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 10 essays across 2 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 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 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 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

Named alongside it

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

Leakage inductanceCoupling coefficientDesign tradeoffMutual inductanceConduction angleCrest factorInterleavingInterwinding capacitanceLoadingMagnetising inductanceRectificationReservoir capacitor

All concepts