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

Magnetising inductance — where it appears

The inductance a winding presents with nothing drawn from the other, and what steals current from a current transformer's reading at low frequency. It sets the low-frequency edge of a transformer's band, and for a current transformer it is the term that steals the reading below a few hertz.

Named by 10 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 A part in ten thousand of ratio, and half a degree that costs 18% of a power reading. computed by solving, not by drawing. The two errors of a 1000:1 current transformer into a 10 Ω burden, read off one solve of a netlist driven by a current source. The magnetising inductance puts a high-pass under the measurement with a corner at 0.456 Hz, so the ratio error falls as the frequency rises — and stops falling at 100 parts per million, which is 1 − k and is a coupling rather than a frequency. The phase error falls too and is worth far more: 0.522° at 50 Hz is 0.0142% of the current and 18.2% of the power at a power factor of 0.05. What bounds it from above is not on this plot: the burden voltage is integrated by the core, so the largest primary current is 916 A at 50 Hz and proportionally more at 400.

The ammeter that is not in the circuit

A shunt measures a current by putting a resistance in the circuit, and every objection to it follows from that. A current transformer puts nothing in the circuit at all — a thousand-turn secondary reflects twelve microhms into the primary — and charges for it in a different currency: no response at direct current, a ratio error that stops falling at one minus the coupling, and a phase error of half a degree at fifty hertz that costs eighteen per cent of a power reading at a power factor of 0.05.

instruments · Current sensing With a leak of a hundred periods the flux settles to a spread of 1.22% of the swing at β = 0.5 — 0.759 of what independent duty errors would give. Seeded: three runs of sixty thousand periods of the event map, 5 mV of threshold noise, β = 0.5, driving a magnetising branch whose flux leaks with τ = L/R. The stationary spread of the period-end flux in volt-nanoseconds against τ in periods, quoted here as a percentage of one half cycle's swing V·T/2 = 5493 V·ns: 1 → 0.146% (closed form 0.145%), 3 → 0.233% (closed form 0.232%), 10 → 0.401% (closed form 0.397%), 30 → 0.685% (closed form 0.673%), 100 → 1.25% (closed form 1.22%), 300 → 2.16% (closed form 2.10%). The closed form linearises one period, keeping the leak inside each half and the covariance −AB between consecutive periods' imbalances; dropping that covariance gives the upper curve, 1.61% at a hundred periods. For a slow leak the spread is V·σ·|B − A|·√(τ/T), the drawn asymptote.

The spread a winding resistance holds

A relaxation oscillator's threshold noise walks an ideal core's flux without limit. Put the winding's resistance back and the walk stops at a spread: 1.22% of a half cycle's swing when L/R is a hundred periods, growing as the square root of L/R, while a comparator skew's offset grows as L/R itself. The anticorrelation between consecutive duty errors is worth 24% of that spread at β = 0.5 and 62% at β = 0.2 — but only in a core slow enough to remember several periods. And the resistance that bounds the walk is a boundary in its own right: with a millihenry and a fifth of the swing to spare, an ohm tolerates 12.9 mV of noise, and a hundred picoseconds of skew needs at least 0.14 Ω.

applied · Hysteresis 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 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.

magnetics · Saturation 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 The second harmonic in each cycle of a transformer's inrush, as a share of the fundamental. computed by solving, not by drawing, the harmonics taken per cycle of the marched current. The first cycle carries 79% second harmonic and the second 93%. The share falls below the 15% a differential relay is commonly set to restrain at only in cycle 30, when the peak is 205 mA against a first peak of 88 A — the harmonic outlasts the inrush it identifies.

The harmonic that names an inrush

A transformer's differential relay sees an inrush as current going in and nothing coming out, which is what an internal fault looks like, and it tells them apart by the second harmonic. Marched cycle by cycle, an inrush starts at 79% second harmonic and stays above the usual 15% restraint for thirty cycles — long after its current has fallen from 88 A to 205 mA, because the current unwinds in a few cycles and the flux offset under it takes seconds. A fault's offset is a smooth exponential and peaks at 14.6% only when X/R is as low as three.

power · Inrush 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.

magnetics · Saturation 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

Named alongside it

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

Model rangeTransformerVolt-secondsCoupling coefficientSaturationFlux linkageInrush currentLeakage inductanceVolt-second limitBandwidthDesign tradeoffFlux density

All concepts