Full-Wave Approximation – Single Frequency Tone Formulation

In this book we will generally look at these equations not as a function of time but as a function of frequency. We get there by simply assuming the time dependence scales as e^jωtejωt where we follow the convention in most engineering books. By doing this we get

∇×H=jωϵE+σE+Ji=jωϵ1+σjωϵE+Ji.(4.18)

From the continuity equation for the conduction component we get

∇⋅σE=−∂ρc∂t=−jωρc(4.19)

This together with the charge equation (4.17) gives

∇⋅ϵE=ρ=ρc+ρi=∇⋅σE−jω+ρi→∇⋅ϵE+∇⋅σEjω=ρi(4.20)

We can now define an effective permittivity

ϵ′=ϵ1+σjωϵ(4.21)

and we have

Using equations (4.6)–(4.8) we get

∇⋅B=∇⋅∇×A=0∇×E=∇×−∇φ−jωA=−jω∇×A=−jωB∇×H=∇×Bμ=1μ∇×∇×A=1μ∇∇⋅A−∇2A=jωϵ′−∇φ−jωA+Ji.

We use the Lorenz gauge which in frequency domain looks like

and get

After rewriting we find

We also have from equation (4.23)

∇⋅ϵ′E=ϵ′∇⋅E=ϵ′∇⋅−∇φ−jωA=−ϵ′Δφ+jωϵ′jϵ′μωφ=−ϵ′Δφ−ω2ϵ′ϵ′μφ=ρi

After rewriting we find

Δφ+ω2ϵ′μφ=−ρiϵ′(4.26)

Equations (4.25) and (4.26) are Maxwell’s equation in yet another form. Knowing A,   φA,φ will give us B   and  EBandE through equations (4.7) and (4.8). We will use these in numerous examples in this and the following chapters.

Long Wavelength Approximation

In the long wavelength approximation, λ ≫ lλ≫l, where λ = 2πc/ωλ=2πc/ω is the wavelength and ll is a length scale of the model, we find the second term on the left-hand side of (4.25), (4.26) disappears and we are left with:

We can also write this directly from the fields as

where (4.27) follows if we use gauge ∇ ⋅ A = 0∇⋅A=0 and

Δφ=−ρiϵ′.Thisfollowsalsofrom∇⋅ϵ′E=ρiifweusegauge∇⋅A=0(4.29)

For integrated circuits the long wavelength approximation is often appropriate since the dimensions are much smaller than any wavelength.

General Solution

We will start discussing the general solution of the one-dimensional case and follow with the two- and three-dimensional versions in the following subsections. There is a rich literature describing these methods and also a list of references at the end of the chapter.

1D – Solution

Let us consider an equation similar to (4.26) in free space

d2φxdx2+k2φx=−δx−x0.(4.30)

This is known as Helmholz’s equation in one-dimensional free space subject to the boundary condition at infinity, φ(±∞) = 0φ±∞=0. The response at xx is due to the delta source at x₀x0. Let us consider the homogeneous equation

d2φxdx2+k2φx=0

This is the same as (4.30) when x ≠ x₀x≠x0. The solution to this equation that satisfies the boundary conditions at infinity is

φx=Aejkxx>x0Be−jkxx<x0.

The unknown constants AA, BB can be determined by the boundary condition at x = x₀ ± Δx=x0±Δ where ΔΔ denotes an infinitesimally small interval. Integrating (4.30) from x = x₀ − Δx=x0−Δ to x = x₀ + Δx=x0+Δ we find

dφdxx=x0−Δx=x0+Δ+∫x=x0−Δx=x0+Δk2φxdx=−1.

Since φ(x)φx is continuous the last term on the left-hand side disappears when Δ → 0Δ→0.

We get

dφdxx=x0−Δx=x0+Δ=−1φx+Δ=φx−ΔikAejkx+Be−jkx=−1Aejkx=Be−jkx.

Solving for AA and BB gives

φx=j2kejkx−x0x>x0j2ke−jkx−x0x<x0=i2kejkx−x0.(4.31)

Long Wavelength Approximation

When kx₀ ≪ 1kx0≪1 (long wavelength approximation) we find

φlwx=j2k1+jkx−x0x>x0j2k1−jkx−x0x<x0=j2k1+jkx−x0=j2k−x−x02.

Since in the long wavelength approximation Helmholz equation reduces to Poisson equation where φ_lwφlw is defined with an arbitrary constant factor, we can simply relabel the constant term for φ_lwφlw and end up with

φlwx=−x−x02+C.(4.32)

Let us verify by examining the Poisson equation

d2φlwxdx2=−δx−x0.

When x ≠ x₀x≠x0 we see ∂2φlwx∂x2≡0. Let us integrate around the singularity as we did earlier

∫x0−Δx0+Δ∂2φlwx∂x2dx=−∫x0−Δx0+Δδx−x0dxdφlwxdxx0−Δx0+Δ=−12−12=−1=r.h.s.

Indeed, in one dimension and the long wavelength approximation (4.32) solves the Poisson equation.

2D – Solution

In two dimensions equation (4.30) becomes

∂2φxy∂x2+∂2φxy∂y2+k2φxy=−δy−y0δx−x0(4.33)

We can now use the Fourier transform

φxy=∫−∞∞φ˜ejβx−x0dβ.

