6.013 Electromagnetics and Applications, Chapter 3: Electromagnetic Fields in Simple Devices and Circuits

Chapter 3: Electromagnetic Fields in Simple Devices and Circuits 3.1 Resistors and capacitors 3.1.1 Introduction One important application of electromagnetic field analysis is to simple electronic components such as resistors, capacitors, and inductors, all of which exhibit at higher frequencies characteristics of the others. Such structures can be analyzed in terms of their: 1) static behavior, for which we can set ∂/∂t = 0 in Maxwell’s equations, 2) quasistatic behavior, for which ∂/∂t is non-negligible, but we neglect terms of the order ∂2/∂t2, and 3) dynamic behavior, for which terms on the order of ∂2/∂t2 are not negligible either; in the dynamic case the wavelengths of interest are no longer large compared to the device dimensions. Because most such devices have either cylindrical or planar geometries, as discussed in Sections 1.3 and 1.4, their fields and behavior are generally easily understood. This understanding can be extrapolated to more complex structures. One approach to analyzing simple structures is to review the basic constraints imposed by symmetry, Maxwell’s equations, and boundary conditions, and then to hypothesize the electric and magnetic fields that would result. These hypotheses can then be tested for consistency with any remaining constraints not already invoked. To illustrate this approach resistors, capacitors, and inductors with simple shapes are analyzed in Sections 3.1–2 below. All physical elements exhibit varying degrees of resistance, inductance, and capacitance, depending on frequency. This is because: 1) essentially all conducting materials exhibit some resistance, 2) all currents generate magnetic fields and therefore contribute inductance, and 3) all voltage differences generate electric fields and therefore contribute capacitance. R’s, L’s, and C’s are designed to exhibit only one dominant property at low frequencies. Section 3.3 discusses simple examples of ambivalent device behavior as frequency changes. Most passive electronic components have two or more terminals where voltages can be measured. The voltage difference between any two terminals of a passive device generally depends on the histories of the currents through all the terminals. Common passive linear twoterminal devices include resistors, inductors, and capacitors (R’s, L’s. and C’s, respectively), while transformers are commonly three- or four-terminal devices. Devices with even more terminals are often simply characterized as N-port networks. Connected sets of such passive linear devices form passive linear circuits which can be analyzed using the methods discussed in Section 3.4. RLC resonators and RL and RC relaxation circuits are most relevant here because their physics and behavior resemble those of common electromagnetic systems. RLC resonators are treated in Section 3.5, and RL, RC, and LC circuits are limiting cases when one of the three elements becomes negligible. 3.1.2 Resistors Resistors are two-terminal passive linear devices characterized by their resistance R [ohms]: - 65 - v = iR (3.1.1) where v(t) and i(t) are the associated voltage and current. That is, one volt across a one-ohm resistor induces a one-ampere current through it; this defines the ohm. The resistor illustrated in Figure 3.1.1 is comprised of two parallel perfectly conducting endplates between which is placed a medium of conductivity σ, permittivity ε, permeability μ, and thickness d; the two end plates and the medium all have a constant cross-sectional area A [m2] in the x-y plane. Let’s assume a static voltage v exists across the resistor R, and that a current i flows through it. (a) (b) v=0 σ, ε, μ i(t) x +v E ρs ++++++++++++ d area A z y - - - - - - - - - - - - d +v -ρs v=0 i(t) Figure 3.1.1 Simple resistor. Boundary conditions require the electric field E at any perfectly conducting plate to be perpendicular to it [see (2.6.16); E × n̂ = 0 ], and Faraday’s law requires that any line integral of E from one iso-potential end plate to the other must equal the voltage v regardless of the path of integration (1.3.13). Because the conductivity σ [Siemens/m] is uniform within walls parallel to ẑ , these constraints are satisfied by a static uniform electric field E = ẑE o everywhere within the conducting medium, which would be charge-free since our assumed E is non-divergent. Thus: d ∫0 E • ẑ dz = Eod = v (3.1.2) where E o = v d ⎡⎣ Vm -1 ⎤⎦ . Such an electric field within the conducting medium induces a current density J , where: J = σE ⎡⎣Am-2 ⎤⎦ (3.1.3) - 66 - The total current i flowing is the integral of J • ẑ over the device cross-section A, so that: i = ∫∫ J • ẑ dxdy = ∫∫ σE • ẑ dxdy = ∫∫ σE o dxdy = σE o A = vσA d A A A (3.1.4) But i = v/R from (3.1.1), and therefore the static resistance of a simple planar resistor is: R = v i = d σA [ ohms ] (3.1.5) The instantaneous power p [W] dissipated in a resistor is i2R = v2/R, and the time-average 2 2 power dissipated in the sinusoidal steady state is I R 2 = V 2 R watts. Alternatively the local instantaneous power density Pd = E • J ⎡⎣ W m-3 ⎤⎦ can be integrated over the volume of the resistor to yield the total instantaneous power dissipated: 2 p = ∫∫∫ E • J dv = ∫∫∫ E • σE dv = σ E Ad = σAv 2 d = v 2 R [ W ] V V (3.1.6) which is the expected answer, and where we used (2.1.17): J = σE . Surface charges reside on the end plates where the electric field is perpendicular to the perfect conductor. The boundary condition n̂ • D = ρs (2.6.15) suggests that the surface charge density ρs on the positive end-plate face adjacent to the conducting medium is: ρs = εE o ⎡⎣ Cm -2 ⎤⎦ (3.1.7) The total static charge Q on the positive resistor end plate is therefore ρsA coulombs. By convention, the subscript s distinguishes surface charge density ρs [C m-2] from volume charge density ρ [C m-3]. An equal negative surface charge resides on the other end-plate. The total stored charged Q = ρsA = CV, where C is the device capacitance, as discussed further in Section 3.1.3. The static currents and voltages in this resistor will produce fields outside the resistor, but these produce no additional current or voltage at the device terminals and are not of immediate concern here. Similarly, μ and ε do not affect the static value of R. At higher frequencies, however, this resistance R varies and both inductance and capacitance appear, as shown in the following three sections. Although this static solution for charge, current, and electric field within the conducting portion of the resistor satisfies Maxwell’s equations, a complete solution would also prove uniqueness and consistency with H and Maxwell’s equations outside the device. Uniqueness is addressed by the uniqueness theorem in Section 2.8, and approaches to finding fields for arbitrary device geometries are discussed briefly in Sections 4.4–6. - 67 - Example 3.1A Design a practical 100-ohm resistor. If thermal dissipation were a problem, how might that change the design? Solution: Resistance R = d/σA (3.1.5), and if we arbitrarily choose a classic cylindrical shape with resistor length d = 4r, where r is the radius, then A = πr2 = πd2/16 and R = 16/πdσ = 100. Discrete resistors are smaller for modern low power compact circuits, so we might set d = 1 mm, yielding σ = 16/πdR = 16/(π10-3×100) ≅ 51 S m-1. Such conductivities roughly correspond, for example, to very salty water or carbon powder. The surface area of the resistor must be sufficient to dissipate the maximum power expected, however. Flat resistors thermally bonded to a heat sink can be smaller than air-cooled devices, and these are often made of thin metallic film. Some resistors are long wires wound in coils. Resistor failure often occurs where the local resistance is slightly higher, and the resulting heat typically increases the local resistance further, causing even more local heating. 