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 4 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
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
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 boundary this ladder's second rung measured. 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 rung 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

Named alongside it

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

Coupling coefficientLeakage inductanceModel rangeTransformerVolt-second limitBandwidthBurden resistanceCurrent sensingCurrent transformerFixed pointFlux linkageInrush current

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