Statics, Quasistatics, and Transmission Lines

Transcription

1 CHAPTER 6 Statics, Quasistatics, an Transmission Lines In the preceing chapters, we learne that the phenomenon of wave propagation is base upon the interaction between the time-varying or ynamic electric an magnetic fiels. In this chapter, we shall use the threa of statics-quasistatics-waves to bring out the frequency behavior of physical structures. Static fiels are stuie by setting the time erivatives in Maxwell s equations equal to zero. We will introuce the lumpe circuit elements familiar in circuit theory, through the ifferent classifications of static fiels. For a nonzero frequency, the fiels are ynamic. The exact solutions are solutions to the complete Maxwell s equations for time-varying fiels. However, a class of fiels, known as quasistatic fiels, can be stuie as low-frequency extensions of static fiels. They are approximations to the exact solutions. We will learn that for quasistatic fiels, the circuit equivalent for the input behavior of a physical structure is essentially same as the lumpe circuit equivalent for the corresponing static case. As the frequency is increase beyon the quasistatic approximation, the lumpe circuit equivalent is no longer vali an the istribute circuit equivalent comes into play, leaing to the transmission line. We begin the chapter with electric potential, a scalar that is relate to the static electric fiel intensity through a vector operation known as the graient. We shall introuce the graient an the electric potential an then consier two important ifferential equations involving the potential, known as Poisson s equation an Laplace s equation. Beginning with static fiel involving the solution of the Laplace s equation, we shall then embark on the stuy base on the threa of statics-quasistatics-waves. 6. GRADIENT AND ELECTRIC POTENTIAL For static fiels, 0/0t = 0, an Maxwell s curl equations given for time-varying fiels by :E =- 0B 0t :H = J+ 0D 0t (6.) (6.2) 86

2 6. Graient an Electric Potential 87 reuce to :E = 0 :H = J (6.3) (6.4) respectively. Equation (6.3) states that the curl of the static electric fiel is equal to zero. If the curl of a vector is zero, then that vector can be expresse as the graient of a scalar, since the curl of the graient of a scalar is ientically equal to zero. The graient of a scalar, say, enote (el ) is given in Cartesian coorinates by = a a x 0 0x + a y 0 0y + a z 0 0z b = 0 0x a x + 0 0y a y + 0 0z a z (6.5) The curl of is then given by : = a x a y a z 0 0x 0 0y 0 0z ( ) x ( ) y ( ) z a x a y a z = 5 0x 0y 0z 5 = 0 0 0x 0 0y 0 0z (6.6) To iscuss the physical interpretation of the graient, we note that # l = a 0 0x a x + 0 0y a y + 0 0z a zb # (x ax + y a y + z a z ) = 0 0x = x + 0 0y y + 0 0z z (6.7) Let us consier a surface on which is equal to a constant, say 0, an a point P on that surface, as shown in Figure 6.(a). If we now consier another point Q on the same surface an an infinitesimal istance away from P, between these two points is zero since is constant on the surface. Thus, for the vector l rawn from P to Q, [ ] P # l = 0 an hence [ ] P is perpenicular to l. Since this is true for all points Q, Q 2, Q 3,... on the constant surface, it follows that [ ] P must be normal to all possible infinitesimal isplacement vectors l, l 2, l 3,... rawn at P an hence

3 88 Chapter 6 Statics, Quasistatics, an Transmission Lines 0 P l 3 Q 3 Q l l 2 l Q 2 a n a P 0 Q 0 a n (a) (b) FIGURE 6. For iscussing the physical interpretation of the graient of a scalar function. is normal to the surface. Denoting a n to be the unit normal vector to the surface at P, we then have [ ] P = ƒ ƒ P a n (6.8) Let us now consier two surfaces on which is constant, having values 0 an 0 +,as shown in Figure 6.l(b).Let P an Q be points on the = 0 an = 0 + surfaces, respectively, an l be the vector rawn from P to Q. Then from (6.7) an (6.8), =[ ] P # l = ƒ ƒ P a n # l = ƒ ƒ P l cos a (6.9) where a is the angle between at P an l.thus, a n ƒ ƒ P = l cos a (6.0) Since l cos a is the istance between the two surfaces along a n an hence is the shortest istance between them, it follows that ƒ ƒ P is the maximum rate of increase of at the point P. Thus,the graient of a scalar function at a point is a vector having magnitue equal to the maximum rate of increase of at that point an is irecte along the irection of the maximum rate of increase, which is normal to the constant surface passing through that point. This concept of the graient of a scalar function is often utilize to fin a unit vector normal to a given surface. We shall illustrate this by means of an example.

4 6. Graient an Electric Potential 89 Example 6. Let us fin the unit vector normal to the surface y = x 2 at the point (2, 4, ) by using the concept of the graient of a scalar. Writing the equation for the surface as we note that the scalar function that is constant on the surface is given by The graient of the scalar function is then given by = (x 2 - y) = 0(x2 - y) 0x = 2xa x - a y x 2 - y = 0 (x, y, z) = x 2 - y a x + 0(x2 - y) 0y a y + 0(x2 - y) 0z The value of the graient at the point (2, 4, ) is 2(2)a x - a y = 4a x - a y.thus, the require unit vector is a z a n = ; 4a x - a y ƒ4a x - a y ƒ = ;a 4 27 a x - 27 a yb Returning to Maxwell s curl equation for the static electric fiel given by (6.3), we can now express E as the graient of a scalar function, say,. The question then arises as to what this scalar function is. To obtain the answer, let us consier a region of static electric fiel. Then we can raw a set of surfaces orthogonal everywhere to the fiel lines, as shown in Figure 6.2. These surfaces correspon to the constant V V A V V B A B E Equipotential Surfaces FIGURE 6.2 A set of equipotential surfaces in a region of static electric fiel.