3.1.3 Capacitors Capacitors are two-terminal passive linear devices storing charge Q and characterized by their capacitance C [Farads], defined by: Q = Cv [Coulombs] (3.1.8) where v(t) is the voltage across the capacitor. That is, one static volt across a one-Farad capacitor stores one Coulomb on each terminal, as discussed further below; this defines the Farad [Coulombs per volt]. The resistive structure illustrated in Figure 3.1.1 becomes a pure capacitor at low frequencies if the media conductivity σ → 0. Although some capacitors are air-filled with ε ≅ εo, usually dielectric filler with permittivity ε > εo is used. Typical values for the dielectric constant ε/εo used in capacitors are ~1-100. In all cases boundary conditions again require that the electric field E be perpendicular to the perfectly conducting end plates, i.e., to be in the ±z direction, and Faraday’s law requires that any line integral of E from one iso-potential end plate to the other must equal the voltage v across the capacitor. These constraints are again satisfied by a static uniform electric field E = ẑE o within the medium separating the plates, which is uniform and charge-free. We shall neglect temporarily the effects of all fields produced outside the capacitor if its plate separation d is small compared to its diameter, a common configuration. Thus Eo = v/d [V m-1] (3.1.2). The surface charge density on the positive end-plate face adjacent to the conducting medium is σs = εEo [C m-2], and the total static charge Q on the positive end plate of area A is therefore: Q = Aσs = AεE o = Aεv d = Cv [ C ] - 68 - (3.1.9) Therefore, for a parallel-plate capacitor: C = εA d [ Farads ] (parallel-plate capacitor) (3.1.10) Using (3.1.2) and the fact that the charge Q(t) on the positive plate is the time integral of the current i(t) into it, we obtain the relation between voltage and current for a capacitor: v ( t ) = Q ( t ) C = (1 C ) ∫ t −∞ i ( t ) dt (3.1.11) i ( t ) = C dv ( t ) dt (3.1.12) When two capacitors are connected in parallel as shown in Figure 3.1.2, they are equivalent to a single capacitor of value Ceq storing charge Qeq, where these values are easily found in terms of the charges (Q1, Q2) and capacitances (C1, C2) associated with the two separate devices. (a) i(t) (b) + Q1 v(t) i(t) Q2 Qeq v(t) C2 C1 + - Ceq = C1 + C2 Figure 3.1.2 Capacitors in parallel. Because the total charge Qeq is the sum of the charges on the two separate capacitors, and capacitors in parallel have the same voltage v, it follows that: Qeq = Q1 + Q2 = (C1 + C2)v = Ceqv (3.1.13) Ceq = C1 + C2 (a) i(t) (capacitors in parallel) (b) v1(t) + v(t) + Q(t) C1 v2(t) + Q(t) - i(t) + v(t) C2 - Figure 3.1.3 Capacitors in series. - 69 - Q(t) Ceq-1 = C1-1 + C2-1 (3.1.14) When two capacitors are connected in series, as illustrated in Figure 3.1.3, then their two charges Q1 and Q2 remain equal if they were equal before current i(t) began to flow, and the total voltage is the sum of the voltages across each capacitor: Ceq-1 = v/Q = (v1 + v2)/Q = C1-1 + C2-1 (capacitors in series) (3.1.15) The instantaneous electric energy density We [J m-3] between the capacitor plates is given by 2 Poynting’s theorem: We = ε E 2 (2.7.7). The total electric energy we stored in the capacitor is the integral of We over the volume Ad of the dielectric: 2 2 ( ε E 2 ) dv = εAd E 2 = εAv2 2d = Cv 2 2 V w e = ∫∫∫ [J] (3.1.16) The corresponding expression for the time-average energy stored in a capacitor in the sinusoidal steady state is: we = C V J 2 4 [ ] (3.1.17) The extra factor of two relative to (3.1.9) enters because the time average of a sinsuoid squared is half its peak value. To prove (3.1.16) for any capacitor C, not just parallel-plate devices, we can compute t t v w e = ∫ 0iv dt where i = dq dt and q = Cv . Therefore w e = ∫ 0C ( dv dt ) v dt = ∫ 0 Cv dv = Cv 2 2 in general. We can also analyze other capacitor geometries, such as the cylindrical capacitor illustrated in Figure 3.1.4. The inner radius is “a”, the outer radius is “b”, and the length is D; its interior has permittivity ε. D b a + v Figure 3.1.4 Cylindrical capacitor. The electric field again must be divergence- and curl-free in the charge-free regions between the two cylinders, and must be perpendicular to the inner and outer cylinders at their perfectly conducting walls. The solution can be cylindrically symmetric and independent of φ. A purely radial electric field has these properties: - 70 - E ( r ) = r̂E o r (3.1.18) The electric potential Φ(r) is the integral of the electric field, so the potential difference v between the inner and outer conductors is: () bE v = Φ a − Φ b = ∫ o dr = E o ln r ab = E o ln b [ V ] a r a (3.1.19) This capacitor voltage produces a surface charge density ρs on the inner and outer conductors, where ρs = εE = εEo/r. If Φa > Φb, then the inner cylinder is positively charged, the outer cylinder is negatively charged, and Eo is positive. The total charge Q on the inner cylinder is then: Q = ρs 2πaD = εE o 2πD = εv2πD ⎡⎣ln ( b a ) ⎤⎦ = CCv [ ] (3.1.20) Therefore this cylindrical capacitor has capacitance C: C = ε2πD ⎡⎣ln ( b a )F⎤⎦ [ ] (cylindrical capacitor) (3.1.21) In the limit where b/a → 1 and b - a = d, then we have approximately a parallel-plate capacitor with C → εA/d where the plate area A = 2πaD; see (3.1.10). Example 3.1B Design a practical 100-volt 10-8 farad (0.01 mfd) capacitor using dielectric having ε = 20εo and a breakdown field strength EB of 107 [V m-1]. Solution: For parallel-plate capacitors C = εA/d (3.1.10), and the device breakdown voltage is EBd = 100 [V]. Therefore the plate separation d = 100/EB = 10-5 [m]. With a safety factor of two, d doubles to 2×10-5, so A = dC/ε = 2×10-5 × 10-8/(20 × 6.85×10-12) ≅ 1.5×103 [m2]. If the capacitor is a cube of side D, then the capacitor volume is D3 = Ad and D = (Ad)0.333 = (1.5×10-3 × 2×10-5)0.333 ≅ 3.1 mm. To simplify manufacture, such capacitors are usually wound in cylinders or cut from flat stacked sheets. 3.2 Inductors and transformers 3.2.1 Solenoidal inductors All currents in devices produce magnetic fields that store magnetic energy and therefore contribute inductance to a degree that depends on frequency. When two circuit branches share magnetic fields, each will typically induce a voltage in the other, thus coupling the branches so they form a transformer, as discussed in Section 3.2.4. - 71 - Inductors are two-terminal passive devices specifically designed to store magnetic energy, particularly at frequencies below some design-dependent upper limit. One simple geometry is shown in Figure 3.2.1 in which current i(t) flows in a loop through two perfectly conducting parallel plates of width W and length D, spaced d apart, and short-circuited at one end. (a) contour C2 contour C1 i(t) d (b) W x da H=0 contour C2 Ho i(t) x D z y J H=0 y z=0 Figure 3.2.1 Parallel-plate inductor. To find the magnetic field from the currents we can use the integral form of Ampere’s law, which links the variables H and J : ∫C H • ds = ∫∫A ( J + ∂ D ∂t ) • da (3.2.1) The contour C1 around both currents in Figure 3.2.1 encircles zero net current, and (3.2.1) says the contour integral of H around zero net current must be zero in the static case. Contour C2 encircles only the current i(t), so the contour integral of H around any C2 in the right-hand sense equals i(t) for the static case. The values of these two contour integrals are consistent with zero magnetic field outside the pair of plates and a constant field H = H o ŷ between them, although a uniform magnetic field could be superimposed everywhere without altering those integrals. Since such a uniform field would not have the same symmetry as this device, such a field would have to be generated elsewhere. These integrals are also exactly consistent with fringing fields at the edges of the plate, as illustrated in Figure 3.2.1(b) in the x-y plane for z > 0. Fringing fields can usually be neglected if the plate separation d is much less than the plate width W. It follows that: ∫C2 H • ds = i ( t ) = Ho W (3.2.2) H = ŷ Ho = ŷ i ( t ) W ⎡⎣A m-1 ⎤⎦ and H ≅ 0 elsewhere. - 72 - ( H between the plates) (3.2.3) contour C i(t) i(t) +v(t) - 2 d E x (t,z) 1 D da H(t) x z z=0 Figure 3.2.2 Voltages induced on a parallel-plate inductor. The voltage v(t) across the terminals of the inductor illustrated in Figures 3.2.1 and 3.2.2 can be found using the integral form of Faraday’s law and (3.2.3): ∂ ∫C E • ds = − ∂t ∫∫A μH • da = − 2 μDd di ( t ) = ∫ E x ( t,z ) dx = −v ( t,z ) 1 W dt (3.2.4) where z = D at the inductor terminals. Note that when we integrate E around contour C there is zero contribution along the path inside the perfect conductor; the non-zero portion is restricted to the illustrated path 1-2. Therefore: v(t) = ( ) μDd di ( t ) = L di t W dt dt (3.2.5) where (3.2.5) defines the inductance L [Henries] of any inductor. Therefore L1 for a single-turn current loop having length W >> d and area A = Dd is: L1 = μDd μA [ ] = H W W (single-turn wide inductor) (3.2.6) To simplify these equations we define magnetic flux ψm as8: ψ m = ∫∫ μH • da [Webers = Vs] (3.2.7) A Then Equations (3.2.4) and (3.2.7) become: v(t) = dψm (t)/dt (3.2.8) ψm (t) = L i(t) 8 (single-turn inductor) (3.2.9) The symbol ψm for magnetic flux [Webers] should not be confused with Ψ for magnetic potential [Amperes]. - 73 - Since we assumed fringing fields could be neglected because W >> d, large single-turn inductors require very large structures. The standard approach to increasing inductance L in a limited volume is instead to use multi-turn coils as illustrated in Figure 3.2.3. W (a) i(t) +v(t) (b) d μ μ μo D i(t) - Figure 3.2.3 N-turn inductor: (a) solenoid, (b) toroid. The N-turn coil of Figure 3.2.3 duplicates the current flow geometry illustrated in Figures 3.2.1 and 3.2.2, but with N times the intensity (A m-1) for the same terminal current i(t), and therefore the magnetic field Ho and flux ψm are also N times stronger than before. At the same time the voltage induced in each turn is proportional to the flux ψm through it, which is now N times greater than for a single-turn coil (ψm = NiμA/W), and the total voltage across the inductor is the sum of the voltages across the N turns. Therefore, provided that W >> d, the total voltage across an N-turn inductor is N2 times its one-turn value, and the inductance LN of an N-turn coil is also N2 greater than L1 for a one-turn coil: ( ) ( ) v ( t ) = L N di t = N 2 L1 di t dt dt H2 LN = N (3.2.10) μA [ ] W (N-turn solenoidal inductor) where A is the coil area and W is its length; W >> (3.2.11) A > d. Equation (3.2.11) also applies to cylindrical coils having W >> d, which is the most common form of inductor. To achieve large values of N the turns of wire can be wound on top of each other with little adverse effect; (3.2.11) still applies. These expressions can also be simplified by defining magnetic flux linkage Λ as the magnetic flux ψm (3.2.7) linked by N turns of the current i, where: Λ = Nψm = N(NiμA/W) = (N2μA/W)i = Li (flux linkage) (3.2.12) This equation Λ = Li is dual to the expression Q = Cv for capacitors. We can use (3.2.5) and (3.2.12) to express the voltage v across N turns of a coil as: (any coil linking magnetic flux Λ) v = L di/dt = dΛ/dt - 74 - (3.2.13) The net inductance L of two inductors L1 and L2 in series or parallel is related to L1 and L2 in the same way two connected resistors are related: L = L1 + L2 L-1 = L1-1 + L2-1 (series combination) (3.2.14) (parallel combination) (3.2.15) For example, two inductors in series convey the same current i but the total voltage across the pair is the sum of the voltages across each – so the inductances add. Example 3.2A Design a 100-Henry air-wound inductor. Solution: Equation (3.2.11) says L = N2μA/W, so N and the form factor A/W must be chosen. Since A = πr2 is the area of a cylindrical inductor of radius r, then W = 4r implies L = N2μπr/4. Although tiny inductors (small r) can be achieved with a large number of turns N, N is limited by the ratio of the cross-sectional areas of the coil rW and of the wire πrw2, and is N ≅ r2/rw2. N is further limited if we want the resistive impedance R << jωL. If ωmin is the lowest frequency of interest, then we want R ≅ ωminL/100 = d/(σπrw2) [see (3.1.5)], where the wire length d ≅ 2πrN. These constraints eventually yield the desired values for r and N that yield the smallest inductor. Example 3.2B carries these issues further. 3.2.2 Toroidal inductors The prior discussion assumed μ filled all space. If μ is restricted to the interior of a solenoid, L is diminished significantly, but coils wound on a high-μ toroid, a donut-shaped structure as illustrated in Figure 3.2.3(b), yield the full benefit of high values for μ. Typical values of μ are ~5000 to 180,000 for iron, and up to ~106 for special materials. Coils wound on high-permeability toroids exhibit significantly less flux leakage than solenoids. Consider the boundary between air and a high-permeability material (μ/μo >>1), as illustrated in Figure 3.2.4. B1⊥ = B2⊥ H2⊥ H 2 , B2 μo μ >> μo B1 H 2 // = H1// << B1// B1// >> H1// Figure 3.2.4 Magnetic fields at high-permeability boundaries. - 75 - The degree to which B is parallel or perpendicular to the illustrated boundary has been diminished substantially for the purpose of clarity. The boundary conditions are that both B⊥ and H // are continuous across any interface (2.6.5, 2.6.11). Since B = μH in the permeable core and B = μo H in air, and since H // is continuous across the boundary, therefore B// changes across the boundary by the large factor μ/μο. In contrast, B⊥ is the same on both sides. Therefore, as suggested in Figure 3.2.4, B2 in air is nearly perpendicular to the boundary because H // , and therefore⎯B2//, is so very small; note that the figure has been scaled so that the arrows representing⎯H2 and⎯B2 have the same length when μ = μ0. i(t) N turns + v(t) - B μ >> μo, e.g. iron Cross-sectional area A Figure 3.2.5 Toroidal inductor. In contrast, B1 is nearly parallel to the boundary and is therefore largely trapped there, even if that boundary curves, as shown for a toroid in Figure 3.2.5. The reason magnetic flux is largely trapped within high-μ materials is also closely related to the reason current is trapped within high-σ wires, as described in Section 4.3. The inductance of a toroidal inductor is simply related to the linked magnetic flux Λ by (3.2.12) and (3.2.7): L= Λ = i μN ∫∫ H • da A (toroidal inductor) i (3.2.16) where A is any cross-sectional area of the toroid. Computing H is easier if the toroid is circular and has a constant cross-section A which is small compared to the major radius R so that R >> A . From Ampere’s law we learn that the integral of H around the 2πR circumference of this toroid is: ∫C H • ds ≅ 2πRH ≅ Ni (3.2.17) where the only linked current is i(t) flowing through the N turns of wire threading the toroid. Equation (3.2.17) yields H ≅ Ni/2πR and (3.2.16) relates H to L. Therefore the inductance L of such a toroid found from (3.2.16) and (3.2.17) is: L≅ μNA Ni μN 2 A [ Henries ] = i 2πR 2πR - 76 - (toroidal inductor) (3.2.18) The inductance is proportional to μ, N2, and cross-sectional area A, but declines as the toroid major radius R increases. The most compact large-L toroids are therefore fat (large A) with almost no hole in the middle (small R); the hole size is determined by N (made as large as possible) and the wire diameter (made small). The maximum acceptable series resistance of the inductor limits N and the wire diameter; for a given wire mass [kg] this resistance is proportional to N2. area A = πr2 i(t) 2R μo μ gap d[m] Figure 3.2.6 Toroidal inductor with a small gap. The inductance of a high-permeability toroid is strongly reduced if even a small gap of width d exists in the magnetic path, as shown in Figure 3.2.6. The inductance L of a toroid with a gap of width d can be found using (3.2.16), but first we must find the magnitude of Hμ within the toroid. Again we can use the integral form of Ampere’s law for a closed contour along the axis of the toroid, encircling the hole. ∫C H • ds ≅ ( 2πR − d ) Hμ + Hg d ≅ Ni (3.2.19) where Hg is the magnitude of H within the gap. Since B⊥ is continuous across the gap faces, μoHg = μHμ and these two equations can be solved for the two unknowns, Hg and Hμ. The second term Hgd can be neglected if the gap width d << 2πRμo/μ. In this limiting case we have the same inductance as before, (3.2.18). However, if A0.5 > d >> 2πRμo/μ, then Hg ≅ Ni/d and: L = Λ i ≅ Nψ m i ≅ Nμo H g A i ≅ N 2μo A d [ H ] (toroid with a gap) (3.2.20) Relative to (3.2.18) the inductance has been reduced by a factor of μo/μ and increased by a much smaller factor of 2πR/d, a significant net reduction even though the gap is small. Equation (3.2.20) suggests how small air gaps in magnetic motors limit motor inductance and sometimes motor torque, as discussed further in Section 6.3. Gaps can be useful too. For example, if μ is non-linear [μ = f (H)], then L ≠ f (H) if the gap and μo dominate L. Also, inductance dominated by gaps can store more energy when H exceeds saturation (i.e., B2 2μ o 2 BSAT 2μ ). - 77 - 3.2.3 Energy storage in inductors The energy stored in an inductor resides in its magnetic field, which has an instantaneous energy density of: Wm ( t ) = μ H 2 2 ⎡⎣ J m-3 ⎤⎦ (3.2.21) Since the magnetic field is uniform within the volume Ad of the rectangular inductor of Figure 3.2.1, the total instantaneous magnetic energy stored there is: w m ≅ μAW H 2 2 2 ≅ μAW ( i W ) 2 ≅ Li 2 2 [ J ] (3.2.22) That (3.2.22) is valid and exact for any inductance L can be shown using Poynting’s theorem, which relates