free space: = r, where is the permittivit of free space = r, where is the permeabiliti of free space and = F/m = H/m 9 Representation of Flu Densit B The Strength of the field = the Number of flu lines per unit area The Direction of the field = the Direction of the flu lines 93

12 II.1. The laws of electromagnetism II.1..-Useful Maths A z a z a A A X a Y ector Notation X AA= A a A a A a z z 94 Curl A: A A A A A A A z z z z a a a z Divergence of A (div A) A A A z A z A Adl. circulation integral c s A. ds flu ds A dl 95

13 Stokes Theorem c Adl. ( A). ds S Divergence Theorem S AdS Ad 96 II.1..1 Mawell s aws Integral Form Mawell-Farada: Edl. BdS. c Mawell-Ampere: where B B t c s Hdl. J D. ds J Conduction current densit [A/m ] D D t Displacement current densit [A/m ] s 97

14 Differential Form E B H J D Differential form linked to integral form b Stokes theorem c E. dl ( E). ds S but E. dl BdS. c E s B Stokes Mawell Farada 98 The Mawell-Farada aw E. dl BdS. a changing magnetic flu B c has rings of field E E s implies that around it B H. dl J D. ds The Mawell-Ampere aw implies that a stead current has rings of H field around it. c s J or a changing electric flu D DEPARTMENT J OF ENGINEERING 99 J H D H. D

15 II.1.. Gauss s aws Differential Form B D where charge densit [C/m 3 ] Integral Form BdS S DdS Q where Q total charge [C] S We can get the integral form from the differential, b integrating over a volume and appling the divergence theorem For eample But B Bd S Bd S BdS S Divergence 1 The divergence gives a measure of the difference between the number of flu lines entering a volume and leaving it, i.e.: X is positive X DEPARTMENT OF + ENGINEERING X is negative Hence, Gauss laws impl that: 1)The B flu lines are continuous, i.e. the are never broken. A flu line eiting from the north pole of a magnet will return to that magnet at the south pole ) The D flu lines are continuous ecept when broken b point charges lines of D begin and end on point charges D - B - 11

16 II.1..3 Displacement current densit It is one of Mawell s ke contributions and eplains electromagnetic waves propagation In Ampere s law the term D is missing. Thus: c H. dl J. ds I s This applies in a wire Mawell added the term D to take into account situations such as a wire with a break in it carring an a.c. current The displacement current allows us to take into account the effect of the gap formed b the break Consider a capacitor with an applied voltage (t) 1 I A B S d When the voltage reverses I A Et () B The charges on plate A move to B, thus creating a current I(t) However, the charges of opposite sign on the two plates create a varing electric field E(t), thus a varing electric field densit D(t)= E(t) 13

17 D t But, in a parallel plate capacitor I(t) in the circuit= ( E) ( / d) t t d t C d S and D t d t t C S S I C t Hence: D t I S J D Displacement current densit 14 II. Electromagnetic Waves Aims To derive the equations for wave propagation in free space, as well as dielectric and conducting media Objectives At the end of this section ou should be able to describe the propagation of a plane wave, derive its velocit and intrinsic impedance in an medium 15

18 II..1 Derivation of Wave Equation Consider an infinite plane z = in which, at all points j t E E,, e j t B, B, e and Hence E and B are perpendicular and uniform B E z B B z E z E z z z In the plane zcambridge z, the UNIERSITY fields will have varied EECTRONIC NANOMATERIAS b DEICES the AND rates of change of B and E with z 16 We can use the diagram to evaluate Mawell s equations and derive the wave equation. c E. dl BdS. E z s E z E Mawell-Farada t zt t We ignore second order terms. Hence we have E B z t Note that the differential form is: E B B B z z E z EECTRONIC NANOMATERIAS DEICES AND E E z DEPARTMENT E E E z OF ENGINEERING E z a z a az AND SPECTROSCOPY BMATERIAS a B GROUP a Bz az B and not B B t z t B 17

19 B E ds B B z z E E z z dl z 18 If we then repeat the calculation but this time using H and D and taking the plane z From c Hdl. J D. ds s H D z t H z H z H z D t No currents (Mawell-Ampere) Note that the differential form is: H J D H H z H H H z H z a z a az D a D a Dz az EECTRONIC NANOMATERIAS DEICES AND H z D D t 19

20 H dl H H z z D ds D D z z z B and H are directl related b the permeabilit : B H D and E are directl related b the permittivit : D E 11 Summarising: E B z t H z D t The net step is to eliminate B from the first equation and D from the second. Since BH D E We get the following equations in E and H: E H z t H z E t (5.1) (5.) 111

21 These are similar to the telegrapher s equations: I (1.1) t I C (1.) t Appling the same technique of differentiating eq. 5.1 and 5. with respect to z, and substituting in from 5., we end up with the equations for electromagnetic waves in free space (or pure dielectric medium, with J=). E H E z H t z t E CAMBRIDGE z UNIERSITY t z EECTRONIC NANOMATERIAS t DEICES AND H 11 E z E t H z H t These have the same form and therefore similar solutions to the equations for waves in transmission lines Wave velocit is defined b: velocit 1 analogous to 1 C 113

22 In free space Remember [ F] and [ H ] F 7 H m m s s 8 m s m s Speed of light All the results obtained from the telegrapher s equations can now be reused for the equations for electromagnetic waves j z j z j E t e EFe EBe e H e H e H e e jz jz jt F B 114 E X H z 115