This gives

∂2φβy∂y2−β2φ˜βy+k2φ˜βy=−δy−y0

The solution to this equation is the one-dimensional free-space Green’s function

φ˜βy=j2κejκy−y0

where κ=k2−β2. We find

φxy=∫−∞∞j2κejκy−y0ejβx−x0dβ=j4H01kx−x02+y−y02

where the last equality relates the Hankel function of order 0, H01 to the integral. We see the solution is simply a composition of a continuous spectrum of plane waves.

Long Wavelength Approximation

The two-dimensional solution in the long wavelength approximation can be found using similar techniques as in the one-dimensional case we noted earlier. Here we will show the solution for the special case of cylindrical symmetry which we will take advantage of later in this chapter when we discuss inductance and current elements.

Helmholz equation in cylindrical symmetry becomes, with a delta function at x = 0x=0,

∇ ⋅  ∇ φ = Cδ(x).

∇⋅∇φ=Cδx.(4.34)

Outside of x = 0x=0 we have

∇⋅∇φ=1r∂∂rr∂φrθ∂r+1r2∂2φrθ∂θ2=0.

To further simplify we will assume there is no θθ-dependence and we find

1rddrrdφrdr=d2φrdr2+1rdφrdr=0.

This has the general solution

φ(r) = D ln r + B.

φr=Dlnr+B.

To find out the constants we need to integrate (4.34) around x = 0x=0. Let us choose a sphere centered at x = 0x=0 with radius ΔΔ as an integration volume. We find for the left-hand side using the divergence theorem (see Appendix B)

∫∇⋅∇φdV=∫∇φ⋅rrda=∫dφdrrdθ=dφdrr2π=2πD.

The right-hand side of (4.35) becomes, as before, CC. Putting all this together we have

2πD=C→D=C2π.

We have for the long wavelength solution to the Helmholz equation in two dimensions:

φr=C2πlnr+B.(4.35)

3D – Solution

We finally present the 3D solution. We will use it in Chapter 6. Equation (4.30) becomes with a source at r = r₀r=r0

∂2φxyz∂x2+∂2φxyz∂y2+∂2φxyz∂z2+k2φxyz=−δr−r0.

Let us first do a change of variables ρ = r − r₀ρ=r−r0. We find we have created a spherically symmetric model. By realizing

∫δ(r − r₀)dV =  ∫ δ(ρ)4πρ²dρ,

∫δr−r0dV=∫δρ4πρ2dρ,

we find

δr−r0=δρ4πρ2.

By substituting

φρ=uρρ,

we find for the Helmholz equation

∇2+k2uρ=−δρ4πρ.

The solution for u(ρ) = Ae^jkρuρ=Aejkρ and we find

φρ=Aejkρρ.

The boundary condition at ρ = 0ρ=0 must be used to determine AA. The Helmholz equation is

∇2+k2φρ=−δρ4πρ2.

After integrating over a small volume we find, as with the one-dimensional case,

φ0+=φ0−,∫∇2φρdV=−1.

After applying the divergence theorem we get

∫∇2φρdV=∮∇φρ⋅ds=Ajkejkρρ−ejkρρ24πρ2ρ→0=−4πA.

So

A=14π.

The general solution is

φxyz=14πr−r0ejκ⋅r−r0+14πr−r0e−jκ⋅r−r0,κ2=κx2+κy2+κz2=ω2ϵ′μ.

Similarly, for the vector potential we get

Ar=μ4π∫Jir′r−r′ejκ⋅r−r′+e−jκ⋅r−r′dr′.

Long Wavelength Approximation

The solution to the long wavelength equations (4.27) and (4.29) in three dimensions are well known. We see the equations are essentially identical and the general solution that vanishes at infinity is

Ar=μ4π∫Jir′r−r′dr′.(4.36)

The solution for φφ is similarly

φr=14πϵ′∫ρir′r−r′dr′.(4.37)

For a point charge at the origin, ρ_i(r^′) = qδ(r^′)ρir′=qδr′ we find

φr=q4πϵ′r.

which is the familiar electrostatic potential from a point charge.

Boundary Conditions

Identifying the relevant equations and how to solve them is a helpful exercise. It is often the easiest part of any investigation. The real problem comes when taking into account what happens at the boundaries, either in time and/or in space. To the novice, the opposite often appears true. To help us there are many great examples in the literature on how to handle the boundaries, and here we will go through the basic methods and leave it to the reader to explore more if needed.

Fundamentally, what one does is to put a “pill box” at the boundary that extends a little into each material with a large surface area. In Figure 4.1 it extends ε/2ε/2 into each region. The volume is thus infinitesimal, while the area is macroscopic. They key thing to note here is that the equations are still valid in this volume and to find out what the boundary conditions are one simply integrates the equations over the small volume. For some entities, portions of the equation will be proportional to the volume and thus small while other entities will be proportional to the area and thus large. We will go through Maxwell’s equations with this method to show explicitly how the conditions work out.

Figure 4.1 Boundary conditions pill box.

Let us look at Ampere’s law (4.28)

∇ × H = J.

∇×H=J.

At the boundary between two media we put a small pill box of height εε which is much smaller than any other dimension in the problem. We integrate Ampere’s law over this volume

∫ ∇  × HdV =  ∫ JdV.

∫∇×HdV=∫JdV.