5 90 Chapter 6 Statics, Quasistatics, an Transmission Lines surfaces. Since on any such surface E# l = 0, no work is involve in the movement of a test charge from one point to another on the surface. Such surfaces are known as the equipotential surfaces. Since they are orthogonal to the fiel lines, they may physically be occupie by conuctors without affecting the fiel istribution. Movement of a test charge from a point, say A,on one equipotential surface to a point, say B,on another equipotential surface involves an amount of work per unit charge equal to # B A E l to be one by the fiel. This quantity is known as the electric potential ifference between the points A an B an is enote by the symbol [V ] B A. It has the units of volts. There is a potential rop from A to B if work is one by the fiel an a potential rise if work is one against the fiel by an external agent. The situation is similar to that in the earth s gravitational fiel for which there is a potential rop associate with the movement of a mass from a point of higher elevation to a point of lower elevation an a potential rise for just the opposite case. It is convenient to efine an electric potential associate with each point. The potential at point A, enote V A,is simply the potential ifference between point A an a reference point, say O.It is the amount of work per unit charge one by the fiel in connection with the movement of a test charge from A to O,or the amount of work per unit charge one against the fiel by an external agent in moving the test charge from O to A.Thus, an V A [V ] B A = L B = L O O A = L E# l = -L E# l A A A = V A - V B O B E# l = L E# l + L E# l - L O (6.) (6.2) If we now consier points A an B to be separate by infinitesimal length l from A to B, then the incremental potential rop from A to B is E A # l, or the incremental potential rise V along the length l is given by A B O E# l O E# l V = -E A # l Writing V = [ V ] A # l in accorance with (6.7), we then have [ V ] A # l = - EA # l (6.3) (6.4) (6.5) Since (6.5) is true at any point A in the static electric fiel, it follows that E = - V (6.6)

6 6. Graient an Electric Potential 9 Thus, we have obtaine the result that the static electric fiel is the negative of the graient of the electric potential. Before proceeing further, we note that the potential ifference we have efine here has the same meaning as the voltage between two points, efine in Section 2.. We, however, recall that the voltage between two points A an B in a time-varying fiel is in general epenent on the path followe from A to B to evaluate # B A E l, since, accoring to Faraay s law, E# l =- B# S C t LS (6.7) is not in general equal to zero. On the other han, the potential ifference (or voltage) between two points A an B in a static electric fiel is inepenent of the path followe from A to B to evaluate # B A E l, since, for static fiels, 0>0t = 0, an (6.7) reuces to E # l = 0 CC (6.8) Thus, the potential ifference between two points in a static electric fiel has a unique value. Fiels for which the line integral aroun a close path is zero are known as conservative fiels. The static electric fiel is a conservative fiel. The earth s gravitational fiel is another example of a conservative fiel, since the work one in moving a mass aroun a close path is equal to zero. Returning now to the iscussion of electric potential, let us consier the electric fiel of a point charge an investigate the electric potential ue to the point charge. To o this, we recall from Section.5 that the electric fiel intensity ue to a point charge Q is irecte raially away from the point charge an its magnitue is Q>4pP 0 R 2, where R is the raial istance from the point charge. Since the equipotential surfaces are everywhere orthogonal to the fiel lines, it then follows that they are spherical surfaces centere at the point charge, as shown by the cross-sectional view in Figure 6.3. If we now consier two equipotential surfaces of raii R an R + R, the potential rop from the surface of raius R to the surface of raius R + R is Q>4pP 0 R 2 2 R, or, the incremental potential rise V is given by V =- = a Q 4pP 0 R 2 R Q 4pP 0 R + Cb (6.9) where C is a constant. Thus, V (R) = Q 4pP 0 R + C (6.20)

7 92 Chapter 6 Statics, Quasistatics, an Transmission Lines E Equipotentials FIGURE 6.3 Cross-sectional view of equipotential surfaces an electric fiel lines for a point charge. We can conveniently set C equal to zero by noting that it is equal to V (q) an by choosing R = q for the reference point. Thus, we obtain the electric potential ue to a point charge Q to be V = (6.2) We note that the potential rops off inversely with the raial istance away from the point charge. Equation (6.2) is often the starting point for the computation of the potential fiel ue to static charge istributions an the subsequent etermination of the electric fiel by using (6.6). Q 4pP 0 R 6.2 POISSON S AND LAPLACE S EQUATIONS In the previous section, we learne that for the static electric fiel, :Eis equal to zero, an hence Substituting this result into Maxwell s ivergence equation for D,an assuming P to be uniform, we obtain or E = - V # D = # PE =P # E =P # (- V ) = r # V =- r P The quantity # V is known as the Laplacian of V, enote 2 V (el square V). Thus, we have 2 V =- r P (6.22)

8 6.2 Poisson s an Laplace s Equations 93 This equation is known as the Poisson s equation. It governs the relationship between the volume charge ensity r in a region an the potential in that region. In Cartesian coorinates, 2 V = # V = a a x 0 0x + a y 0 0y + a z 0 0z b # a 0V 0x a x + 0V 0y a y + 0V 0z a zb = 02 V 0x V 0y V 0z 2 (6.23) an Poisson s equation becomes 0 2 V 0x V 0y V 0z 2 = -r P (6.24) For the one-imensional case in which V varies with x only, 0 2 V >0y 2 an 0 2 V >0z 2 are both equal to zero, an (6.24) reuces to 0 2 V 0x 2 = 2 V x 2 = -r P (6.25) We shall illustrate the application of (6.25) by means of an example. Example 6.2 Let us consier the space charge layer in a p-n junction semiconuctor with zero bias, as shown in Figure 6.4(a), in which the region x 6 0 is ope p-type an the region x 7 0 is ope n-type. To review briefly the formation of the space charge layer, we note that since the ensity of the holes on the p sie is larger than that on the n sie, there is a tenency for the holes to iffuse to the n sie an recombine with the electrons. Similarly, there is a tenency for the electrons on the n sie to iffuse to the p sie an recombine with the holes. The iffusion of holes leaves behin negatively charge acceptor atoms, an the iffusion of electrons leaves behin positively charge onor atoms. Since these acceptor an onor atoms are immobile, a space charge layer, also known as the epletion layer, is forme in the region of the junction,with negative charges on the p sie an positive charges on the n sie. This space charge gives rise to an electric fiel irecte from the n sie of the junction to the p sie so that it opposes iffusion of the mobile carriers across the junction, thereby resulting in an equilibrium. For simplicity, let us consier an abrupt junction, that is, a junction in which the impurity concentration is constant on either sie of the junction. Let N A an N D be the acceptor an onor ion concentrations, respectively, an p an n be the withs in the p an n regions, respectively, of the epletion layer. The space charge ensity r is then given by r = e - ƒeƒn A for - p 6 x 6 0 ƒeƒn D for 0 6 x 6 n (6.26)

