Circuit Response
every model has an edge, and the figure carries it
Concept

Mutual inductance — where it appears

The flux one winding links into another, entering a solve as an off-diagonal term rather than as a current or a dissipation. It is what makes a two-port out of two one-ports, and the coupling it implies is recoverable from two terminal measurements that do not contain it.

Named by 14 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

magnetics · Transformer 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.

magnetics · Mutual inductance Two tracks, and a far end that cancels exactly when the field is all in one material. computed by solving, not by drawing, on 12 coupled sections of a 100 mm pair terminated in 50 Ω at all four ends. A mutual capacitance injects a current proportional to dV/dt and splits it towards both ends of the quiet track; a mutual inductance injects a voltage proportional to dI/dt and drives the two ends in opposite directions. So the near end goes as Cm/Ct + Lm/Lt and the far end as their difference, with the same constant in front of both — measured here as 1.048e-2 either way, over a slider that moves the ratio by five times. The consequence is that the far end is not a smaller effect but a cancellation: at a ratio of one it is 7.52e-19 of the drive, which is zero to the last bits of a double, while the near end is 1.048e-3. That is why a stripline has no far-end crosstalk and a microstrip has some — what shows up there measures the field that is in air, not the spacing. The model is lumped and stops where it says: a section is one degree long at 50.0 MHz.

The far end that cancels

Two mechanisms couple two parallel tracks: a mutual capacitance injecting a current and a mutual inductance injecting a voltage. They add at the near end of the quiet track and subtract at the far end, with the same constant in front of both — measured here as 1.048 times ten to the minus two either way, across a slider that moves their ratio by five times. So the far end is not a smaller effect: when the two couplings are equal it is 3.5 times ten to the minus nineteen of the drive, which is zero to the last bits of a double.

lines · Crosstalk 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.

magnetics · Four-terminal 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.

networks · Reciprocity One part, two corners: 3.98 kHz to the common mode and 7.86 MHz to the signal. computed by solving, not by drawing. Two windings on one core with a coupling of 0.999, driven twice from the same netlist — once with the two conductors in opposition, which is the signal, and once with them in parallel, which is everything the cable picked up. The mode that goes the same way round both windings meets (1+k)L and is down three decibels by 3.98 kHz; the mode that goes opposite ways meets the leakage, (1−k)L, and is untouched until 7.86 MHz. The ratio is 1975, which is 2/(1−k) and contains no inductance at all. Neither number is computed here: both modes are driven and the answer is read.

The inductor one mode cannot see

Two windings on one core present a millihenry to a current that goes the same way round both and a microhenry to one that goes opposite ways, so the same component has a corner at 3.98 kHz and another at 7.86 MHz — a ratio of two thousand, which is 2/(1−k) and contains no inductance at all. It is bought to remove the conversion the previous essay measured, and five picofarads across each winding turn it over at 1.59 MHz and leave it worth 0.02 decibels by ten gigahertz.

lines · Common-mode choke The null is a V and not a bowl: one per cent of ratio error is 5.0e-3 of the near end. computed by solving, not by drawing. The same twelve coupled sections read at 10.0 MHz, with the ratio of the two couplings swept across the null rather than sat at one setting. The far end divided by the near end is |1 − r|/(1 + r) at every point — a straight-sided V through zero, first order in the departure with a coefficient of one half, and not a rounded minimum with a flat bottom. So there is no tolerance band: a ratio one per cent off gives 4.98e-3 of the near end and ten per cent off gives 4.76e-2, and the exchange rate between them is fixed. The upper trace is the near end over the same sweep, which moves by 11 per cent while the lower one moves through 14 decades — the two ends are the same coupling read as a sum and as a difference, which is why one of them has a zero in it and the other cannot.

How wide a null is

A far end at 3.5×10⁻¹⁹ of the drive is a statement about arithmetic until somebody asks how far the two couplings may differ before it comes back. The answer has no flat bottom in it: the far end divided by the near end is |1−r|/(1+r) exactly, so the null is a V and a ratio one per cent off returns 4.98×10⁻³ of the near end. On the axis a board is built to that is a difference of 0.0081 between the two modes' effective permittivities, out of 3.99 — two parts in a thousand, and 0.675 picoseconds of mode skew over a hundred millimetres.

lines · Crosstalk 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.

networks · Reciprocity 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.

lines · Return path 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.

lines · Return path 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.

magnetics · Mutual inductance 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.

lines · Crosstalk 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.

lines · Crosstalk 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.

magnetics · Mutual inductance

Named alongside it

The objects these essays reach for when they reach for this one.

CrosstalkTransformerCoupling coefficientLeakage inductanceMutual capacitanceTransmission lineHomogeneous dielectricModel rangeReciprocityClosed formCommon-impedanceControlled source

All concepts