We can here use Stoke’s theorem for the left-hand side and denoting by A_rea = A_reanArea=Arean where A_reaArea is the area and nn is the outward normal to the area segment

∫ ∇  × HdV =  ∮ H × A_readA_rea = A_rea(H_{t, +} − H_{t, −}) + εH_n → A_rea(H_{t, +} − H_{t, −}),    ε → 0.

∫∇×HdV=∮H×AreadArea=AreaHt,+−Ht,−+εHn→AreaHt,+−Ht,−,ε→0.

Volume Current

For the right-hand side without delta functions, which we call volume current, we get

∫JdV~ε → 0    when   ε → 0.

∫JdV∼ε→0whenε→0.

Putting it all together we get at the boundary

∫ ∇  × HdV = A_rea(H_{t, +} − H_{t, −}) =  ∫ JdV = 0.

∫∇×HdV=AreaHt,+−Ht,−=∫JdV=0.

or

H_{t, +} = H_{t, −}.

Ht,+=Ht,−.(4.38)

Surface Current

We also see when we have a delta function on the right-hand side, a surface current,

∫JdV =  ∫ J_sδ(y)dV = J_sA_rea    when   ε → 0.

∫JdV=∫JsδydV=JsAreawhenε→0.

Putting this together with the left-hand side we get

A_rea(H_{t, +} − H_{t, −}) = J_sA_rea

AreaHt,+−Ht,−=JsArea

or

(H_{t, +} − H_{t, −}) = J_s

Ht,+−Ht,−=Js(4.39)

Similarly, we can use the charge equation (4.23)

∇ ⋅ (ϵ^′E ) = ρ_i.

∇⋅ϵ′E=ρi.

We integrate over volume

∫ ∇  ⋅ (ϵ^′E )dV =  ∫ ρ_idV

∫∇⋅ϵ′EdV=∫ρidV

For the left-hand side we use Gauss’ law

∫∇⋅ϵ′EdV=∮ϵ′E⋅ndArea=Areaϵ+′En+−ϵ−′En−+ε…→Areaϵ+′En+−ϵ−′En−,ε→0

Volume Charge

For the case of volume charge, the right-hand side is now treated the same way as for the volume current earlier so we end up with

ϵ+′En+=ϵ−′En−(4.40)

Finally, for the case of surface charges we have

Surface Charge

ϵ+′En+−ϵ−′En−=ρsArea(4.41)

This way of treating the boundary is standard and can be found in many textbooks.

First Principle Calculation of Capacitance of Two-Plate System

Let us know look at the same situation from first principles, in this case Maxwell’s equations. We will follow the estimation analysis method. It will seem a little excessive at first, in particular when comparing with the previous calculation, but we will show we can easily extend what we learn here to other situations with little extra effort.

Simplify

The first thing to do is simplification. Let us imagine the two plates are infinitely extended in all directions in the plane. That will simplify the problem to one dimension. Next the top plate has a voltage, VV while the bottom plate is grounded. Lastly we assume the long wavelength approximation so we need to solve equation (4.26). We find as in Figure 4.2.

Figure 4.2 Two-dimensional projection of two plates.

In free space assuming no x-dependency we have then

Δφ=−Cδy−y0ϵ′

Subject to the boundary conditions

φ(y₀) = V   φ(0) = 0

φy0=Vφ0=0

Solve

This equation is independent of z and is a one-dimensional Helmholz equation in the long wavelength approximation which makes it Poisson’s equation and it has solution:

φ(y) =  − A(y − y₀) + B,    for  y₀ ≥ y

φy=−Ay−y0+B,fory0≥y

Let us plug in the boundary conditions

B=VAy0+B=0

A=−Vy0

We get for

φ(y) =  − A(y − y₀) + B

φy=−Ay−y0+B

The electric field is E =  −  ∇ φE=−∇φ. At the lower perfect electrical conductor (PEC) boundary the electric field will change abruptly from the gradient of the potential to zero. This will result in surface charge and it follows from the boundary condition on ε ∇  ⋅ E = ρε∇⋅E=ρ which gives ϵE₊ = ρϵE+=ρ. In effect what is happening is the voltage on the top is inducing a charge in the bottom plate. At long wavelength this is simply known as a capacitor effect. We will go through the simple derivation now.

We have

E=−∇φ=A=−Vy0.

Since we now know the electric field we can simply plug it into the formula for capacitance (4.43) and recover equation (4.47). Going further we can use our more sophisticated model to recover another formula involving capacitance. Let us put a pill box around the boundary and integrating we find:

ε∫∇⋅EdV=εEy=0+−Ey=0−Area=εEy=0+Area=−εVy0Area=∫ρdV=∫ρ0δy=0dV=Q.

We have

−εVy0Area=Q

where Q < 0Q<0. If we identify εy0Area=C or capacitance we have

C V = Q.

CV=Q.(4.48)

This is another famous relationship involving capacitance and we see how straightforward it is to derive it directly from Maxwell’s equations.

First Principle Calculation of Capacitance with Two Different Dielectric Media

Here we will examine how the field solutions behave when there are different dielectric media in the problem. This situation often shows up in integrated circuits where there are different dielectric layers and instead of implementing all of them in a field solver it is often enough to use the equivalent permittivity. Here we will show how to calculate the effective permittivity and we will again start from Maxwell’s equations and end up with an expression that is quite familiar to most readers.

Simplify

We build on the previous simple model ad simply add another boundary: see Figure 4.3. All will still be one-dimensional.

Figure 4.3 Two-dimensional picture of two plates with two dielectric media.