9 94 Chapter 6 Statics, Quasistatics, an Transmission Lines Space Charge Acceptor Ion Hole E Electron x 0, p-type x 0 x 0, n-type Donor Ion (a) r e N D p 0 n x e N A (b) E x p 0 n x e N A p P (c) e N A 2P 2 ( p + n p ) V 2P 2 e N A p p 0 n x FIGURE 6.4 For illustrating the application of Poisson s equation for the etermination of the potential istribution for a p-n junction semiconuctor. () as shown in Figure 6.4(b), where ƒ e ƒ is the magnitue of the electronic charge. Since the semiconuctor is electrically neutral, the total acceptor charge must be equal to the total onor charge; that is, ƒ e ƒn A p = ƒ e ƒn D n (6.27)

10 6.2 Poisson s an Laplace s Equations 95 We wish to fin the potential istribution in the epletion layer an the epletion layer with in terms of the potential ifference across the epletion layer an the acceptor an onor ion concentrations. Substituting (6.26) into (6.25), we obtain the equation governing the potential istribution to be ƒ e ƒn A for - 2 V x 2 = P p 6 x ƒ e ƒn D for 0 6 x 6 P n To solve (6.28) for V,we integrate it once an obtain (6.28) V x = ƒ e ƒn A P - ƒ e ƒn D P x + C for - p 6 x 6 0 x + C 2 for 0 6 x 6 n where C an C 2 are constants of integration. To evaluate C an C 2, we note that since E = - V = -0V >0x2a x, 0V >0x is simply equal to -E x. Since the electric fiel lines begin on the positive charges an en on the negative charges, an in view of (6.27), the fiel an, hence, 0V >0x must vanish at x = - p an x = n, giving us V x = ƒ e ƒn A P x + p2 for - p 6 x ƒ e ƒn D x - P n 2 for 0 6 x 6 n (6.29) The fiel intensity, that is, -V >x, may now be sketche as a function of x, as shown in Figure 6.4(c). Proceeing further, we integrate (6.29) an obtain V = ƒ e ƒn A 2P x + p2 2 + C 3 for - p 6 x ƒ e ƒn D 2P x - n2 2 + C 4 for 0 6 x 6 n where C 3 an C 4 are constants of integration.to evaluate C 3 an C 4, we first set the potential at x = - p arbitrarily equal to zero to obtain C 3 equal to zero.then we make use of the conition that the potential be continuous at x = 0, since the iscontinuity in V >x at x = 0 is finite, to obtain ƒ e ƒn A 2P p 2 =- ƒ e ƒn D 2P n 2 + C 4 or C 4 = ƒ e ƒ 2P N A p 2 + N D n 2 2

11 96 Chapter 6 Statics, Quasistatics, an Transmission Lines Substituting this value for C 4 an setting C 3 equal to zero in the expression for V, we get the require solution V = ƒ e ƒn A 2P x + p2 2 for - p 6 x ƒ e ƒn D 2P x2-2x n 2 + ƒ e ƒn A 2 2P p for 0 6 x 6 n (6.30) The variation of potential with x as given by (6.30) is shown in Figure 6.4(). We can procee further an fin the with = p + n of the epletion layer by setting V n 2 equal to the contact potential, V 0, that is, the potential ifference across the epletion layer resulting from the electric fiel in the layer.thus, V 0 = V n 2 = ƒ e ƒn D 2P = ƒ e ƒ N A N D 2 2P N A + N D 2 n + ƒ e ƒn A 2P = ƒ e ƒ N D N A + N D 2 n 2 + ƒ e ƒ N A N A + N D 2 2 p 2P N A + N D 2P N A + N D = ƒ e ƒ N A N D 2 n + p n p 2 2P N A + N D where we have mae use of (6.27). Finally, we obtain the result that 2 p = B 2PV 0 ƒ e ƒ a N A + N D b which tells us that the epletion layer with is smaller, the heavier the oping is. This property is use in tunnel ioes to achieve layer withs on the orer of 0-6 cm by heavy oping as compare to withs on the orer of 0-4 cm in orinary p-n junctions. We have just illustrate an example of the application of Poisson s equation involving the solution for the potential istribution for a given charge istribution. Poisson s equation is even more useful for the solution of problems in which the charge istribution is the quantity to be etermine, given the functional epenence of the charge ensity on the potential. We shall, however, not pursue this topic any further. If the charge ensity in a region is zero, then Poisson s equation reuces to (6.3) This equation is known as Laplace s equation. It governs the behavior of the potential in a charge-free region. In Cartesian coorinates, it is given by 0 2 V 0x 2 2 V = V 0y V 0z 2 = 0 (6.32) The problems for which Laplace s equation is applicable consist of fining the potential istribution in the region between two conuctors given the charge

12 6.3 Static Fiels an Circuit Elements 97 istribution on the surfaces of the conuctors or the potentials of the conuctors or a combination of the two. The proceure involves the solving of Laplace s equation subject to the bounary conitions on the surfaces of the conuctors. We shall o this in the following section. 6.3 STATIC FIELDS AND CIRCUIT ELEMENTS In the previous two sections, we consiere static fiels with reference to electric fiel alone. In this section, we shall expan the treatment to all types of static fiels, for the purpose of introucing circuit elements. Thus, for static fiels, 0/0t = 0. Maxwell s equations in integral form an the law of conservation of charge become C E# l = 0 (6.33a) C CS CS H# l = LS J# S D# S = LV r v B# S = 0 (6.33b) (6.33c) (6.33) CS J# S = 0 (6.33e) whereas Maxwell s equations in ifferential form an the continuity equation reuce to :E = 0 :H = J # D = r # B = 0 # J= 0 (6.34a) (6.34b) (6.34c) (6.34) (6.34e) Immeiately, one can see that, unless Jinclues a component ue to conuction current, the equations involving the electric fiel are completely inepenent of those involving the magnetic fiel. Thus, the fiels can be subivie into static electric fiels, or electrostatic fiels, governe by (6.33a) an (6.33c), or (6.34a) an (6.34c), an static magnetic fiels,or magnetostatic fiels, governe by (6.33b) an (6.33), or (6.34b) an (6.34). The source of a static electric fiel is r, whereas the source of a static magnetic fiel is J.One can also see from (6.33e)or(6.34e)that no relationship exists between Jan r. If Jinclues a component ue to conuction current, then, since J c = se, a coupling between the electric an magnetic fiels exists for that part of the total fiel associate with J c. However, the coupling is only one way, since the right sie of (6.33a) or (6.34a) is still zero. The fiel is then referre to as electromagnetostatic fiel.it can