power P = vi at the device terminals to changes in energy storage: wm = ∫ t −∞ v ( t ) i ( t ) dt = ∫ t −∞ i L ( di dt ) i dt = ∫ Li di = Li 2 2 [ J ] 0 (3.2.23) Earlier we neglected fringing fields, but they store magnetic energy too. We can compute them accurately using the Biot-Savart law (10.2.21), which is derived later and expresses H directly in terms of the currents flowing in the inductor: 3 H ( r ) = ∫∫∫ dv ' [ J ( r ') × ( r − r ')] ⎡⎣ 4π r − r ' ⎤⎦ V' (3.2.24) The magnetic field produced by current J ( r ') diminishes with distance squared, and therefore the magnitude of the uniform field H within the inductor is dominated by currents within a distance of ~d of the inductor ends, where d is the nominal diameter or thickness of the inductor [see Figure 3.2.3(a) and assume d ≅ D << W]. Therefore H at the center of the end-face of a semi-infinite cylindrical inductor has precisely half the strength it has near the middle of the same inductor because the Biot-Savart contributions to H at the end-face arise only from one side of the end-face, not from both sides. The energy density within a solenoidal inductor therefore diminishes within a distance of ~d from each end, but this is partially compensated in (3.2.23) by the neglected magnetic energy outside the inductor, which also decays within a distance ~d. For these reasons fringing fields are usually neglected in inductance computations when d << W. Because magnetic flux is nondivergent, the reduced field intensity near the ends of solenoids implies that some magnetic field lines escape the coil there; they are fully trapped within the rest of the coil. The energy stored in a thin toroidal inductor can be found using (3.2.21): ( wm ≅ μ H 2 ) 2 A2πR (3.2.25) - 78 - The energy stored in a toroidal inductor with a non-negligible gap of width d can be easily found knowing that the energy storage in the gap dominates that in the high-permeability toroid, so that: ( ) 2 w m ≅ μo Hg 2 2 Ad ≅ μo ( Ni d ) Ad 2 ≅ Li 2 2 (3.2.26) Example 3.2B Design a practical 100-Henry inductor wound on a toroid having μ = 104μo; it is to be used for ω ≅ 400 [r s-1] (~60 Hz). How many Joules can it store if the current is one Ampere? If the residual flux density Br of the toroid is 0.2 Tesla, how does this affect design? Solution: We have at least three unknowns, i,e., size, number of turns N, and wire radius rw, and therefore need at least three equations. Equation (3.2.18) says L ≅ μN2A/2πRm where A = πr2. A fat toroid might have major radius Rm ≅ 3r, corresponding to a central hole of radius 2r surrounded by an iron torus 2r thick, yielding an outer diameter of 4r. Our first equation follows: L = 100 ≅ μN2r/6. Next, the number N of turns is limited by the ratio of the cross-sectional area of the hole in the torus (π4r2) and the cross-sectional area of the wire πrw2; our second equation is N ≅ 4r2/rw2. Although tiny inductors (small r) can be achieved with large N, N is limited if we want the resistive impedance R << ωL. If ωmin is the lowest frequency of interest, then we obtain our third equation, R ≅ ωminL/100 = 400 = d/(σπrw2) [see (3.1.5)], where the wire length d ≅ 4πrN. Eliminating rw2 from the second and third equation yields N2 ≅ 400σr, and eliminating N2 from the first equation yields r = (600/400σμ)0.5 ≅ 1.5mm, where for typical wires σ ≅ 5×107; the maximum diameter of this toroid is 8r ≅ 1.2 cm. Since N2 ≅ 400σr, therefore N ≅ 5600, and rw ≅ 2r/ N ≅ 40 microns. We might suppose the stored energy wm = Li2/2 = 100×12/2 = 50 joules. However, if 1 ampere flows through 5600 turns, and if H = 5600/2π3r = 5600/0.031 = 1.8×105 [A m-1], then B = μH ≅ 2300 Tesla, well above the limit of Br = 0.2 Tesla where saturation was said to occur. Since the incremental μo applies at high currents, this device is quite non-linear and the computed stored energy of 50J should be reduced by a factor of ~μo/μ to yield ~5 mJ. If linearity and low loss (R<<ωL) are desired, either this toroid must be made much larger so that the upper limit on μH inside the toroid is not exceeded, or the maximum current must be reduced to the ~100 μA level. Moreover, a sinusoidal current of 1 ampere through this small 400-ohm resistance would dissipate 200 W, enough to damage it. Note that if ωmin is increased by a factor of F, then r decreases by F0.5. - 79 - 3.2.4 Transformers Transformers are passive devices used to raise or lower the voltages of alternating currents or transients. The voltage v across two terminals of any coil can be found using Faraday’s law (2.4.14): d ∫C E • ds = − dt ∫∫A μo H • da (3.2.27) which leads to the voltage across any N turns of a coil, as given by (3.2.13): v = dΛ dt (3.2.28) where the flux linkage Λ = Nψm and the magnetic flux ψm within the cross-sectional area A of the coil is defined by (3.2.7): Ψ m = ∫∫ μH • da [ Webers = Vs ] (3.2.29) A Consider the ideal toroidal transformer of Figure 3.2.7. + v2(t) + i1(t) v1(t) i2(t) N2 turns ⎯H - - μ >> μo ; e.g. iron N1 turns Cross-sectional area A Figure 3.2.7 Toroidal transformer. Its high permeability traps the magnetic flux within it so that ψm is constant around the toroid, even though A varies. From (3.2.28) we see that the voltage vk across coil k is therefore: vk = dΛ k dt = N k dΨ m dt (3.2.30) The ratio between the voltages across two coils k = 1,2 is therefore: v 2 v1 = N 2 N1 (3.2.31) where N2/N1 is the transformer turns ratio. If current i2 flows in the output coil, then there will be an added contribution to v1 and v2 due to the contributions of i2 to the original ψm from the input coil alone. Note that current flowing into the “+” terminal of both coils in the figure contribute to⎯H in the illustrated direction; this - 80 - distinguishes the positive terminal from the negative terminal of each coil. If the flux coupling between the two coils is imperfect, then the output voltage is correspondingly reduced. Any resistance in the wires can increment these voltages in proportion to the currents. Figure 3.2.8 suggests traditional symbols used to represent ideal transformers and some common configurations used in practice. The polarity dot at the end of each coil indicates which terminals would register the same voltage for a given change in the linked magnetic flux. In the absence of dots, the polarity indicated in (a) is understood. Note that many transformers consist of a single coil with multiple taps. Sometimes one of the taps is a commutator that can slide across the coil windings to provide a continuously variable transformer turns ratio. As illustrated, the presence of an iron core is indicated by parallel lines and an auto-transformer consists of only one tapped coil. (a) (c) (b) (d) Figure 3.2.8 Transformer configurations: (a) air-core, (b) iron-core, (c) tapped, and (d) auto-transformer. The terminal voltages of linear transformers for which μ ≠ f(H) are linearly related to the various currents flowing through the windings. Consider a simple toroid for which H, B, and the cross-sectional area A are the same everywhere around the average circumference πD. In this case the voltage V1 across the N1 turns of coil (1) is: V1 = jωN1 (3.2.32) Ψ = μHA (3.2.33) H = (N1I1 + N2I2)/πD (3.2.34) V1 = jω[μAN1(N1I1 + N2I2)/πD] = jω(L11I1 + L12I2) (3.2.35) Therefore: where the self-inductance L11 and mutual inductance L12 [Henries] are: L11 = μAN12/πD L12 = μAN1N2/πD (3.2.36) Equation (3.2.35) can be generalized for a two-coil transformer: ⎡ V1 ⎤ ⎡ L11 L12 ⎤ ⎡ I 1 ⎤ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎣ V2 ⎦ ⎣ L21 L22 ⎦ ⎣ I 2 ⎦ (3.2.37) - 81 - Consider the simple toroidal step-up transformer illustrated in Figure 3.2.9 in which the voltage source drives the load resistor R through the transformer, which has N1 and N2 turns on its input and output, respectively. The toroid has major diameter D and cross-sectional area A. ψm + I1 Vs - N1 turns area A μ + D μo I2 V2 R - N2 turns Figure 3.2.9 Toroidal step-up transformer loaded with resistor R. Combining (3.2.33) and (3.2.34), and noting that the sign of I2 has been reversed in the figure, we obtain the expression for total flux: Ψ = μA (N1I1 + N2I2)/πD (3.2.38) We can find the admittance seen by the voltage source by solving (3.2.38) for I1 and dividing by Vs: I1 = (πDΨ/μAN1) + I2 N2/N1 (3.2.39) Vs = jωN1Ψ = V2N1/N2 = I2RN1/N2 (3.2.40) I1/Vs = (πDΨ/μAN1)/jωN1Ψ + I2N22/(N12I2R) (3.2.41) = - jπD/(ωN12μA) + (N2/N1)2/R = 1/jωL11 + (N2/N1)2/R (3.2.42) Thus the admittance seen at the input to the transformer is that of the self-inductance (1/jωL11) in parallel with