Compared with previous calculations, everything is the same – the same equations, the same solutions – but we have an additional boundary condition at the interface between the two dielectrics. From ∇ ⋅ εE = ρ∇⋅εE=ρ we find

ε₁E|₁ = ε₂E|₂

ε1E1=ε2E2

since there is no surface charge in the dielectric media.

Solve

We now have two region and two solutions:

φ1y=−A1y−y0+B1φ1y=−A2y+B2

For boundary conditions we get

B1=VB2=0ε1E1=ε2E2φ1y1=φ2y1

The third one can be expanded:

ε₁E|₁ =  − ε₁ ∇ φ₁ = ε₁A₁ = ε₂E|₂ = ε₂A₂

ε1E1=−ε1∇φ1=ε1A1=ε2E2=ε2A2

or

A2=ε1ε2A1

We find from the fourth boundary condition

−A1y1−y0+V=−ε1ε2A1y1

which gives

A1=−Vy0−y1+ε1ε2y1

and

A2=−ε1ε2Vy0−y1+ε1ε2y1

We find for the electric field at y = 0

E=−∇φ2=−ε1Vy1ε1+ε2y0−y1

As before we have for the charge at the lower PEC

ε2EArea=−ε1ε2y1ε1+ε2y0−y1VArea=Q

We can rewrite this in more familiar form:

−1y1/Areaε2+y0−y1/Areaε1V=−11/C1+1/C2V=Q

This is the well-known serial formula for capacitors. We can furthermore rewrite the equation in a form that makes use of an effective permittivity, ε^′ε′

ε′Vy0Area=ε1ε2y1ε1+ε2y0−y1VArea

which gives

ε′=ε1ε2y1ε1+ε2y0−y1y0=1y1/ε2+y0−y1/ε1y0.

Summary

We have applied the estimation analysis to two situations where the total capacitance was needed. We calculated it directly from Maxwell’s equations using some basic simplifications as directed by the estimation analysis technique.

Simplify

We have already set up a situation that is relatively simple. A straight wire in free space with uniform current distribution. We will model it as a system with no z-dependence, a two-dimensional system with cylindrical symmetry so all entities only depend on the distance, r, from the center of the wire. Seen from the side we find as in Figure 4.4.

Figure 4.4 Cross-section of single wire.

Let us look at this simple wire and dig into the mathematical details. We only need Ampere’s law in the long wavelength approximation

Solve

We can integrate Ampere’s law over a cross-section containing the wire as in Figure 4.5:

We can use Stoke’s theorem on the curl side:

where we have taken advantage of the cylindrical symmetry. On the right-hand side we get from the fact the current, JJ, is a density

Putting the two sides together we find

Hφr=Jtotal2πr

It is interesting to see here this result is independent of the precise radial distribution of the current internal to the conductor. The right-hand side is only sensitive to the total current in conductor.

Figure 4.5 Integration boundary of single wire.

This in turn implies that the inductance from the total magnetic energy is.

12LJtotal2=12∫μH⋅HdV=12μl12π2∫Jtotalr2rdrdθ=122πμlJtotal2lnR.

Here we have excluded the magnetic energy internal to the wire. It makes little difference. We find

L=μl2πlnR

where R is the “extent” of the wire. It has to be a closed loop somehow and RR is simply the size of the loop. We see the inductance will be infinite for a single wire without a loop as we hinted at in the previous section.

As a quick side note we can also calculate the vector potential outside the conductor when we know the magnetic field

∇×A=B=μH→A=Arez=−μJtotal2πlnrR0+Cez.(4.49)

Verify

This is a known solution and can be found in for example [4].

Evaluate

The inductance for a wire scales as the length of the wire times the logarithm of its size.

Simplify

We use the same basic model as previously but with the addition of one wire. The internal current distribution is assumed to be uniform: see Figure 4.6.

Figure 4.6 Cross-section of two wires in two dimensions.

Figure adapted from [16].

Solve

To calculate the inductance of the system we will use equation (4.46) we derived earlier. The equation states that knowing the vector potential and the current distribution inside the conductors we can calculate the inductance through a simple integration over the conductors cross-sections only. We have already assumed the current distribution is uniform and we are left with calculating the vector potential. The vector potential external to the wires we already know from equation (4.49). Let us also calculate AA inside a conductor. We have using the same strategy as in the previous section where we integrated Ampere’s law and used Stoke’s theorem:

2πHrr=∫J⋅dS=JtotalπR02∫0R0r22πr′dr′=JtotalR02

where we have used the current density

J=JtotalπR02

and r ≤ R₀r≤R0.

We get

Hr=Jtotal2πR02r.

From B = μH =  ∇  × AB=μH=∇×A we find

μJtotal2πR02r=−∂A∂r→A=C+μJtotal2π12−μJtotal2πR02r22

where the constant has been chosen to match the external solution (4.49) at the r = R₀r=R0 boundary. We can now put things together to calculate the energy by integrating the self-inductance and the mutual inductance from (4.46). We denote the energy for the self-inductance, F_selfFself, and the energy for the mutual inductance calculation, F_mutualFmutual.

Fself=12∫0R0C+μJtotal2π12−μJtotal2πR02r′22JtotalπR02r′dr′2πZFmutual=−12∫02π∫0R0−μJtotal2πlnrR0+CJtotalπR02r′dr′dφ′Z

where ZZ is the unit length of the wire. The variable r is the distance between the center of conductor 2 and r^′,    φ^′r′,φ′ as indicated in the figure.

