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2416 IEEE TRANSACTIONS ON MAGNETICS, VOL. 31. NO. 4, JULY 1995 Calculation of Self and Mutual Impedances in Planar Magnetic Structures W. G. Hurley, Senior Member IEEE, M. C. Duffy AbstractâThe high frequency operation of magnetic compoÂ¬â¨nents, in applications such as filters, makes them ideal candiÂ¬â¨dates for thick film technology along with resistors and capacÂ¬â¨itors. This in turn leads to distinct advantages over laborâ¨intensive wire wound components: improved reliability, reÂ¬â¨peatability, accuracy and consequential cost reductions. Thisâ¨paper establishes a new set of formulas for the self and mutualâ¨impedances of planar coils on ferromagnetic substrates. Aâ¨planar coil in air is a special case of the generalized formulas.â¨The formulas are derived directly from Maxwell's equationsâ¨and therefore serve as a useful yardstick for simpler approxiÂ¬â¨mations. The formulas take full account of the current densityâ¨distribution in the coil cross-section and the eddy current lossesâ¨in the substrate. Experimental and calculated impedances upâ¨to 100 MHz are presented for a four layer device with threeâ¨turns per layer which is 150 /um thick and 40 mm2 in area. additional mutual impedance due to presÂ¬â¨ence of ferromagnetic substrate,â¨angular frequency (rad/s).â¨electrical conductivity of substrate,â¨permeability of free space (47r x 10 03 o -7 MO H/m). relative permeability of the substrate,â¨defined in equations (21) and (22). Mr <t>, V I. Introduction HPHE momentum towards high density electronic cir-â¨X cuits continues unabated. The effects are obvious inâ¨very large scale integration (VLSI) design: componentâ¨densities are being quadrupled every three years. In theâ¨case of magnetic components, modern microelectronicâ¨techniques such as thick film and thin film technologiesâ¨are being examined with a view to reducing size and costâ¨and to improving reliability. Planar magnetic componentsâ¨can become an integral part of the process, whereby reÂ¬â¨sistors and capacitors are already established components.â¨One of the major drawbacks in establishing planar magÂ¬â¨netic technology is the lack of accurate analytical modelsâ¨for the type of structures encountered. Prototypes are exÂ¬â¨pensive to fabricate and test. One would normally expectâ¨to complete a second cycle of fabrication and testing beÂ¬â¨fore a final design is achieved. While this procedure mayâ¨give a better insight, it does not lead to an establishedâ¨design methodology. The purpose of this paper is to adÂ¬â¨dress this situation. The starting point for all inductance calculations is theâ¨celebrated formula for the mutual inductance between twoâ¨filaments given by Maxwell [1], Coils have a finite cross-â¨section and the standard technique is to integrate the filÂ¬â¨amentary formula over the cross-section, assuming a conÂ¬â¨stant current density. Alternatively, an approximate resultâ¨can be obtained by placing a filament at the center of eachâ¨coil and the mutual inductance can be calculated directlyâ¨from the filament formula. These approaches have workedâ¨well in the past [2]. In the case of planar magnetic comÂ¬â¨ponents, the aspect ratio of height to width of a section isâ¨usually very severe. This paper shows that the currentâ¨density is not constant and when this factor is taken intoâ¨account, accuracy is greatly improved. The simplest configuration of a planar magnetic comÂ¬â¨ponent is the air-cored spiral inductor [3]. Despite itsâ¨physical simplicity, it forms the basis for more advanced Abbreviations filament radii, see Fig. 1.