the admittance of the transformed resistance [(N2/N1)2/R]. The power delivered to the load is |V2|2/2R = |V1|2(N2/N1)2/2R, which is the time-average power delivered to the transformer, since |V2|2 = |V1|2(N2/N1)2; see (3.2.31). The transformer equivalent circuit is thus L11 in parallel with the input of an ideal transformer with turns ratio N2/N1. Resistive losses in the input and output coils could be represented by resistors in series with the input and output lines. Usually jωL11 for an iron-core transformer is so great that only the ideal transformer is important. One significant problem with iron-core transformers is that the changing magnetic fields within them can generate considerable voltages and eddy currents by virtue of Ohm’s Law (⎯J = σ⎯E) and Faraday’s law: - 82 - ∫C E • ds = − jωμ ∫A H • da (3.2.43) where the contour C circles each conducting magnetic element. A simple standard method for reducing the eddy currents⎯J and the associated dissipated power ∫V (σ|J|2/2)dv is to reduce the area A by laminating the core; i.e., by fabricating it with thin stacked insulated slabs of iron or steel oriented so as to interrupt the eddy currents. The eddy currents flow perpendicular to⎯H, so the slab should be sliced along the direction of⎯H. If N stacked slabs replace a single slab, then A, E, and J are each reduced roughly by a factor of N, so the power dissipated, which is proportional to the square of J, is reduced by a factor of ~N2. Eddy currents and laminated cores are discussed further at the end of Section 4.3.3. 3.3 Quasistatic behavior of devices 3.3.1 Electroquasistatic behavior of devices The voltages and currents associated with all interesting devices sometimes vary. If the wavelength λ = c/f associated with these variations is much larger than the device size D, no significant wave behavior can occur. The device behavior can then be characterized as electroquasistatic if the device stores primarily electric energy, and magnetoquasistatic if the device stores primarily magnetic energy. Electroquasistatics involves the behavior of electric fields plus the first-order magnetic consequences of their variations. The electroquasistatic approximation includes the magnetic field⎯H generated by the varying dominant electric field (Ampere’s law), where: ∇ × H = σE + ∂D ∂t (3.3.1) The quasistatic approximation neglects the second-order electric field contributions from the time derivative of the resulting⎯H in Faraday’s law: ∇ × E = −μ o ∂H / ∂t ≅ 0 . One simple geometry involving slowly varying electric fields is a capacitor charged to voltage V(t), as illustrated in Figure 3.3.1. It consists of two circular parallel conducting plates of diameter D and area A that are separated in vacuum by the distance d << D. Boundary conditions require⎯E to be perpendicular to the plates, where E(t) = V(t)/d, and the surface charge density is given by (2.6.15): E • n̂ = ρs / εo = V / d (3.3.2) ρs = εo V / d [C m −2 ] (3.3.3) - 83 - I Area A + V - d ⎯E −ρs φ̂ H r Figure 3.3.1 Quasistatic electric and magnetic fields in a circular capacitor. Since the voltage across the plates is the same everywhere, so are⎯E and ρs, and therefore the total charge is: Q(t) ≅ ρsA ≅ (εoA/d)V = CV(t) (3.3.4) where C ≅ εoA/d is the capacitance, as shown earlier (3.1.10). The same surface charge density ρs(t) can also be found by evaluating first the magnetic field⎯H(r,t) produced by the slowly varying (quasistatic) electric field⎯E(t), and then the surface current⎯Js(r,t) associated with⎯H(r,t); charge conservation then links⎯Js(r,t) to ρs(t). Ampere’s law requires a non-zero magnetic field between the plates where⎯J = 0: ∫C H • ds = εo ∫∫A′ ( ∂E ∂t ) • da (3.3.5) Symmetry of geometry and excitation requires that⎯H between the plates be in the φ̂ direction and a function only of radius r, so (3.3.5) becomes: 2πr H(r) = εoπr2 dE/dt = (εoπr2/d) dV/dt (3.3.6) H(r) = (εor/2d) dV/dt (3.3.7) If V(t) and the magnetic field H are varying so slowly that the electric field given by Faraday’s law for H(r) is much less than the original electric field, then that incremental electric field can be neglected, which is the essence of the electroquasistatic approximation. If it cannot be neglected, then the resulting solution becomes more wavelike, as discussed in later sections. The boundary condition n̂ ×⎯H =⎯Js (2.6.17) then yields the associated surface current⎯Js(r) flowing on the interior surface of the top plate: ⎯Js(r) = r̂ (εor/2d) dV/dt = r̂ Jsr (3.3.8) - 84 - This in turn is related to the surface charge density ρs by conservation of charge (2.1.19), where the del operator is in cylindrical coordinates: ∇ •⎯Js = -∂ρs/∂t = - r-1∂ (r Jsr)/∂r (3.3.9) Substituting Jsr from (3.3.8) into the right-hand side of (3.3.9) yields: ∂ρs/∂t = (εo/d) dV/dt (3.3.10) Multiplying both sides of (3.3.10) by the plate area A and integrating over time then yields Q(t) = CV(t), which is the same as (3.3.4). Thus we could conclude that variations in V(t) will produce magnetic fields between capacitor plates by virtue of Ampere’s law and the values of either ∂ D ∂t between the capacitor plates or⎯Js within the plates. These two approaches to finding⎯H (using ∂ D ∂t or⎯Js) yield the same result because of the self-consistency of Maxwell’s equations. Because the curl of⎯H in Ampere’s law equals the sum of current density⎯J and ∂ D ∂t , the derivative ∂ D ∂t is often called the displacement current density because the units are the same, A/m2. For the capacitor of Figure 3.3.1 the curl of⎯H near the feed wires is associated only with⎯J (or I), whereas between the capacitor plates the curl of⎯H is associated only with displacement current. Section 3.3.4 treats the electroquasistatic behavior of electric fields within conductors and relaxation phenomena. 3.3.2 Magnetoquasistatic behavior of devices All currents produce magnetic fields that in turn generate electric fields if those magnetic fields vary. Magnetoquasistatics characterizes the behavior of such slowly varying fields while neglecting the second-order magnetic fields generated by ∂D ∂t in Ampere’s law, (2.1.6): ∇ × H = J + ∂D ∂t ≅ J (quasistatic Ampere’s law) (3.3.11) The associated electric field⎯E can then be found from Faraday's law: ∇ × E = −∂B ∂t (Faraday’s law) (3.3.12) Section 3.2.1 treated an example for which the dominant effect of the quasistatic magnetic field in a current loop is voltage induced via Faraday’s law, while the example of a short wire follows; both are inductors. Section 3.3.4 treats the magnetoquasistatic example of magnetic diffusion, which is dominated by currents induced by the first-order induced voltages, and resulting modification of the original magnetic field by those induced currents. In every quasistatic problem wave effects can be neglected because the associated wavelength λ >> D, where D is the maximum device dimension. - 85 - We can roughly estimate the inductance of a short wire segment by modeling it as a perfectly conducting cylinder of radius ro and length D carrying a current i(t), as illustrated in Figure 3.3.2. An exact computation would normally be done using computer tools designed for such tasks because analytic solutions are practical only for extremely simple geometries. In this analysis we neglect any contributions to⎯H from currents in nearby conductors, which requires those nearby conductors to have much larger diameters or be far away. We also make the quasistatic assumption λ >> D. θ ⎯H(r) r z ro i(t) D Figure 3.3.2 Inductance of an isolated wire segment. We know from (3.2.23) that the inductance of any device can be expressed in terms of the magnetic energy stored as a function of its current i: L = 2w m i2 [ H ] (3.3.13) Therefore to estimate L we first estimate H and wm. If the cylinder were infinitely long then H ≅ θˆH ( r ) must obey Ampere’s law and exhibit the same cylindrical symmetry, as suggested in the figure. Therefore: ∫C H • ds = 2πr H ( r ) = i ( t ) (3.3.14) and H ( r ) ≅ i 2πr . Therefore the instantaneous magnetic energy density is: Wm = 1 μo H 2 ( r ) = 1 μo ( i 2πr ) 2 2 2 ⎡⎣J/m3 ⎤⎦ (3.3.15) To find the total average stored magnetic energy we must integrate over volume. Laterally we can neglect fringing fields and simply integrate over the length D. Integration with respect to radius will produce a logarithmic answer that becomes infinite if the maximum radius is infinite. A plausible outer limit for r is ~D because the Biot-Savart law (1.4.6) says fields decrease as r2 from their source if that source is local; the transition from slow cylindrical field decay as r-1 to decay as r-2 occurs at distances r comparable to the largest dimension of the source: r ≅ D. With these approximations we find: - 86 - D 0 ( ( ) 2πr dr D Wm 2πr dr ≅ D ∫ 1 μo i ro ro 2 2πr w m ≅ ∫ dz ∫ D 2 ) = μo Di 4π ln r D ro ( 2 ) (3.3.16) = μo Di 4π ln ( D ro ) [ J ] 2 Using (3.3.13) we find the inductance L for this wire segment is: L ≅ ( μo D 2π ) ln ( D ro ) [ Hy] (3.3.17) where the units “Henries” are abbreviated here as “Hy”. Note that superposition does not apply here because we are integrating energy densities, which are squares of field strengths, and the outer limit of the integral (3.3.16) is wire length D, so longer wires have slightly more inductance than the sum of shorter elements into which they might be subdivided. 