13 98 Chapter 6 Statics, Quasistatics, an Transmission Lines also be seen, then, that for consistency, the right sies of (6.33c) an (6.34c) must be zero, since the right sies of (6.33e) an (6.34e) are zero. We shall now consier each of the three types of static fiels separately an iscuss some funamental aspects. Electrostatic Fiels an Capacitance The equations of interest are (6.33a) an (6.33c), or (6.34a) an (6.34c). The first of each pair of these equations simply tells us that the electrostatic fiel is a conservative fiel, an the secon of each pair of these equations enables us, in principle, to etermine the electrostatic fiel for a given charge istribution. Alternatively, the Poisson s equation, equation (6.22), can be use to fin the electric scalar potential, V, from which the electrostatic fiel can be etermine by using (6.6). In a charge-free region, the Poisson s equation reuces to the Laplace s equation, (6.3). The fiel is then ue to charges outsie the region, such as surface charge on conuctors bouning the region. The situation is then one of solving a bounary value problem, as we shall illustrate by means of an example. Example 6.3 Figure 6.5(a) is that of a parallel-plate arrangement in which two parallel, perfectly conucting plates (s = q, E = 0) of imensions w along the y-irection an l along the z-irection lie in the x = 0 an x = planes. The region between the plates is a perfect ielectric (s = 0) of material parameters P an m.the thickness of the plates is shown exaggerate for convenience in illustration. A potential ifference of V 0 is maintaine between the plates by connecting a irect voltage source at the en z = -l. If fringing of the fiel ue to the finite imensions of the structure normal to the x-irection is neglecte, or, if it is assume that the structure is part of one which is infinite in extent normal to the x-irection, then the problem can be treate as oneimensional with x as the variable, an (6.3) reuces to 2 V x 2 = 0 (6.35) We wish to carry out the electrostatic fiel analysis for this arrangement. The solution for the potential in the charge-free region between the plates is given by V (x) = V 0 ( - x) (6.36) which satisfies (6.35), as well as the bounary conitions of V = 0 at x = an V = V 0 at x = 0. The electric fiel intensity between the plates is then given by E = - V = V 0 a x (6.37) as epicte in the cross-sectional view in Figure 6.5(b), an resulting in isplacement flux ensity D = PV 0 a x (6.38)

14 6.3 Static Fiels an Circuit Elements 99 l w y V 0 x 0 z z l P, m x z z 0 (a) x C Pwl V 0 x 0, V V 0 E, D x, V 0 z l z z 0 r s y x z (c) r s (b) FIGURE 6.5 Electrostatic fiel analysis for a parallel-plate arrangement. Then, using the bounary conition for the normal component of D given by (5.94c), we obtain the magnitue of the charge on either plate to be Q = a PV 0 Pwl b (wl) = V 0 (6.39) We can now fin the familiar circuit parameter, the capacitance, C, of the parallel-plate arrangement, which is efine as the ratio of the magnitue of the charge on either plate to the potential ifference.thus, V 0 C = Q V 0 = Pwl (6.40) Note that the units of C are the units of P times meter, that is, faras. The phenomenon associate with the arrangement is that energy is store in the capacitor in the form of electric fiel energy between the plates, as given by W e = a 2 PE2 xb (wl) = 2 a Pwl b V 2 0 = 2 CV 2 0 (6.4) the familiar expression for energy store in a capacitor.

15 200 Chapter 6 Statics, Quasistatics, an Transmission Lines Magnetostatic Fiels an Inuctance The equations of interest are (6.33b) an (6.33), or (6.34b) an (6.34). The secon of each pair of these equations simply tells us that the magnetostatic fiel is solenoial, which as we know hols for any magnetic fiel,an the first of each pair of these equations enables us,in principle, to etermine the magnetostatic fiel for a given current istribution. In a current-free region, J= 0. The fiel is then ue to currents outsie the region, such as surface currents on conuctors bouning the region. The situation is then one of solving a bounary value problem as in the case of (6.3). However, since the bounary conition (5.94b) relates the magnetic fiel irectly to the surface current ensity, it is straightforwar an more convenient to etermine the magnetic fiel irectly by using (6.34b) an (6.34). We shall illustrate by means of an example. Example 6.4 Figure 6.6(a) is that of the parallel-plate arrangement of Figure 6.5(a) with the plates connecte by another conuctor at the en z = 0 an riven by a source of irect current I 0 at the en z = -l.if fringing of the fiel ue to the finite imensions of the structure normal to the x-irection is neglecte, or, if it is assume that the structure is part of one which is infinite in extent normal to the x-irection, then the problem can be treate as one-imensional with x as the variable an we can write the current ensity on the plates to be J S = e (I 0 >w)a z on the plate x = 0 (I 0 >w)a x on the plate z = 0 -(I 0 >w)a z on the plate x = (6.42) We wish to carry out the magnetostatic fiel analysis for this arrangement. l y w z I 0 x 0 x P, m z l x z z 0 L ml w (a) x 0 y z (c) I 0 J S H, B x z l z z 0 x (b) FIGURE 6.6 Magnetostatic fiel analysis for a parallel-plate arrangement.

16 In the current-free region between the plates, (6.34b) reuces to 6.3 Static Fiels an Circuit Elements 20 a x a y a z 0 0x 0 0 = 0 (6.43) H x H y H z an (6.34) reuces to 0B x 0x = 0 (6.44) so that each component of the fiel, if it exists, has to be uniform. This automatically forces H x an H z to be zero, since nonzero value of these components o not satisfy the bounary conitions (5.94b) an (5.94) on the plates, keeping in min that the fiel is entirely in the region between the conuctors.thus, as epicte in the cross-sectional view in Figure 6.6(b), H = I 0 w a y (6.45) which satisfies the bounary conition (5.94b) on all three plates, an results in magnetic flux ensity B = mi 0 w a y (6.46) The magnetic flux, c, linking the current I 0, is then given by c = a mi 0 ml b (l) = a w w b I 0 (6.47) We can now fin the familiar circuit parameter, the inuctance, L, of the parallel-plate arrangement, which is efine as the ratio of the magnetic flux linking the current to the current. Thus, L = c I 0 = ml w (6.48) Note that the units of L are the units of m times meter, that is, henrys. The phenomenon associate with the arrangement is that energy is store in the inuctor in the form of magnetic fiel energy between the plates, as given by W m = a 2 mh2 b (wl) = 2 a ml w b I 0 2 = 2 LI 0 2 (6.49) the familiar expression for energy store in an inuctor.