â¨interlayer capacitance,â¨dielectric thickness above ferromagneticâ¨substrate. a, r C d d |, d2 height of filaments or coil centers above ferromagnetic substrate,â¨coil heights in axial direction,â¨current density at radius r. Bessel function of the first kind, order v. hu h2 J{r) Jv(x) K{f),E{f) Complete Elliptic Integrals of the firstâ¨and second kind respectively,â¨self inductance of coil 1 in air. u additional coil inductance due to subÂ¬â¨strate. mutual inductance between two filaÂ¬â¨ments in air.â¨mutual inductance between two coils,â¨defined in equations (13) and (14). geometric mean, GM = V(r, r2).â¨additional coil resistance due to subÂ¬â¨strate,â¨coil dc resistance,â¨axial separation. mutual impedance between two coils. Ls M M ,2 G, S ro R* z z Manuscript received June 6, 1994; revised January 19, 1995. This workâ¨is supported by Power Electronics Ireland. The authors are with the Department of Electronic Engineering, UniverÂ¬â¨sity College, Galway, Ireland. IEEE Log Number 9410680. 0018-9464/95S04.00 C"i 1995 IEEE HURLEY AND DUFFY: SELF AND MUTUAL IMPEDANCES IN PLANAR MAGNETIC STRUCTURES 2417 configurations such as magnetic substrates [4] and sandÂ¬â¨wich inductors [5], This paper establishes a new formulaâ¨for the mutual inductance between two planar spirals inâ¨air which takes full account of the current density distriÂ¬â¨bution in the planar section. The result can be extendedâ¨to a component with several turns per layer and with sevÂ¬â¨eral layers. The next step is to add a magnetic substrate,â¨which introduces eddy current losses. A frequency deÂ¬â¨pendent mutual impedance formula for this case is deÂ¬â¨rived, which takes the eddy current losses into account. Planar magnetic components are suitable because ofâ¨their small size. This is a direct manifestation of the genÂ¬â¨eral principle that the size of magnetic components is reÂ¬â¨duced as frequency increases. Unfortunately high freÂ¬â¨quency operation gives rise to unwanted skin effect andâ¨proximity effect losses. In multilayer devices the inter-â¨layer capacitance introduces resonance at high frequenÂ¬â¨cies. Experimental results are compared with predictedâ¨values for a 4 layer spiral inductor with 3 turns per layer,â¨measurements are taken up to 100 MHz. a T z r i Fig. 1. Circular concentric filaments in air. coll 2 -1 II. Spiral Coils in Airâ¨The derivation of the general mutual inductance forÂ¬â¨mula for planar structures starts with the mutual inductÂ¬â¨ance between two filaments [ 1]. z Â°2 a1 rt OO Ji(kr)J](ka)e-klzl dk Jo M = jx^-Kar (I) T coil 1 V^r2 where 7, is a Bessel function of the first kind, a, r are theâ¨filament radii shown in Fig. 1 and p0 is the permeabilityâ¨of free space. The solution of (1) can be written in terms of ellipticâ¨integrals & V1" T Fig. 2. Planar coils of rectangular cross-section. f2 2 - 1AT(/) - E(f) M = flQ (2) ar -- / J(r) and the radius r. Since the height of the section isâ¨much smaller than the width we shall assume that there isâ¨negligible variation in current density in the z direction.â¨Given that the total current in the section is /, then where K( f) and E(f) are complete elliptic integrals ofâ¨the first and second kind, respectively and where 4 ar fÂ»r 2 h \ J(r) dr = /â¨Jri / = (3) z: + (a + r)~ Fig. 2 shows the arrangement and dimensions of twoâ¨illustrative circular and concentric planar sections. Inâ¨practice a spiral arrangement would connect two sectionsâ¨in series, which can be accurately modelled by the conÂ¬â¨centric circular coils. The traditional approach involvesâ¨integrating the filamentary formula (1) over each cross-â¨section, assuming the current density is constant in eachâ¨section [2], [6]â[8]. The approach works well when theâ¨width to height ratio of the section approaches 1. HowÂ¬â¨ever in a planar structure this ratio could be 50: 1. EviÂ¬â¨dently the path on the inside edge of the section is muchâ¨shorter than that on the outside edge and therefore theâ¨resistance is reduced on the inside, with