Fself=12JtotalπR022πZ∫0R0C+μJtotal2π12−μJtotal2πR02r′22r′dr′=JtotalZC12+μJtotal22πZ18(4.50)

Fmutual=−12JtotalπR022πZCR022+μ12Jtotal22ππR02Z∫02π∫0R012lnd2+r′2−2dr′cosφ′R02r′dr′dφ′.

The integral in the expression for F_mutualFmutual can be calculated by identifying

Id=∫02π∫0R012lnd2+r′2−2dr′cosφ′R02r′dr′dφ′.

This is straightforward to calculate and we find

Id=πR02lndR0.

Now

Fmutual=−12JtotalπR022πZCR022+μ12Jtotal22ππR02ZπR02lndR0=−JtotalZC12+μ14πJtotal2ZlndR0.(4.51)

The total energy is now

F=Fself+Fmutual=μ12π2Jtotal2Z14+lndR0.

The other conductor combination is now easy to include because of symmetry, we just multiply by 2. By removing Jtotal2/2 we find the inductance is

L=μZπ14+lndR0(4.52)

This is one of our fundamental results and we will use this again and again in various contexts.

Verify

This is a standard calculation in physics literature: see [4, 15], for example. For the reader coming from a microwave background this example might seem familiar. If so, it should, but the microwave literature often assumes the conductors to be ideal and in that case the solution is somewhat different. We leave it as an exercise for the reader to solve the ideal conductor case.

Evaluate

For two inductors on top of each other, if the current is going the same way in both inductors the magnetic field is doubled (with equal current) →→ magnetic energy is quadrupled →→ inductance is four times larger. If currents are going the opposite way →→ magnetic field is nulled →→ inductance is zero.

This effect is often referred to as coupling between the wires. It comes from the high frequency case where currents are induced in (or coupled to) the neighboring conductors.

We will study the high frequency case and induced currents later in this chapter.

Simplify

Using the method of images we can model the situation in exactly the same way where the distance dd is simply

where bb as in the figure is the distance to the ground plane from the wire. The method of images, [1], simply states the field outside a perfect ground plane can be found by removing the ground plane and put mirror conductors with opposite charge/current equidistant from the ground plane border.

Solve

We can now use exactly the same calculations as before with one important difference. The total field energy is now half of what we had before since no field exists in the ground plane which takes up half the volume.

We find

Lground=12μZπ14+lndR0.(4.53)

Verify

This situation has been simulated as a 0.1 → 4 mm long wire (1 µm × 1 µm cross-section) over a ground plane in HFSS: see Figure 4.8.

Figure 4.8 Figure of an HFSS sim setup.

The conductor was a variable height over the ground plane. The excitation is through wave ports at the end of the structure as indicated. The inductance is calculated as

Lsim=im−2Y12+Y211ω

where YY refers to the Y-parameter (admittance) gain. Figure 4.9 shows the simulation comparison to (4.53) as a function of length where the height over the ground plane is 3 µm.

Figure 4.9 Simulated and estimated inductance of single wire over ground plane vs length.

A comparison of simulation vs estimation of inductance vs height over ground plane shows in Figure 4.10.

Figure 4.10 Inductance of single wire over ground plane vs height over ground plane.

Evaluate

When adding a ground plane under a conducting wire the resulting inductance will be reduced compared to a single wire. The inductance scales roughly as the logarithm of the distance to the ground plane. This is for the case of no other conductors nearby.

Simplify

Let us know look at the basic equations in terms of the vector potential A = A(y)e_zA=Ayez, where the source term is a current. We could utilize the same simplifications we used in the capacitance calculation but here we will go one step further and start with the full wave equation and later go to the long wavelength limit so we can demonstrate some new mathematical steps.

In free space assuming no x-dependency we have

We have

∂2∂y2A+ω2ϵ′μA=−μJδy−y0

Solve

We can use

∂2∂y2A+κ2A=−μJδy−y0

This equation has the solution from equation (4.31).

Ay=μj2κejκy−y0+Bejκy=Cejκy−y0+Bejκy=Cejκy0−y+Bejκy,fory0≥y

For the boundary conditions we have

We get

Bxy=∂Ay∂y=−jκCejκy0−y+jκBejκy.

From the boundary conditions we find:

−jκC+jκBejκy0=JArea−jκCejκy0+jκB=JArea

Solving for the constant BB we get:

jκBej2κy0−jκB=JAreaejκy0−1→B=JjκAreaejκy0−1ej2κy0−1

and

jκC=jκBejκy0−JArea→C=JjκAreaejκy0−1ej2κy0−1−1=JjκAreaejκy01−ejκy0ej2κy0−1.

We have

Bxy=∂Ay∂y=−JAreaejκy01−ejκy0ej2κy0−1ejκy0−y+JAreaejκy0−1ej2κy0−1ejκy.

At the long wavelength approximation we can simplify

Bxy≈−JArea−jκy0j2κy01+jκy0−y+JAreajκy01+jκyj2κy0=JArea2+jκy02≈JArea.

By integrating the magnetic energy over the region we find:

∫0y0B2dV=JArea2∫0y0Aready=JArea2Areay0=J2Areay0.