17 202 Chapter 6 Statics, Quasistatics, an Transmission Lines Electromagnetostatic Fiels an Conuctance The equations of interest are CS CS E# l = 0 C H# l = J c # S = s C CS CS E# S D# S = 0 B# S = 0 (6.50a) (6.50b) (6.50c) (6.50) or, in ifferential form, :E = 0 : H = J c = se # D = 0 # B = 0 (6.5a) (6.5b) (6.5c) (6.5) From (6.5a) an (6.5c), we note that Laplace s equation for the electrostatic potential, (6.3), is satisfie, so that, for a given problem, the electric fiel can be foun in the same manner as in the case of the example of Figure 6.6. The magnetic fiel is then foun by using (6.5b), an making sure that (6.5) is also satisfie. We shall illustrate by means of an example. Example 6.5 Figure 6.7(a) is that of the parallel-plate arrangement of Figure 6.5(a) but with an imperfect ielectric material of parameters s, P, an m, between the plates. We wish to carry out the electromagnetostatic fiel analysis of the arrangement. The electric fiel between the plates is the same as that given by (6.37), that is, resulting in a conuction current of ensity E = V 0 a x (6.52) J c = sv 0 (6.53) from the top plate to the bottom plate, as epicte in the cross-sectional view of Figure 6.7(b). Since 0r>0t = 0 at the bounaries between the plates an the slab, continuity of current is satisfie by the flow of surface current on the plates. At the input z = -l, this surface current, which is the current rawn from the source, must be equal to the total current flowing from the top to the bottom plate. It is given by a x I c = a sv 0 swl b (wl) = V 0 (6.54)

18 6.3 Static Fiels an Circuit Elements 203 l w y V 0 x 0 z s, P, m x z l x z z 0 (a) J S r s V 0 I c H, B E, J c, D x 0, V V 0 y z x, V 0 z l z z 0 J S r s x (b) FIGURE 6.7 Electromagnetostatic fiel analysis for a parallel-plate arrangement. We can now fin the familiar circuit parameter, the conuctance, G, of the parallel-plate arrangement,which is efine as the ratio of the current rawn from the source to the source voltage.thus, V 0 G = I c = swl V 0 (6.55) Note that the units of G are the units of s times meter, that is, siemens (S). The reciprocal quantity, R,the resistance of the parallel-plate arrangement,is given by R = V 0 I c = swl (6.56) The unit of R is ohms. The phenomenon associate with the arrangement is that power is issipate in the material between the plates, as given by P = (se 2 )(wl) = a swl b V 0 2 = GV 2 0 = V 0 2 R (6.57) the familiar expression for power issipate in a resistor.

19 204 Chapter 6 Statics, Quasistatics, an Transmission Lines Proceeing further, we fin the magnetic fiel between the plates by using (6.5b), an noting that the geometry of the situation requires a y-component of H,epenent on z,to satisfy the equation.thus, H = H y (z) a y 0H y 0z =-sv 0 H =- sv 0 z a y (6.58a) (6.58b) (6.58c) where the constant of integration is set to zero, since the bounary conition at z = 0 requires H y to be zero for z equal to zero. Note that the magnetic fiel is irecte in the positive y-irection (since z is negative) an increases linearly from z = 0 to z = -l, as epicte in Figure 6.7(b). It also satisfies the bounary conition at z = -l by being consistent with the current rawn from the source to be w[h y ] z =-l = (sv 0 >)(wl) = I c. Because of the existence of the magnetic fiel, the arrangement is characterize by an inuctance, which can be foun either by using the flux linkage concept or by the energy metho. To use the flux linkage concept,we recognize that a ifferential amount of magnetic flux c =mh y (z ) between z equal to (z -z ) an z equal to z, where -l 6 z 60, links only that part of the current that flows from the top plate to the bottom plate between z = z an z = 0,thereby giving a value of (- z >l) for the fraction, N,of the total current linke.thus, the inuctance, familiarly known as the internal inuctance, enote L i, since it is ue to magnetic fiel internal to the current istribution, as compare to that in (6.48) for which the magnetic fiel is external to the current istribution, is given by L i = I c L 0 z =- = ml 3 w N c (6.59) or, >3 times the inuctance of the structure if s = 0 an the plates are joine at z = 0, as in Figure 6.6(b). Alternatively, if the energy metho is use by computing the energy store in the magnetic fiel an setting it equal to 2 L ii 2 c, then we have L i = 0 I 2 (w) c L = ml 3 w z =-l mh y 2 z (6.60) same as in (6.59). Finally, recognizing that there is energy storage associate with the electric fiel between the plates, we note that the arrangement has also associate with it a capacitance C, equal to Pwl>. Thus, all three properties of conuctance, capacitance, an inuctance are associate with the structure. Since for s = 0 the situation reuces to that of Figure 6.5, we can represent the arrangement of Figure 6.7 to be equivalent to the circuit shown in Figure 6.8. Note that the capacitor is charge to the voltage an the current through it is zero (open circuit conition). V 0