consequentialâ¨higher current density. It is reasonable then to assume thatâ¨an inverse relationship exists between the current density (4) K J(r) = (5) r Solving (4) and (5) gives / m = (6) r2 h.r. In n In the following analysis, the current is sinusoidal, jolt t) = J(r)e (7) where oj is the angular frequency. The voltage induced in a filament at (r, n) in coil 1 dueâ¨to the current in an annular section da x dr2 at radius a 2418 IEEE TRANSACTIONS ON MAGNETICS, VOL. 31, NO. 4, JULY 1995 in coil 2 is A. Numerical Calculations Equation (16) appears rather formidable, however, it isâ¨perfectly amenable to numerical evaluation. In the past, the filament formula (2) has been used forâ¨coils with a filament placed at the center of the sectionâ¨and with z replaced by the Geometric Mean Distanceâ¨(GMD) between the coils [2]. In the case of self inducÂ¬â¨tance, z is replaced by the GMD of the coil from itself,â¨[GMD = 0.2235 (w + h)]. The central filament is placedâ¨so that the current is divided equally on either side of theâ¨filament. dV = jojMJ(a) da dr2 where M is the mutual inductance between the filamentsâ¨at (r, t|) and (a, z + t2). The total voltage at (r, tx) dueâ¨to all the current in coil 2 is obtained by integrating (8)â¨over the cross-section of coil 2 (8) n F(r) = jup0 yrr -hl/2 I aJ(a)J](kr)Jx(ka)e Jai â k\z + T2 â Tl\ da dr2 dk. As a further improvement it seems reasonable that anâ¨equivalent filament could be obtained provided that theâ¨filament is so placed that the correct current density isâ¨taken into account. Integration of (6) shows that equalâ¨current division occurs at the radius given by the geoÂ¬â¨metric mean (GM) of the inside and outside radii [r0 = (9) The power transferred to the annular segment at (r, r,)â¨due to coil 2 is dP = V(r)J(r) dr dr GO) V (rx â¢ r2)]. Finally, the total power transferred to coil 1 is found byâ¨integrating (10) over its cross-section There are three cases (see Fig. 3): 1) Self Inductance Lx Replace z in (3) by GMD of the coil from itself.â¨Place filament at the center or at the GM of theâ¨cross-section, 2) Mutual Inductance M12, MI3, z =Â£ 0 z is replaced by the GMD between sections. Forâ¨sections with different radial dimensions, such as 1â¨and 3 in Fig. 3, it is sufficiently accurate to takeâ¨GMD = z. Place filament at the center or at the GMâ¨of the cross-section, 3) Mutual Inductance M14, z â 0. In this case a single filament is not sufficiently accurateâ¨Lyle's Method [2] is used here with two filaments replacÂ¬â¨ing each section. The radial dimensions are given by AOO p/ll/2 Nhl/2 fÂ»CI2 P = )<J>H ox \ rJ(r)Jx(kr) JO v-h\12 J-fo/2 Jri Jai â aJ(a)J](ka)e~k^z + 72~7i<' da dr drx dr2 dk. (11) The internal integrals are readily solved, with the aid ofâ¨(6), to give r Â» 00 Jo hh P = yoj/L0ir (2 02 hx In h2 In Â«i â¢ S(kr2, krx)S(ka2, kax)Q(khx, kh2)e dk (12) h2 w2 - h1 where G.2 = R( 1 + (17) + 24 R2 12 Q(kx, ky) = p 2 2 cosh k where R is taken at the center or the GM of the cross-â¨section. The total mutual inductance between the two sectionsâ¨is the sum of the individual mutual inductances betweenâ¨the equivalent filaments of each section, each carrying halfâ¨the total current. h\ + h2 z > 2 -kh 2 , e~ = -ih + - - 1 z = 0,x = y = h k k (13) M, 4 â (Mac + Mad + Mhc + Mhd)l 4 where a and b represent the filaments in one cross-sectionâ¨and c and d represent the filaments in the other cross-secÂ¬â¨tion. B. Experimental Validation An experimental device was constructed with the diÂ¬â¨mensions shown in Fig. 3. Table I summarizes the results for the following conÂ¬â¨ditions: I) Measurement; this was carried out at 10 kHz toâ¨avoid high frequency effects which shall be disÂ¬â¨cussed later. (18) J0(kx) - JQ(ky) (14) S(kx, ky) = k But P â v2i2 â jioM\2l\I2 (15) where Mn is the mutual inductance between the two coils. Equating (12) and (15): ft 00 I S(kr2, kr{)S(ka2, ka,) Jo " Mo Tt M\2 = rJ, 02 hxh2 In In ax r\ â¢ Q(khx, kh2)e dk (16) HURLEY AND DUFFY: SELF AND MUTUAL IMPEDANCES IN PLANAR MAGNETIC STRUCTURES 2419 o 6.9mm R 2.3mm 1 c tSJ LL T L L â âi O 1 I 14! 