The collection

Every essay — page 4

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence. Page 4 of 6.

Lines, where a wire has a length

On the far side of the frequency at which Kirchhoff's laws give out. What a source drives into is decided by geometry before the load has any say; what comes back one delay later decides the rest. The wave picture is checked against a lumped ladder that has never heard of a wave — and the useful result is how badly the ladder does.

A track needs about three heights of copper beside it, and it is the resistance that says so. computed by solving, not by drawing at 100 MHz, each point a strip solve of its own on a 50 mm plane of the same area, moved sideways. The horizontal axis is where the track's centre sits relative to the plane's edge, in units of the track's height above it; negative is a track hanging past the edge with no copper beneath it. With the centre directly over the edge the loop's inductance is 1.161 times its centred value and its resistance 2.90 times, because the return has to crowd into the last few hundred micrometres of copper. Three heights in, the inductance is 1.006 times and the resistance 1.09; ten heights in, both are within 0.7 per cent. Three heights past the edge the inductance is 1.82 times. At direct current every point on this axis is exactly one, because the copper has been moved and not removed.

Where the plane runs out

The corner that is three decades wide solved a return current over a plane that extends well past the track on both sides. Where it does not, the two costs arrive at opposite ends of the band: at direct current a plane that ends under the track costs 27 per cent of inductance and not one part in a million of resistance, and above the band it costs 16 per cent of inductance and 199 per cent of resistance. Three track-heights of copper beside the track removes almost all of both, and the number three has no millimetres in it — sixteen times the whole cross-section gives the same ratios to a part in a billion.

8 figures
1 pF across a 50 Ω line: a dip of 0.320 V and an area of 25 ps. computed by solving, not by drawing as a cascade of two-ports, with a raised-cosine edge of 59 ps sent into it. The incident edge is the faint curve; what comes back is the shaded dip and what goes on is the third. The dip reaches -0.3202 V and its area is 25 ps, which is Z₀C/2 to a part in ten thousand. Driven by an edge fifty times faster the same cascade returns the single exponential the closed form gives, to 9.3e-6 of a volt. The transmitted edge leaves at 81.1 ps, against 59 ps arriving.

The dip whose area is fixed

A picofarad across a 50 Ω line makes a dip in what comes back. Its depth is 0.833 volts to a six-picosecond edge and 0.0196 volts to a nanosecond one, forty-two times less; its area is 25 picoseconds to both, to six parts in a hundred thousand, because the area is Z₀C/2 and contains nothing about the edge. Two half-picofarad discontinuities too close to tell apart read as exactly one picofarad, and so do two far enough apart to be separate — the area is additive where the depth is not. And what a reflectometer calls the capacitance of an impedance step is the step's real excess capacitance times 1 + Z/Z₀.

7 figures
A permittivity quoted as one number falls 0.645 across five decades. computed by solving, not by drawing. The real part of the relative permittivity against frequency for FR-4 (ε′ = 4.4 at 1 GHz, tanδ = 0.02), from a continuum of relaxations spread uniformly in log-frequency — the arrangement that makes the loss tangent flat. The dashed line is the single number a datasheet quotes. FR-4: 4.787 at 1 MHz and 4.142 at 100 GHz, a fall of 14.7 per cent. Nothing here is fitted: the slope is what a flat loss tangent forces.

The permittivity a loss forbids

A datasheet quotes a relative permittivity and a loss tangent as two independent numbers, and they are not two numbers. A material that dissipates has a permittivity that falls logarithmically with frequency at a rate its own loss fixes — 0.129 of permittivity a decade for FR-4, so the 4.4 quoted at a gigahertz is 4.79 at a megahertz and 4.14 at a hundred. Three hundred millimetres of track loses 62.9 picoseconds of delay between 100 MHz and 10 GHz, which a constant permittivity puts at 1.3; and the constant-permittivity model smears an edge backwards, taking 180 picoseconds to reach half height and 133 more to reach nine tenths.

7 figures
A via's area changes sign at 44.7 Ω, and two impedances give its 0.5 pF and 1 nH back. computed by solving, not by drawing, as a cascade of two-ports: a via of 0.25 pF, 1 nH and 0.25 pF, met by an edge of 59 ps from reference lines of 20 to 150 Ω. The area under the reflection is −Z₀C/2 + L/2Z₀ at every impedance: a bump below 44.7 Ω, where the inductance's term is the larger, nothing at it, and a dip above. From 50 Ω the area is 2.5 ps of dip, which a single-capacitance reading calls 0.100 pF. From 50 and 75 Ω together the two areas give 0.5000 pF and 1.0000 nH. An error of 50 fs on each area moves them by up to 0.8% and 1.5%.

The via two lines can weigh

The area under a reflection is a property of the discontinuity rather than of the edge, and for a via it is one number made of two: −Z₀C/2 from its pads and +L/2Z₀ from its barrel, with opposite signs. A via of half a picofarad and a nanohenry, seen from fifty ohms, leaves 2.5 picoseconds of dip — which a reading that assumes a capacitor calls 0.100 pF, a fifth of what is there. From fifty and seventy-five ohms together the two areas give 0.5000 pF and 1.0000 nH back, and fifty femtoseconds of error on each costs 0.8 per cent of the capacitance and 1.5 of the inductance. A second line at fifty-five ohms costs five times as much.

6 figures
At 44.7 Ω the via's area is zero; its reflection is a doublet falling as the 1.99 power of the edge, not the first. computed by solving, not by drawing, as a cascade. A via of 0.25 pF, 1 nH and 0.25 pF on a line of √(L/C) = 44.72 Ω, met by an edge of 59 ps, against the same capacitance alone on the same line. The via's reflection is a doublet — a dip and a bump of equal area — whose largest excursion is 24.4 mV against 166 mV for the capacitance alone. Over edges from 5.9 ps to 295 ps the capacitance's reflection falls as the −0.97 power of the edge and the via's as the −1.99 power: 24.4 mV at 59 ps, 994 µV at 295 ps. The via delays the edge going past it by 22.4 ps, against √(LC) = 22.4 ps.

The via that is a piece of line

A via of half a picofarad of pad and a nanohenry of barrel puts no area under its reflection from a line of √(L/C) = 44.72 ohms, from any edge. It has not vanished. It delays the edge going past it by 22.4 picoseconds, which is √(LC) exactly, and it still reflects: a doublet whose largest excursion falls as the −1.99 power of the edge where a lone capacitance's falls as the −0.97 — 24.4 millivolts for a 59-picosecond edge against 166 for the pads alone, and 994 microvolts for a 295-picosecond one against 35. The balanced via is a short piece of line, and the reflection it leaves is the reflection of its length rather than of its size.

6 figures
Two tracks 1 mm apart share all of the plane's resistance at direct current and 14.5% of it above the band. computed by solving, not by drawing, across a 50 mm plane cut into 239 strips, with two tracks 200 µm above it and 1 mm apart — 5.0 heights. One track carries the current and the voltage along the other's loop is measured. The shared resistance, as a fraction of the driven loop's own, is 1 at direct current, where both returns spread across the whole plane and share its 10 mΩ/m; it falls through a half at 183 kHz and settles at 0.1454 above 10.0 MHz, where each return has gathered under its own track. The shared inductance is 0.398 of the loop's own at direct current and 0.0341 above the band.

Two returns in one plane

Two tracks over one plane share the whole of its resistance at direct current, however far apart they are routed: 10 milliohms a metre on a fifty-millimetre plane, from tracks a millimetre apart or ten. The sharing ends across the same band a single return gathers over, and it ends sooner the farther apart the tracks are — through a half at 525 kilohertz for tracks three heights apart and at 9.88 kilohertz for fifty. Above the band what is left is the overlap of two image distributions, 4h²/(4h² + d²): 14.5 per cent of the resistance at a millimetre, 0.68 at five. And in the middle of the band the two loops' mutual inductance changes sign.

6 figures
20 sections imitating a 1 m line: its delay is 1% long at 318 MHz and its group delay at 185 MHz, and it passes nothing above 1.32 GHz. computed by solving, not by drawing, as a chain of 20 series inductors and shunt capacitors carrying the inductance and capacitance of a metre of 50 Ω line of delay 4.83 ns, terminated in 50 Ω at both ends, beside the Bloch phase of an endless chain, 2·arcsin of ω over the cutoff, a section. The chain's cutoff is the cutoff 2/√(LₛCₛ), 1.32 GHz. Its phase delay is too long by arcsin(x)/x − 1 with x = f over that cutoff, 1% at 318 MHz (x = 0.2417); its group delay by 1/√(1 − x²) − 1, 1% at 185 MHz (x = 0.1404). Well below cutoff the solved chain's delay follows the closed form; nearer it the fifty-ohm terminations, which are not the LC ladder's own impedance there, add a ripple. Ten sections per wavelength is 414 MHz for this chain.

The sections a wavelength needs

A ladder of inductors and capacitors is a line only below its own cutoff, 2/√(LC) of one section, and in the frequency domain how far below can be written down exactly: its delay is too long by arcsin(x)/x − 1 and its group delay by 1/√(1 − x²) − 1, where x is π over the number of sections per wavelength. One per cent of delay needs 13.0 sections per wavelength; one per cent of group delay, 22.4. The rule of ten per wavelength is 1.72 per cent slow in phase and 5.33 in group delay. Twenty sections imitating a metre of cable are a line to a per cent of group delay up to 185 megahertz and pass nothing at all above 1.32 gigahertz.

5 figures
A 10 ps stub under a balanced via, and the pad it has to come out of. computed by solving, not by drawing, as a cascade. The reflection from a via of 0.5 pF of pad and 1 nH of barrel on its balancing line of 44.72 Ω, met by an edge of 59 ps, four ways: with no stub, whose largest excursion is 24.4 mV; with a 10 ps open stub below its bottom pad, 90.5 mV and an area of 5 ps; with the stub's 0.224 pF taken out of both pads, 56.1 mV; and taken out of the bottom pad alone, 24.6 mV. Both trims balance the area again. Only the second restores the doublet.

The pad the stub comes out of

A via used to a middle layer leaves barrel hanging below the connection: an open stub. Under a via balanced on its √(L/C) line, a 10 ps stub is 0.224 pF of extra capacitance, 5 ps of area, and a reflection of 90.5 mV where the balanced via gave 24.4. Taking that capacitance out of the pads balances the area again whichever pad it comes from. Only one restores the doublet: out of both pads the via still reflects 56.1 mV, out of the bottom pad the stub hangs from, 24.6. The trim works for every stub whose capacitance fits inside that pad, delays up to Z·C/2 = 11.2 ps, whose quarter wave is above 22.4 GHz. A longer stub has nothing left to come out of. At 40 ps it is a notch in the transmission at 6.25 GHz.

6 figures
At the same rise time a lossy edge draws 2.9–4.0 times more from a balanced via than a clean one, and the square law survives. computed by solving, not by drawing. The largest excursion of the reflection from a balanced via and from a 0.5 pF capacitive via against the 10–90 rise time of the edge that reaches them: behind 5 to 50 cm of lossy line, with a 20 ps edge sent in, and from clean raised-cosine edges of the same rise times on a lossless line. Behind the loss the balanced via's reflection falls as the −1.97 power of the rise time and the capacitive one's as the −0.82. Against clean edges of the same rise time the lossy edge draws 2.93 to 3.97 times as much from the balanced via and 1.07 to 1.54 times as much from the capacitive one.

The edge the loss delivers

A balanced via's reflection falls as the square of the edge's rise time, so the loss in front of it should hide it. A 20 ps edge sent down 20 cm of lossy line arrives with a 10–90 rise time of 194 ps, and the via's reflection falls to 9.1 mV. But a clean edge with that same 194 ps rise time draws only 2.29 mV from the same via. The lossy edge excites it 3.97 times as much, because loss makes an edge steep in the middle and slow at both ends, 3.07 times steeper at its centre than a raised cosine of the same 10–90 time. Across 10 to 50 cm the square law survives, an exponent of −1.97. It survives shifted: the 10–90 time overstates, by a factor of three or four, how slow the edge is to anything that reads its curvature.

6 figures
A 0.4 mm slot between the tracks leaves all of the resistance shared at DC and raises the in-band share from 4.03% to 4.98%. computed by solving, not by drawing, across the plane in strips. Two tracks 200 µm above a 50 mm plane and 2 mm apart, one driven and the other listened to: the share of the driven loop's resistance that the quiet loop's return also carries, against frequency, over an uncut plane (dashed) and with a slot 0.4 mm wide cut along the plane midway between the tracks (solid). The slot's two halves are one conductor, joined where the run ends, so at direct current the return still spreads across both and the whole of the plane's resistance is shared, 1.0000 of it. Above the band each return has gathered under its own track, and the share is 4.98% with the slot against 4.03% without: the slot is where the copper nearest the other track used to be.

The slot that does not separate

Cutting a slot in the plane between two tracks is the common response to coupling between their returns, and in a cross-section it looks as if it separates them. It does not, if the plane is one ground: the halves stay joined where the run ends, so at direct current both returns still spread across the whole plane and share all of its resistance — 1.0000 with a 0.4 mm slot — and above the band the slot makes the sharing worse, 4.98% against 4.03%, rising to 15.7% as the slot widens to 1.6 mm, because it takes away the copper between the tracks and pushes each return towards the other. Only a split into two grounds removes the shared resistance at DC, at the price of each loop's resistance multiplied by 2.10, and it leaves the in-band coupling 29% lower, not gone: most of two tracks' coupling at 100 MHz is their own magnetic field, which the plane's copper does not carry.

5 figures
Three imbalances, three slopes: a 2% amplitude imbalance is 1% common mode at every frequency, 1 mm of skew overtakes it above 455 MHz, and a 5 ps rise-time mismatch climbs at forty decibels a decade. computed by solving, not by drawing, from the transforms of the two conductors' edges and checked against the sampled edges transformed by FFT to two per cent: the common mode at the far end of a pair, as a fraction of the differential, for three separate imbalances between its halves. An amplitude imbalance of 2% gives 1.00% at every frequency. A skew of 1 mm on FR-4, 7.00 ps, gives sin(ωΔτ/2), rising at twenty decibels a decade: 2.20% at 1 GHz. Edges of 50 and 55 ps, perfectly timed and equal in size, give a common mode rising at forty decibels a decade: 0.043% at 1 GHz and 5.38% at 10 GHz. The skew passes the amplitude imbalance near 455 MHz, which is a/(2πΔτ).

