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

Transformer — where it appears

Two windings sharing a magnetic path, which presents the turns ratio to signals inside a band and something else outside it. The band has a bottom set by the magnetising inductance and a top set by the leakage and the winding capacitance, and what a designer buys is the distance between them.

Named by 17 essays across 5 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 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 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 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.

magnetics · Winding The crest factor is 13.4 and the winding is sized by 3.01. computed by solving, not by drawing. Three ratios of the same settled march, against the reservoir. The crest factor — the peak diode current over the load's direct current — runs 6.43 to 24.25. The form factor, which is the root-mean-square current over the same direct current and is what a winding heats by, runs 2.114 to 4.077. Its square is the copper loss against a winding carrying the direct current alone, and that runs 4.47 to 16.62. The first ratio is 3.04 times the second at 220 µF and 5.95 times at 4700, so quoting one of them tells a reader nothing about the other.

The current that sizes the transformer

A reservoir's crest factor is 13.374 at a thousand microfarads and the winding is not sized by it. The root-mean-square of the same marched current is 3.0069 times the load's direct current, so the copper dissipates 9.0417 times what it would carrying the direct current alone — and the two ratios diverge, from 3.04 apart at 220 microfarads to 5.95 apart at 4700. The expression for the mean output is wrong in three places whose signs differ, and at 313.9 microfarads they cancel to six microvolts while the ripple expression inside it is still 32.3 per cent high.

applied · Unregulated supply The same 100 mΩ in the winding instead of the capacitor: 1.317 V of ripple against 1.335 V, at the same crest factor of 11.15. computed by solving, not by drawing, marched with the diodes in the netlist: a 1000 µF reservoir behind a centre-tapped rectifier, with a series resistance from 1 mΩ to 1 Ω placed either in the capacitor or in each half-winding. The crest factor is the same in both places to two parts in a thousand at every resistance — 13.34, 13.26, 13.03, 12.49, 11.15, 9.069, 6.548 — because both limit the charging current alike. The ripple is not: in the capacitor it has a minimum and rises to 1.711 V at an ohm; in the winding it falls throughout, to 1.172 V, against 1.331 V with no resistance. At an ohm the winding costs 534 mV of mean output and the capacitor 380 mV. The diode's root-mean-square current at an ohm is 0.3449 A with the resistance in the winding and 0.3484 A with it in the capacitor.

The resistance that belongs in the winding

A reservoir capacitor's series resistance lowers the ripple to a minimum of 1.3303 V at 17 mΩ and raises it past that. The same resistance moved into the transformer's winding limits the peak current by the same amount — the crest factor agrees to two parts in a thousand at every value from a milliohm to an ohm — and the minimum is gone: the ripple falls throughout, to 1.172 V at an ohm against 1.711 V in the capacitor. The two resistances each carry a current the other does not, and that one asymmetry decides where a deliberate one should go.

applied · Unregulated supply 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 With a 0.30 µF coupling capacitor and a stiff driver the transformer's damping is 0.289: the pulse's tail crosses zero 8 times in sixty units of L/Rₗ. computed by solving, not by drawing, marched and checked against the closed form to 2.3e-4: a pulse two L/Rₗ long through a series coupling capacitor into a 1 mH magnetising inductance with 100 Ω reflected across it, from a stiff driver. The capacitor is 0.30 µF, x = CRₗ²/L = 3. The poles are those of 3.000e-10·s² + 1.000e-5·s + 1, a damping of 0.2887 and a natural frequency of 9.19 kHz. The top ends at 0.128 of the high-frequency gain, and after the pulse the tail crosses zero 8 times, with lobes above a thousandth of the pulse, in the sixty units drawn. The slider is the capacitor.

The ring a blocking capacitor buys

A series capacitor in front of a transformer stops its flux from walking: a 55%-duty drive that walks a bare core by 0.40 units of volt-time a period settles within a few dozen periods once the capacitor takes the drive's average. The price is a second-order high-pass with the magnetising inductance, and an inductance lets its poles go complex where two resistive coupling stages never could. Between equal resistances the damping cannot fall below 1/√2, and the poles are complex only for a capacitor between 0.17 and 5.8 times L/R². From a stiff driver the damping is √(L/C)/2R and has no floor: a capacitor large enough to make the droop small — the droop is 1/RC, the capacitor's and not the transformer's — makes the tail ring, from one zero crossing at a tenth of L/R² to thirty-five at a hundred.