1 C I â¡ IâI I Fig. 4. Lumped parameter model of the test device. 600^ 15 M 5000 r 250fi Fig. 3. Layout of experimental device. 4000 - TABLE I cT Experimental and Calculated Results 3000 - Calculated Z c Measured Z aâ¨â a L, Mâ S. 2000 - Measurement 620 E FEA (ANSOFT) 4 350 3.942 2 210 628 1000 - New Mode! s 16) 4.365 3 95! 2 22'9 627 GM Approximation 4.38! 4017 2 2! 7 627 Ofc â t- 0 10 20 Center Approximation 4.451 4 145 2.287 635 30 40 50 60 70 80 90 100â¨Frequency (MHz) L. is the self inductance of the prototype device All values m nH Fig. 5. Input impedance of the test device. I , is the self inductance of section I in figure 3 and M12 is the mutual inductance between sections 1 and 2 etc. Clearly the GM approximation provides very accurate estimate of the approach. The equivalent resistance is a function of theâ¨skin effect losses and proximity effect losses and the efÂ¬â¨fective ac resistance can be found from finite elementâ¨analysis simulation of the device [9], overall inductance The approximation for filaments at the center of sections is quite good in this case, however, as the ratio w/h increases the error grows rapidly 2) Finite Element Analysis; the finite element analysisâ¨(FEA) [9] was carried out at 10 kHz. 3) Model; Numerical evaluation of (16) [10]. 4) GM Approximation; Equation (2) with filaments atâ¨the geometric means. 5) Center Approximation; Equation (2) with filamentsâ¨at the center of sections. III. Spiral Coil on a Ferromagnetic Substrate The presence of a ferromagnetic substrate in the vicinÂ¬â¨ity of the planar coil in Fig. 2 enhances its self inducÂ¬â¨tance. If the half plane (z < 0) were replaced by an idealâ¨magnetic material (a = 0, p.r = oo) the self inductanceâ¨would be doubled as compared with the air case [4]. Theâ¨presence of coil currents gives rise to eddy current effectsâ¨in a ferrite with finite conductivity. The ultimate appliÂ¬â¨cation of these devices necessarily means high frequencyâ¨operation and therefore a general impedance equation isâ¨required, which takes frequency dependent eddy currentâ¨losses in the substrate into account. In this section such aâ¨generalized impedance formula is derived for magneticâ¨substrates, similar to (16) in Section II, which takes fullâ¨account of eddy currents in the substrate. The startingâ¨point of the analysis is the mutual impedance between twoâ¨filaments placed above a magnetic substrate as shown inâ¨Fig. 6. The substrate is assumed to be infinite in the âzâ¨direction. In practice, the substrate should be at least fiveâ¨skin depths thick to ensure the validity of this assumption.â¨The lower filament in Fig. 6 is at a height d above theâ¨substrate, so that a dielectric layer can be accounted forâ¨later. Maxwell's Equations are solved from first principles forâ¨the configuration in Fig. 6 and the details are given in the C. High Frequency Effects A general lumped-parameter model of the prototype deÂ¬â¨vice is shown in Fig. 4. The capacitance C consists of theâ¨three interlayer capacitances of the device connected inâ¨series. The input impedance of the device was measuredâ¨on a HP network analyzer from 10 kHz to 100 MHz andâ¨the results are shown in Fig. 5. The input impedance ofâ¨the equivalent circuit of Fig. 4 was calculated for L = 628 nH as calculated, R = 15.7 fl and C = 10.8 pF. The resonant frequency is 61 MHz. The capacitance C can beâ¨estimated by calculating the interlayer capacitance in Fig.