Lines, where a wire has a length

The corner that is three decades wide

Where the current comes back put the change in a return current's path at 106 kilohertz, from a low-frequency path assumed three track-widths wide and an inductance taken from a parallel-plate formula. Solved across a plane cut into a hundred and twenty strips, the loop's resistance equals its reactance at 1.59 kilohertz, where the current has not moved at all, and the return then gathers beneath the track over three decades — half of the way by 283 kilohertz, nine tenths by 1.42 megahertz. The single corner is a point about a quarter of the way through a band.

Assumes: Where the current comes back · Kirchhoff's own frequency

Where the current comes back asks the right question of a ground plane — where does the return current run, and when does that change — and answers it with one number. For ten centimetres of track two hundred micrometres above a plane of half a milliohm a square, the return takes the path of least resistance below 106 kilohertz and the path of least inductance above it.

The number comes from two paths chosen before anything is solved. One is a strip of plane three times as wide as the track, which gives a resistance of 83.3 milliohms. The other is the strip directly beneath the track, given the inductance of a parallel-plate line, 125.7 nanohenries. The corner is where the first path’s resistance equals the second path’s reactance. That essay says plainly that the factor of three is not a derivation, and everything that depends on it carries it.

This essay does the derivation. What it finds is not a better corner. It finds that there is no corner to find: the return gathers under the track across three decades of frequency, the frequency at which resistance and reactance are equal is where the gathering has not yet started, and the estimated corner is a point about a quarter of the way through.

The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres. — Fig. 1 The estimate this essay replaces, as it was drawn: an assumed wide path of 83.3 mΩ, the parallel-plate inductance µ₀h/w of 125.7 nH, and the frequency at which the two cost the same, 106 kHz. It says where two chosen paths would be equal, and nothing about where the current is.

Two paths that were never solved for

Each of the two assumptions is reasonable on its own, and each is doing more work than it looks.

The inductance µ₀h/w is the inductance of two parallel plates of width w a distance h apart, with the field confined between them. That is the right answer for a track much wider than its height above the plane. The track here is exactly as wide as it is high, so most of its magnetic field is outside the strip the formula assumes, and the formula has no way to say how much.

The resistance of a strip three track-widths wide is not measured from anything. At direct current a return uses whatever copper is connected, in proportion to its conductance, and in a plane fifty millimetres wide that is the whole plane. A width tied to the track’s own width also has a consequence nobody chose: it makes the corner independent of the track’s width by construction, since the resistance and the inductance both scale as one over it.

And the corner itself, R = ωL, is a statement about a loop’s impedance: the frequency at which its phase angle reaches forty-five degrees. Reading it as the frequency at which the current changes path adds a claim — that the current’s position follows the phase of the impedance — which nothing in the estimate tests. The solve below tests it, and it fails by nearly two decades.

A plane cut into strips

The solve is a cross-section. The plane, fifty millimetres wide, is cut into a hundred and twenty strips running parallel to the track, narrow beneath it — the finest is a sixth of the track’s height — and widening towards the plane’s edges. Every strip has a resistance, which is the sheet resistance over its width, and a partial inductance to every other strip and to the track. The partial inductance of two long parallel conductors is µ₀/2π times the logarithm of one over their distance, averaged across both widths, which has a closed form for two strips in one plane and for a strip at a height above another.

The strips are joined together at both ends, so they all see the same voltage per metre, and the return divides itself between them however the impedance matrix says. Nothing tells it where to go.

That makes two limits available as checks, and the solve reaches both unaided. At direct current the inductances drop out and the current is the same in every strip, to a part in a billion, so the resistance is the sheet resistance over the plane’s width: 1.000 milliohm over ten centimetres, eighty-three times less than the assumed wide path. Far above any frequency of interest the resistances drop out instead and the current takes the distribution of least inductance, which for a plane much wider than the height is the image-current distribution, and that has a closed form of its own. Solved, 48.7, 79.3 and 93.7 per cent of the return lies within one, three and ten heights of the point under the track; the closed form for a strip of this width says 48.7, 79.4 and 93.7, with a worst gap of 1.8×10⁻⁴ of the return. The table Where the current comes back quoted — fifty per cent inside one height — was the same distribution for a track of no width at all.

Two edges come with the method. A cross-section has no length in it, so it describes a run long compared with the width of plane the return spreads over, and it cannot say anything about the ends. The plane is thin: the current is taken as uniform through the copper’s thickness, which holds until the skin depth reaches it. Half a milliohm a square is thirty-four micrometres of copper, and the resistance that grows with frequency puts the skin depth there at 3.68 megahertz.