The common mode that has three slopes

A differential pair turns into common mode wherever its two halves differ, and they can differ in three ways — in size, in timing and in edge rate — each with its own slope. A 2 per cent amplitude imbalance is 1 per cent common mode at every frequency. A millimetre of skew rises at twenty decibels a decade and overtakes it above 455 MHz, a/(2πΔτ). Edges of 50 and 55 picoseconds, perfectly timed, rise at forty: 0.043 per cent at a gigahertz and 5.4 at ten. In time they are a step that stays, a pulse with an area, and a doublet with none — each slope is a moment that vanishes. And an amplitude imbalance and an edge-rate mismatch of the right signs cancel at one frequency, down 74 dB, which half a picosecond of skew fills back to −25.

5 figures
Routed along the glass weave, a length-matched pair with one conductor over a bundle and one over a gap builds 63.6 ps of skew in 20 cm — the skew nobody routed. computed by solving, not by drawing by integrating each conductor's delay along its route: a length-matched pair of 0.25 mm pitch over a glass weave of 0.5 mm pitch whose effective permittivity swings between 4.2 and 4.6 across it, positioned so that one conductor runs over a bundle and the other over the gap. Routed along the weave (dashed) the skew grows in proportion to the length, 0.318 ps a millimetre, to 63.62 ps in 200 mm. The slider rotates the route.

The skew in the glass

Match a differential pair's two lengths exactly and it can still have tens of picoseconds of skew, because a circuit board is woven glass in resin and the two conductors need not sit over the same one. Over a weave of half a millimetre whose effective permittivity swings from 4.2 to 4.6, a pair of quarter-millimetre pitch with one conductor over a bundle and the other over a gap builds 0.318 ps of skew every millimetre — 63.6 ps in 20 cm, the equivalent of 13.6 mm of routed mismatch by 30 cm, and 68 per cent common mode at 2.5 GHz. Where the pair falls across the weave is a lottery from nothing to that. Rotate the route by ten degrees and the skew never exceeds 0.146 ps, however long the pair: each conductor crosses bundle and gap in turn, and the difference averages out.

5 figures
A 0.6 mm guard between the tracks leaves all the resistance shared at DC and moves the in-band share from 4.03% to 3.50%. computed by solving, not by drawing, across the plane and the guard in strips. Two tracks 200 µm above a 50 mm plane and 2 mm apart, one driven and the other listened to, with a grounded trace of 0.2, 0.6, 1.2 mm centred between them at the tracks' height and joined to the plane at the ends of the run: the share of the driven loop's resistance that the quiet loop's return also carries, against frequency. The guard is part of the return net, so at direct current it changes nothing — the whole plane's resistance is still shared. Above the band a 0.6 mm guard moves the share from 4.03% to 3.50%.

The guard that is part of the return

Where a slot between two tracks makes their returns share more of the plane, adding copper instead — a grounded trace between them — makes them share less. But not much, and not at every width. At direct current a guard joined to the plane at the ends of the run is part of the same return and changes nothing: the whole plane's resistance is still shared. At 100 MHz a guard 1.4 mm wide in the 2 mm gap cuts the shared resistance from 4.03 to 2.55 per cent and the mutual inductance by half; a guard 0.1 mm wide raises the sharing to 4.44 per cent, since a thin trace beside both tracks carries both returns through its own small cross-section. The guard works by taking return current the quiet track's plane would have carried — 7 per cent of the neighbour's for a 0.6 mm guard — and it is bounded by exactly that.

5 figures
With the weft counted as well as the warp, the skew a pair can reach is least at 45.0° to the warp — 128 fs — and turning further off the warp makes it worse again. computed by solving, not by drawing, to first order in the permittivity's swing, the summed bound held against an integration over 64 registrations of the pair to the cloth at six angles. A 0.25 mm pair over a cloth whose warp has a 0.5 mm pitch and whose weft has a 0.5 mm pitch, each swinging the permittivity by ±0.2 about 4.4. The warp's contribution falls as the route turns off it, as the earlier figures found; the weft's rises, because the pair's projection across the weft grows as d·sin θ while its yarns are crossed less often. The worst skew is least at 45.00°, 128.3 fs. At 10°, where the warp alone allows 291.4 fs over every registration — twice the earlier figure's 146 fs, which was for a route starting over bundle and gap — the weft adds 13.8 fs. Integrated: 5° 587.6 fs, 15° 215.1 fs, 30° 139.3 fs, 40° 126.3 fs, 60° 139.2 fs, 80° 304.9 fs. The slider is the weft's pitch.

The angle two weaves allow

Turning a differential pair a few degrees off a board's glass weave stops its skew growing with length — but a cloth has two weaves, and turning off the warp turns onto the weft. Counting both, the worst skew over every registration of the artwork to the cloth is a sum of two bounds, one falling as the route leaves the warp and one rising as it meets the weft, and it is least at an angle the cloth decides: 45° and 128 fs for two identical half-millimetre weaves and a quarter-millimetre pair, against 305 at 10°; 52° for a finer weft; 67° for a flatter one. Doing the count properly also finds the earlier bound a factor of two low, and finds the obvious answer a trap: on a square weave's exact diagonal the two weaves' product stops averaging and the skew grows again, 2.2 fs a millimetre, while a degree either side it stays at 128.

6 figures
The delay of 50 Ω FR-4 track against frequency, with nothing dispersing, the dielectric causal, and the conductor causal as well. computed by solving, not by drawing at 108 frequencies on one track: 4.4 and a loss tangent of 0.02 at a gigahertz, 2 Ω/m of copper rising as √f above 1 MHz. Held constant (faint), the delay is flat above the resistance's corner. With the permittivity allowed to disperse (dashed), it falls 209.6 fs/mm from 100 MHz to 10 GHz. With the conductor's internal reactance put beside its resistance (solid), it falls 464.4 fs/mm over the same two decades: 7.389 ps/mm at 100 MHz, 7.094 at 1 GHz, 6.925 at 10 GHz.

The conductor made causal too

A dielectric that loses energy must disperse, and the essay before this one made FR-4's permittivity do so: the track's delay then falls 209.6 fs/mm from 100 MHz to 10 GHz. The copper has the same obligation. A conductor whose resistance rises as √f has an internal impedance (1 + j) times that rise, and putting the reactance beside the resistance more than doubles the fall, to 464.4 fs/mm — the conductor's share largest at low frequency, 418 of 584 fs/mm between 10 and 100 MHz, and the dielectric's constant at about 100 a decade. On a 30 ps edge after 300 mm the toe sharpens from 74.5 to 55.6 ps and the tail stretches from 221 to 398: the rise time is 453 ps against 295, and the tail-over-toe ratio a scope can read is 8.0 at 400 mm against 2.6.

6 figures
A pair with one conductor over glass-rich laminate (4.6, tan δ 0.015) and one over resin-rich (4.2, tan δ 0.025): the skew per millimetre against frequency. computed by solving, not by drawing: each conductor as its own causal line — permittivity dispersing with its own loss tangent, copper with its internal reactance — and the skew as the difference of their delays. With both permittivities held at their 1 GHz values the skew would be 0.318 ps/mm at every frequency (level line). Dispersing, it is 0.272 ps/mm at 100 MHz, 0.319 at 1 GHz and 0.366 at 10 GHz: the resin, lossier, loses permittivity faster with frequency, and the gap between the two halves widens.

The skew that moves with frequency

A pair with one half over glass-rich laminate and one over resin-rich builds 0.318 ps of skew a millimetre when each permittivity is held at its 1 GHz value. But the two regions lose differently — tan δ 0.015 against 0.025 — so their permittivities fall at different rates, and the skew is a curve: 0.272 ps/mm at 100 MHz, 0.319 at 1 GHz, 0.366 at 10 GHz. A deskew set at one frequency leaves ±2.35 ps in 50 mm at the other two. The resin half also loses 3.7 dB more by 10 GHz over 200 mm, so its edge arrives 63.7 ps early and rises in 257 ps against 202; the halves differ in timing and in edge rate at once, and leave a common-mode pulse of 151 mV for each volt.

6 figures
Near-end crosstalk from a 200 ps edge, 100 mm of coupled track against 10 mm. computed by solving, not by drawing, the pair solved mode by mode and returned to time from its harmonics. Over 100 mm the near end rises with the edge, holds a plateau of 12.51 mV per volt for twice the line's delay, 1.33 ns, and falls. Over 10 mm the round trip is shorter than the edge and the pulse never reaches the plateau: 8.34 mV. The plateau is Kb, a quarter of the sum of the two couplings, on the 0.5 V the driven line carries: 12.50 mV.

The near end that stops growing

Near-end crosstalk is a length only until the coupled run's round trip equals the edge's rise time; past that it is a constant. Solved mode by mode and returned to time, a 200 ps edge on 100 mm of coupled track gives a plateau of 12.51 mV per volt — Kb, a quarter of the sum of the two couplings, on the driven line's half-volt — held for the 1.33 ns round trip. The knee is at v·tr/2: 7 mm for a 100 ps edge, 75 mm for 1 ns. Past it a longer run adds width and not height, and the pulse's area keeps growing as Kb times the round trip.

5 figures
Both ends of the quiet line, 100 mm of microstrip-like coupling, edge 200 ps. computed by solving, not by drawing. The near end (pale) is the saturated plateau, 16.28 mV per volt. The far end (dark) is a single pulse arriving with the edge after one delay, 0.676 ns, as wide as the edge and 25.16 mV per volt deep — negative, because the inductive coupling exceeds the capacitive. Its depth is the coupling difference, the delay and the edge's slope multiplied: 24.96 predicted.

The far end that is a derivative

Where the two couplings are unequal the far end does not cancel, and in time it is the derivative of the edge: a single pulse as wide as the rise time, arriving after one delay, its height the coupling difference times the delay times the edge's slope. On 100 mm of microstrip-like coupling a 200 ps edge gives −25.16 mV per volt against 24.96 predicted, and the height falls as one over the rise time with a fitted slope of −1.000 while the saturated near end sits still. It grows with length without a knee until the two modes separate by a whole edge — 249 mm for a 50 ps edge — and then stops at half the driven line's voltage.

5 figures
A 3.3 V edge from a 10 Ω driver down a 50 Ω line of 1 ns, received through 50 Ω in series with four capacitors, beside a plain 50 Ω and an open end. computed by solving, not by drawing: the line marched by its characteristics. The open end rings up to 5.50 V; the plain resistor holds 2.750 V, the launched level, and never reaches the swing. With the capacitor the far end arrives at the same 2.75 V and then climbs as the capacitor charges and stops terminating: with a time constant of half a round trip (20.0 pF) it overshoots to 4.48 V; with one, 3.83 V; with three (120 pF) it rises without overshoot to 3.30 V by 24 ns; with ten, 3.13 V by 24 ns, still on its way.

The termination that runs out

A 50 Ω resistor at the end of a line terminates every edge and draws current for as long as the line is held high: 75.6 mW on an alternating pattern from a 3.3 V, 10 Ω driver. Put a capacitor in series with it and the standing current is gone, and so, eventually, is the termination: the far end reflects nothing as an edge arrives and the whole of it once the capacitor has charged, by one less the exponential of −t/2Z₀C exactly. Under two round trips of time constant the receiver overshoots — 54.8 per cent at a quarter of one — and above it the receiver waits on the capacitor to reach the swing. On a pseudo-random pattern the levels are tightest at three round trips, and on a fast pattern the capacitor holds the line's mean and the resistor still burns half of what a plain one does.

5 figures
Three matches of one length, half a wavelength at f₀: two quarter-wave sections are bad again at 2f₀ and 4f₀; an exponential and a Klopfenstein taper are not bad again at all. computed by solving, not by drawing, as cascades of exact line input impedances. 50 Ω into 200 Ω through three matching sections each half a wavelength long at f₀: a two-section binomial transformer (dashed), an exponential taper (solid) and a Klopfenstein taper designed for 5% ripple (dotted), the tapers as four hundred thin slices. The stepped transformer is a band-pass: matched at f₀, and at 2f₀, where each section is half a wavelength and transparent, |Γ| is the bare mismatch, 0.600, again — and at 4f₀, and every even multiple. The tapers are high-pass: the exponential is 0.0194 at f₀ and under 0.131 everywhere above 1.6f₀; the Klopfenstein holds 0.050 from 1.1f₀ up.

The match with no upper edge

A stepped quarter-wave transformer is a band-pass in disguise: at twice its design frequency every section is half a wavelength and transparent, and a 50-to-200 Ω match reflects the bare 0.600 again. Let the steps multiply over a fixed length and the staircase of n steps returns the mismatch at n·f₀; in the limit the upper edge has gone, and what is left is a taper — a high-pass match. An exponential taper half a wavelength long is 0.019 at f₀ and never above 0.131 past 1.6f₀. The Klopfenstein profile, the Chebyshev design's continuous limit, holds 0.050 from 1.1f₀ up, needs half the exponential's length for |Γ| ≤ 0.1 and an eighth of it for 0.02, and is not continuous: it starts with a step from 50 to 52.6 Ω. And the small-reflection formula that defines all of them as a Fourier transform overstates a 10:1 taper's reflection by 41% — the line gives tanh of it.

5 figures
A 4:1 load on a cable of 3 dB matched loss costs the cable 1.53 dB more: the standing wave makes the line lossier, by up to the mismatch loss. computed by solving, not by drawing. The extra loss — power into the line over power reaching the load, beyond the matched line's — against the matched loss, for loads giving a standing-wave ratio of 1, 1.5, 2 and 4: 10·log((a² − |Γ|²)/(a(1 − |Γ|²))) − 10·log a (lines), with a 200 Ω load on a 50 Ω line solved exactly as a line whose loss is all series resistance (dots). At a ratio of 4 the extra is 0.109 dB on a 0.1 dB line, 0.82 on a 1 dB line, 1.53 on 3 dB and 1.94 on 20 dB — rising towards the mismatch loss 10·log(1/(1 − |Γ|²)) = 1.94 dB, which it reaches when the reflected wave is burnt before it can return.

The heat a standing wave leaves

A cable improves a mismatch's apparent return loss by twice its own loss because the reflection is burnt on the way back, and the burning is not free: it is extra loss in the cable. For one unit delivered to a load reflecting |Γ| through a line of matched loss a, the line must be fed (a² − |Γ|²)/(a(1 − |Γ|²)). A 4:1 load on a 3 dB cable costs 1.53 dB more than a matched one; on a 20 dB cable, 1.94 dB, which is exactly the mismatch loss, because by then every reflected watt is heat. Short lines multiply their dissipation by (1 + |Γ|²)/(1 − |Γ|²), long ones by 1/(1 − |Γ|²). Solved exactly on a line whose loss is in its conductor, the heat stands where the current does: the hottest point runs at the standing-wave ratio times a matched line's, 4.04 against 4.00, so a power rating limited by temperature falls by the ratio, not by the average.