â¨3 and treating each layer as a parallel plate capacitor, theâ¨dielectric constant of the insulating material is 7. Thisâ¨gives a predicted value of C = 13.4 pF which in turnâ¨gives a resonant frequency of 55 MHz. In the prototypeâ¨device the turns are spiralled and do not overlap in exactâ¨concentric circles, however, a reasonable estimate of resÂ¬â¨onant frequency is obtainable using this straightforward 242(1 IEEE TRANSACTIONS ON MAGNETICS, VOL. 31, NO. 4, JULY 1995 Iâ z"^2 medium 1 medium 1 "Â°2 1*5 dielectric layer/â¨magnetic substrata-â¨medfum 2 l dielectric layerâ¨mogrtetic substrateâ¨medium 2 hi: fli-0 1 Fig. 6. Circular concentric filaments on a magnetic substrate. Fig. 7. Planar coils on a magnetic substrate. Appendix. The mutual impedance between the two filaÂ¬â¨mentary circular concentric turns of Fig. 6 is Z = juM + Z, where M is the mutual inductance which would exist in â Â£â¨the absence of the substrate and is the same as (1). Zs is ^â¨the additional impedance due to the presence of the sub- Â§â¨strate. 1500 r â 140 1300 - (19) j 120â¨4100 gâ¨â j 80 gâ¨60 e 1100 - / 900 - Â« Calculated L (riH) o Simulated L (nH) Calculated R (ohms) x Simulated R (ohms) I I 700 - / n oo Jo Â£ 500 - e i 7s = Rx + ./col. = join o -war 40 *> ui 300 - 20 â¢ Ji(kr)Jl(ka)0(/t)e""M</| + d2) dk (20) 100 " â A f Q 10Â« 107 10" KM 102 102 10* 10s _ 1 Mr k Fig. 8. Self-impedance with a magnetic substrate. (j){k) = (21) 4- Vâ¨Mr + -k y= t joifiQ ixro. Fig. 3 on a magnetic substrate of transformer steel (a =â¨2 X 106 (fl â m) nr = 1000, d = 0) using finite eleÂ¬â¨ment analysis. The self impedance results are shown inâ¨Fig. 8. The calculated results were obtained using theâ¨impedance formula (23) in conjunction with (16) to acÂ¬â¨count for the air term. The dc resistance of the coil (0.686â¨(1) is included in the calculated resistance. Skin and proxÂ¬â¨imity effects in the winding are not included. There is veryâ¨good agreement between the simulated and calculated reÂ¬â¨sults which establishes the validity of the proposed forÂ¬â¨mula in predicting the effect of a magnetic substrate onâ¨the inductance and on the losses in a planar magnetic deÂ¬â¨vice. It is noteworthy that at 40 MHz skin and proximityâ¨effect losses contributed less than 5% to the total losses,â¨evidently at very high frequencies the eddy current lossesâ¨in the substrate dominate. The most salient feature of Fig.â¨8 is that the inductance remains essentially flat up to 1â¨MHz. Clearly with a less lossy substrate such as ferriteâ¨(a = 1 (Q â m)~') the frequency where the inductanceâ¨falls off is several orders of magnitude above 1 MHz. (22) A filament placed directly on an ideal magnetic subÂ¬â¨strate (d = 0, a = 0, fir -* oo) means 17 = k and Ls = Mâ¨giving a doubling of the inductance as expected. In air fxrâ¨â 1 and $(&) = 0 giving Ls = 0 as expected. Fig. 7 showsâ¨two circular concentric planar sections on a magnetic subÂ¬â¨strate. The current density is taken as being inversely proÂ¬â¨portional to the radius as described in Section II. ApplyÂ¬â¨ing the procedure outlined in Section II to the substrateâ¨term Zs in (20) gives yojMi) 7T /Â» OO 1 S(kr2, krl)S(ka2, kas)â¨Jo 7S = a2 ["l h\h2 In In at r 1 â¢ Q{khx, kh2)f(k)e ku/["(h) dk. (23) A. Validation of the New Formula The frequency dependent mutual impedance formulaâ¨(23) takes full account of eddy current losses in the subÂ¬â¨strate. The resistive component of Zv represents the subÂ¬â¨strate losses and the reactive component of Zs representsâ¨the enhanced inductance due to the reflected field of theâ¨magnetic substrate. Simulations were carried out for the device shown in IV. Conclusionsâ¨A new set of formulas has been established for calcuÂ¬â¨lating self and mutual impedances of planar coils on hoÂ¬â¨mogeneous ferromagnetic substrates. The formula forâ¨planar spiral coils in air is a special case of the generalâ¨formula. HURLEY AND DUFFY: SELF AND MUTUAL