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24 II.. Intrinsic Impedance For electromagnetic waves we define a quantit: intrinsic impedance This is a function of permeabilit and permittivit in the same wa that, the characteristic impedance, was a function of inductance and capacitance per unit length. (.1) C (5.3) 118 links E & H in the same wa that linked & I. E H F F E H F B Similar to I I F Note: The vectors E and H are orthogonal to one another hence the subscripts and in our epression for B B B In free space =

25 II..3 Wave propagation in conducting media So far we have considered EM wave propagation in air or in a pure dielectric medium: = conductivit = If then J J E The Mawell-Ampere law is then.. Hdl J D ds E E. ds c s s This changes the wave equation to: E E E z t t Helmholtz equation 1 We then proceed in a similar fashion to what done in loss transmission lines For simplicit we assume E f( z) e jt E t E t ( ) j jf z e t je j E E z j j E E ( j) j j propagation constant (5.4) 11

26 Re E E e E e e (5.5) j z j z j t F B Re H H e H e e (5.6) j z j z j t F B Then E H F F E H B B Where j j i e is a comple impedance (5.7) 1 Note that if ( i.e. infinite resistivit) then j j Having a finite conductivit leads to a comple impedance. This, in turn, leads to the E and H waves having a real deca term in the eponent. This signifies a wave with decaing amplitude as it travels into a conductive medium. The deca constant is e - 13

27 II..4 Eample Characteristic Impedance A printed circuit board is one millimetre thick, has an earthing plane on the bottom and r =.5 & r = 1 Estimate the characteristic impedance of a track mm wide. l d=1mm W=mm Capacitance C Area Wl ' r r d d ' C AND SPECTROSCOPY 44 MATERIAS r GROUP l d m Capacitance per unit length C W pf 14 Wave velocit 1 1 C r r Hence :.631 C r r 6 H m Then 1 C 15

28 II..4 Eample Termination of a Coaial Cable BNC Connector (Baonet Neill Concelmann) 16 The characteristic impedance is given b: C And the voltage reflection coefficient is: In order to avoid unwanted reflections, we need a to terminate our coaial cable which has the same impedance as the characteristic impedance of the cable. 17

29 b Capacitance per unit length r l b outer radius a inner radius C a q Q with q= charge per unit lenght l DdS Q D E S Erl ql 18 But Then b q E r d E dr a a q q a q b Edr dr ln ln r b a b C b ln a 19

30 We can work out the inductance per unit length of the cable from Ampere s law: H. D J C. Hdl. JD. ds S Where S is an surface bounded b a closed curve C. In the absence of a changing electric field we can simplif the equation to: Hdl. J. ds DEPARTMENT COF ENGINEERING S. D 13 Since J is the current densit then: J. ds I S for the surface shown below r H I Surface (S) Closed curve (C) Note: since the field strength comes from the total current flowing, this eplains how coaial cables shield fields. The inner conductor carries current flowing in one direction. The outer in the opposite direction. Hence, summing the currents for a surface which includes both inner and outer conductors gives a total current of, hence a field strength of 131

31 The field strength H in the cable is then: a H H I r For a length of coa the total magnetic flu is derived b integrating the field strength between inner and outer conductors. Since b I ds=drdl the inductance per unit length is : NOT rdr Radial b I b Hdr ln a a b ln a 13 Hence C b ln a b ln a If we consider a tpical b/a~3-3.5 b ln a And a tpical r ~ (e.g. Teflon, PTFE) We get a tpical value for ~5. Thus, we can use a 5 resistor to terminate our coa So, for eample, if we have an input impedance of 5 for a television aerial socket, then the aerial lead should also have a characteristic impedance of 5 133

32 II..6 The Ponting vector and power in EM waves et us consider a parallel plate transmission line: I I d w E H 134 From the Mawell-Ampere law: c Hdl. JD. ds s But D is in the same direction as to d S dda z Thus: E. Therefore it is orthogonal s D ds Hdl c. JdS. s 135

33 I I d w E H ds=dda z 136 We consider W>>d and neglect fringe effects Solving for the top plate: H W Solving for the bottom plate HW I I Since the current is I on the top plate and I on the bottom, the H fields sum inside the transmission line, while the cancel outside: H Inside I I I W W W H Outside I I W W 137

34 I H W The electric field and the voltage are related b: Ed But the transmission line average power is: PAv 1 Re 1 Re I * E * H Wd=Area of the line Wd The average power densit of the electromagnetic wave is then PAv Area 1 Re = 1 R * * CAMBRIDGE E H UNIERSITY e E H W AND SPECTROSCOPY m MATERIAS GROUP * S EH Comple Ponting ector This gives the direction of power flow, which is is perpendicular to both EandH For a plane wave E E,, and H, H, 1 S E H a Thus * Since z E H E Average Power= F C.f. in a transmission line= H E S z J. H. Ponting 1884 z 139

35 S EH Ponting vector points in the direction of propagation (note different definition with respect to Comple Ponting ector) Wavefront is locus of points having the same phase E H S z 14 II..6.1 Eample Duck a la microwave A duck with a cross-sectional area of.1m is heated in a microwave oven. If the electromagnetic wave is: jtz j tz E Re 75 e m, H Re e Am What power is delivered to the duck? 1 * P PAv Area Re E H Area (75).1 75W 141