3.3.3 Equivalent circuits for simple devices Section 3.1 showed how the parallel plate resistor of Figure 3.1.1 would exhibit resistance R = d/σA ohms and capacitance C = εA/d farads, connected in parallel. The currents in the same device also generate magnetic fields and add inductance. Referring to Figure 3.1.1 of the original parallel plate resistor, most of the inductance will arise from the wires, since they have a very small radius ro compared to that of the plates. This inductance L will be in series with the RC portions of the device because their two voltage drops add. The R and C components are in parallel because the total current through the device is the sum of the conduction current and the displacement current, and the voltages driving these two currents are the same, i.e., the voltage between the parallel plates. The corresponding first-order equivalent circuit is illustrated in Figure 3.3.3. + L C R Figure 3.3.3 Equivalent RLC circuit of a parallel-plate capacitor. Examination of Figure 3.3.3 suggests that at very low frequencies the resistance R dominates because, relative to the resistor, the inductor and capacitor become approximate short and open circuits, respectively. At the highest frequencies the inductor dominates. As f increases from zero beyond where R dominates, either the RL or the RC circuit first dominates, depending on whether C shorts the resistance R at lower frequencies than when L open-circuits R; that is, RC dominates first when R > L/C . At still higher frequencies the LC circuit dominates, followed by L alone. For certain combinations of R, L, and C, some transitions can merge. - 87 - Even this model for a resistor is too simple; for example, the wires also exhibit resistance and there is magnetic energy stored between the end plates because ∂D dt ≠ 0 there. Since such parasitic effects typically become important only at frequencies above the frequency range specified for the device, they are normally neglected. Even more complex behavior can result if the frequencies are so high that the device dimensions exceed ~λ/8, as discussed later in Section 7.1. Similar considerations apply to every resistor, capacitor, inductor, or transformer manufactured. Components and circuits designed for very high frequencies minimize unwanted parasitic capacitance and parasitic inductance by their very small size and proper choice of materials and geometry. It is common for circuit designers using components or wires near their design limits to model them with simple lumped-element equivalent circuits like that of Figure 3.3.3, which include the dominant parasitic effects. The form of these circuits obviously depends on the detailed structure of the modeled device; for example, R and C might be in series. Example 3.3A What are the approximate values L and C for the100-Ω resistor designed in Example 3.1A if ε = 4εo, and what are the three critical frequencies (RC)-1, R/L, and (LC)-0.5? Solution: The solution to 3.1A said the conducting caps of the resistor have area A = πr2 = π(2.5×10-4)2, and the length of the dielectric d is 1 mm. The permittivity ε = 4εo, so the capacitance (3.1.10) is C = εA/d = 4×8.85×10-12×π(2.5×10-4)2/10-3 ≅ 7×10-15 farads. The inductance L of this device would probably be dominated by that of the connecting wires because their diameters would be smaller and their length longer. Assume the wire length is D = 4d = 4×10-3, and its radius r is 10-4. Then (3.3.17) yields L ≅ (μoD/16π)ln(D/r) = (1.26×10-6 × 4×10-3/16π)ln(40) = 3.7×10-10 [Hy]. The critical frequencies R/L, (RC)-1, and (LC)-0.5 are 2.7×1011, 6.2×1011, and 1.4×1012 [r s-1], respectively, so the maximum frequency for which reasonably pure resistance is obtained is ~10 GHz (~R/2πL4). 3.4 General circuits and solution methods 3.4.1 Kirchoff’s laws Circuits are generally composed of lumped elements or “branches” connected at nodes to form two- or three-dimensional structures, as suggested in Figure 3.4.1. They can be characterized by the voltages vi at each node or across each branch, or by the currents ij flowing in each branch or in a set of current loops. To determine the behavior of such circuits we develop simultaneous linear equations that must be satisfied by the unknown voltages and currents. Kirchoff’s laws generally provide these equations. Although circuit analysis is often based in part on Kirchoff’s laws, these laws are imperfect due to electromagnetic effects. For example, Kirchoff’s voltage law (KVL) says that the voltage drops vi associated with each lumped element around any loop must sum to zero, i.e.: ∑ i vi = 0 (Kirchoff’s voltage law [KVL]) - 88 - (3.4.1) branches - nodes i2 v1 current loop i1 v2 Figure 3.4.1 Circuit with branches and current loops. which can be derived from the integral form of Faraday’s law: ∫C E • ds = − ( ∂ ∂t ) ∫∫ A B • da (3.4.2) This integral of E • ds across any branch yields the voltage across that branch. Therefore the sum of branch voltages around any closed contour is zero if the net magnetic flux through that contour is constant; this is the basic assumption of KVL. KVL is clearly valid for any static circuit. However, any branch carrying time varying current will contribute time varying magnetic flux and therefore voltage to all adjacent loops plus others nearby. These voltage contributions are typically negligible because the currents and loop areas are small relative to the wavelengths of interest (λ = c/f) and the KVL approximation then applies. A standard approach to analyzing circuits that violate KVL is to determine the magnetic energy or inductance associated with any extraneous magnetic fields, and to model their effects in the circuit with a lumped parasitic inductance in each affected current loop. The companion relation to KVL is Kirchoff’s current law (KCL), which says that the sum of the currents ij flowing into any node is zero: ∑ ji j = 0 (Kirchoff’s current law) (3.4.3) This follows from conservation of charge (2.4.19) when no charge storage on the nodes is allowed: ( ∂ ∂t ) ∫∫∫V ρ dv = − ∫∫ A J • da (conservation of charge) (3.4.4) If no charge can be stored on the volume V of a node, then ( ∂ ∂t ) ∫∫∫ ρ dv = 0, and there can be V no net current into that node. For static problems, KCL is exact. However, the physical nodes and the wires connecting those nodes to lumped elements typically exhibit varying voltages and D , and therefore have - 89 - capacitance and the ability to store charge, violating KCL. If the frequency is sufficiently high that such parasitic capacitance at any node becomes important, that parasitic capacitance can be modeled as an additional lumped element attached to that node. 3.4.2 Solving circuit problems To determine the behavior of any given linear lumped element circuit a set of simultaneous equations must be solved, where the number of equations must equal or exceed the number of unknowns. The unknowns are generally the voltages and currents on each branch; if there are b branches there are 2b unknowns. Figure 3.4.2(a) illustrates a simple circuit with b = 12 branches, p = 6 loops, and n = 7 nodes. A set of loop currents uniquely characterizes all currents if each loop circles only one “hole” in the topology and if no additional loops are added once every branch in the circuit is incorporated in at least one loop. Although other definitions for the loop currents can adequately characterize all branch currents, they are not explored here. Figure 3.4.2(b) illustrates a bridge circuit with b = 6, p = 3, and n = 4. b (a) (b) 1 2 a b = 12, p = 6, n = 7 d 5 3 c 4 6 I Figure 3.4.2 12-branch circuit and bridge circuit. The simplest possible circuit has one node and one branch, as illustrated in Figure 3.4.3(a). (a) (b) (c) b=1 p=1 n=p=1 b=n=2 b=3 n=p=2 