5 figures
A line ten rises long stood in for by 20 LC sections, matched at both ends: the edge overshoots 17.5% and takes 1.66 rises to climb, where the line delivers it unchanged. computed by solving, not by drawing: an edge of 10–90% rise tr at the far end of a chain of 20 LC sections with a total delay of 10 tr, matched at both ends, marched (solid), against the same edge delayed by exactly 10 tr, which is what a line delivers (dashed). Overshoot 17.5%, rise 1.66 tr. With no reflections at either end, everything here is the chain's own dispersion: a section has a group delay that rises with frequency, so the fast end of the edge arrives late and rings behind it.

The sections an edge needs

A chain of LC sections rings on an ideal step however many sections it has, because a step has energy all the way to the chain's cut-off. An edge has a top frequency, and for a matched chain carrying one the section count has a clean answer: for one per cent of overshoot, 6 sections on a line one rise long, 183 on ten rises and 5,760 on a hundred, 5.8 times the length in rises to the three-halves power, within a few per cent from end to end. So sections per rise is not a constant. A section's phase error goes as the cube of frequency and the inverse square of the count, and it accumulates along the line. A 1 ns edge on a metre of cable asks for 62 sections, a 100 ps edge for 2,048, and the delay is right long before the edge is.

5 figures

Two windings, and the band between them

The first two-sided model on this site. Everything else here is right below a number or above one; an ideal transformer is wrong at both ends and right in the middle, and what a designer buys is the distance between the two — measured on the solve, not taken from a T-model. Beside it, a core whose energy is almost all in its air gap, a saturation limit that is an integral rather than a frequency, and a flux that walks to it however small the imbalance. And a core that can get warm: a single-valued curve has no area, so it cannot dissipate, and giving the material a second branch turns its loss into an area, its Steinmetz exponents into local slopes, and its inductance into two numbers at one current.

A 1:1 transformer at k = 0.99, and the band it is a turns ratio over. computed by solving, not by drawing. Two 10 mH windings coupled at 0.99, driven from 50 Ω into 50 Ω, with 0.5 Ω of winding resistance and 100 pF across the secondary. The response is flat at 0.4901 — which is 98.02% of the 0.5000 an ideal transformer of this ratio would give, and that shortfall is the coupling itself: the flat part is k times the turns ratio, times what the two winding resistances leave of the loop, to four figures at every k on the slider — between 400 Hz and 81.3 kHz, which is 2.31 decades. Both edges are bisected on the solved network. Below the first, the magnetising inductance is a short across the source; above the second, the leakage inductance is in series with the load. The slider moves the coupling, and it moves the upper edge only.

The band a turns ratio holds over

Most circuit models are right below a number or above one. A transformer is wrong at both ends and right in the middle, and the flat part is not the turns ratio either — measured on the solve it is the turns ratio times the coupling, times what the two winding resistances leave of the whole loop — and the first version of that last factor was a coincidence that held for every coupling and broke at a different load.

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What coupling buys: the upper edge only. computed by solving, not by drawing. Six couplings from 0.8 to 0.999, each transformer solved and both its edges bisected. The lower edge moves by 1.083× across the whole range — it is set by the magnetising inductance against the source and the reflected load, and the coupling barely enters it. The upper edge moves by 168×, from 4.84 kHz to 814 kHz, because it is set by the leakage — which is what the coupling is. Winding a better transformer widens the band at the top and does nothing at the bottom, where the answer is more inductance or a smaller load.

What coupling buys, and where it does not

Winding a transformer better is winding it more tightly coupled, and the coupling coefficient is the number a maker works on. Measured across six designs from k = 0.8 to k = 0.999, it moves the upper band edge by 168 times and the lower one by 1.083 — so every hour spent on the winding buys bandwidth at one end of the band and, to within eight per cent, nothing at all at the other.

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Which mechanism sets the upper edge, against the load. computed by solving, not by drawing. Two candidate upper edges drawn against the measurement. The one every textbook names is a resonance between the leakage inductance and the winding capacitance; the one that actually binds at ordinary loads is the leakage in series with the load, a first-order corner at R/2πL. At 50 Ω they are 80.4 kHz and 1.13 MHz — a factor of 14 apart — and the measurement follows the first, to 19.5% at worst across nine loads. They swap at about 1500 Ω, above which the resonance is the binding one and the usual picture is right — which is why a transformer feeding a high impedance behaves as the textbooks say and one feeding fifty ohms does not.

Which picture sets the upper edge

Every account of a transformer's high-frequency limit names the same mechanism: the leakage inductance resonating with the winding capacitance. At fifty ohms that resonance is at 1.13 MHz and the measured edge is at 81.3 kHz — a factor of fourteen away — because what actually binds is the leakage in series with the load, a first-order corner with no resonance in it at all. The two swap at about 1500 Ω, and both accounts are current because both are sometimes right.

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The load at which a transformer becomes a resonant circuit, at k = 0.99. computed by solving, not by drawing. The peak output of the same 1:1 transformer against its load, as a multiple of what the turns ratio would give. Below about a kilohm the load damps the leakage resonance, the peak is the plateau, and the ratio is one: there is a band, and it is what the rest of this field measures. Above it the damping goes and the response peaks — 1.18× at 832 kHz into 1500 Ω, rising to 8.80× at the light end. A transformer with voltage gain is not a transformer behaving badly; it is a resonant circuit, and asking for "the band" of one returns the skirts of a resonance. So the measurement reports that the response is peaked, rather than returning two edge frequencies in the wrong order.

Where the band goes entirely

The three essays before this one measure a transformer's band, and all three assume there is one. Past a few hundred ohms of load there is not: the leakage that sets the upper edge is also what damps the resonance behind it, and a lightly loaded, well-coupled transformer peaks at 3.03 times its own turns ratio. A passive component with voltage gain is not a transformer behaving badly. It is a resonant circuit, and asking for its band returns the skirts of a resonance.

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The coupling coefficient, from two measurements that do not know it. computed by solving, not by drawing. Two windings in series, connected one way and then the other, each solved as a netlist and its inductance read out of the impedance. The two differ by four times the mutual inductance, so k comes out of the difference and the geometry never enters. The recovered value matches the one stamped into the coupling to 2.4e-15 at eight couplings from 0.1 to 0.99 — which is the second route the new element needed, since neither current law nor the energy balance can see a mutual inductance at all. Their sum stays at L₁ + L₂ throughout, which is the check that the two measurements are of one pair.

One number from two measurements

Neither of the two standing checks on a solved circuit can see a mutual inductance. A coupling adds no current anywhere, so Kirchhoff's law is unmoved by it; a coupled pair dissipates nothing, so the energy balance is unmoved too. A coupling stamped into the wrong row would produce a well-formed solution to a different circuit and both checks would pass — so the field needed a third route, and the one it uses is the bench method: connect the windings in series one way, then the other, and the difference is four times the mutual inductance.

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A coupled pair as a two-port, at k = 0.8. computed by solving, not by drawing. Each port driven in turn with the other open, four solves, and the four impedance parameters read out. The diagonal terms measure each winding's own inductance — 10.0000 mH and 40.0000 mH against 10 and 40 — and both transfer terms measure the mutual inductance, 16.0000 mH against k√(L₁L₂) = 16.0000. The two transfer terms agree to 2.83e-16, which is reciprocity — a property of the device rather than of the measurement, and the first thing a coupling stamped into the wrong row would break. Neither of this site's two standing checks can see it: a coupling adds no current and dissipates nothing.

Two ports from two one-ports

An inductor is a one-port: one impedance, one number. Two of them coupled is a two-port, and the four impedance parameters that describe it are recovered here by four solves — each port driven with the other open, the definition read literally. Two of the four come out equal to 2.8 × 10⁻¹⁶, which is reciprocity, and is the first property a coupling stamped into one row instead of two would break.

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A gapped core: where the inductance goes, and where the energy is. computed by solving, not by drawing. Reluctance in series — the gap's lg/µ₀Ae and the core's le/µ₀µᵣAe — with the inductance N²/ℛ and the share of the stored energy in each proportional to its share of the reluctance. At two hundred microns on a µᵣ = 2000 core the inductance has fallen from 41.89 mH to 5.366, and 87.2% of the energy is in the gap — which is air. The share is lg/(lg + le/µᵣ), so it contains neither the turns nor the area and what decides it is µᵣ·lg against the path length; the slider shows the same gap holding 40% at µᵣ = 200 and 98% at 15,000. That is the reason a gap is a design parameter: it is the part of the magnetic circuit whose properties do not drift, do not saturate and do not depend on temperature.

The energy is in the gap

A ferrite core is chosen for its permeability and then deliberately cut, and the cut is not a compromise. Reluctance adds in series, so a two-hundred-micron gap in a µᵣ = 2000 core holds 87.0% of the stored energy while the ferrite holds thirteen — and the share is lg/(lg + le/µᵣ), a ratio of two lengths, containing neither the turns nor the area. The material chosen for its permeability holds almost none of what the component stores.

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The voltage a winding may carry, which is a volt-second limit read at a frequency. computed by solving, not by drawing. The dots are bisections on a marched flux — the voltage integrated sample by sample until the peak excursion reaches 0.35 T — and the line is N·Ae·Bsat·2πf. They agree to 0.001% over three decades, and the fitted slope is 1.000000: exactly proportional, because flux is the integral of voltage and nothing else. The quantity that belongs to the core is the 3.500 mWb-turn, which has no frequency in it. A transformer "rated for 50 Hz" is a transformer whose volt-second product was divided by 2π × 50 once.

A boundary in volt-seconds

A core saturates on the integral of the voltage applied to it, not on the current through it and not at a frequency. The quantity that belongs to the core is N·Ae·Bsat — 3.500 mWb-turn here — and it has no frequency in it at all. Everything a data sheet says about a transformer's frequency rating is that one number divided by 2πf once: measured on a marched flux, the voltage a winding may carry is proportional to frequency to a fitted exponent of 1.000000.

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A 1% imbalance, and the 64 cycles it survives. computed by solving, not by drawing. The upper panel is the peak flux density, marched cycle by cycle, under a square drive whose positive half is 1% larger in area than its negative half. It does not settle. It walks, by the same area every cycle, and reaches 0.35 T after 64 cycles — 1280 ms at 50 Hz — against a closed form of 63.7. The lower panel is the count against the imbalance, and it rises without bound and never becomes infinite. Halving the drive gives 128 cycles, which is exactly twice: reducing the amplitude buys time and not safety, and there is no amplitude at which this design is inside a limit.

The flux that walks

The previous essay's saturation limit is an amplitude, and an amplitude can be respected. This one cannot. A drive whose two half-cycles differ in volt-seconds by one per cent adds the same small area to the flux every cycle, so it reaches saturation after 64 cycles — and halving the drive gives 128, and a tenth of it gives 637. Reducing the amplitude buys time in exact proportion and removes nothing. There is no amplitude at which the design is inside a limit.

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A 0.5 mm conductor's resistance against frequency, exact and asymptotic. computed by solving, not by drawing. The exact ratio is computed from the Kelvin functions by their series; the dashed curve is the asymptote everybody quotes, which treats the current as flowing in one skin depth of the rim and is drawn only where that annulus is inside the wire. At 17.4 kHz, where the skin depth equals the radius and the rule of thumb says the effect "starts", the asymptote says 1.0000 — no effect at all — and the exact answer is already 1.0208. The rule of thumb names a frequency the effect has passed, which is the same shape as the tenth-of-a-wavelength criterion marking a point at which the lumped model is already 30% wrong. Two decades above, the two agree to 0.00%, which is what makes it an asymptote rather than a formula.

The resistance that grows with frequency

The rule of thumb names the frequency at which the skin depth equals the conductor's radius as the point where the effect begins. Computed exactly from the Kelvin functions, the resistance is already 2.05% up there — and the rule's own asymptote says 1.0000, no effect at all. Two decades higher the two agree to 0.01%, which is what makes it an asymptote rather than a formula, and what makes the frequency it names the wrong one to design at.

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A 10 mH inductor with 8 pF across it, and where ωL stops being its impedance. computed by solving, not by drawing. The dashed line is ωL, which is what an inductor is supposed to be; the solid one is the impedance of the same inductor with 8 pF of winding capacitance across it. They part company at 170 kHz, which is ten per cent, and the impedance peaks at 563 kHz and falls thereafter — above which the component is a capacitor. The ratio between the two is 3.317, and the slider shows it is the same ratio at every capacitance: the shape of the departure belongs to the resonance rather than to either part. This is the capacitor essay with the components exchanged, and it comes out with the same structure and a different number.

The inductor that is a capacitor

The frequency field's second essay measures where a capacitor stops being one, because its own leads are an inductance. This is the same measurement with the components exchanged, and it comes out with the same structure and a different number: the ten-per-cent departure from ωL sits at f₀/3.317, and it sits at f₀/3.317 at every winding capacitance and every inductance tried. The shape of the departure belongs to the resonance rather than to either part.

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Alternating-current resistance against foil thickness, 4 layers. computed by solving, not by drawing. The falling dashed curve is the direct-current resistance, which is what more copper buys. The solid curve is the alternating-current resistance at 100 kHz for a portion of 4 layers, and it turns over: past ξ = 0.663 skin depths, thicker foil has MORE resistance, not less. The minimum sits at 1.3368 times the direct-current resistance of the same foil, which is four thirds and is the same number for every layer count above one. The resistance per turn there is 2.016 against √m = 2.000, which is the law the layer count obeys.

The copper that makes it worse

The rung below measured one conductor pushing its own current to its rim, and there is nothing to optimise in it: thicker wire is always less resistance. Stack the conductors and the quantity changes character. Each layer sits in the field of the ones below it, the loss that field drives has no upper bound in the thickness, and the product turns over — so a portion of four layers has a best foil thickness, and above it more copper is more resistance. The best thickness is the fourth root of three over the square root of the layer count, in skin depths, and the penalty at it is four thirds for every layer count above one.

8 figures
One part, three saturation currents: 1.90 A, 2.40 A, 2.69 A. computed by solving, not by drawing. B(H) is μ₀H plus a saturating magnetisation, written as a flux linkage, and the inductance drawn here is dλ/di — the slope of that flux, which is what a small signal on a direct current actually meets. It is 18.92 µH at no current, 16.73 µH at two amps and 10.43 µH at three, and it never reaches zero: the vacuum is still there, so the part falls to its air-core 0.05 µH and stays. The three marks are the ten, twenty and thirty per cent drops different manufacturers print as the saturation current — 1.896, 2.395, 2.694 amps, a spread of 42 per cent on one part.