IMPEDANCES IN PLANAR MAGNETIC STRUCTURES 2421 The formulas have been derived from Maxwell's equaÂ¬â¨tions and therefore they can be fully expected to representâ¨practical planar devices accurately. As such, the formulasâ¨serve as a useful gauge for simpler approximations. Whileâ¨simple approximations are useful in initial design, itâ¨should be borne in mind that mathematical software packÂ¬â¨ages have reached such a maturity that they are reliable,â¨readily available and straightforward to use. Comparisons between experimental and calculated data,â¨for a 4 layer 12 turn device, show that the new formulaâ¨represents physical planar devices accurately. The workâ¨presented in this paper lays the groundwork for future deÂ¬â¨velopments, such as sandwich inductors where anotherâ¨substrate is added above the coil winding. BE <t> â- = jwfirij.0Hr dz 1 d(rEJ _ -juHoflrH. (A4) Z' r dr E has a ^-component only and we shall drop the <Â£ subÂ¬â¨script. Eliminating H gives the following result for meÂ¬â¨dium 1 ^ + p + = - a)8(z - d). (A5) Applying the Fourier-Bessel Integral Transformation [11],â¨noting that: V. Acknowledgmentâ¨The assistance of the National Microelectronic ReÂ¬â¨search Center, University College, Cork and Pulse EnÂ¬â¨gineering, Tuam, Co. Galway, Ireland, is gratefully apÂ¬â¨preciated. /Â» oo 1 5(r â a)Jx{kr)r dr = aJ^ka) Jo gives the transformed version of (A5) d2E* = k2E* + joi/jioI^aJ^k^diz - d). (A6) dz1 Appendix For a magnetoquasistatic system, the following formsâ¨of Maxwell's Equations hold in a linear homogeneousâ¨isotropic medium Equation (A6) has a solution of the form E* = Ae~kz + Bekz. (A7) V x H = J At this point we must distinguish between the region ofâ¨medium 1 above the filament i.e. z > d and the area beÂ¬â¨tween the filament and the surface of the substrate i.e. 0â¨< z < d. Clearly for z > d, the field decays at infinityâ¨and E* is given by E* = Ae~kz q> BB V X E = - (Al) dt The filamentary turn at z = d\ = d, in Fig. 6, carries aâ¨sinusoidal current /$(/) = Medium 1 refers to z > 0 in air and medium 2 refers to the magnetic substrate z z > d E* = Be+kz + Ce'kz 0 < z < d. (A8)â¨Eliminating H in (A3) gives d2E d2E 1 BE E . ^ 7T + ~z~2 + ~ a 2 = joiHrHooE. (A9) Bz Br r Br r Applying the Fourier-Bessel Integral Transform d2E* < 0. On the basis of cylindrical symmetry the followingâ¨identities apply to the electric field intensity E and theâ¨magnetic field intensity H Er = 0 E: = 0 = 0 B<k H.= 0 ^ = 0 BH; â (k2 + j<j)nrfiQo)E* = 0. = 0. (A2) (A 10) </> 2 B<$> B<)> dz Maxwell's Equations reduce toâ¨Medium 1 (z S 0): The electric field in medium 2 must tend to zero at infinityâ¨(z -* oo) and therefore the solution of (A 10) is of the form E* = Dev BHr BHZ (All) = /0i5(r - a) B{z ~ d) Bz Br where BEÂ± = 1] = \Jk2 + joifloHrO- The following boundary conditions apply:â¨(a) Â£ is continuous at the boundary between the two mediaâ¨at z = 0 /oj/xo Hr Bz 1 B(rE0) (A3) - = -y'w/i 0HZ r Br Medium 2 (z < 0) (A 12) B + C = D BHr BH: ââ = at (b) The radial component of H is continuous at the boundÂ¬â¨ary between the two media at z = 0 From (Al) <t> Bz Br 2422 IEEE TRANSACTIONS ON MAGNETICS. VOL. 31, NO. 4, JULY 1995 dE Thus ~T~ = juftoV-rH, (A13) rm dz 2txrE{r, d2) Z = - (A21) Application to (A7) and (All), noting that pr = 1 in meÂ¬â¨dium 1 with rearranging gives h The mutual impedance is k D = n, - (B â C) (A14) Z = juM + Z5 (A22) n where M is the component of mutual inductance whichâ¨would exist in the absence of the substrate and correÂ¬â¨sponds to (1) and (c) E is continuous in the plane of the filament at z â d (A 15) (d) The boundary condition on the H field in the plane ofâ¨the filament at z = d is given by Ae'kd = Beu + Ce -kd 00 Zs â jo)fx0irar \ Jx(kr)Jx{ka)<t>{k)e~â¨' Jo k(d[ 4- d2) dk. (A23) n X (H+ - H ) = K, References where n