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 8 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

Every model this collection has drawn is right below a number or above one. A transformer is the first that 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 this site's two standing checks 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 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 rung below's 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

The rung below 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

Named alongside it

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

Model rangeReciprocityTransformerControlled sourceCoupling coefficientCrosstalkHomogeneous dielectricLeakage inductanceMode conversionModified nodal analysisMutual capacitanceSymmetric matrix

All concepts