The inductance the current decides

The rung below bounded the flux and the boundary is exact: the volt-seconds decide the flux swing whatever the material does, and a cycle whose current ripple runs from 519 mA to 75 A has the same flux excursion to better than two per cent. What saturation breaks is the relationship between that flux and the current — so a ripple the design expression puts at 514 mA is 1,022 mA at 2.56 A of load, the peak reaches 3.57 A where the part is at a fifth of its nameplate inductance, and the same part has three saturation currents depending on which per cent it was quoted at.

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A major hysteresis loop at 9.0 A/m of coercivity, and the anhysteretic curve it closes onto. computed by solving, not by drawing. The B–H loop of a core driven sinusoidally to ±400 A/m, marched through a superposition of twenty-four play operators and drawn over the single-valued magnetisation curve with no hysteresis in it. The loop encloses 12.481 joules per cubic metre per cycle, which is the core loss and which no single-valued model can produce, because a curve has no area. The coercivity is 8.96 amperes per metre and the remanence 22.3 millitesla; both are read off the marched descending branch rather than handed in. The slider takes the threshold spread down to zero, where the two branches become one, the area falls to 9.8e-15 J/m³, and the object is exactly the core the field already had.

The area a curve cannot have

Every magnetic model in this collection is a single-valued B(H), and a single-valued B(H) cannot dissipate: ∮H dB around a curve is zero, so the transformers and inductors here have all run cold. Giving the material a second branch costs one object — a play operator, which lags the field by a threshold and is otherwise nothing — and a superposition of twenty-four of them produces a loop with 12.481 joules per cubic metre in it, a coercivity of 8.958 amperes per metre and a remanence of 22.3 millitesla, none of which was handed in. At zero threshold the whole thing collapses onto the curve the field already had, to fifteen figures.

8 figures
The Steinmetz exponent is a local slope, and how far it moves is a property of the material. computed by solving, not by drawing. Loss per cycle against peak flux density over three decades, marched on a play-operator core, with the local exponent d ln W / d ln B drawn across the top of the same frame. It is not a constant anywhere: 2.797 at 5.5 millitesla, heading for the three that Rayleigh's law gives, and 1.462 near saturation where the material has run out of magnetisation to give — a range of 1.420. How wide that range is is itself a property of the material: over the same amplitudes a soft core's exponent moves by 1.73 and a hard one's by 0.21. A single power law fitted across the whole range returns β = 2.518 and misses by 72.2 per cent; the same law fitted over the quarter of it from 9.7 to 24 millitesla returns 2.743 and misses by 0.97. Below 0.58 millitesla this discretisation has no loss at all, which is the finite operator count showing and not the material; the sweep starts above it.

The exponent nobody put in

A catalogue prints core loss as a coefficient times the frequency raised to one power and the flux to another, and the two exponents look like material constants. Neither is. On a loop built from play operators the frequency exponent is exactly one — a theorem, not a fit, because a rate-independent locus has the same area however fast it is traced — and the flux exponent is a local slope that runs from 2.94 at half a millitesla to 1.46 near saturation, so five windows on one measured curve give β from 1.58 to 2.84 and predictions three times apart at a hundred and fifty millitesla.

9 figures
A 4.0:1 flux slope ratio and a sinusoid enclose the same loop to 0.00%. computed by solving, not by drawing. A core driven in flux rather than in field — the way a winding drives it, by integrating a rectangular voltage — around a triangle of ±100 millitesla at a duty cycle of 0.2, whose two slopes differ by 4.00 to one. The loop it traces encloses 2.1006 joules per cubic metre, against 2.1007 for a symmetric triangle and 2.1006 for a sinusoid of the same peak: the same number to 0.001 per cent. That is not an approximation, it is a theorem about the model — a rate-independent locus depends on where the flux went and not on how fast — and it is the prediction that real cores disagree with by tens of per cent. The disagreement is the measurement of what the model has left out.

The duty cycle that costs nothing

A converter drives its core with a rectangular voltage, so the flux is a triangle whose two slopes differ by nineteen to one at a five per cent duty. The play-operator core charges exactly the same for all of them — 2.1006 joules per cubic metre at every duty and for a sinusoid of the same peak, to three parts in ten thousand — because a rate-independent locus depends on where the flux went and not on how fast. Real cores charge tens of per cent more, and the standard correction hides its entire waveform dependence in α − 1, which is the one term a rate-independent model has none of.

8 figures
The same core at the same current has two inductances, 1.80 times apart. computed by solving, not by drawing. The small-signal inductance of a sixty-turn winding on a core walked down from 400 amperes per metre, measured by pushing the excitation up and by pushing it down at each bias. They are never the same: 14.899 millihenries against 8.297 at zero bias, a factor of 1.796, and up to 1.796 across the sweep. The mechanism is that an operator held inside its own backlash contributes nothing to dB/dH, so a reversal is measured by whichever operators are still moving, and that is a different set in each direction. The single-valued curve the field already had is drawn above both, which is what it is: an optimistic reading of an object with two answers.

Two inductances at one current

A data sheet prints one L(i) curve and there are two. An operator sitting inside its own backlash contributes nothing to dB/dH, so a small excitation sees only the operators still moving — every one at the tip of a loop, none just after a reversal — and the same core at zero bias measures 14.90 millihenries pushed downward and 8.30 pushed upward, a factor of 1.80. The same split decides what a converter's ripple costs: held at the flux swing volt-seconds actually fix, a twenty-millitesla ripple costs twelve times more at a hundred and seventy-five amperes per metre of bias than at none, and above a hundred and eighty-three the question has no answer at all.

8 figures
The eddy term is f² below the skin-depth frequency and f^1.5 above it, both exactly. computed by solving, not by drawing. Eddy-current loss in a 0.35 millimetre lamination held at a mean flux of 1 tesla, against frequency, with the classical uniform-flux expression drawn beside it and the local exponent across the top. Below the frequency at which the sheet is two skin depths thick — 465 Hz here — the two agree and the exponent is 2.0000. Above it the flux is confined to a layer whose thickness falls as one over the square root of the frequency, and the exponent is 1.5000: three halves, exactly. It does not arrive there monotonically — it undershoots to 1.4847 at ξ = 3.28 and comes back up, which is the bounded cosine term the asymptotic statement drops. At 1.00 kHz the classical term is already 1.74 times the truth. The solve and the closed form agree to 7.5e-3 per cent across six decades.

The current inside the iron

Every core-loss law has an eddy term of the form d²f²B²/6ρ, and it is derived by assuming the flux is uniform across the lamination — an assumption that is a frequency and that the expression does not carry. Solved instead as a diffusion, the exponent is exactly 2 below the frequency at which the sheet is two skin depths thick, exactly 1.5 above it, and it undershoots to 1.485 on the way. For a 0.35 mm sheet the crossing is 465 hertz, so at a kilohertz the classical term is 1.74 times the truth and at ten kilohertz it is forty-seven times.

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The winding current a hysteretic core asks for, and the current the same circuit's single-valued core asks for. computed by solving, not by drawing. One netlist — a 22.6 volt peak sinusoid at 5.00 kHz through 4 ohms into a sixty-turn winding — marched twice, once with the core as a superposition of play operators and once with the single-valued saturating curve every earlier figure in this field used. The two currents differ in shape and not only in size: the hysteretic one leads the flux by an angle that is not ninety degrees, which is the whole of the core loss, and its peak is 59.45 milliamperes against 50.12. The energy the source delivers over a complete cycle is 17.19 microjoules for the loop and 2.5e-5 for the curve, which is zero to the resolution of the march. The slider takes the flux the drive demands from a twentieth of the material's saturation to most of it.

The core the solver has to remember

A saturating inductor's state is one number, because the current is a function of the flux linkage. A hysteretic core's is not: half an amp is one flux on the way up and a different flux on the way down, and there is no function of the current that returns the flux. So the march's state vector gains twenty-four more numbers, one per play operator, and the Newton loop is forbidden to touch them. The circuit and the bench then agree on the loss to three parts in ten million — one integral taken at two wires, the other inside the material.

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The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance.

The loss that depends on what it causes

Every thermal figure in this collection has had the power handed in. A ferrite's has no business being: its saturation flux falls with temperature, its permeability rises, and both move the loss. Closing that loop makes the temperature a fixed point rather than a product — and the fixed point has a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists.

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Four threshold densities, all matched to the same measured coercivity. computed by solving, not by drawing. Every hysteresis figure below this one used a uniform density of play-operator thresholds, and the reason given was that a uniform density makes the small-signal loss law come out cubic — Rayleigh's law, which is what soft ferrites do. That is a real constraint and it is a weak one. These four are all bisected to a measured coercivity of 9 amperes per metre on a deeply driven loop, and they need threshold spreads from 30.1 to 76.0 amperes per metre to get there. The slider moves the coercivity they are all matched to.

The distribution the bench cannot see

Six rungs of this ladder assumed a uniform density of operator thresholds, and the reason given was that a uniform density makes the small-signal law come out cubic. That reason is weaker than it looks: every density finite at the origin does the same. Matched on a measured coercivity, four distributions whose spreads differ by 2.5 times give the same remanence to 0.6 per cent, the same loop area to 1.0, and the same branch-slope ratio to 0.54 — and small-signal exponents from 2.81 to 3.81. The major loop cannot see the shape and the small signal cannot see anything else.

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The winding window solved in two dimensions, copper filling 100% of it. computed by solving, not by drawing. The grey frame is iron of infinite permeability, which in this formulation is a Neumann boundary — flux enters it at right angles and pays nothing. The thin curves are flux lines, which are contours of the vector potential, so equal spacing is equal flux. The copper is shaded by its own share of the loss. At 100 per cent fill the solved ratio is 16.280 against Dowell's 16.382, and the difference is entirely the flux that curls round the ends of the foils — which the one-dimensional model has no way to hold.

The assumption that is a geometry

Every alternating-resistance number this collection has computed for a winding rests on one sentence — the field is parallel to the layers everywhere — and the sentence has never been tested, because testing it needs a field. Solved as one, a portion of foils that fills its window returns Dowell's expression to 0.155 per cent; the same copper filling a quarter of it returns 9.00 against the expression's 16.38, and dissipates 0.528 watts a metre against 0.232. The ratio falls by 45 per cent and the loss more than doubles.

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The fringing field, and the turns standing in it. computed by solving, not by drawing. The slot on the left is the gap, cut through the centre leg to the core's own symmetry plane where the potential is zero. Flux crossing it does not stay in the slot: it bulges into the window and crosses the copper at right angles to the layers, which is the one direction Dowell's expression and every one-dimensional winding model assumes has no field in it. The turns are shaded by their own loss. The worst is turn 4, level with the gap, at 32.5 times its direct-current dissipation; the best is 1.39 times. Same wire, same current, same winding, and a spread of 23.4 between them.

The turns nearest the gap

A gapped inductor's flux does not turn a corner into the iron on its way out of the gap; it bulges into the window and crosses the copper at right angles to the layers. Four tenths of a millimetre from a one-millimetre gap, the worst turn of an eight-turn winding dissipates 37.5 times its direct-current loss and the winding as a whole 14.1 times. Move the same winding three millimetres further out and those become 2.9 and 2.4 — and the distance that governs it is 0.60 millimetres, which is not the gap length and does not scale with it.

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A porosity of 0.50, with the field the substitution smooths away. computed by solving, not by drawing. The flux lines between the conductors are the whole difference. The porosity substitution replaces this layer with a foil of the same direct-current resistance spread over the full breadth, in which the field is parallel to the layers by construction; here it is not, and it crowds between the turns. The solved ratio is 5.816 against the substitution's 6.212, 6.4 per cent apart. The copper is shaded by its own loss, which is what says the turns inside a layer are not alike either.

The wire that is not a foil

Almost no winding is made of foil, and the closed form for a winding's alternating-current resistance is about foils. The bridge between them is a substitution — squeeze the layer's conductors together, spread the result back across the breadth, divide the conductivity by the porosity — and it replaces a two-dimensional geometry with a one-dimensional one. Solved as a field it is exact where it must be, at a porosity of one, and 7.2 per cent high at a porosity of 0.40. It errs on the safe side, which is the half of the answer nobody could have assumed.

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How much larger a gap is than its own length says. computed by solving, not by drawing. A magnetic circuit prices a gap as g/(µ₀A) and everybody knows that is low, because the flux bulges out of the sides. The usual repair is to add one gap length to each dimension of the gap's area, which is the dashed line. The measurement is the solid one: at a 0.3 mm gap in a 6 mm leg the true correction is 1.100 and the rule offers 1.050, so the rule supplies 50 per cent of a correction worth 10 per cent of the inductance; at 1.7 mm it supplies 85 per cent. The rule is not wrong so much as it is a rule whose accuracy depends on the thing it is correcting.

The gap that is bigger than it is

A magnetic circuit prices a gap as g/µ₀A and everybody knows that is low, because the flux bulges out of the sides. The usual repair — add one gap length to each dimension of the gap's area — supplies half the correction at a 0.3 mm gap and 85 per cent at 1.7 mm, on a correction worth 10 per cent of the inductance at the first and 33 at the second. It is not a rule that is right or wrong; it is a rule whose accuracy is a function of the very thing it is correcting.

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Where a transformer's leakage inductance actually is. computed by solving, not by drawing. Both windings carry the same ampere-turns in opposite directions, which is the short-circuit condition a leakage measurement is made under, so the flux drawn here is the flux that fails to link the two — the leakage field, and nothing else. It is largest in the insulation between the portions, where the magnetomotive force is at its full value and there is no copper to be in. The energy in this window is 2.058 microjoules per metre, which is 4.116 microhenries per metre referred to the primary against a closed form of 5.213.

The inductance that is a shape

Leakage inductance is the one transformer parameter that belongs to the geometry rather than to the material: twice the magnetic energy in the window under equal and opposite ampere-turns, divided by the square of the current. Solved as a field it is 3.086 microhenries a metre against a closed form's 3.128 when the copper fills the window, and 4.608 against 6.255 when it fills half of it. Interleaving is worth 3.11 times and not the four it is quoted as, and the missing 0.89 is the insulation nobody puts in the formula.