Fig. 2 The share of the return inside one track-height of the point under the track, and inside ten, against frequency on a 50 mm plane. At direct current 0.8 per cent is within one height; far above the band 48.7 per cent is, which is the closed form. A tenth of the gathering is done by 26.3 kHz, half by 283 kHz and nine tenths by 1.42 MHz, shaded; resistance equals reactance at 1.59 kHz, and the estimated corner at 106 kHz is 28 per cent of the way in.

Where resistance equals reactance, nothing has moved

The solved loop has one resistance and one inductance at each frequency, and they are equal — R = ωL, the estimate’s own criterion — at 1.59 kilohertz.

At that frequency 0.98 per cent of the return is within one height of the track, against 0.80 per cent at direct current, and 8.9 per cent is within ten heights against 8.0. The loop’s impedance has turned to forty-five degrees and the current is where it was at direct current to within a fifth of a percentage point. Whatever R = ωL marks, it is not the frequency at which the return changes path.

What happens instead is gradual, and the reason it is gradual is geometric. A strip of plane at a distance x from the track is cheap in resistance and expensive in inductance in a proportion that depends on x, and the strips are at every distance from a sixth of a height to twenty-five millimetres. So each strip gives up its share at its own frequency, and the return narrows continuously rather than switching. Measured as the share within one height, going from its direct-current value to its final one, the return has gathered a tenth of the way by 26.3 kilohertz, half by 283 kilohertz and nine tenths by 1.42 megahertz.

From the frequency at which resistance equals reactance to nine tenths gathered is 2.95 decades. From a tenth to nine tenths is 1.73. Either way it is a band and not a corner, and the band is wide on the scale of anything else in this field: Kirchhoff’s own frequency for the same ten centimetres of copper is 3.97 megahertz, so on this board the return has done nine tenths of its gathering before the track stops being a lumped element at all.

The estimated corner lands inside the band. At 106 kilohertz 14.0 per cent of the return is within one height, against 48.7 per cent when the gathering is complete, which is 27.5 per cent of the way through. The sentence the estimate supports — that above the corner the return runs directly beneath the track — is false at the corner itself.

Fig. 3 Current per millimetre of plane against distance sideways from the point beneath the track, one side only, for one ampere in the track. The faint curves are 100 Hz, 10 kHz, 1 MHz and 100 MHz; the heavy one is 100 kHz, where 13.5 per cent of the return is within one height and 65.3 per cent within ten. The dashed curve is the image-current closed form. The slider is the frequency.

One distribution, not two paths

The profile is the picture the two-path estimate stands in for, and it does not contain two paths.

At a hundred hertz the current per millimetre is flat all the way to the plane’s edge, twenty-five millimetres out. As the frequency rises the density under the track lifts and the far strips give theirs up, first at the edge of the plane and then progressively nearer. By a megahertz the curve lies close to the closed form within a few heights and is still carrying current tens of millimetres out; by a hundred megahertz it lies on the closed form everywhere drawn.

The tail is the part a designer should notice. At a hundred kilohertz — the estimate’s “above the corner” — 34.7 per cent of the return is more than ten heights, two millimetres, from the track. A second track two millimetres away is running in that current. The consequence of sharing copper with somebody else’s return is what the millivolts in the wire measures on a lumped conductor, and what the far end that cancels assumes has been arranged when it takes the two couplings from one field.

The path's resistance rises 38-fold as the return gathers, and its inductance falls to 44 nH. computed by solving, not by drawing, on the same cross-section. The solid curves are the resistance and the reactance of the return loop over 100 mm of track, from the strip solve; the dashed pair is the estimate — a resistance of 83.3 mΩ from a path three track-widths wide, and a reactance from the parallel-plate inductance µ₀h/w, 125.7 nH. Solved, the resistance is 1.000 mΩ at direct current, where the current uses the whole plane, and rises to 38.31 mΩ above the band, where it is packed under the track; the inductance falls from 103.2 to 44.3 nH. The solved curves cross at 1.59 kHz and the estimated ones at 106 kHz. — Fig. 4 The resistance and reactance of the return loop over 100 mm of track, solved (solid) and estimated (dashed). Solved, the resistance is 1.000 mΩ at direct current and 38.31 mΩ above the band, and the inductance falls from 103.2 to 44.3 nH; the estimate holds 83.3 mΩ and 125.7 nH throughout. The solved pair cross at 1.59 kHz and the estimated pair at 106 kHz.

