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

Leakage inductance — where it appears

The part of a winding's inductance whose flux does not link the other winding, appearing in series with the ideal transformer. It sets the upper edge of the band and, at switch-on, turns the first charge of a reservoir into a resonance rather than an exponential.

Named by 13 essays across 4 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 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.

magnetics · Transformer 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.

magnetics · Transformer 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.

magnetics · Transformer The inductance divides the current by 7 and leaves the capacitor 29% above the mains peak. computed by solving, not by drawing. The first conduction of a rectifier whose transformer has a leakage inductance as well as a winding resistance, marched from an empty capacitor at the worst instant of the mains. The peak falls from 36.4 A at 20 µH to 5.3 A at 5000, and — unlike the winding resistance measured on the first cycle alone, which limits the current and leaves ∫i²dt exactly where it was — the inductance takes the energy down with it, from 0.370 to 0.073 A²s. What it costs is the second curve: the inductor's current cannot stop at the instant the two voltages are equal, so the capacitor overshoots to 21.86 V at 1000 µH — 28.6% above the 17 V peak of its own supply — and the diodes will not let the charge back out. The overshoot has an interior maximum, because past it the mains reverses before the ring has finished.

The inductance that limits, and lifts

Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

power · Inrush 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 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.

magnetics · Transformer 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.

magnetics · Winding capacitance 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.

magnetics · Winding capacitance 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 The supply's mean output against its load, with no leakage inductance, 0.3 mH and 1 mH in the winding. computed by solving, not by drawing, marched with the diodes in the netlist: 17 V peak into 1000 µF, 50 mΩ of winding throughout. Without inductance the output falls 1.69 V from 200 Ω to 25 Ω, an average slope of 3.43 Ω. With 0.3 mH it falls 0.62 V, 1.14 Ω, from a higher start; with 1 mH, 0.42 V, 0.77 Ω. An element that carries the load's current and drops no mean voltage makes the regulation better, not worse.

The inductance that flattens the supply

An unregulated supply's output resistance counts each resistance by the mean current it carries, and an inductance in the winding carries the whole mean current and drops no mean voltage — so by that rule it should count for nothing. Marched, it counts for less than nothing: 0.3 mH of leakage takes the output resistance at 100 Ω from 3.88 Ω to 0.67, and the mean output falls 0.62 V from 200 Ω to 25 Ω instead of 1.69. The effect has an interior optimum — 0.03 mH makes it slightly worse, 0.3 mH best, 3 mH back to 2.86 Ω — because the inductance first rings the reservoir up and then turns the rectifier toward choke input.

applied · Unregulated supply 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%.

magnetics · Transformer 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.

TransformerCoupling coefficientDesign tradeoffEnergy-storageInterwinding capacitanceModel rangeMutual inductanceClosed formField solutionMagnetising inductanceBandwidthLoading

All concepts