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The temperature a part cannot come back from, and how long it takes to leave. computed by solving, not by drawing. The same fixed-point equation as the rung below, marched in time with a thermal capacitance rather than solved for its steady states: C dT/dt = P(T) − (T − T_a)/R_th, stepped adaptively on the temperature change because dT/dt goes through zero at each fixed point. Every trajectory starting below 191.1 °C returns to 88.8, however far above the operating point it began; every one starting above it leaves the material's range entirely, the closest in 0.2 minutes. The two nearest starts are 3.0 kelvin apart. The ignition temperature is a boundary in the STARTING CONDITION, and no steady-state analysis contains one.

The boundary that is a starting point

A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes.

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The one number the window does not move. computed by solving, not by drawing. Loss against foil thickness, at three window fills, with each curve's minimum located by a parabola through its three lowest points rather than read off the grid. The optimum sits at 0.654 skin depths at full fill, 0.708 at forty per cent, against the closed form's 0.663 — a drift of 8.3 per cent while the ratio the same winding carries moves by eighty. The alternating-current resistance at the optimum is 1.340, 1.351, 1.406, against four thirds. What did move is the loss it costs: 0.0555 watts a metre at full fill and 0.1383 at forty per cent, for the same current in the same number of layers.

The optimum that does not move

A foil winding has a best thickness — past it, more copper is more resistance — and at that thickness the alternating-current resistance is four thirds of the direct-current resistance, whatever the layer count. Both of those are one-dimensional results, and this ladder has spent three rungs finding that the one-dimensional picture is 82 per cent wrong about the resistance ratio. Solved as a field, the optimum drifts by 8.3 per cent between a full window and a quarter-full one, and four thirds becomes 1.34, 1.35, 1.41. The trade barely moves while everything it is made of moves a great deal.

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3 gaps of 0.40 mm, where one of 1.20 would do. computed by solving, not by drawing. The same total gap, the same turns, the same core, and very nearly the same inductance — cut into 3 instead of one. Each gap now drops one part in 3 of the magnetomotive force, so the field it throws into the window is that much weaker; and because the field's own energy goes as its square, the loss it causes falls faster than the field does. The winding dissipates 3.00 times its direct-current loss here against 6.52 with a single gap, and its worst turn 3.7 against 13.3.

The gap that is three gaps

Reluctances in series add, so one gap of 1.2 millimetres and three of 0.4 are the same magnetic circuit: the same inductance, the same saturation current, the same energy in the air. They are not the same field. Solved in two dimensions, the winding beside the single gap dissipates 6.52 times its direct-current loss and its worst turn 13.3; beside three gaps those are 3.00 and 3.7. And there is a best number of gaps rather than a monotone gain — past three, spreading them along the leg brings each one close to a different part of the winding.

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The winding window solved electrostatically, in two portions. computed by solving, not by drawing. The same cross-section the loss solve reads, read with ∇·(ε∇φ) = 0 instead. Two things are the opposite way round from the magnetic problem and both are the whole difference. The iron is now a Dirichlet boundary rather than a Neumann one — an earthed core is an equipotential, so the field meets it at right angles instead of running along it — and a conductor carries a prescribed potential rather than a prescribed current. The thin curves are equipotentials, which are contours of φ, so equal spacing is equal potential step and crowded curves are a strong field. The copper is shaded by the potential each foil sits at, which rises along the winding rather than being one number. Winding to winding this window is 926.9 picofarads a metre, and 89 per cent of the energy is inside insulation that occupies a fraction of the window.

The other half of the same window

The two-dimensional solve that settled what a winding's alternating-current resistance really is computed one of the window's two parameters and never mentioned the other. Read with Laplace instead of the vector potential, the same cross-section returns 926.9 picofarads a metre — and 89 per cent of that energy sits inside films that occupy 14.1 per cent of the window. The instrument agrees with a layered slab to three parts in ten thousand billion and converges on a real winding at order 1.34, and the reason for the shortfall is not the arithmetic but the corner of a conductor.

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The trade, in the plane where both halves of it live. computed by solving, not by drawing. Leakage inductance across, interwinding capacitance up, both solved on the same cross-section with the same cells. The faint diagonals are lines of constant leakage-times-capacitance, so a design action that runs along one of them has bought nothing and only moved where the energy is kept. Interleaving runs at slope -0.78, which is nearly along them: six sections cut the leakage 21.3 fold and multiply the capacitance 11.0 fold, and the product moves by 1.94. What it does change is the winding's characteristic impedance, 95 ohms down to 6.2 — a factor of 15. Thickening the interlayer instead runs at -4.8, steeply across the diagonals, and moves the product 5.2 fold over the same sweep. It is the cheaper action by that measure and it is not free either: the millimetre it spends is a millimetre of window that is not copper.

Interleaving is a choice, not an improvement

Splitting a transformer's windings into six sections divides its leakage inductance by 21.4 and multiplies its winding-to-winding capacitance by 11.0. The product of the two — which is what sets the frequency the part stops being a transformer at — moves by 1.94, and the resonance it decides goes from 1.804 to 2.515 megahertz for all that work. What interleaving really changes is the winding's characteristic impedance, 95.2 ohms down to 6.2, and nobody quotes it.

7 figures
The best foil thickness for 4 layers, for three currents with the same fundamental. computed by solving, not by drawing. The loss of a portion of 4 layers against foil thickness, with the loss weighted by the current in each harmonic rather than computed for one frequency. A sinusoid wants 0.6631 skin depths and lands at 1.3368 times the direct-current resistance — four thirds, the single-frequency constant, reproduced. A triangular ripple wants 0.6432, which is the same answer to within 3.0 per cent, so a winding carrying one needs none of this. A square current of the same fundamental wants 0.3838 — thinner by a factor of 1.728 — and lands at 1.8313, which is not four thirds and is not any constant the geometry knows. Building to the sinusoid's answer costs 16.8 per cent more loss.

The optimum a spectrum moves

The best foil thickness for a winding is derived for one sinusoid and quoted as a property of the geometry: a minimum at four thirds of the direct-current resistance, whatever the layer count. Weight the loss by the current in each harmonic instead and a square current of the same fundamental wants foil 1.728 times thinner and lands at 1.83, and a narrow pulse wants it 3.68 times thinner. Four thirds is a property of the current. The constant that replaces it for an ideal square edge is exactly two, and a real winding sits between them at a place its edge rate decides.

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A core walked from lossless to lossy, and the three straight lines it walks along. computed by solving, not by drawing. Loop area, measured coercivity and measured remanence against the threshold spread the model was handed, over five decades of it, driven sinusoidally to ±400 A/m. None of the three is an input: the area is ∮H dB round the marched loop, the coercivity is interpolated where the descending branch crosses zero, and the remanence is read at zero field. Over the lowest four decades all three are exactly proportional to the spread — 0.625961 joules per cubic metre per ampere-metre of spread, a coercivity 0.4499 of it and a remanence of 1.1304 millitesla per ampere-metre — and at 80 A/m the area is 2.69 per cent below the line and the remanence 16.42 per cent, because the pinned operators have reached the flat of the magnetisation curve. At zero the area is 9.8e-15 J/m³, which is the single-valued core of the anhysteretic model.

One dissipation, two exponents

The core this ladder built takes two numbers — a threshold spread and the fraction of the magnetisation that follows the field with no threshold — and seven rungs moved the first and left the second at 0.55 without ever saying why. The first decides how much loss there is: area, coercivity and remanence are all exactly proportional to it over four decades, at 0.625961 joules per cubic metre, 0.449775 and 1.130408 millitesla per ampere-metre of spread, the last two of which are closed forms. The second decides nothing about the loss at all — it is single-valued, so it contributes exactly zero to the loop area, to twelve digits — and it moves the Steinmetz exponent from 1.5042 to 2.9860.

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The same imbalance at five winding resistances, and the fixed point three of them reach. computed by solving, not by drawing. The peak flux density in each cycle, against the cycle, under a square drive whose positive half carries 1 per cent more volt-seconds than its negative one, for winding resistances of 0, 0.05, 0.15, 0.4, 1.5 ohms. With none the flux walks to 0.35 T in 15 cycles, which is the walk an earlier measurement of this drive found. With 1.5 Ω it settles at an offset of 5.67 millitesla and stays there, because the offset draws a direct magnetising current and that current's drop across the winding opposes the imbalance. The resistance at which the two outcomes change places is 0.1014 ohms, bisected on whether saturation is reached at all. Every curve carries the identical drive; the resistance is the only difference between them.

The walk that stops

A drive whose two half-cycles differ in volt-seconds walks the flux to saturation in a count of cycles, and the essay that measured it concluded that no amplitude puts the design inside a limit. That model has a stiff source and a winding of no resistance. With resistance in the loop the walk has a fixed point, held by an identity the material is not in — the mean magnetising current is the drive's direct component divided by the resistance, to seven parts in 10¹³ — and the boundary becomes a resistance rather than a time: 0.1014 ohms bisected, at a one per cent imbalance and half the volt-second limit.

7 figures
A 10 mH winding read at 1 kHz while a 40 mH winding coupled to it at k = 0.99 is loaded from a short to an open. computed by solving, not by drawing at 61 loads for each coupling. Open, the first winding reads 10.0000 mH, its own inductance. Shorted, it reads 199.00 µH — L₁(1 − k²), the leakage seen from this side, because the second winding's current cancels all of the flux the two share. The change happens around a load equal to the second winding's reactance, 251 Ω. At k = 0.5, 0.9, 0.99 and 0.999 the shorted readings are 7500 µH, 1900 µH, 199.00 µH, 19.99 µH.

The leakage a series reading cannot see

The series-aiding and series-opposing readings recover a coupling coefficient to fifteen figures on a netlist, and on a bench with a 0.1% meter they recover it to about 0.001 at every coupling — which is useless exactly where transformers live. At k = 0.999 the leakage, L₁(1 − k²), is two thousandths of the winding, and the series route leaves it 133% uncertain: 88 of 400 seeded benches report a negative leakage. Shorting the other winding reads the leakage directly, to the meter's own 0.1%, and recovers k 943 times more precisely at 0.999. The short is a short only where the shorted winding's reactance dwarfs its resistance by 10k/√(1 − k²), and the winding to short is the one with the longer time constant.

6 figures
The Q a 10 mH coil with 1.2 Ω of winding and 8 pF across it shows, against frequency. computed by solving, not by drawing, the coil's impedance at each frequency and its reactance over its resistance. The straight ωL/R line (pale) rises without limit; the coil's Q follows it at first, peaks at 11340 at 325 kHz — 0.5773 of the 563 kHz self-resonance, which is 1/√3 — and falls to zero at the resonance. The peak is 0.3849 of what the line reaches at resonance, 2/(3√3).

The Q that peaks before resonance

A coil's quality factor is quoted as ωL/R and drawn rising with frequency, and its own winding capacitance turns it over: Q peaks at the self-resonance over √3 and falls to zero at the resonance. Eighteen coils from 1 mH with 300 pF to 100 mH with 1 pF all peak at 0.5774 of their resonance, at 2/(3√3) of the line's value there. Skin effect moves the peak down toward the resonance over √5. And the falling Q is a property of the coil read alone: in a tank tuned at 0.8 of its self-resonance, the same coil gives the full ωL/R, 23,570 against the 8,485 a meter reads off the bare part.

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The Q of a 10 mH coil with 1.2 Ω of winding and 8 pF across it, with cores whose own Q is 1000, 300, 100 and 50. computed by solving, not by drawing: the core's loss a resistance ωL·Qc across the inductance, constant loss tangent, the coil solved at each frequency. Without a core loss the Q peaks at 11340 at 325 kHz, the self-resonance over √3. With the cores: 825.6 at 136 kHz for Qc = 1000, 274.9 at 94.0 kHz for Qc = 300, 95.9 at 66.2 kHz for Qc = 100, 48.7 at 52.8 kHz for Qc = 50. The peak falls to just under the core's own Q for a poor core and further under it for a good one, and moves down from the self-resonance over √3.

The Q a core allows

A coil's Q, limited by its winding and its own capacitance, peaks at the self-resonance over √3. Put a core with a loss of its own across the inductance and the peak falls and moves: with a core whose own Q is 100, the 10 mH coil that peaked at 11,340 near 325 kHz peaks at 95.9 near 66 kHz. The peak sits where x³ = a/2 — x the frequency over the self-resonance, a the core's Q over the winding's — and reaches the core's Q less three times the two-thirds power of a/2 of it. At that peak the core carries 97 per cent of the loss, so a better winding hardly moves it. A core whose loss is a fixed resistance instead puts the peak at the geometric mean of two corners, as high as half the square root of the core's resistance over the winding's; a loss tangent that rises with frequency drags the peak down to the material's corner.

5 figures
A power ferrite driven one way to 0.3 T: ungapped, every pulse starts at 90.8 mT; with a 0.1 mm gap, at 22.0 mT. computed by solving, not by drawing, marched on the hysteretic core. A ferrite with 90.6 mT of remanence and a coercive field of 38 A/m, 100 turns on 100 mm² and 60 mm of path, its flux raised to 0.3 T by a pulse and then released by letting the winding current fall to zero, cycle after cycle until the starting flux stops moving. Ungapped (solid) the release ends at H = 0 on the descending branch, 90.8 mT: each pulse can add only 209.2 mT. With a 0.1 mm gap (dashed) the current is zero where le·H + lg·B/μ₀ = 0 — the shear line (dotted) — at H = −29.0 A/m and 22.0 mT, under the bound μ₀·Hc·le/lg = 30.2 mT; each pulse can add 278.0 mT.

The flux the reset leaves behind

A transformer driven one way only — a forward converter, a single-ended pulse transformer — is reset each cycle by letting its magnetising current fall to zero, and zero current is not zero flux. Marched on the hysteretic core, a power ferrite with 90.6 mT of remanence driven to 0.3 T starts every pulse but the first at 90.8 mT, and keeps 34.9% of the swing a symmetric drive would get, where half is the most a one-way drive could; a square-loop core starts at 212 mT and keeps 14.7%. A gap shears the loop: the current is zero where H = −lg·B/(μ₀·le), so the start is bounded by μ₀·Hc·le/lg and falls onto the bound as the gap grows — 22.0 mT at 0.1 mm, recovering 69 mT of swing for 239 mA more magnetising current. The first pulse from a demagnetised core is the only one that starts at zero, and it is the one a bench test sees.

