Series

Transformer — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · magnetics
  2. 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.

    part 2 · magnetics
  3. 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.

    part 3 · magnetics
  4. 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.

    part 4 · magnetics
  5. 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.

    part 5 · magnetics
  6. 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%.

    part 6 · magnetics

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