20 6.4 Low-Frequency Behavior via Quasistatics 205 V 0 C Pwl R swl L i 3 ml w FIGURE 6.8 Circuit equivalent for the arrangement of Figure 6.7. The voltage across the inuctor is zero (short circuit conition) an the current through it is V 0 >R. Thus, the current rawn from the voltage source is V 0 >R an the voltage source views a single resistor R, as far as the current rawn from it is concerne. 6.4 LOW-FREQUENCY BEHAVIOR VIA QUASISTATICS In the preceing section, we introuce circuit elements via static fiels. A class of ynamic fiels for which certain features can be analyze as though the fiels were static are known as quasistatic fiels.in terms of behavior in the frequency omain,they are low-frequency extensions of static fiels present in a physical structure, when the frequency of the source riving the structure is zero, or low-frequency approximations of time-varying fiels in the structure that are complete solutions to Maxwell s equations. In this section, we consier the approach of low-frequency extensions of static fiels. Thus, for a given structure, we begin with a time-varying fiel having the same spatial characteristics as that of the static fiel solution for the structure, an obtain fiel solutions containing terms up to an incluing the first power (which is the lowest power) in v for their amplitues. Depening on whether the preominant static fiel is electric or magnetic, quasistatic fiels are calle electroquasistatic fiels or magnetoquasistatic fiels. We shall now consier these separately. Electroquasistatic Fiels For electroquasistatic fiels, we begin with the electric fiel having the spatial epenence of the static fiel solution for the given arrangement. We shall illustrate by means of an example. Example 6.6 Figure 6.9 shows the cross-sectional view of the arrangement of Figure 6.5(a) excite by a sinusoially time-varying voltage source V g (t) = V 0 cos vt instea of a irect voltage source. We wish to carry out the electroquasistatic fiel analysis for the arrangement. From (6.37), we write E 0 = V 0 cos vt a x (6.6)

21 206 Chapter 6 Statics, Quasistatics, an Transmission Lines V g (t) V 0 cos vt I g (t) J S r s H P, m E 0 x 0 x z l z z 0 y x z FIGURE 6.9 Electroquasistatic fiel analysis for the parallel-plate structure of Figure 6.5. where the subscript 0 enotes that the amplitue of the fiel is of the zeroth power in v.this results in a magnetic fiel in accorance with Maxwell s equation for the curl of H, given by (3.28). Thus, noting that J= 0 in view of the perfect ielectric meium, we have for the geometry of the arrangement, 0H y 0z =- 0D x0 0t H = vpv 0z = vpv 0 sin vt a y sin vt (6.62) where we have also satisfie the bounary conition at z = 0 by choosing the constant of integration such that [H y ] z = 0 is zero, an the subscript enotes that the amplitue of the fiel is of the first power in v. Note that the amplitue of H y varies linearly with z, from zero at z = 0 to a maximum at z = -l. We stop the solution here, because continuing the process by substituting (6.62) into Maxwell s curl equation for E,(3.7),to obtain the resulting electric fiel will yiel a term having amplitue proportional to the secon power in v.this simply means that the fiels given as a pair by (6.6) an (6.62) o not satisfy (3.7), an hence are not complete solutions to Maxwell s equations. They are the quasistatic fiels. The complete solutions are obtaine by solving Maxwell s equations simultaneously an subject to the bounary conitions for the given problem. Proceeing further, we obtain the current rawn from the voltage source to be I g (t) = w[h y ] z =-l = -va Pwl b V 0 sin vt = C V g(t) t (6.63a) or, Ī g = jv CV g (6.63b) where C = (Pwl>) is the capacitance of the arrangement obtaine from static fiel consierations. Thus, the input amittance of the structure is jvc, such that its low frequency input behavior is essentially that of a single capacitor of value same as that foun from static fiel

22 6.4 Low-Frequency Behavior via Quasistatics 207 analysis of the structure. Inee, from consierations of power flow, using Poynting s theorem, we obtain the power flowing into the structure to be P in = w3e x0 H y 4 z = 0 = -a Pwl b vv 2 0 sin vt cos vt = t a 2 CV 2 g b (6.64) which is consistent with the electric energy store in the structure for the static case, as given by (6.4). Magnetoquasistatic Fiels For magnetoquasistatic fiels, we begin with the magnetic fiel having the spatial epenence of the static fiel solution for the given arrangement. We shall illustrate by means of an example. Example 6.7 Figure 6.0 shows the cross-sectional view of the arrangement of Figure 6.6(a), excite by a sinusoially time-varying current source I g (t) = I 0 cos vt instea of a irect current source. We wish to carry out the magnetoquasistatic fiel analysis for the arrangement. From (6.45) we write H 0 = I 0 w cos vt a y (6.65) where the subscript 0 again enotes that the amplitue of the fiel is of the zeroth power in v. This results in an electric fiel in accorance with Maxwell s curl equation for E, given by (3.7). Thus, we have for the geometry of the arrangement, 0E x 0z =- 0B y 0 0t E = vmi 0z w = vmi 0 w sin vt a x sin vt (6.66) r s E I g (t) I 0 cos vt x 0 V g (t) P, m H 0 J x S z l z z 0 y x z FIGURE 6.0 Magnetoquasistatic fiel analysis for the parallel-plate structure of Figure 6.6.

23 208 Chapter 6 Statics, Quasistatics, an Transmission Lines where we have also satisfie the bounary conition at z = 0 by choosing the constant of integration such that [E x ] z = 0 is equal to zero, an again the subscript enotes that the amplitue of the fiel is of the first power in v. Note that the amplitue of E x varies linearly with z, from zero at z = 0 to a maximum at z = -l. As in the case of electroquasistatic fiels, we stop the process here, because continuing it by substituting (6.66) into Maxwell s curl equation for H,(3.28),to obtain the resulting magnetic fiel will yiel a term having amplitue proportional to the secon power in v.this simply means that the fiels given as a pair by (6.65) an (6.66) o not satisfy (3.28), an hence are not complete solutions to Maxwell s equations. They are the quasistatic fiels. The complete solutions are obtaine by solving Maxwell s equations simultaneously an subject to the bounary conitions for the given problem. Proceeing further, we obtain the voltage across the current source to be or V g (t) = [E x ] z =-l = -va ml w b I 0 sin vt = L I g(t) t V g = jvlī g (6.67a) (6.67b) where L = (ml>w) is the inuctance of the arrangement obtaine from static fiel consierations. Thus, the input impeance of the structure is jvl, such that its low frequency input behavior is essentially that of a single inuctor of value same as that foun from static fiel analysis of the structure. Inee, from consierations of power flow, using Poynting s theorem, we obtain the power flowing into the structure to be P in = w3e x H y 0 4 z =-l = -a ml w b vi2 0 sin vt cos vt = t a 2 LI2 gb (6.68) which is consistent with the magnetic energy store in the structure for the static case, as given by (6.49). Quasistatic Fiels in a Conuctor If the ielectric slab in an arrangement is conuctive, then both electric an magnetic fiels exist in the static case, because of the conuction current, as iscusse uner electromagnetostatic fiels in Section 6.3. Furthermore, the electric fiel of amplitue proportional to the first power in v contributes to the creation of magnetic fiel of amplitue proportional to the first power in v,in aition to that from electric fiel of amplitue proportional to the zeroth power in v.we shall illustrate by means of an example.