From the definition of inductance, equation (4.4) we find

L=μ1Areay0.(4.54)

The steps we followed here were more complicated than absolutely necessary to get to this answer. However, we have demonstrated a full solution to this version of Maxwell’s equations, in this case Helmholz equation, and we will look more into this kind of calculation in the rest of this chapter.

Verify

We can now see an interesting relationship with the calculation of capacitance between two plates. If we multiply equations (4.47) and (4.54):

LC=μ1Areay0ϵy0Area=μϵ(4.55)

Evaluate

This turns out to be a general relationship for two dimensions, any two shapes have an inductance per unit length times a capacitance per unit length that is equal to μϵμϵ in two dimensions. The proof of this goes beyond the scope of this book but can be found in many references. It is a very useful rule to keep in mind. In practice it also holds up well for planar geometries in three dimensions where one has long skinny conductors. This is an example of a situation where estimation analysis yields a result that turns out to be quite general. If you find some simple relationship like the one just descrbied try to see if it is more general than your simplifications imply. Perhaps you have discovered something fundamental?

Summary

We have applied the estimation analysis to a few situations where the total inductance was needed. We calculated it directly from Maxwell’s equations using some basic simplifications as directed by the estimation analysis technique.

Simplify

We assume again cylindrical symmetry and we are using a spherical conductor with conductance σσ which we assume is high in the sense the effective permittivity is that of a good conductor: see Figure 4.12. Outside the conductor we assume again the long wavelength approximation since this is more common in circuit design. We have inside the conductor

Solve

Outside the conductor we have again

∇2A=0withsolutionAzr=C+BlnrR0.

The magnetic flux density is

Bφr=−Br

Inside the conductor the general solution is

A=Djκe−jκr+Ejκejκrez

representing an outgoing and an incoming wave. We will assume the outgoing wave is zero (E = 0)E=0. Above

κ=ω2μϵ′

With

ϵ′=−jσω

We have

−j=1−j2

We find

κ=ω2μϵ′=−ω2μjσω=ωσμ1−j2

We find for the solution

A=Djκe−jκrez=Djκe−jωσμ1−j2rez=Djκe−jωσμ2r−ωσμ2rez.

We see a term in the exponent of the solution that is real and shows a decline as a function of rr with a length scale

δ=2ωσμ.(4.56)

Figure 4.12 Cross-section of single wire with a finite conductance.

Verify

This is known as the skin depth of the conductor.

Evaluate

The depth to which the electromagnetic fields can penetrate a conductor, skin depth, is dependent on frequency, conductivity and permeability according to (4.35). In typical small geometry CMOS processes the metal layers are really thin – 90 nm is not unusual. For a 6 GHz frequency in copper with a conductivity of 1.67 ∙ 10⁻⁸10−8 ohmm we find a skin depth of 800 nm, 9× the thickness of some metal layers. So, when creating metal shields beware that there are enough metal layers stacked to make an efficient shield given the particular skin depth. Also, the resistance of the metal layers can be substantial – something to keep in mind for large currents.

Another interesting point here is the fact that when the skin depth is less than the conductor size the magnetic field is excluded from part of the conductor. Outside the conductor the field behaves independently of the current distribution inside the conductor and is not affected, as we noted earlier. This exclusion of the magnetic field thus lowers the total magnetic field over all space and the resulting inductance is reduced. This effect can be observed in simulators and while often not very big can reduce the inductance by a small percentage.

Simplify

We examined this situation previously where we noted the resulting inductance is less when the ground plane is present. Here we will use the same approximations as before but we will look at the field solution near the ground plane. This situation is very common in the circuit world and it is worthwhile to explore this more.

Solve

The fields above the ground plane can be described as a sum of the field generated from the source current and a mirror field generated from a location equidistant from the ground plane but on the other side of the ground plane. The ground plane itself is removed. The phase of the mirror current is 180 degrees opposite the source current. This is known as the method of images and we described this earlier. We have from Figure 4.14:

Hsumrr’=Jtotal2πrcosθex+sinθey−Jtotal2πr′cosθ′ex+sinθ′eywherer=x2+y−y02,r′=x2+y+y02.

At the ground plane we have from the boundary conditions that H_{sum, x} |_y = 0Hsum,xy=0 is equal to the surface current in the ground plane. From symmetry we have r = r^′r=r′ and we find from HH above

J=Jtotal2πx2+y022cosθez=Jtotal2πx2+y022y0rez=Jtotalx2+y022y02πez.

We see the induced ground current varies with a scale corresponding to the distance of the current above the ground plane: see Figure 4.15. This is as expected. Note that in this long wavelength approximation no other length scales exist with the exception of skin depth which here is zero. We explore this further in the next section.

Figure 4.14 Coordinate system of current over ground plane.

Figure 4.15 Current distribution plot.

Let us integrate this current density in the x-direction

∫−∞∞Jdx=∫−∞∞Jtotalx2+y022y02πdx=2y0Jtotal∫−∞∞1×2+y02dx=2y0Jtotal2πy02∫−∞∞1x2y02+1dx=2y0Jtotal2πy0∫−∞∞1x2y02+1dxy0=2Jtotal2πtan−1xy0−∞∞=2Jtotalπ2π=Jtotal

so the total induced current is the same as the forced current. This makes sense since the magnetic field far away will then be canceled out by the two opposing currents.

Verify

This is a well-known calculation and can be found in for example [17].

Evaluate

The integrated induced current density is equal to the total imprinted current when no current loops are present. If this was not the case there would net magnetic fields at far distances and thus infinite impedance.