instruments · AC coupling A 50 Ω source over 20.0 Hz–20.0 kHz: the best turns ratio is 10.7, giving 1.30 dB; the white optimum's 11.5 gives 1.31, and no transformer at all 13.2. computed by solving, not by drawing, the band integrals checked by quadrature. The drifting part of the essay before — 4 nV/√Hz and 0.6 pA/√Hz white, flicker corners at 10 Hz and 1 kHz — fed from a 50 Ω source through an ideal step-up transformer of ratio n, so it sees n²·50 Ω. The noise figure over 20.0 Hz to 20.0 kHz against n: least at n = √(√(E/I)/Rₛ) = 10.73, where the amplifier sees 5.76 kΩ, at 1.298 dB. The ratio that matches the white optimum, 11.55, gives 1.310 dB; the source connected directly, 13.23 dB. The slider is the band.

One turns ratio for a band

A 50 Ω source and an amplifier whose optimum is 6.67 kΩ want a step-up transformer, and the ratio that matches the white optimum, 11.5, is the one a design note computes. Over the audio band it is nearly right: 1.31 dB against 1.30 at the band's own best ratio, 10.7. Over 1 Hz to 100 Hz it is badly wrong — 9.2 dB where a ratio of 4.8 gives 5.4 — because the band's noise lives in its bottom decade, where the flicker has moved the optimum down. One ratio for a two-decade band costs at most 0.24 dB against a ratio that could follow the optimum. The transformer's winding resistance multiplies the whole noise factor by one plus the winding's resistance over the source's, not just its own share: a tenth of the source costs 0.41 dB where its own noise alone would cost 0.31. And 115 amplifiers in parallel are an ideal 10.7:1 transformer, noise for noise.

noise · Device noise 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 A 230 V transformer's primary switched on at 0° of the mains, secondary open. computed by solving, not by drawing, marched with the core's saturating flux law and checked at the voltage zero against the netlist march. Settled, the primary draws a magnetising current of 111 mA peak. Switched on at 0° it draws 87.6 A in the first three cycles — 791 times the settled peak. At the voltage zero the first peak is 87.6 A, 791 times; at the voltage peak there is no inrush at all.

The instant a transformer is worst at

A capacitor-input rectifier draws its worst inrush when it is switched on at the peak of the mains. A transformer's own magnetising branch is the other way round: switched on at the voltage zero, a 230 V primary worked at 1.4 T draws 87.6 A in its first half-cycle, 791 times its settled magnetising current, and at the voltage peak it draws nothing extra at all. The core's flux has to start from zero, the voltage does not, and how close the working flux sits to half of saturation decides the size.

power · Inrush 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 Paralleling a 4 nV/√Hz amplifier on a 50 Ω source, summed through 100 Ω from each: the best is 1.19 dB at 140 parts, against 1.14 dB at 134 parts with a perfect summing node. computed by solving, not by drawing: the noise figure from a 50 Ω source against the number of amplifiers in parallel, each 4 nV/√Hz and 0.6 pA/√Hz, with a perfect summing node (dashed) and with their outputs averaged through 100 Ω each into one node (solid), which adds a thermal 4kTR/N at the node. The summing resistors' noise falls as 1/N exactly as the amplifiers' voltage noise does, so their ratio is the same at every N, and a summing resistor comparable with the resistance an amplifier's voltage noise is worth is a constant factor on that term.

The resistors that sum a hundred amplifiers

A hundred amplifiers in parallel across a 50 Ω source are an ideal transformer of ratio ten, and the summing network between them and the output is where the comparison was left. Its resistors turn out to be the least of it. N resistors averaging N outputs make the noise of R/N, falling exactly as the amplifiers' voltage noise does, so the resistor equal to a 4 nV/√Hz amplifier is about a kilohm at every N, and at 133 parts 200 Ω costs only a tenth of a decibel. The stage that was forgotten is whatever reads the sum. Averaging leaves it alone, so a following amplifier as good as one of the parts costs 12.14 dB, and each part needs a gain of 26 to bring that under a tenth. Parallel amplifiers then match a transformer with one ohm of winding.

noise · Device noise 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.

Leakage inductanceCoupling coefficientDesign tradeoffMagnetising inductanceMutual inductanceClosed formWindingBandwidthConduction angleCrest factorInterleavingInterwinding capacitance

All concepts