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A 100 µs pulse through the transformer whose band runs 400 Hz to 81.3 kHz: it rises in 4.03 µs, droops 21.6% along its top, and swings 21.6% below zero after it. computed by solving, not by drawing, twice. A 1 V, 100 µs pulse into the 1:1 transformer of the band essays — 10 mH windings, coupling 0.99, 50 Ω each side — from residues (solid) and from a trapezoidal march of the netlist (dots, within 0.34% of the plateau). The upper edge, 81.3 kHz, is a rise of 4.03 µs from 10 to 90%; the lower edge, 400 Hz, is a droop with time constant 1/(2πfₗ) = 398 µs, and the top ends at 0.7842 of the 0.4901 V plateau against exp(−T/τₗ) = 0.7778. When the pulse ends the output falls by the full plateau and goes negative by what the top lost, then recovers with the same time constant.

The band counted in rise times

A transformer's band is two frequencies and its pulse response is two times, and they are the same two facts. Through the 1:1, k = 0.99 transformer whose band runs 400 Hz to 81.3 kHz, a 100 µs pulse rises in 4.03 µs — 0.35 over the upper band edge — and droops 21.6% along its top with a time constant of one over 2π times the lower edge, 398 µs; solved by residues and by a march of the netlist, they agree to a third of a per cent. Coupling buys rise time and nothing else: from 30.8 µs at k = 0.9 to 422 ns at 0.999, while the droop's time constant moves from 414 to 396 µs. The longest pulse that droops no more than 10%, over the rise, is 0.048 times the ratio of the upper band edge to the lower — the band ratio counted in rise times, 10.4 at k = 0.99 and 98.8 at 0.999. And the droop is not lost: a transformer has no gain at direct current, so the output swings below zero afterwards by exactly what the top lost, and the two areas match to 0.08%.

5 figures
A shorted reading at k = 0.99: within 1% of the leakage from 279 Hz to 112 kHz, too high below through the winding's resistance and above through the leakage's resonance with 100 pF, at 1.13 MHz. computed by solving, not by drawing, in closed form, and by a netlist with a coupling element at three frequencies. The reading Im(Z)/ω at the driven terminals of a 10 mH and 40 mH pair, with the other winding shorted through its own 1 Ω, 0.5 Ω in the driven winding and 100 pF across it, against frequency, as a fractional distance from the leakage L₁(1 − k²) = 199.00 µH. Below the band the short does not cancel the shared flux. Above it the leakage and the capacitance resonate, at 1.13 MHz, and the reading rises to meet them. Band: 279 Hz to 112 kHz, a ratio of 402.1.

The window that moves

A shorted winding reads its own leakage only above the frequency at which the short really cancels the shared flux, and only below the leakage's resonance with whatever capacitance is across it. A tighter coupling raises the first edge as one over the root of 1 − k², and it was expected to pull the second down and close the window. It raises the second as well, by the same factor. So the band is 82 Hz to 36 kHz at k = 0.9, 279 Hz to 112 kHz at 0.99 and 2.81 kHz to 1.12 MHz at 0.9999, about four hundred times wide throughout. Its width is 0.01·L₂/(R₂k√(L₁C)), with the leakage nowhere in it, so what closes the window is the winding and not the coupling. Reading the real part as well is exact on paper and limited by phase.

5 figures

Networks, and how a solve is checked

A circuit has one answer and a matrix finds it. What matters is not that the answer exists but that it can be checked: the branch currents are rebuilt from the element laws and summed at every node, and the energy is counted twice. Networks with no answer are refused by name rather than returned as a large plausible number.

A network solved, and checked: a bridge, which no series-parallel reduction reaches. Node potentials from modified nodal analysis. The branch currents are then recomputed from each element's own law and summed at every node; the residual is 2.7e-16 of the largest current in the circuit, which is floating-point rounding and nothing else.

What a network answers, and how the answer is checked

A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number.

4 figures
A 10 kΩ + 10 kΩ divider, solved with its load. The unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does.

The divider, and the thing it does not know about

A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

6 figures
A 9 V source with 500 mΩ inside it. The ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

6 figures
The equivalent of six elements, and the heat it does not account for. computed by solving, not by drawing at 73 loads. The network reduces to 8.56032 V behind 599.768 Ω, obtained twice and agreeing to twelve figures. Into every load on the axis the two put the same voltage to 2.2e-16 of it and deliver the same power to 7.8e-16 — there is no load that can tell them apart. Inside, at a load of 5.62 kΩ, the real network is dissipating 49.168 mW and the equivalent claims 1.1349 mW, a factor of 43.3. With the port open the equivalent says nothing at all is being burned and the network is burning 48.03 mW. The slider moves the resistor bridging the two sources, which joins two ideal sources and so cannot change anything the port can reach: the equivalent stays 8.56032 V behind 599.768 Ω to the last bit at every setting, while the heat inside moves by two and a half times.

Exact outside and wrong within

Six elements reduce to one source and one resistor that no load can distinguish from them: the same voltage into every load across six decades, to the last bit of a double. The reduction is wrong about the heat by a factor of forty-three, and with nothing connected it says the network is dissipating nothing while it burns 48 milliwatts.

8 figures
A network the solver will answer, and should not be asked, into 0.01 Ω. computed by solving, not by drawing at 61 spreads. A hundredth of an ohm either side of a wire whose resistance is made smaller and smaller, into 0.01 Ω — an element written where the right answer is no element at all. The solution stays exact: 0.499999998422607 at the last spread before the refusal, against 0.500000000000000. What grows is the current-law residual, as the 0.96 power of the spread. The solve is refused at a spread of 2.5e+8, where the smallest pivot falls under 1e-12; the residual tolerance of 1e-7 would have been reached at about 3.3e+9. Two guards written for unrelated reasons, arriving within a decade of one another.

The answer that is perfect and absurd

A network with a wire written into it as a small resistance returns the right node voltage to fifteen figures, passes both verifications with a residual of two parts in ten to the sixteenth, and reports two hundred thousand amperes. The solver refuses it one decade further on, and by then it has been answering for eight decades.

8 figures
Superposition holds for the solution and not for its square. computed by solving, not by drawing, at 81 settings of the second source. Two ten-volt sources reach one hundred-ohm load through a hundred ohms each. Every node voltage and every branch current is the sum of the two single-source solves to 3.8e-16 of itself — superposition, exactly. The power is not: the difference is 2·Re(I₁·conj(I₂))·R to 1.0e-15, and at 0° between the sources and equal size it is 444.4 mW against the 222.2 mW that adding gives. Adding the two powers is within one per cent of the truth only when one source is below 0.00505 of the other. The slider is the angle between them: at 90° the two curves coincide to the last bit, and at 180° the true power falls to zero while the sum does not.

Two solves that add, and the one that does not

Every node voltage and every branch current in a linear network is the sum of the per-source solves, here to the last bit of a double at eighty-one settings. The power is not, and the gap is not a correction: two equal sources in antiphase put nothing at all into a load while adding their powers gives 222 milliwatts, and the sum is within one per cent of the truth only when one source is two hundred times the other.

8 figures
A passive network, read from each end — and the two readings are one number. computed by solving, not by drawing. A five-element RLC ladder with a current injected at one port and the voltage read at the other, then the two exchanged. With no controlled source the two readings agree to 4.9e-14 of themselves over four decades, which is the arithmetic's noise rather than a physical difference — the network cannot tell which way round it is being used. A mutual inductance keeps that: a 1 mH and a 4 mH winding at k = 0.7 give 0.0e+0, because the coupling puts the same entry in both halves of the matrix. A transconductance does not, and the departure is proportional to it with a fitted exponent of 1.000 over four decades — so there is no small amount of gain that is harmless. It passes the arithmetic's own floor at 0.781 femtosiemens, and the smallest transistor in this collection is nine orders above that.

The reading that does not care which way round it is

Inject a current at one end of a network and read the voltage at the other; then swap the two. A passive network gives back the same number — not a similar one, but the same to five parts in ten to the fourteenth over four decades of frequency, and for a coupled pair of windings the same double. Put a transconductance of eight tenths of a femtosiemens anywhere in it and the two readings part company.

8 figures
A divider of 4 equal 1.0% resistors, solved 3000 times. computed by solving, not by drawing. Every resistor drawn from its tolerance band and the divider solved, 3000 times. The worst case is ±1.000% — the part tolerance itself, and it does not improve when the divider is built from more parts — while the measured spread is 0.2944% and the worst of 3000 draws reached 82% of the bound. The root-sum-square, offered as though it were a standard deviation, is 1.70 of one here: for uniformly distributed parts it is √3 σ, a coverage of about 92%.

The tolerance that is not on any part

Four one per cent resistors in a divider give an answer whose worst case is one per cent, whose measured spread is 0.29 per cent, and whose root-sum-square bound — offered everywhere as though it were a standard deviation — is 1.70 of one. Adding parts does not move the worst case at all and shrinks the spread as one over their root, so the gap between the promise and the fact widens with every resistor. And the same arithmetic draws a boundary in tolerance rather than in frequency: an R–2R ladder is a twelve-bit converter only while its resistors are inside 0.14 per cent.

8 figures
Fourteen decades of imbalance, fourteen digits gone, and a matrix in perfect health. computed by solving, not by drawing. A Wheatstone bridge walked towards balance, with the relative error of the solved output against a closed form that cannot lose digits. The condition number of the nodal matrix is 505.0 at every imbalance and the smallest pivot is 2.0e-3 of the matrix norm — orders above the 1e-12 at which this solver refuses to answer at all. Neither number moves, and the answer still loses one digit per decade of imbalance, reaching 33% at δ = 1e-15. The bound drawn over it is the round-off divided by the imbalance, which the measurement stays under at every point. The third curve is the same closed form written as ½ − 1/(2+δ) — algebraically identical, and it loses its digits at the same rate, which is where the loss lives: in the subtraction of two nearly equal numbers, not in the matrix.

The matrix that is ill, and the answer that is not

Every other essay here treats the solve as exact, and it is not. A bridge walked towards balance loses one digit per decade of imbalance and has none left at a part in 10¹⁵ — on a matrix whose condition number never moves and whose smallest pivot stays four orders above the threshold this solver refuses at. A feedback amplifier does the opposite: the solver declines to answer at a gain of 10⁹, one decade after it returned an answer that was exact to the last bit.

8 figures
A difference quotient is best at a step of 1e-5, and is 1e+6 times worse at 10⁻¹¹. computed by solving, not by drawing. The worst disagreement between the adjoint network's derivatives and a central difference quotient of the same quantities, against the fractional step the quotient is taken with, on a 7-element LC ladder filter at 1000 Hz. The curve has a minimum because two errors pull opposite ways: the curvature the quotient neglects falls as the square of the step, and the digits its subtraction destroys rise as one over the step. The best it reaches is 1.4e-9, against the 2.3e-11 that the two-thirds power of the machine epsilon predicts. The exact route costs 2 solves against 15, and has neither error term.

Every derivative, and the one that is zero

How much does this response move if that capacitor is one per cent out? A difference quotient answers it one component at a time, in two solves each, and its best possible accuracy is four parts in a hundred million. Transposing the matrix and solving once more answers it for every component at once, exactly. Pointed at a claim the essays on ladder filters have made from the start and never tested directly — that a doubly terminated ladder filter's response is stationary in every element at its passband maxima — it returns two parts in ten billion, where the cascade realising the identical response returns 0.72.

7 figures
Sixteen more digits move the boundary by sixteen decades and leave it exactly where it was. computed by solving, not by drawing. The earlier essay's bridge, walked towards balance and solved twice: once in double precision and once with a pair of doubles carrying about 31 decimal digits, against a closed form that cannot lose any. The 33 per cent error at an imbalance of 10⁻¹⁵ becomes 7.0e-18 — so that loss was the arithmetic's and not the network's, which is what the earlier essay could not say. Each arithmetic's error is its own round-off divided by the imbalance, drawn as the two straight lines, so the second boundary is the first one moved by exactly the extra digits. The condition number is 505 in both cases and at every point, which is the diagnostic being blind twice over.

The digits the arithmetic did not have

Bounding the arithmetic of double-precision circuit solving finds two boundaries that cannot be attributed to it: a bridge with no correct figures left at an imbalance of 10⁻¹⁵, and a filter synthesis that stalls at order 14. An ill-conditioned problem stays ill-conditioned however many digits are used, and a well-conditioned one computed badly gets better — so adding digits is the experiment that tells them apart. The bridge's loss is entirely the arithmetic's. The synthesis's is mostly the data's, and doubling the digits makes it worse.

8 figures
Where one pole goes when each component is 5% high. computed by solving, not by drawing. A series R–L–C, its poles recovered by rooting the determinant, and the derivative of the upper one with respect to each element taken exactly from the two null vectors at the pole. The dashed lines are the first-order prediction for a 5 per cent change; the filled circles are where the root actually goes when the element is changed and the determinant re-rooted. The three directions are the argument: the resistance moves the pole along a circle of constant radius, because the natural frequency does not contain it — its normalised sensitivity has a real part of 3.1e-16. The inductance and the capacitance each carry exactly −½ of the radius, and imaginary parts that are exact negatives. At 5 per cent the prediction is out by 0.122 per cent of the pole's own magnitude.

The derivative of a root

An earlier essay turned one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder filter stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder filter's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.

8 figures
At a stationary point a tolerance has a mean, not a spread. computed by solving, not by drawing. Six hundred ladder filters with every reactance drawn independently from ±1 per cent, measured at the ripple peak and at a frequency between the peaks. Away from the peak the distribution is centred on nominal and 325 of 600 are above it. At the peak none is: the whole distribution lies below, with a mean of -0.0030 per cent and a worst case of -0.0140. A yield calculation that assumes a symmetric spread at a frequency that has a stationary point is wrong in both directions at once — it allows parts above a limit that cannot exist, and it misses that the whole batch has moved.

The tolerance that can only take away

At a doubly terminated ladder filter's ripple peak the first derivative of the magnitude with respect to every reactance is zero to ten digits, which an earlier essay measured and which says nothing about how much the response moves. This says how: every second derivative is negative, so of six hundred ladder filters built from one per cent components not one is above nominal, the mean has shifted rather than the spread having grown, and doubling the tolerance quadruples the damage instead of doubling it.

8 figures
Five tolerances, and the response moves in two directions. computed by solving, not by drawing. The eigenvalues of the relative second-derivative matrix of a fifth-order 0.5 dB Chebyshev ladder's magnitude at its lower ripple peak, over its five reactances. Two are of order one — -0.9473 and -0.8051, both negative — and the other three are 3.9e-9, which is zero at the precision the arithmetic has. So the quadratic form is negative semi-definite of rank two, and there is a three-dimensional subspace of component errors that the peak cannot see. The open circles are the same matrix computed by four re-solves per pair, sharing no adjoint arithmetic with the filled ones: they agree to parts in ten thousand on the two that are there and place the three zeros about two decades higher, which is the price of differencing a difference.