The tight path is the resistive one

In the two-path picture the wide path is the resistive one and the tight path is the inductive one, and a rising frequency trades one for the other. Solved, there is one path, and its resistance and inductance move together.

The resistance rises thirty-eight-fold, from 1.000 milliohm to 38.3, because the same current is being crowded into less copper. The inductance falls from 103.2 nanohenries to 44.3, because crowding it under the track shrinks the field it has to establish. Both directions are required rather than observed: in any network of resistances and inductances the resistance seen at a pair of terminals never falls as the frequency rises and the inductance never rises, and the solve is held to that at every frequency it computes.

Against that, the estimate’s two numbers are wrong in opposite directions. Its wide-path resistance, 83.3 milliohms, is eighty-three times the resistance the return actually meets while it is spread, and more than twice what it meets when it is packed under the track. Its inductance, 125.7 nanohenries, is 2.8 times the solved loop above the band — the parallel-plate formula applied to a track as wide as it is high, which is outside the limit it describes. Neither error is a tolerance. Both come from choosing the paths before solving for the current.

One of the estimate’s results does survive, and it is worth separating from the rest. The radiated field it computes above the band uses a loop area of the track’s length times its height, and that is an identity rather than a measurement: the plane’s current flows in the plane’s own surface, so wherever across the plane it runs, and provided it is spread symmetrically, the magnetic moment of the loop is the track’s current times its height times its length. The width of the distribution changes the inductance and the resistance. It does not change that product.

The plane’s width moves the bottom of the band

Twenty times the plane’s width moves the frequency at which resistance equals reactance thirty-seven-fold and moves the top of the band by five per cent.

The bottom follows the plane because it is set by the direct-current resistance, and the direct-current resistance is the sheet resistance over the plane’s width: widen the plane and the resistance falls, so the reactance catches up with it sooner. The top does not follow the plane, because by the time nine tenths of the return is within a few heights of the track the strips at the far edges are carrying almost nothing and removing them changes nothing.

That is why the bottom of the band cannot be read off a stack-up. In a cross-section the width of plane available to the return is the plane’s width. On a real board, where the track leaves the plane through a via at each end, what limits the spreading at low frequency is a three-dimensional question — the length of the run is presumably part of it — and a cross-section cannot answer it. That is left unproved here rather than guessed.

The track's height moves the top of the band 19-fold and the bottom by 5 per cent. computed by solving, not by drawing at 6 heights, each a strip solve of its own. The shaded band runs from a tenth to nine tenths of the return gathered; the line through it is half; the line below it is where the path's resistance equals its reactance; the dashed line is the single corner estimated from a three-width path and a parallel-plate inductance. At 50 µm: R = ωL at 1.60 kHz, half gathered by 752 kHz (0.59 of Rs/2πµ₀h), nine tenths by 3.69 MHz. At 100 µm: R = ωL at 1.59 kHz, half gathered by 511 kHz (0.81 of Rs/2πµ₀h), nine tenths by 2.61 MHz. At 200 µm: R = ωL at 1.59 kHz, half gathered by 283 kHz (0.89 of Rs/2πµ₀h), nine tenths by 1.42 MHz. At 400 µm: R = ωL at 1.57 kHz, half gathered by 149 kHz (0.94 of Rs/2πµ₀h), nine tenths by 729 kHz. At 800 µm: R = ωL at 1.55 kHz, half gathered by 79.0 kHz (1.00 of Rs/2πµ₀h), nine tenths by 373 kHz. At 1.6 mm: R = ωL at 1.51 kHz, half gathered by 44.0 kHz (1.11 of Rs/2πµ₀h), nine tenths by 196 kHz. — Fig. 6 The band against the track’s height above a 50 mm plane. Nine tenths of the gathering is done by 3.69 MHz at 50 µm and by 196 kHz at 1.6 mm, nineteen times lower, while resistance equals reactance between 1.60 and 1.51 kHz throughout. Half the gathering falls at 0.59 of Rs/2πµ₀h at 50 µm and at 1.11 of it at 1.6 mm.

The height and the sheet move the top

The top of the band is the stack-up’s, as Where the current comes back said the corner was. Raising the track from fifty micrometres to 1.6 millimetres lowers nine-tenths-gathered nineteen-fold and moves the bottom of the band by five per cent.