Figure 3.4.3 Simple circuit topologies; n, p, and b are the numbers of nodes, loops, and branches, respectively. It is easy to see from the figure that the number b of branches in a circuit is: b=n+p–1 (3.4.5) - 90 - As we add either nodes or branches to the illustrated circuit in any sequence and with any placement, Equation (3.4.5) is always obeyed. If we add voltage or current sources to the circuit, they too become branches. The voltage and current for each branch are initially unknown and therefore any circuit has 2b unknowns. The number of equations is also b + (n – 1) + p = 2b, where the first b in this expression corresponds to the equations relating voltage to current in each branch, n-1 is the number of independent KCL equations, and p is the number of loops and KVL equations; (3.4.5) says (n – 1) + p = b. Therefore, since the numbers of unknowns and linear equations match, we may solve them. The equations are linear because Maxwell’s equations are linear for RLC circuits. Often circuits are so complex that it is convenient for purposes of analysis to replace large sections of them with either a two-terminal Thevenin equivalent circuit or Norton equivalent circuit. This can be done only when that circuit is incrementally linear with respect to voltages imposed at its terminals. Thevenin equivalent circuits consist of a voltage source VTh(t) in series with a passive linear circuit characterized by its frequency-dependent impedance Z(ω) = R + jΧ, while Norton equivalent circuits consist of a current source INo(t) in parallel with an impedance Z(ω). An important example of the utility of equivalent circuits is the problem of designing a matched load ZL(ω) = RL(ω) + jXL(ω) that accepts the maximum amount of power available from a linear source circuit, and reflects none. The solution is simply to design the load so its impedance ZL(ω) is the complex conjugate of the source impedance: ZL(ω) = Z*(ω). For both Thevenin and Norton equivalent sources the reactance of the matched load cancels that of the source [XL(ω) = - X(ω)] and the two resistive parts are set equal, R = RL. One proof that a matched load maximizes power transfer consists of computing the timeaverage power Pd dissipated in the load as a function of its impedance, equating to zero its derivative dPd/dω, and solving the resulting complex equation for RL and XL. We exclude the possibility of negative resistances here unless those of the load and source have the same sign; otherwise the transferred power can be infinite if RL = -R. Example 3.4A The bridge circuit of Figure 3.4.2(b) has five branches connecting four nodes in every possible way except one. Assume both parallel branches have 0.1-ohm and 0.2-ohm resistors in series, but in reverse order so that R1 = R4 = 0.1, and R2 = R3 = 0.2. What is the resistance R of the bridge circuit between nodes a and d if R5 = 0? What is R if R5 = ∞? What is R if R5 is 0.5 ohms? Solution: When R5 = 0 then the node voltages vb = vc, so R1 and R3 are connected in parallel and have the equivalent resistance R13//. Kirchoff’s current law “KCL” (3.4.3) says the current flowing into node “a” is I = (va - vb)(R1-1 + R3-1). If Vab ≡ (va - vb), then Vab = IR13// and R13// = (R1-1 + R3-1)-1 = (10+5)-1 = 0.067Ω = R24//. These two circuits are in series so their resistances add: R = R13// + R24// ≅ 0.133 ohms. When R5 = ∞, R1 - 91 - and R2 are in series with a total resistance R12s of 0.1 + 0.2 = 0.3Ω = R34s. These two resistances, R12s and R34s are in parallel, so R = (R12s-1 + R34s-1)-1 = 0.15Ω. When R5 is finite, then simultaneous equations must be solved. For example, the currents flowing into each of nodes a, b, and c sum to zero, yielding three simultaneous equations that can be solved for the vector V = [ v a , v b , v c ] ; we define vd = 0. Thus (va - vb)/R1 + (va - vc)/R3 = I = va(R1-1 + R3-1) - vbR1-1 -vcR3-1 = 15va - 10vb - 5vc. KCL for nodes b and c similarly yield: -10va + 17vb - 2vc = 0, and -5va -2vb + 17vc = 0. If we define the current vector I = [I, 0, 0], then these three equations can be written as a matrix equation: ⎡ 15 −10 −5 ⎤ Gv = I , where G = ⎢ −10 17 −2 ⎥ . ⎢ ⎥ ⎢⎣ −5 −2 17 ⎥⎦ Since the desired circuit resistance between nodes a and d is R = va/I, we need only −1 solve for va in terms of I, which follows from v = G I , provided the conductance matrix G is not singular (here it is not). Thus R = 0.146Ω, which is intermediate between the first two solutions, as it should be. 3.5 Two-element circuits and RLC resonators 3.5.1 Two-element circuits and uncoupled RLC resonators RLC resonators typically consist of a resistor R, inductor L, and capacitor C connected in series or parallel, as illustrated in Figure 3.5.1. RLC resonators are of interest because they behave much like other electromagnetic systems that store both electric and magnetic energy, which slowly dissipates due to resistive losses. First we shall find and solve the differential equations that characterize RLC resonators and their simpler sub-systems: RC, RL, and LC circuits. This will lead to definitions of resonant frequency ωo and Q, which will then be related in Section 3.5.2 to the frequency response of RLC resonators that are coupled to circuits. (a) C i(t) (b) + v(t) - R L R L C Figure 3.5.1 Series and parallel RLC resonators. The differential equations that govern the voltages across R’s, L’s, and C’s are, respectively: vR = iR (3.5.1) - 92 - vL = L di dt (3.5.2) vC = (1 C ) ∫ i dt (3.5.3) Kirchoff’s voltage law applied to the series RLC circuit of Figure 3.5.1(a) says that the sum of the voltages (3.5.1), (3.5.2), and (3.5.3) is zero: d 2i dt 2 + ( R L ) di dt + (1 LC ) i = 0 (3.5.4) where we have divided by L and differentiated to simplify the equation. Before solving it, it is useful to solve simpler versions for RC, RL, and LC circuits, where we ignore one of the three elements. In the RC limit where L = 0 we add (3.5.1) and (3.5.3) to yield the differential equation: di dt + (1 RC ) i = 0 (3.5.5) This says that i(t) can be any function with the property that the first derivative is the same as the original signal, times a constant. This property is restricted to exponentials and their sums, such as sines and cosines. Let's represent i(t) by Ioest, where: { i ( t ) = R e Io est } (3.5.6) where the complex frequency s is: s ≡ α + jω (3.5.7) We can substitute (3.5.6) into (3.5.5) to yield: { } st Re ⎡s ⎣ + (1 RC ) ⎦⎤ Io e = 0 (3.5.8) Since est is not always zero, to satisfy (3.5.8) it follows that s = - 1/RC and: − 1 RC ) t i ( t ) = Io e ( = Io e − t τ (RC current response) (3.5.9) where τ equals RC seconds and is the RC time constant. Io is chosen to satisfy initial conditions, which were not given here. A simple example illustrates how initial conditions can be incorporated in the solution. We simply need as many equations for t = 0 as there are unknown variables. In the present case we need one equation to determine Io. Suppose the RC circuit [of Figure 3.5.1(a) with L = 0] was at - 93 - rest at t = 0, but the capacitor was charged to Vo volts. Then we know that the initial current Io at t = 0 must be Vo/R. In the RL limit where C = ∞ we add (3.5.1) and (3.5.2) to yield di/dt + (R/L)i = 0, which has the same form of solution (3.5.6), so that s = -R/L and: − RL t i ( t ) = I o e ( ) = Io e − t τ (RL current response) (3.5.10) where the RL time constant τ is L/R seconds. In the LC limit where R = 0 we add (3.5.2) and (3.5.3) to yield: d 2i dt 2 + (1 LC ) i = 0 (3.5.11) Its solution also has the form (3.5.6). Because i(t) is real and ejωt is complex, it is easier to assume sinusoidal solutions, where the phase φ and magnitude Io would be determined by initial conditions. This form of the solution would be: i ( t ) = Io cos ( ωo t + φ ) (LC current response) (3.5.12) where ωo = 2πfo is found by substituting (3.5.12) into (3.5.11) to yield [ωo2 – (LC)-1]i(t) = 0, so: ωo = 1 ⎡⎣ radians s-1 ⎤⎦ LC (LC resonant frequency) (3.5.13) We could alternatively express this solution (3.5.12) as the sum of two exponentials using the identity cos ωt ≡ ( e jωt + e− jωt ) 2 . RLC circuits exhibit both oscillatory resonance and exponential decay. If we substitute the generic solution Ioest (3.5.6) into the RLC differential equation (3.5.4) for the series RLC resonator of Figure 3.5.1(a) we obtain: ( s2 + sR L + 1 LC) Ioest = ( s − s1 )( s − s2 ) Ioest = 0 (3.5.14) The RLC resonant frequencies s1 and s2 are solutions to (3.5.14) and can be found by solving this quadratic equation9 to yield: 2 si = −R 2L ± j ⎡⎣(1 LC ) − ( R 2L ) ⎤⎦ 0.5 (series RLC resonant frequencies) When R = 0 this reduces to the LC resonant frequency solution (3.5.13). 