Simplify

Using the full Maxwell equations above we now assume with the same geometry as in Figure 4.13 but the perfect ground plane is replaced a resistive plane:

where J_iJi is the impressed current. With the chosen coordinate system we find

We have

Solve

Since x is unbounded we can Fourier transform:

δx=12π∫ejxβdβAxyz=12π∫A˜zβyzejxβdβ.

We have

∫∂2∂y2A˜z−β2A˜z+ω2ϵ′μA˜zejxβdβ=−∫μJ0δy−y0ejxβdβ.

We can just look at the integrand:

∂2∂y2A˜z−β2A˜z+ω2ϵ′μA˜z=−μJ0δy−y0.

In general this equation cannot be solved exactly but needs a numerical solution.

For our case the solution has the general form:

A˜zβy=Cj2κejκy−y0+Bejκy

With

Inside the lower material the solution is:

A˜zβy=Ae−jκ2y

where

−β2+ω2ϵ′μ=κ2−β2+ω2ϵ2′μ=κ22.

The boundary conditions are now

Ez1=Ez2Bz1=Bz2

Also

We get

Bx=∂A˜z∂yBy=∂A˜z∂x=jβA˜zBz=0

From E_z,    B_zEz,Bz is continuous we get

Cj2κejκy0+B=A−κCj2κejκy0+κB=−κ2A

We see continuity of B_yBy is guaranteed by the E_z = E_zEz=Ez condition. Multiplying first by κ₂κ2 and adding.

Ci2κeiκy0κ2−κ+Bκ2+κ=0B=κ−κ2κ2+κCj2κejκy0

and

A=Cj2κejκy01+κ−κ2κ2+κ=Cj2κejκy02κκ2+κ

We have

A˜zβy=Cj2κejκy−y0+κ−κ2κ2+κCj2κejκy0ejκy=Cj2κejκy−y0+κ−κ2κ2+κejκy0ejκyAxy=12π∫Cj2κejκy−y0+κ−κ2κ2+κejκy0ejκyejxβdβ.

The first term is the definition of Hankel function of degree zero:

H0kρ=12π∫j2κejκy−y0ejxβdβ.

Inside the lower material the solution is:

A˜zβy=Cj2κejκy02κκ2+κe−jκ2yA2xy=12π∫Cj2κejκy02κκ2+κe−jκ2yejxβdβ(4.57)

This is a general solution and needs to be evaluated numerically for a given configuration.

Verify

However we can learn a few things by looking at extreme cases of key parameters. Let us use conductivity as such an example. Let us assume we have situation described in the previous section with static field and a ground plane with high conductance. We then have

or

The solution in media 2 is then

A2xy=12π∫C12jβe−βy02jβκ2+jβe−jκ2yejxβdβ=12π∫Ce−βy01−β2−jωσμ+jβe−j−β2−jωσμyejxβdβ.

To get the current in the medium we need to take the curl of this twice. This corresponds to multiplying by −κ22. In this extreme case we can limit ourselves to y = 0. So we find.

Jxy=0=12π∫Ce−βy0β2+jωσμ−β2−jωσμ+jβejxβdβ.

For high σσ we see the β²β2 will not dominate the sum for a long while. We can assume this happens when the first exponent |β| |y₀|βy0 is very large so as to dampen out the integrand. We then get

Jxy=0≈12π∫Ce−βy0jωσμejxβdβ

This is a well-known Fourier inverse function and we have

Jxy=0∼2y0x2+y02.

We see we come back to our old solution, so this is encouraging. What about the other extreme where σσ is very small? We can still keep the long wavelength approximation both in medium 1, 2 and we find:

Jxy=12π∫Ce−βy0β2+jωσμ−β2−jωσμ+jβe−j−β2−jωσμyejxβdβ.

In the extreme case where σ = 0σ=0 we simply recover our old static solution where J = 0J=0 outside the current carrying conductor. What if σσ is simply really small so that when we look at the integrand that term containing σσ is only relevant for small ββ? By analogy with a more common physical situation where have frequency and time and not wavelength and distance we would say only the low frequency spectrum is affected and thus only long timescales will show any impact of σσ. Similarly, we can here say only long distances will show the impact of a situation where the conductivity is low. In other words, for high resistance substrates the current distribution under the current carrying conductor will be of the order 1/σωμ1/σωμ, which is large.

Evaluate

Although we have not provided a formal proof of the following Key Concept it is really just a reworking of Ohm’s law: More currents flow where the impedance is small.

Simplify

Let us consider the current distribution inside a metal slab. The general solution for a rectangular cross-section is difficult to solve analytically and has not been done to this author’s knowledge. However, for an ellipsoidal shape there is a known exact solution: [16]. Here we will just outline the model and do some simple scaling estimations to arrive at one of the asymptotic solutions. We will use the long wavelength approximation in two dimensions, with the current, J_zJz, (and vector potential, A_zAz) going in the z-direction, which states that externally we need to solve Laplace’s equation:

Imagine the conductor has the shape as described in Figure 4.16.

Figure 4.16 Ellipsoidal coordinate system.

Adapted from [17].

As a function of coordinates x, y the shape is described by

yx=±h1−x/a2

We will here go to the limit of h→0andσ→”large” such that h ≪ δ ≪ ah≪δ≪a and

For this situation we then naturally have the current in the slab is only dependent on xx.