The three tolerances that do nothing

An earlier essay computed every second derivative of a ladder filter's magnitude at a ripple peak, found them all negative, and built six hundred ladder filters to argue that no combination of tolerances could raise the response. The whole matrix says so outright — and says something six hundred samples could not have found, because a sample of a five-dimensional box never lands on a three-dimensional subspace: two of the five eigenvalues are of order one and the other three are nine decades down.

8 figures
The most sensitive direction at one ripple peak is not the one at the next. computed by solving, not by drawing. The eigenvector of the largest eigenvalue of the relative curvature matrix, at each of the 2 ripple peaks of a 5th-order 0.5 dB Chebyshev ladder, drawn as its components over the reactances in order along the ladder filter. Every one is symmetric under the filter's own reversal and no two of them are the same direction: the largest overlap between any pair is 0.0603, which is 86.5 degrees apart. The eigenvalue that belongs to them grows from -0.9473 at 554.9 Hz to -12.005 at 897.9 Hz, a factor of 12.7, so the peak nearest the band edge is both the most sensitive place in the passband and sensitive to a different combination of parts. Each direction is confirmed by a second computation of the whole matrix — four re-solves per pair, sharing no adjoint arithmetic — which returns the same direction to 7.6e-7 radians.

The direction a response is most sensitive to

An earlier essay diagonalised a ladder filter's curvature at one ripple peak and read only the eigenvalues: rank two of five, three directions of nothing. The eigenvectors say what the two directions are, and doing it at every peak rather than one changes the conclusion. The most sensitive combination at 554.9 Hz and the one at 897.9 Hz are 86.5 degrees apart, the curvature that belongs to them grows from −0.556 to −75.4 across a ninth-order passband, and exactly one combination survives the whole band at every order — which turns out to be the ripple depth.

8 figures
Inside the passband the matrix is worst at the edge, and at no ripple peak. computed by solving, not by drawing. The condition number of a 5th-order Chebyshev filter's nodal matrix at every frequency from a hundredth of its cutoff to ten times it. It is 4.004e+3 at direct current, rises to 1.5868e+7 at 964.0 Hz — located by golden section rather than read off the sweep — falls to 4.970e+6 at 1520 Hz and rises again through the stopband. The maximum is above every ripple peak (the highest is at 897.9 Hz) and within 3.6 per cent of the half-power frequency: the worst-conditioned place in the passband is where the filter stops passing and starts blocking, which is a place the response curve has no feature at. It is a local maximum: above the band κ climbs again, reaching 2.702e+7 at ten times the cutoff, because the susceptances in the matrix grow with the frequency and κ counts them. Nothing about any of these numbers says whether a digit is actually lost anywhere.

Where the matrix is worst, and where the answer is not

When a solve stops being exact has been answered with two networks at direct current. Swept along the frequency axis, a resistive chain's nodal matrix has a condition number of 4.00×10³ at direct current and 1.59×10⁷ at 964 Hz — a maximum at the band edge, at no ripple peak and at no feature the response has. It predicts nothing. The solution vector is right to three units of round-off everywhere, the response taken out of it loses four decades into the stopband, and the same filter written at 400 kΩ instead of 10 Ω has a condition number 1.4×10⁹ times larger and returns the same twelve digits.

8 figures
The ratio of the two readings, drawn where it lives. computed by solving, not by drawing. The same five-element RLC ladder at 10 kHz, read from each end, with the quotient of the two readings plotted in the complex plane. A reciprocal network sits at the point 1. One transconductance moves it along a straight ray, 1 − gm·Z, whose direction is the phase of the single branch between the source's control node and its output node and whose length is gm|Z| — 62.83 Ω for the 1 mH inductor, 120 Ω for a resistor, 72.34 Ω for 220 nF, all at 10 kHz. A mirrored pair of transconductances stays at 1 to 2.3e-15; reversing one of them runs the ratio around the unit circle to 8.9e-16, where the two readings are the same size and differ only in phase. The straight-ray law needs the two nodes joined by exactly one branch, and a second path between them takes it away by a factor rather than by a percentage.

The reading that does care which way round

An earlier essay measured a network's departure from reciprocity and left its size as a constant — 62.8 per siemens, for that network at that frequency. It is not a constant and it is not the network's: it is 2πfL for the single inductor between the controlled source's control node and its output node, 62.832 ohms at ten kilohertz, and the ratio of the two readings is 1 − gm·Z to 8.8 parts in 10¹⁴ over ninety-nine readings. So 4.5455 millisiemens across a 220 ohm branch makes a ladder that transmits a hard zero forwards at every frequency at once, and 206.13 ohms back at ten kilohertz.

8 figures
The load that takes the most power is 3.67 Ω, and the open-circuit voltage over the short-circuit current is 3.97 Ω. computed by solving, not by drawing. The terminal characteristic of a photocurrent with a junction across it, with the power along it drawn on the same axes and scaled to fill them. The most power, 80.99 W, is delivered at 17.244 V and 4.696 A, which is a load of 3.67 Ω. The incremental resistance of the source there — the negative of the characteristic's own slope — is 3.67 Ω, the same number to four figures. A straight line drawn between the two end points has a resistance of 3.97 Ω and would promise 24.81 W. The ratio of the true maximum to that promise is 3.264, and its reciprocal is the fill factor, 0.8160.

The load a curve recommends

The maximum power theorem says to match the load to the source's internal resistance, and for a straight-line source the internal resistance, the slope and the open-circuit voltage over the short-circuit current are one number. On a photovoltaic panel they are three: the slope at the maximum is 3.6717 ohms, the load there is 3.6717, and the open-circuit voltage over the short-circuit current is 3.9709 — and a quarter of the open-circuit voltage times the short-circuit current, which is what a straight line would deliver, is 24.81 watts against the 80.99 that is there. The theorem survives with the incremental resistance in place of the internal one, and it survives only where the characteristic has a slope: on a supply whose current limit folds back, the most power is delivered at a corner.

7 figures
One gain takes the equivalent resistance from 5 kΩ through infinity to negative. computed by solving, not by drawing. A 5 V source drives a node through 10 kΩ; the node also reaches 10 kΩ whose far end is held at A times the node's own voltage. The Thévenin resistance looking into that node is r1 in parallel with r2/(1 − A), which the solve returns to a part in a billion without being told: 5 kΩ at no gain, 10 kΩ at unity where r2 takes no current at all, and unbounded at A = 2.00 where the two conductances cancel. Above that it is negative. The open-circuit voltage follows it, because the short-circuit current is 500.0 µA at every gain — a short across the controlling node leaves the dependent source nothing to be controlled by — so the open-circuit voltage is simply the short-circuit current times whatever the resistance is, and reaches 150.0 volts from a five-volt source inside the range drawn.

The resistor that is not made of the resistors

Exact outside and wrong within reduced six elements to one source and one resistor and found the resistor two ways that agreed to the last bit. Put a dependent source in the network and one of those routes stops working, because setting the sources dead kills the independent ones and leaves the dependent one where it is. On a bootstrap of two ten-kilohm resistors the Thévenin resistance runs from five kilohms through infinity to minus ten, the open-circuit voltage of a five-volt source reaches 225, and above one gain the equivalent's resistor is negative — which the netlist refuses to stamp, correctly, because a negative resistance is a controlled source and not a resistor.

6 figures
The circuit does not care which node is called zero, and the matrix does. computed by solving, not by drawing. The condition number of the nodal matrix for a 12-section chain, against which of its nodes was taken as the reference. The network, its elements and its physics are identical in every case — only a label has moved — and every branch voltage and branch current comes back the same to 1.1e-13. The condition number runs from 5.25e+4 at "n6" to 1.72e+5 at "n12", a factor of 3.28, which is 0.52 decimal digits of the arithmetic's own margin.

The node that is not in the circuit

Nodal analysis needs a node to call zero and no circuit contains one. Moving it changes every node voltage by the same amount and no branch voltage or branch current at all — to a part in ten to the fourteenth on a well-behaved chain. What it does change is the matrix: the condition number of a twelve-section chain moves by a factor of 3.3 with the reference, and on a star whose resistances span nine decades by 8.0. Measured against the same matrix solved in twice the precision, the worst reference costs four decimal digits of the answer, and it is the reference the condition number named before the error was looked at.

6 figures
Three dividers in a row, and where the error actually is. computed by solving, not by drawing. Three two-resistor dividers cascaded with nothing between them. The product of their ratios is 0.1250 and the solved output is 0.076923, 38.5% low at a staircase of ×1. Decomposed stage by stage with the rest of the chain in place — and the product of those three is the answer exactly — they are 0.3846, 0.4000, 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is at the front, which is the opposite of where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion, fitted exponent -0.993, so a decade a stage is within 4.88% and two decades within 0.50% — which is why a chain that has to be right is built as a staircase and not out of one value repeated.

The stage that is wrong is the far one

Three identical ten-kilohm dividers in a row give 0.076923 rather than the product of their ratios, 0.125 — 38.5 per cent low. Decomposed stage by stage with the rest of the chain in place, and the product of those three is the answer exactly, they are 0.3846, 0.4000 and 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is entirely at the front, which is the opposite end from where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion — fitted exponent −0.993.

6 figures
The current divider, and the resistance that is not in the branch. computed by solving, not by drawing. A current source into two parallel branches, the metered one 10.0 kΩ and the other 1.00 kΩ. The metered branch takes 0.090909 of the current, which is the OTHER branch's resistance over the sum; writing the subscripts the way a voltage divider writes them gives 0.90909, a different number at every ratio but one. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with the roles exchanged: one per cent of error at 111.1 Ω, which is that resistance over ninety-nine to 1.7e-12%. And the headline of the loaded divider holds in the dual too — two current dividers of identical ratio read 0.04762 and 0.09090 into one hundred-ohm meter — while a perfect ammeter reads them identically.

The branch the other resistance decides

A voltage divider's output is set by the resistance the output is taken across; a current divider's is set by the resistance the current does not go through. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with one word changed: one per cent at a meter resistance of R/99 where R is what the meter looks back into — and removing an ideal current source means OPENING it, so that R is the two branches in series, 11 kΩ here, not the 909 Ω of their parallel combination — a factor of twelve in the same construction on the same network.

6 figures
One rms value, and a peak that moves 41 per cent with a phase. A 10 V fundamental and a 3.33 V 3th harmonic, summed, with the harmonic's relative phase swept. The two are orthogonal over a period, so the root-mean-square value is the quadrature sum 7.45356 V at every phase — to 6.2e-15, which is the one case superposition allows and is a theorem rather than an approximation. The peak is not a sum of anything: it runs from 9.428 V to 13.33 V, a factor of 1.414, against a bound of 13.33 V that it reaches exactly at 180° — where this harmonic's own crest lands on the fundamental's, which for n ≡ 3 (mod 4) is that end of the sweep and not the other. At the far end it flattens the crest to 9.428 V, and there the peak has no closed form at all. So one signal has one heating and a headroom requirement that varies by 41 per cent, decided by a phase that no magnitude spectrum carries.

The peak that only has a bound

Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.

5 figures
Zeroing the dependent source too: an answer that does not depend on the gain. computed by solving, not by drawing. Two independent sources reach a node through resistors; a voltage-controlled source of gain −10 is driven from that node and fed back to it through 10 kΩ. Solved whole, the output is -25.0000 V. Solved by superposition with the dependent source LEFT IN each of the two solves it is -25.0000 V — the same number to 1.1e-16, so superposition is exact here as it is everywhere else in a linear network. Zeroing the dependent source along with each independent one gives nought volts at the output and 4.06250 V at the node, against the node's true 2.50000 V — 62.5 per cent out — because a controlled voltage source set to zero is a short circuit and the three solves are then of three different circuits. What makes it unmistakable is that the wrong answer is 4.06250 V at EVERY gain from one to three hundred: the route has deleted the only element that knows what the gain is, while the true answer moves by more than a factor across the same range.

The source that must not be zeroed

Superposition says to zero all the sources but one, and the care usually taken over that instruction is that zeroing is not deleting — a voltage source with no voltage is a short circuit and it stays in the netlist. A dependent source needs the opposite care: it must not be zeroed at all, because it is an element whose law involves another node rather than a source the theorem is about. Zeroed, it is a short circuit, and the answer that comes back is 4.0625 volts at every gain from one to three hundred — while the true answer moves from 3.82 volts to 0.206.

5 figures
Adding powers is wrong by the number of sources, when they are coherent. computed by solving, not by drawing, at five source counts. 8 ten-volt sources reach one hundred-ohm load through a hundred ohms each. With every source in phase the load takes 790.1 mW against the 98.77 mW that adding their separate powers gives — exactly 8 times, at every count, to a part in 10¹², because all 28 cross terms are positive and equal. With seeded random phases it takes 0.9865 times the sum over 64 runs, against a standard error of 0.125 — the cross terms average to nothing, which is the reason noise powers add and a root-sum-square is the right arithmetic for uncorrelated interference. The factor between the two limits is the source count itself, so it grows without bound: adding powers is not wrong by a bounded amount, it is wrong by how many things are being added.

The cross terms that outnumber the sources

With two sources the power in a load has one cross term, and adding the separate powers is wrong by a factor of two. With thirty-two there are four hundred and ninety-six of them, and the factor is thirty-two: 940 milliwatts delivered against the 29.4 that adding gives. With the same thirty-two sources at seeded random phases it is 26.9 milliwatts — the sum itself, inside the standard error of sixty-four runs, because every cross term averages to nothing. One identity, two limits, and the two are a phased array's gain and a noise sum's.

5 figures
4 branches read with 10 Ω ammeters: one at a time every reading is low, all at once they add up and the small branches are high. computed by solving, not by drawing. A 1 A source into 4 parallel branches of 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ, each read with an ammeter of 10 Ω. With one ammeter moved from branch to branch, each reading is low — by −0.989%, −0.902%, −0.099%, −0.010% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 990.275 mA rather than 1000.000. With an ammeter in every branch at once the readings add to 1000.000 mA and are wrong by −0.892%, +7.939%, +8.910%, +9.008%: the meter in the largest branch pushes current out of it, and the small branches take it.

The readings that add up and are wrong

Four branches of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ share a 1 A source. Read them one at a time with a single 10 Ω ammeter moved from branch to branch, and every reading is within one per cent of the truth, but the four add to 990.275 mA. Put a 10 Ω ammeter in every branch at once and the four readings add to 1000.000 mA exactly, satisfying Kirchhoff's current law to the last digit. Yet three of them are 7.9 to 9.0 per cent high. The check a reader would run on the readings is passed by the wrong set and failed by the right one.