is the unit vector normal to the plane of the filÂ¬â¨ament and Kjas the surface current density at the boundaryâ¨which is given by 11 ] J. C. Maxwell, A Treatise on Electricity and Magnetism. Oxford Clarendon Press, 1873. [2] F. W. Grover, Inductance Calculations: Working Formulas and TaÂ¬â¨bles. New York: Dover Publications, 1946. '</+ /^(r - a)Â§(z - d) dz. = /0 6(r â a) K, = |3] R. Rodriquez, J. M. Dishman, F. T. Dickens, and E. W. Whelan,â¨"Modeling of two-dimensional spiral inductors," IEEE Trans. Com- f ponents, Hybrids and Mfg. Technology, vol. CHMT-3, no. 4. pp.â¨535-541, Dec. 1980 and in terms of transformed variables [4] W. A. Roshen, and D. E. Turcolte, "Planar inductors on magnetic K* â l$aj\(ka) substrates," IEEE Trans. Magn.. vol. MAG-24, no. 6, pp. 3213Â¬â¨3216, Sept. 1988. H+ and //_ are found from (A13) and the boundary conÂ¬â¨dition now becomes [5] W. A. Roshen, "Analysis of planar sandwich inductors by current images," IEEE Trans. Magn., vol. MAG-26, no. 5. pp. 2880-2887,â¨Sept. 1990. [6) A. Gray, Absolute Measurements in Electricity and Magnetism, vol.â¨11, pt. I. London: Macmillan, 1893. [7) H. B. Dwight, "Some new formulas for reactance coils," AIEE - k(Behl - Ce kd) = j<j3fJioI0aJl(ka). (A16) The four unknowns A, B, C and D are now solved using Trans., vol. 38, pt. 2, pp. 1675-1696, 1919. (A12), (A14), (A15) and (A16) to give E* in medium 1 [KJ T. R. Lyle, "On the self inductance of circular coils of rectangular for z > d section," Phil. Trans., vol. 2L3A, pp. 421-435. 1914. |9| User's Manual, Ansoft Corporation, Maxwell 2D Field Simulator, yojpo L Â« (version 4.33), Sept. 1991. " + 4>(k)e kA:~d)]Jx{ka) E* = - 11User's Manual, Matlab. The MaihWorks, Inc. (version 4.0), Feb. 2k 1993. [11] P. Morse, H. Feshback, Methods of Theoretical Physics. New York:â¨McGraw-Hill, ch. 6, pt. 1, 1953. [12) J. A. Tegopoulos and E. E. Kriezis, Eddy Currents in Linear ConÂ¬â¨ducting Media, ch. 5 Amsterdam: Elsevier, 1985. (A17) where _ 1 V-r k <t>(k) = (A 18) William Gerard Hurley (M"77, SM'90) was born in Cork, Ireland in 1952. He graduated from the National University of Ireland, Cork in 1974 withâ¨a first class honors degree in Electrical Engineering. He received the MasÂ¬â¨ter's degree in electrical engineering at the Massachusetts Institute of TechÂ¬â¨nology in 1976. He received the Ph.D. on Transformer Modelling at theâ¨National University of Ireland, Galway, in 1988. He worked for Honeywell Controls in Canada as a product engineer fromâ¨1977 to 1979 and as a development engineer in transmission lines at OnÂ¬â¨tario Hydro from 1979 to 1983. He lectured in electronic engineering atâ¨the University of Limerick, Ireland from 1983 to 1991 and is currently aâ¨senior lecturer in the Department of Electronic Engineering at Universityâ¨College, Galway, Ireland. He is the director of the Power Electronics ReÂ¬â¨search Center there. 1 dr + k Applying the inverse transform of the Fourier-Bessel InÂ¬â¨tegral 00 E = 5 \e-k\z-d\ + 0 â¢ Jx{ka) Ji(kr) dk. (A19) Dr. Hurley is a member of the Administrative Committee of the Powerâ¨Electronics Society of the IEEE and a member of Sigma Xi. An alternative derivation of (A 19) using the method ofâ¨separation of variables is given in reference [12], Mutual impedance Z, between the source at (a, d\) andâ¨the circular filament at (r, d2) gives the induced voltageâ¨V = Z/^. This voltage is Maeve C. Duffy was born in Monaghan, Ireland, in 1971. She graduatedâ¨from University College, Galway with a first class honors degree in ElecÂ¬â¨tronic Engineering in 1992. She is presently studying for the Ph.D. degreeâ¨in high frequency planar magnetics under Dr. W. G. Hurley in the Powerâ¨Electronics Research Center at University College, Galway. She is a memÂ¬â¨ber of the IEE. 2x f Jo E(r, d2)r dt}> = -2xrÂ£(r, d2) (A20) V = -