Solve

We have from the definitions of the electric field in terms of the gauge potentials (4.49) and our assumption of E_z = 0Ez=0

Ez=−∂φ∂z−jωAz=0→−jωAz=∂φ∂z.

Let us take the derivative w.r.t xx of the last expression. With our approximations we assume the only current flowing flows in the zz-direction. For this to be consistent the voltage drop along the segment in zz has to be constant over the surface of the conductor so we expect the derivative w.r.t xx to be zero.

∂−jωAz∂x=−jω∂Az∂x=−jωBy=0(4.60)

which becomes our boundary condition at the slab.

The problem amounts to solving Laplace’s equation for A_zAz with the boundary condition (4.60) at the conductor. For large distances we must have from (4.49)

Az=−lμ2πlnx2+y2R(4.61)

corresponding to a current filament with a large return radius RR. This is our second boundary condition.

We now introduce the conformal coordinate transformation

yielding

∂x∂v=asinhvcosu→0,v→0∂x∂u=−acoshvsinu→−asinu,v→0

∂y∂v=acoshvsinu→asinu,v→0∂y∂u=asinhvcosu→0,v→0

The transformation is one-one with the restrictions

In the (x, y)-plane the curves of constant v are homofocal ellipses; the segment y = 0,    |x| ≤ ay=0,x≤a is the infinitely flat ellipse v = 0v=0 and vv increases outwards to infinity: see Figure 4.17. The curves of constant u are the hyperbolae but there is a cut along the segment v = 0, so that u is positive in the upper-half plane Re y > 0Rey>0 and negative in the lower half-plane. The semi-infinite segment y = 0,    x ≥ ay=0,x≥a corresponds to u = 0u=0, whereas the segment y = 0,   x ≤  − ay=0,x≤−a corresponds to u =  ± πu=±π.

Figure 4.17 Conformal coordinate transformation.

Adapted from [17].

With these new coordinates we have for our equation to solve

With the boundary conditions

∂Az∂xy→0=∂Az∂u∂u∂xv→0+∂Az∂v∂v∂xv→0=∂Az∂u1asinu=0→∂Az∂u=0

The beauty of conformal transformations is if the solution solves Laplace’s equation in the new coordinates it also solves it in the old coordinates: see [17, 18]. The main advantage of this transformation is that if the new coordinates follow the basic geometry the equations are easier to solve.

We will further limit ourselves to the situation where u ≠ 0,    ∓ πu≠0,∓π in other words we avoid the endpoints of the ellipsoid. We know the behavior of A_zAz for large values of v and we also know when v → 0v→0 A_zAz cannot have any dependence on xx (or uu in our new coordinates at the limit v → 0v→0). To meet Laplace’s equation we know a solution of the form

will work. To find C,   BC,B we can attempt a look at the far distance requirement, v → ∞v→∞, equation (4.61) and see if we get a self-consistent solution. We have

x2+y2R≈coshv2R=coshvR≈evR

Taking the logarithm of this we find

lnx2+y2R≈v−lnR

We can now identify the constants B,   CB,C from (4.61) and (4.63) and we find

Az=−lμ2πln2Ra−v(4.64)

This equation solves Laplace’s equation and as v → 0v→0 it has no dependence on uu which we need for the boundary condition at the slab and has the right scaling for large vv. In other words, it solves the conditions at the boundaries and the equation inside the domain. It is thus a solution to the problem. We will confirm this later when comparing to a more general solution.

We are interested in calculating the current across the thin slab with this solution. It can be calculated from the boundary condition discussed in section “Solutions to Maxwell’s Equations: Boundary Conditions,” in particular (4.39) where we here find

where H_{x, 0}Hx,0 refers to the size of the magnetic field at y = 0+y=0+. The magnetic field is given by (4.6) and (4.7).

Hxy=0+=1μ∂Az∂yy=0+=1μ∂Az∂yy=0+=1μ1asinu∂Az∂vv=0+.

We then find the current

Jzu=−21μ1asinu∂Az∂vv=0+=−1μ1asinulμ2π=−Iasinu12π

And in terms of xx

Jzx=−I1−x2/a212π,x<a(4.66)

Verify

This problem has been solved more generally in [17]. The σ → ∞σ→∞ solution in that paper is the same as the one derived here. In the field solver we have analyzed the situation with various thicknesses. See Figure 4.18. As the thickness of the metal slab increase it is getting much better at expelling the field, just as we would expect from the simplified model. For thin wires the prediction fits reasonably well, within 10% except close to the end points (the actual edge is excluded because of the singularity in expression (4.66)). As the thickness increases the effect of the expulsion becomes clear and we see the skin depth start to dominate.

Figure 4.18 Simulation results and comparison wtih estimation analysis. The pictures to the left shows a cross section of the conductor with the current densities indicated in gray. The right pictures shows the current densities through the middle of the conductor. (a, b) corresponds to a thickness of 0.1 µm, (c, d) has a thickness of 0.5 µm, (e, f) shows a slab with 1 µm thickness, (g, h) has 2 µm thickness and finally (i, j) shows 5 µm thickness. The estimated curve has been normalized such that the current density in the middle of the conductor is the same for the simulated and estimated responses.

Evaluate

The lateral skin effect for wide conductors much thinner than the skin depth has a current length scale variation that is dependent on width only, equation (4.66). As the thickness increases the normal skin effect starts to manifest itself.