Two laws here are exact rather than fitted. Every frequency in the band is proportional to the plane’s sheet resistance: four times the sheet resistance gives four times each frequency, to 4×10⁻¹⁶, because the strip equations contain the frequency only as ω divided by the sheet resistance. And the band has no length scale but the geometry’s own: the whole cross-section made three times larger — height, width and plane together — divides every frequency by three, to 2×10⁻¹³.

Those two together give the natural frequency of the problem, the sheet resistance over 2πµ₀h, which is 317 kilohertz at two hundred micrometres. The estimate’s corner was that expression divided by its spreading factor of three. The solve puts half the gathering at 0.89 of it with no factor, and the ratio is not a constant: it is 0.59 at fifty micrometres and 1.11 at 1.6 millimetres, and at two hundred micrometres of height it is 0.91 for a track a hundred micrometres wide and 0.77 for one five hundred wide.

So the track’s width is not absent from the answer. It enters the top of the band through its ratio to the height, by as much as the difference between 0.91 and 0.77 over the widths drawn. The claim that width does not appear was a property of the assumed spreading width, which was defined as a multiple of the track’s.

Fig. 7 The same track 1.6 mm above the plane, which is a two-layer board. At direct current 6.4 per cent of the return is within one height, because a height is now a larger share of the plane; a tenth of the gathering is done by 7.14 kHz, half by 44.0 kHz and nine tenths by 196 kHz. The estimated corner, 13.2 kHz, is 20 per cent of the way in.

A two-layer board, and what survives

On a two-layer board the estimate’s errors grow, because the track is now an eighth as wide as it is high and the parallel-plate formula is further outside its limit. It gives 1,005 nanohenries; the solved loop above the band is 85.5, twelve times less. Its resistance is still 83.3 milliohms; the solved return meets 5.02 once it has gathered.

What survives is the part of the earlier essay’s argument that did not depend on the numbers. Length does not appear, in the only sense a cross-section can test. The stack-up sets the top of the band, through the height and the sheet resistance. A plane is there to give the return somewhere tight, and on a four-layer board it has done so by a megahertz and a half, on a two-layer board by two hundred kilohertz. And the loop’s moment above the band is length times height, so the radiated estimates built on it stand.

What does not survive is a single frequency. Every sentence of the form “above the crossover the return runs directly beneath the track” names a point in a band three decades wide, and the useful sentence names the band: on this stack-up, nothing has moved below a couple of kilohertz, half of the return has gathered by three hundred, and almost all of it by a megahertz and a half.

That changes one practical reading in particular. A quarter wave, and the path the current takes back treats the return as undecided near a hundred kilohertz, and the solve agrees and widens it: a board carrying signals anywhere from ten kilohertz to a megahertz has a return partly spread and partly gathered, and the proportion moves with frequency across the whole of that range. It also moves the edges that are lengths one entry: the two hundred microns between a track and its plane is still the length that decides where the current goes, and it now decides a band rather than a corner.

Still open: where the plane runs out

The solve here has a plane that extends well beyond the track on both sides. The next question is the geometry in which it does not: a track near the edge of its plane, or over a slot running alongside it.

A first cut of the same cross-section, with the plane ending beside the track, puts numbers on how much of that matters to the inductance. At a hundred megahertz the loop inductance is 1.160 times its centred value when the track’s centre is directly over the plane’s edge, 1.038 times at one height in, 1.006 times at three and 1.001 times at ten. The resistance triples at the edge, because the return crowds into the last few hundred micrometres of copper. A track three heights past the edge, with no plane beneath it at all, has 1.82 times the inductance; a slot two heights wide running directly under the track costs 1.15 times, and one ten heights wide 1.73.

So the inductance barely notices an edge a few heights away, while the current density at the edge and the resistance notice it strongly — which is a distinct argument from this one, about a boundary rather than about a frequency, and it deserves its own measurement rather than a paragraph.

That question has two limits to state before it can state anything else. A cross-section can only represent an edge or a slot running along the track. A slot that crosses the track, which is the usual case of a split plane, is three-dimensional, and so are the vias at each end of a real run. And what an edge radiates is a question about the field leaving the board, which a solve for the currents in a plane does not contain. The inductance and the current density are what the cross-section can honestly report.

A second question is available from the tail measured above. Below the band two tracks share the whole plane, and their returns overlap however far apart they are routed; as the returns gather the shared fraction falls, at a rate the same solve with two tracks in it would give frequency by frequency. That turns the common-impedance coupling the millivolts in the wire measures on one conductor into a property of a plane, and the transition it would find is the same three-decade band rather than a corner.