6 figures
The R–2R ladder is the chain designed for the rest of itself: every node halves exactly, in place and only in place. computed by solving, not by drawing. A current source into an 8-bit R–2R ladder, each bit's 2R returned to a virtual earth. With the rest of the ladder in place every node splits the current arriving at it exactly in half, so bit k carries 2 to the −k of the source and the step at the major carry is exactly one least significant bit. A stage alone, into a short, would send a third on rather than a half; it is right because of what is connected to it. The same ladder with every resistor equal — the three-divider chain's habit — gives 0.61803, 0.23607 and then falls away by 2.6182 a stage, which is the square of the golden ratio, 2.6180, to the fourth figure.

The R–2R ladder, built for the rest of itself

Three equal voltage dividers in a row deliver 0.076923 of their input instead of 0.125. Their error sits in the front stage, at 0.3846, 0.4000 and 0.5000 per stage. The dual chain, three current dividers fed by a current source, gives the same three numbers to the last digit, and its repair runs the other way: each stage's resistance must fall, not rise. The R–2R ladder is the chain whose stages are wrong alone and right in place, halving exactly at every node. It still keeps its error at the front: a one per cent error in its first resistor makes an eight-bit ladder step backwards at the major carry, and the same error in its last shunt moves that step by 0.0034 LSB.

6 figures
With each source 0.8 of the one before, 32 coherent sources cost 8.99 times the sum of their powers, and never more than 9.00. computed by solving, not by drawing, at five counts and five amplitude ratios. Sources of 10 V, 10q V, 10q² V … each reach one 100 Ω load through 100 Ω; with all in phase the load's power over the sum of each source's power alone is (Σ|a|)²/Σa² to ten figures. Equal sources (q = 1) give the count itself, 2, 4, 8, 16, 32. Falling amplitudes give less, and a limit: for q = 0.8, 1.98, 3.77, 6.41, 8.51, 8.99, tending to (1 + q)/(1 − q) = 9; for q = 0.5, 3.000 against 3. The coherent penalty counts sources by their participation, not their number.

How many sources the cross terms count

N equal coherent sources on one load deliver N times the sum of their separate powers. Sources of unequal size deliver (Σ|a|)²/Σa² times it — solved on the network to ten figures — which is the number of equal sources that would carry the same penalty: a participation ratio. Amplitudes falling by 0.8 a source cap it at (1 + q)/(1 − q) = 9 however many there are; one large source and thirty-one at a tenth of its amplitude give 12.8, not 32, and for small sources the excess is 2(N − 1)ε — their amplitudes, not their powers, so thirty-one sources carrying a hundredth of the large one's power each turn 1.31 times its power into 16.8 times it when they line up. Random amplitudes take a fixed fraction off the count, (mean)²/(mean square): 0.75 for uniform ones. And with random phases none of it matters: the load takes the sum.

5 figures
Four 1% parts spread evenly across their band: the divider's error has a spread of 0.289%, and 8.6% of builds are beyond ±0.5%. computed by solving, not by drawing: a divider of four equal 10 kΩ resistors tapped at the middle, built 6000 times from seeded parts uniform across ±1%, each copy's output error taken from the ratio in closed form and the first four hundred also solved on the netlist, agreeing to 7e-16. The inset is the parts' own distribution; the histogram is the divider's error. Its standard deviation is 0.2894%, half the parts' 0.5780%, because each part moves the output by a quarter of its own error and four quarters add in quadrature to a half. 8.60% of builds fall beyond ±0.5%, which is half the worst case the printed tolerance allows.

The shape a tolerance does not print

A resistor marked 1% promises only that its value lies within one per cent. How the values fill that band is not printed, and it decides almost everything a design does with the part. Four populations that all honour ±1% — spread evenly, Gaussian with the band at three standard deviations, screened from a wider population, and sorted with the best parts removed — give a four-resistor divider spreads of 0.289, 0.168, 0.267 and 0.378 per cent, a factor of 2.26 under one label. Specify the divider at ±0.5 per cent and the yield is 91.4, 99.8, 93.7 and 85.3 per cent. The shape's signature fades as one over the number of parts, so it bites hardest in the circuits with fewest.

5 figures
Parts sharing 90% of their error: the divider's ratio spreads 0.091% and the chain's total resistance 0.559% — one from the same four parts, cancelling, and one adding. computed by solving, not by drawing, over 20,000 seeded builds: four 1% resistors whose errors share a fraction ρ = 0.9 of their variance — as parts from one reel, one wafer or one temperature do — built into a divider tapped at the middle. The ratio's error (solid bars) has a standard deviation of 0.0912%, half the parts' 0.577% times √(1 − ρ); the total resistance of the same four in series (outline bars) has 0.5587%, the parts' spread times √(ρ + (1 − ρ)/4). Correlation that makes a ratio better makes a sum worse, by the same arithmetic.

What a ratio does not see

Parts from one reel, one wafer or one temperature share part of their error, and a circuit that uses them as a ratio cancels exactly the part they share. Four 1% resistors sharing ninety per cent of their error make a divider whose ratio spreads 0.091 per cent and a series chain whose total spreads 0.559 — the same four parts, one answer six times better than independent parts would give and one nearly twice as bad. The ratio's spread falls as √(1 − ρ), the total's rises as √(ρ + (1 − ρ)/4). From −40 to 85 °C the chain moves by up to 0.76 per cent and the ratio by 0.135, inside the ±25 ppm/K the coefficients differ by. And a resistor array whose parts share 99.5 per cent of their error puts a divider inside ±0.1 per cent every time that four discrete parts manage it one time in four.

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The best reference is the middle of the chain below 2.37 kHz and ground above it — the node the capacitors return to. computed by solving, not by drawing. The condition number of the nodal matrix of an 8-section RC ladder — 1 kΩ in series, 10 nF and 1 MΩ to ground at each node — solved at s = jω with each of its 10 nodes in turn called zero, against frequency. Drawn: ground, the input, the middle node and the far end; the others lie between. At direct current n4 is best, near the middle as the earlier resistor chain found; from 2.37 kHz ground is — well below a single section's 15.9 kHz corner, where each capacitor's admittance is already 0.15 of a series resistor's conductance, and all 8 together about 1.2 of one.

The reference a sweep should keep

At direct current the best node to call zero in a resistor chain is its middle. Put a capacitor to ground at each node and solve at s = jω, and the best reference moves: the middle of an eight-section RC ladder below 2.37 kHz, ground above it, where every capacitor returns. So a solver that chooses its reference once for a whole sweep is choosing wrongly somewhere, as the earlier essay suspected — but the two wrong choices are not alike. Kept for the whole sweep, ground is never worse than 1.27 times the best condition number, a tenth of a digit. The middle node, best at direct current and kept, is 143 times worse at a megahertz, 2.2 digits. A fifth-order LC filter says the same: ground within 2.18 times throughout.

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Two 1% parts sharing 90% of their error: the divider's output spreads 0.129% at a half, 0.236% at ten to one and 0.258% at a hundred to one. computed by solving, not by drawing, over 20,000 seeded builds at each of seven ratios, every ratio computed exactly: a divider of two 1% parts, each with a uniform part's 0.577% spread and sharing a fraction ρ = 0.9 of it, with the top resistor N times the bottom. The output ratio's standard deviation (dots) against N, beside N/(N + 1)·√2·σ·√(1 − ρ) (line) and the same law for the other shared fractions (faint). N 1: 0.1286%; N 2: 0.1712%; N 5: 0.2145%; N 10: 0.2356%; N 20: 0.2477%; N 50: 0.2536%; N 100: 0.2581%. What is shared cancels at every ratio; what is not reaches the output with a leverage of N/(N + 1), which is a half for a halving divider and nearly one for a large ratio — so the same two parts make a ten-to-one divider 1.83 times worse.

The small leg of a large ratio

A divider of two 1% parts sharing ninety per cent of their error spreads 0.129 per cent at a half and 0.236 at ten to one — the same two parts, nearly twice as bad, because what they do not share reaches the output with a leverage of N/(N + 1) rather than one half. Correlation cancels at every ratio; the leverage does not. Built from equal units of one array, a divider of N to one spreads as σᵤ·√(N/M), so at any one area a ten-to-one ratio costs √10 against a half and needs ten times the units to match it, and the units it needs are the bottom leg's. And a ratio far from a half heats its own parts unequally: at 50 V across 11 kΩ the top of a two-part ten-to-one divider runs 46 K hotter than the bottom and the ratio moves 1,055 ppm with perfectly matched coefficients, where a chain of eleven equal units moves at most 29.

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The bootstrap with 1 nF at its node: one natural frequency, against the gain. computed by solving, not by drawing: the natural frequency of the network found as the root of its determinant at each of 31 gains. It is −G/C to a part in a million at every gain, where G is the Thévenin conductance the reduction gives: −200 thousand per second at A = 0, zero at A = 2 where the resistance passes through infinity, and +100 thousand at A = 3 — a growth rather than a decay. The gain at which the pole crosses into the right half-plane is the gain at which the resistance changes sign, and the capacitor's size moves the pole's speed and nothing else.

The pole that crosses with the resistance

A bootstrap's Thévenin resistance passes through infinity and turns negative at a gain of two. Put one nanofarad at its node and the network has one natural frequency, −G/C to a part in a million: −200 thousand per second with no gain, zero at A = 2, +100 thousand at A = 3. The gain at which the pole crosses is the gain at which the resistance changes sign — for this network. Loaded by a resistor it crosses at one plus R₂ times the sum of the conductances to ground, so a negative resistance is stable across a smaller load and not a larger one; a lag in the source slows the growth and does not move the crossing; a tank on the node turns the runaway into a 30.5 kHz oscillation once its coil is below about 1.5 Ω; and with an inverting source the resistance is positive at every gain and the node still oscillates beyond A = −16.

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The bootstrap reduced both ways: Thévenin's resistance diverges at A = 2, Norton's conductance crosses zero there. computed by solving, not by drawing at 121 gains, R₁ = R₂ = 10 kΩ. The Norton current is Vs/R₁, 0.500 mA, at every gain to a part in a billion (dashed, at 1), because shorting the node also shorts the bootstrap's input. The Norton conductance times R₁ is a straight line, 2 at A = 0 through zero at A = 2 to −1 at A = 3 (solid). The Thévenin resistance over R₁ is its reciprocal (faint, clipped at ±3): 0.5, 1 at A = 1, infinite at A = 2, −1 at A = 3. One network, two coordinate systems, and only one of them is singular exactly where the network is interesting.

The reduction that stays finite

The bootstrap's Thévenin pair diverges at A = 2: its resistance and its open-circuit voltage both pass through infinity and the solve that finds them refuses. Its Norton pair does not. The short-circuit current is Vs/R₁, 0.5 mA, at every gain to a part in a billion, and the conductance is a straight line through zero. A bench that drives a current and reads a volt loses the network for every gain from 1.992 to 2.008; a bench that drives a volt and reads a current sees it to a nanosiemens throughout. A 0.1% error in the gain moves the conductance by a constant 0.1 µS and the resistance by an amount that diverges at the pole — and four hundred bootstraps built from 0.1% parts to sit on it have conductances within ±0.3 µS of zero and resistances that are a median 10.6 MΩ, 196 of them negative.

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Five readings of one passive ladder network, each taken forwards and backwards. computed by solving, not by drawing; every reading built as its own netlist, never converted from another. Transimpedance (current in, open-circuit voltage out), transadmittance (voltage in, short-circuit current out) and the reverse voltage gain against minus the forward current gain all give a ratio of one, to 1.0e-13. The voltage gain forwards against backwards does not: 1.000 at 100 Hz, 2.129 at 1 MHz. Nor does the current gain.

The readings that come back the same

Reciprocity is usually stated as 'swap the source and the meter and the reading is unchanged', and that is true of some readings and false of others. On one passive ladder network the transimpedance, the transadmittance, and the reverse voltage gain against the forward current gain each come back the same to 10⁻¹³, while the voltage gain forwards and backwards differ by 13.1 times at 82.5 kHz — exactly z₂₂/z₁₁. A 600-to-50-ohm pad has voltage gains of 0.0426 and 0.511 and one power gain both ways. A controlled source breaks all three reciprocal pairs by the same factor.

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A terminated ladder network's output noise three ways: one solve per resistor, one solve from the output, and 4kT·Re{Z}. computed by solving, not by drawing, at 290 K. The line is 4kT·Re{Zout}. The dots sum each resistor's 4kT/R through its own transfer to the output, one solve per resistor. The rings get the same transfers from a single solve with an ampere injected at the output — reciprocity — and agree to 4.4e-16; the sum agrees with Nyquist's expression to 4.4e-16. 4.60 nV/√Hz at 100 Hz, 3.82 at 1 MHz.

The resistance colder than the room

In a passive network the noise at a port is 4kT times the real part of its impedance, and reciprocity is why: the transfer from each resistor to the port equals the transfer from the port to that resistor, so one solve from the output gives every resistor's share and the shares add to Nyquist's expression, here to 4.4 × 10⁻¹⁶. With gain in the network both fail. A 50-ohm termination made by feedback around an amplifier of gain A presents 50.000 Ω and has the noise of a 50-ohm resistor at 290/(1 + A) kelvin — 2.90 K at a gain of 99 — but only where the amplifier still has its gain.

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Three trackers finding a panel's maximum power from open circuit, against the number of readings each takes. computed by solving, not by drawing: the single-diode panel of 81.0 W at 17.244 V, every tracker starting at 0.995 of the 19.85 V open-circuit voltage. Perturb-and-observe with 0.2 V steps walks down and then oscillates about the maximum, within two parts in a thousand of it from reading 39. Newton on I + V·dI/dV, two readings a step for the current and its slope, is within a part in a million from reading 10. The fractional open-circuit rule needs one reading and lands exactly, because its fraction, 0.8686, was taken at this light.

How many readings a maximum takes

A panel's maximum power sits where I + V·dI/dV is zero, and a tracker finds it by reading the panel. From open circuit, perturb-and-observe in 0.2 V steps takes 39 readings to come within two parts in a thousand and then oscillates about the maximum forever, giving up 7|P''|δ²/24 of it; Newton's method on the condition itself takes 10 readings to come within a part in a million, the digits doubling at each step. Newton converges only from a window: above 15.82 V on a panel whose maximum is at 17.24, because below it the condition is nearly the short-circuit current and nearly flat, and one step leaves the characteristic. And the one-reading rule — a fixed fraction of the open-circuit voltage — loses nothing at the light it was set at and 17 per cent at a fiftieth of it.

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