9 A quadratic equation in x has the form ax2 + bx + c = 0 and the solution x = (-b ± [b2 - 4ac]0.5)/2a. - 94 - (3.5.15) The generic solution i ( t ) = Io′est is complex, where Io′ ≡ Ioejφ: { } { } − R 2L ) t jωt − R 2L ) t i ( t ) = R e Io′es1t = R e Io e jφe ( e cos ( ωt + φ ) = Io e ( (3.5.16) where ω = [(LC)-1 + (R/2L)2]0.5 ≅ (LC)-0.5. Io and φ can be found from the initial conditions, which are the initial current through L and the initial voltage across C, corresponding to the initial energy storage terms. If we choose the time origin so that the phase φ = 0, the instantaneous magnetic energy stored in the inductor (3.2.23) is: ( ) ( ) w m (t) = Li 2 2 = LIo 2 2 e− Rt L cos2 ωt = LIo 2 4 e− Rt L (1 + cos 2ωt ) (3.5.17) Because wm = 0 twice per cycle and energy is conserved, the peak electric energy we(t) stored in the capacitor must be intermediate between the peak magnetic energies stored in the inductor (eRt/LLIo2/2) during the preceding and following cycles. Also, since dvC/dt = i/C, the cosine variations of i(t) produce a sinusoidal variation in the voltage vC(t) across the capacitor. Together these two facts yield: we(t) ≅ (LIo2/2)e-Rt/L sin2ωt. If we define Vo as the maximum initial voltage corresponding to the maximum initial current Io, and recall the expression (3.1.16) for we(t), we find: w e (t) = Cv 2 2 ≅ ( CVo 2 2 ) e−Rt L sin 2 ωt = ( CVo 2 4 ) e−Rt L (1− cos 2ωt ) (3.5.18) Comparison of (3.5.17) and (3.5.18) in combination with conservation of energy yields: Vo ≅ ( L C ) 0.5 Io (3.5.19) Figure 3.5.2 illustrates how the current and energy storage decays exponentially with time while undergoing conversion between electric and magnetic energy storage at 2ω radians s-1; the time constant for current and voltage is τ = 2L/R seconds, and that for energy is L/R. One useful way to characterize a resonance is by the dimensionless quantity Q, which is the number of radians required before the total energy wT decays to 1/e of its original value, as illustrated in Figure 3.5.2(b). That is: w T = w Toe−2αt = w Toe−ωt Q [ J ] (3.5.20) The decay rate α for current and voltage is therefore simply related to Q: α = ω/2Q (3.5.21) - 95 - (a) Stored energy (b) i(t) wTo -(R/2L)t Io e -αt wm(t) = Io e wT = wTo e-(R/L)t = wm(t) + we(t) we(t) t 0 wTo/e 0 Q radians, Q/ω seconds t Figure 3.5.2 Time variation of current and energy storage in RLC circuits. If we find the power dissipated Pd [W] by differentiating total energy wT with respect to time using (3.5.20), we can then derive a common alternative definition for Q: Pd = −dw T dt = ( ω Q ) w T (3.5.22) Q = ωw T Pd (one definition of Q) (3.5.23) For the series RLC resonator α = R/2L and ω ≅ (LC)-0.5, so (3.5.21) yields: Q = ω 2α = ωL R ≅ ( L C ) 0.5 R (Q of series RLC resonator) (3.5.24) Figure 3.5.1(b) illustrates a parallel RLC resonator. KCL says that the sum of the currents into any node is zero, so: C dv dt + v R + (1 L ) ∫ v dt = 0 (3.5.25) d 2 v dt 2 + (1 RC ) dv dt + (1 LC ) v = 0 (3.5.26) If v = Voest, then: ⎡⎣s 2 + (1 RC ) s + ( L C ) ⎤⎦ = 0 (3.5.27) - 96 - 2 s = − (1 2RC ) ± j ⎡⎣(1 LC ) − (1 2RC ) ⎤⎦ 0.5 (parallel RLC resonance) (3.5.28) Analogous to (3.5.16) we find: { } v(t) = R e Vo′es1t = Vo e−(1 2RC ) t cos ( ωt + φ ) (3.5.29) where V o′ = Vo e jφ . It follows that for a parallel RLC resonator: −1 −2 ω = ⎡⎣( LC ) − ( 2RC ) ⎤⎦ 0.5 Q = ω 2α = ωRC = R ( C L ) ≅ ( LC ) −0.5 0.5 (3.5.30) (Q of parallel RLC resonator) (3.5.31) Example 3.5A What values of L and C would give a parallel resonator at 1 MHz a Q of 100 if R = 106/2π? Solution: LC = 1/ωo2 = 1/(2π106)2, and Q = 100 = ωRC = 2π106(106/2π)C so C = 10-10 [F] and L = 1/ωo2c ≅ 2.5×10-4 [Hy]. 3.5.2 Coupled RLC resonators RLC resonators are usually coupled to an environment that can be represented by either its Thevenin or Norton equivalent circuit, as illustrated in Figure 3.5.3(a) and (b), respectively, for purely resistive circuits. (a) (b) I(ω) RTh + VTh - + V(ω) - I(ω) R I No GNo L + V ( ω) G L C - C Norton equivalent source Thevenin equivalent source Figure 3.5.3 Series and parallel RLC resonators driven by Thevenin and Norton equivalent circuits. A Thevenin equivalent consists of a voltage source VTh in series with an impedance ZTh = R Th + jXTh , while a Norton equivalent circuit consists of a current source INo in parallel with an admittance Y No = G No + jU No . The Thevenin equivalent of a resistive Norton - 97 - equivalent circuit has open-circuit voltage VTh = INo/GNo, and RTh = 1/GNo; that is, their opencircuit voltages, short-circuit currents, and impedances are the same. No single-frequency electrical experiment performed at the terminals can distinguish ideal linear circuits from their Thevenin or Norton equivalents. An important characteristic of a resonator is the frequency dependence of its power dissipation. If RTh = 0, the series RLC resonator of Figure 3.5.3(a) dissipates: Pd = R I 2 2 [W] (3.5.32) 2 2 2 2 Pd = ⎡⎣ R V Th 2 ⎦⎤ R + Ls + C−1s −1 = ⎡⎣ R V Th 2 ⎤⎦ s L ( s − s1 )( s − s2 ) 2 (3.5.33) where s1 and s2 are given by (3.5.15): 2 si = −R 2L ± j ⎡⎣(1 LC ) − ( R 2L ) ⎤⎦ 0.5 = −α ± jω′o (series RLC resonances) (3.5.34) The maximum value of Pd is achieved when ω ≅ ω′o : Pd max = VTh 2 2R (3.5.35) This simple expression is expected since the reactive impedances of L and C cancel at ωo, leaving only R. If (1 LC ) ( R 2L ) so that ωo ≅ ω′o , then as ω - ωo increases from zero to α, | s − s1 | = | jωo − ( jωo + α) | increases from α to 2 α. This departure from resonance approximately doubles the denominator of (3.5.33) and halves Pd. As ω departs still further from ωo and resonance, Pd eventually approaches zero because the impedances of L and C approach infinity at infinite and zero frequency, respectively. The total frequency response Pd(f) of this series RLC resonator is suggested in Figure 3.5.4. The resonator bandwidth or half-power bandwidth Δω is said to be the difference between the two half-power frequencies, or Δω ≅ 2α = R/L for this series circuit. Δω is simply related to ωo and Q for both series and parallel resonances, as follows from (3.5.21): Q = ωo/2α = ωo/Δω (Q versus bandwidth) (3.5.36) (parallel RLC resonances) (3.5.37) Parallel RLC resonators behave similarly except that: 2 si = −G 2L ± j ⎣⎡(1 LC ) − ( G 2L ) ⎦⎤ 0.5 = −α ± jω′o where R, L, and C in (3.5.34) have been replaced by their duals G, C, and L, respectively. - 98 - Pd Pd max Pd max 2 Δω 0 ωo ω Figure 3.5.4 RLC power dissipation near resonance. Resonators reduce to their resistors at resonance because the impedance of the LC portion approaches zero or infinity for series or parallel resonators, respectively. At resonance Pd is maximized when the source Rs and load R resistances match, as is easily shown by setting the derivative dPd/dR = 0 and solving for R. In this case we say the resonator is critically matched to its source, for all available power is then transferred to the load at resonance. This critically matched condition can also be related to the Q’s of a coupled resonator with zero Thevenin voltage applied from outside, where we define internal Q (or QI) as corresponding to power dissipated internally in the resonator, external Q (or QE) as corresponding to power dissipated externally in the source resistance, and loaded Q (or QL) as corresponding to the total power dissipated both internally (PDI) and externally (PDE). That is, following (3.5.23): QI ≡ ωw T PDI (internal Q) (3.5.38) QE ≡ ωw T PDE (external Q) (3.5.39) (loaded Q) (3.5.40) Q L ≡ ωw T ( PDI + PDE ) Therefore these Q’s are simply related: Q L −1 = QI −1 + Q E −1 (3.5.41) It is QL that corresponds to Δω for coupled resonators ( QL = ωo Δω) . For example, by applying Equations (3.5.38–40) to a series RLC resonator, we readily obtain: Q I = ωo L R (3.5.42) - 99 - QE = ωo L R Th (3.5.43) Q L = ωo L ( R Th + R ) (3.5.44) For a parallel RLC resonator the Q’s become: Q I = ωo RC (3.5.45) Q E = ωo R Th C (3.5.46) Q L = ωo CR Th R ( R Th + R ) (3.5.47) Since the source and load resistances are matched for maximum power dissipation at resonance, it follows from Figure 3.5.3 that a critically coupled resonator or matched resonator results when QI = QE. These expressions for Q are in terms of energies stored and power dissipated, and can readily be applied to electromagnetic resonances of cavities or other structures, yielding their bandwidths and conditions for maximum power transfer to loads, as discussed in Section 9.4. - 100 -