24 6.4 Low-Frequency Behavior via Quasistatics 209 Example 6.8 Let us consier that the ielectric slab in the arrangement of Figure 6.9 is conuctive, as shown in Figure 6.(a), an carry out the quasistatic fiel analysis for the arrangement. Using the results from the static fiel analysis from the arrangement of Figure 6.7, we have for the arrangement of Figure 6.(a), E 0 = V 0 cos vt a x J c0 = se 0 = sv 0 H 0 =- sv 0z cos vt a x cos vt a y (6.69) (6.70) (6.7) as epicte in the figure. Also, the variations with z of the amplitues of E x0 an H y0 are shown in Figure 6.(b). V g (t) V 0 cos vt I g E 0, J c0 H 0 x 0 s, P, m x z l z z 0 y x z (a) E x0 H y0 z l z z 0 (b) H y H yc E x FIGURE 6. z l z z 0 (c) (a) Zero-orer fiels for the parallel-plate structure of Figure 6.7. (b) Variations of amplitues of the zero-orer fiels along the structure. (c) Variations of amplites of the first-orer fiels along the structure.

25 20 Chapter 6 Statics, Quasistatics, an Transmission Lines The magnetic fiel given by (6.69) gives rise to an electric fiel having amplitue proportional to the first power in v, in accorance with Maxwell s curl equation for E, (3.7).Thus, 0E x 0z =- 0B y0 0t E x =- vmsv 0 2 =- vmsv 0z sin vt (z 2 - l 2 ) sin vt (6.72) where we have also mae sure that the bounary conition at z = -l is satisfie.this bounary conition requires that E x be equal to V g > at z = -l. Since this is satisfie by E x0 alone, it follows that E x must be zero at z = -l. The electric fiel given by (6.69) an that given by (6.72) together give rise to a magnetic fiel having terms with amplitues proportional to the first power in v, in accorance with Maxwell s curl equation for H, (3.28).Thus, 0H y 0z 0E x0 = -se x -P 0t = vms2 V 0 2 H y = vms2 V 0 (z 3-3zl 2 ) 6 (z 2 - l 2 ) sin vt + vpv 0 sin vt + vpv 0z sin vt sin vt (6.73) where we have also mae sure that the bounary conition at z = 0 is satisfie. This bounary conition requires that H y be equal to zero at z = 0, which means that all of its terms must be zero at z = 0. Note that the first term on the right sie of (6.73) is the contribution from the conuction current in the material resulting from E x, an the secon term is the contribution from the isplacement current resulting from E x 0. Denoting these to be H yc an H y, respectively, we show the variations with z of the amplitues of all the fiel components having amplitues proportional to the first power in v in Figure 6.(c). Now, aing up the contributions to each fiel, we obtain the total electric an magnetic fiels up to an incluing the terms with amplitues proportional to the first power in v to be or H y =- sv 0z E x = V 0 cos vt - vmsv 0 (z 2 - l 2 ) sin vt 2 cos vt + vpv 0z E x = V g sin vt + vms2 V 0 (z 3-3zl 2 ) 6 ms + jv 2 (z2 - l 2 )V g H y =- sz V g - jv Pz V g - jv ms2 (z 3-3zl 2 ) V g 6 sin vt (6.74a) (6.74b) (6.75a) (6.75b) Finally, the current rawn from the voltage source is given by I g = w[h y] z =-l = a swl + jv Pwl - jv ms2 wl 3 3 b V g (6.76)

26 6.5 The Distribute Circuit Concept an the Parallel-Plate Transmission Line 2 The input amittance of the structure is given by Y in = I g V Pwl = jv g + swl a - jv msl 2 3 b Pwl L jv + swl a + jv msl 2 3 b (6.77) where we have use the approximation further, we have [ + jv(msl 2 >3)] - L [ - jv(msl 2 >3)]. Proceeing Y Pwl in = jv + swl + jv ml 3w = jvc + R + jvl i (6.78) where C =Pwl> is the capacitance of the structure if the material is a perfect ielectric, R = >swl is the resistance of the structure, an L i = ml>3w is the internal inuctance of the structure, all compute from static fiel analysis of the structure. The equivalent circuit corresponing to (6.78) consists of capacitance C in parallel with the series combination of resistance R an internal inuctance L i, the same as in Figure 6.8. Thus, the low-frequency input behavior of the structure is essentially the same as that of the equivalent circuit of Figure 6.8, with the unerstaning that its input amittance must also be approximate to first-orer terms. Note that for s = 0, the input amittance of the structure is purely capacitive. For nonzero s, a critical value of s equal to 23P>ml 2 exists for which the input amittance is purely conuctive. For values of s smaller than the critical value, the input amittance is complex an capacitive, an for values of s larger than the critical value, the input amittance is complex an inuctive. 6.5 THE DISTRIBUTED CIRCUIT CONCEPT AND THE PARALLEL-PLATE TRANSMISSION LINE In the preceing section, we have seen that, from the circuit point of view, the parallelplate structure of Figure 6.5 behaves like a capacitor for the static case an the capacitive character is essentially retaine for its input behavior for sinusoially time-varying excitation at frequencies low enough to be within the range of valiity of the quasistatic approximation. Likewise, we have seen that, from a circuit point of view, the parallel-plate structure of Figure 6.6 behaves like an inuctor for the static case an the inuctive character is essentially retaine for its input behavior for sinusoially time-varying excitation at frequencies low enough to be within the range of valiity of the quasistatic approximation. For both structures, at an arbitrarily high enough frequency, the input behavior can be obtaine only by obtaining complete (wave) solutions to Maxwell s equations, subject to the appropriate bounary conitions. Two questions to ask at this point are () whether there is a circuit equivalent for the structure itself, inepenent of the termination, that is representative of the phenomenon taking place along the structure an vali at any arbitrary frequency, to the

27 22 Chapter 6 Statics, Quasistatics, an Transmission Lines extent that the material parameters themselves are inepenent of frequency, an (2) what the limit on frequency is beyon which the quasistatic approximation is not vali. The answer to the first question is, yes, uner a certain conition, giving rise to the concept of the istribute circuit, which we shall evelop in this section by consiering the parallel-plate structure, to be then known as the parallel-plate transmission line. The conition is that the waves propagating along the structure be the so-calle transverse electromagnetic or TEM waves, meaning that the irections of the electric an magnetic fiels are entirely traverse to the irection of propagation of the waves. The answer to the secon question is that for the quasistatic approximation to hol, the length of the physical structure along the irection of propagation of the waves must be very small compare to the wavelength corresponing to the frequency of the source, in the ielectric region between the plates. While this can be obtaine by extening the solution for the quasistatic case beyon the terms of the first power in v by successive solution of Maxwell s equations (as in Section 4.3) an fining the conition uner which the term of the first power in v is preominant, it is more straightforwar to obtain the exact solution by resorting to simultaneous solution of Maxwell s equations an fining the conition for which it approximates to the quasistatic solution. We shall o this in Section 7. by consiering the structure of Figure 6.0 as a shortcircuite transmission line an fining its input impeance. Now, to evelop an iscuss the concept of the istribute circuit, we consier the parallel-plate arrangement of Figure 6.7(a) excite by a sinusoially time-varying source of arbitrary frequency, as shown in Figure 6.2(a). Then, for an exact solution, the equations to be solve are :E =- 0B 0t = -m0h 0t (6.79a) :H = J c + 0D 0t = se +P0E 0t (6.79b) For the geometry of the arrangement, neglecting fringing of the fiels at the eges or assuming that the structure is part of a much larger-size configuration, E = E x (z, t)a x an H = H y (z, t)a y, so that (6.79a) an (6.79b) simplify to 0E x 0z = -m0h y 0t (6.80a) 0H y 0z = -se x -P 0E x 0t (6.80b) The situation is one of uniform plane electromagnetic waves propagating in the z-irection as though the conuctors are not present, being guie by them, since all the bounary conitions are satisfie. We then have the simple case of a parallel-plate transmission line. Now, since E z an H z are zero in a given constant-z plane, that is, aplane transverse to the irection of propagation of the wave, as shown in Figure 6.2(b), we can uniquely efine a voltage between the plates in terms of the electric fiel intensity in that plane, an a current crossing that plane in one irection on the top plate an in the opposite irection on the bottom plate in terms of the magnetic

28 6.5 The Distribute Circuit Concept an the Parallel-Plate Transmission Line 23 w x 0 y x z x Transverse Plane (a) x 0 J S E H z y x w (b) J S x FIGURE 6.2 (a) Parallel-plate transmission line. (b) A transverse plane of the parallel-plate transmission line. fiel intensity in that plane. These are given by V(z, t) = E x (z, t) x = E x (z, t) x = E x (z, t) L L x = 0 x = 0 w w w I(z, t) = J S (z, t) y = H y (z, t) y = H y (z, t) y L L L y = 0 y = 0 y = 0 = wh y (z, t) (6.8a) (6.8b) Proceeing further, we can fin the power flow own the line by evaluating the surface integral of the Poynting vector over a given transverse plane. Thus, P(z, t) = Ltransverse plane (E : H)# S = L = L w x = 0 Ly = 0 w x = 0 Ly = 0 E x (z, t)h y (z, t) a z # x y az V(z, t) I(z, t) w x y = V(z, t)i(z, t) (6.82) which is the familiar relationship employe in circuit theory.

29 24 Chapter 6 Statics, Quasistatics, an Transmission Lines From (6.8a) an (6.8b), we have E x = V H y = I w (6.83a) (6.83b) E x H y Substituting for an in (6.80a) an (6.80b) from (6.83a) an (6.83b), respectively, we now obtain two ifferential equations for voltage an current along the line as 0 0z a V b = -m 0 0t a I w b 0 0z a I w b = -sa V b -P0 0t a V b (6.84a) (6.84b) or 0V 0z = -a m w b 0I 0t 0I 0z = -a sw b V - apw b 0V 0t (6.85a) (6.85b) We now recognize the quantities in parentheses in (6.85a) an (6.85b) to be the circuit parameters L, G,an C,ivie by the length l of the structure in the z-irection. Thus, these are the inuctance per unit length, capacitance per unit length, an conuctance per unit length, of the line, enote to be l, g, an c, respectively, an we can write the equations in terms of these parameters as where 0I 0z 0V 0z = -l0i 0t = -gv - c0v 0t (6.86a) (6.86b) l = m w c = Pw g = sw (6.87a) (6.87b) (6.87c) We note that l, c, an g are purely epenent on the imensions of the line an lc = mp g c = s P (6.88a) (6.88b)

30 6.5 The Distribute Circuit Concept an the Parallel-Plate Transmission Line 25 Equations (6.86a) an (6.86b) are known as the transmission line equations. They characterize the wave propagation along the line in terms of the circuit quantities instea of in terms of the fiel quantities. It shoul, however, not be forgotten that the actual phenomenon is one of electromagnetic waves guie by the conuctors of the line. It is customary to represent a transmission line by means of its circuit equivalent, erive from the transmission-line equations (6.86a) an (6.86b). To o this, let us consier a section of infinitesimal length z along the line between z an z + z. From (6.86a), we then have or, for z : 0, V(z + z, t) - V(z, t) Lim z:0 z 0I(z, t) = -l 0t V(z + z, t) - V(z, t) = -l z 0I(z, t) 0t (6.89a) This equation can be represente by the circuit equivalent shown in Figure 6.3(a), since it satisfies Kirchhoff s voltage law written aroun the loop abca.similarly,from (6.86b), we have I(z + z, t) - I(z, t) Lim z:0 z or, for z : 0, + z, t) = Lim c -gv(z + z, t) - c0v(z z:0 0t I(z + z, t) - I(z, t) = -g z V(z + z, t) - c z 0V(z + z, t) 0t (6.89b) This equation can be represente by the circuit equivalent shown in Figure 6.3(b), since it satisfies Kirchhoff s current law written for noe c.combining the two equations,we then obtain the equivalent circuit shown in Figure 6.3(c) for a section z of the line. It b z I(z, t) c I(z, t) I(z z, t) c z V(z, t) V(z z, t) z z V(z z, t) z z a z (a) (b) (c) z z FIGURE 6.3 Development of circuit equivalent for an infinitesimal length z of a transmission line.