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

Flux density — where it appears

The flux in a core divided by its cross-sectional area, in teslas, and the quantity that saturates rather than the flux itself. It is what a volt-second limit is really a limit on, and a core is chosen by area for exactly that reason.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

The voltage a winding may carry, which is a volt-second limit read at a frequency. computed by solving, not by drawing. The dots are bisections on a marched flux — the voltage integrated sample by sample until the peak excursion reaches 0.35 T — and the line is N·Ae·Bsat·2πf. They agree to 0.001% over three decades, and the fitted slope is 1.000000: exactly proportional, because flux is the integral of voltage and nothing else. The quantity that belongs to the core is the 3.500 mWb-turn, which has no frequency in it. A transformer "rated for 50 Hz" is a transformer whose volt-second product was divided by 2π × 50 once.

A boundary in volt-seconds

A core saturates on the integral of the voltage applied to it, not on the current through it and not at a frequency. The quantity that belongs to the core is N·Ae·Bsat — 3.500 mWb-turn here — and it has no frequency in it at all. Everything a data sheet says about a transformer's frequency rating is that one number divided by 2πf once: measured on a marched flux, the voltage a winding may carry is proportional to frequency to a fitted exponent of 1.000000.

magnetics · Saturation A 1% imbalance, and the 64 cycles it survives. computed by solving, not by drawing. The upper panel is the peak flux density, marched cycle by cycle, under a square drive whose positive half is 1% larger in area than its negative half. It does not settle. It walks, by the same area every cycle, and reaches 0.35 T after 64 cycles — 1280 ms at 50 Hz — against a closed form of 63.7. The lower panel is the count against the imbalance, and it rises without bound and never becomes infinite. Halving the drive gives 128 cycles, which is exactly twice: reducing the amplitude buys time and not safety, and there is no amplitude at which this design is inside a limit.

The flux that walks

The previous essay's saturation limit is an amplitude, and an amplitude can be respected. This one cannot. A drive whose two half-cycles differ in volt-seconds by one per cent adds the same small area to the flux every cycle, so it reaches saturation after 64 cycles — and halving the drive gives 128, and a tenth of it gives 637. Reducing the amplitude buys time in exact proportion and removes nothing. There is no amplitude at which the design is inside a limit.

magnetics · Saturation Twenty nanoseconds of skew is 0.28% of duty, not 0.40. computed by solving, not by drawing. The duty cycle of a relaxation oscillator whose comparator takes longer to go one way than the other, marched, against the expression that lengthens each half cycle by its own delay. The two part company immediately and by a constant factor of about 1.42: during a delay the capacitor keeps charging past the threshold it already crossed, so the next half cycle starts further out and takes longer, and the two halves partly cancel. At 200 ns of skew on a 2585 ns period the duty is 52.805 per cent where the expression says 54.004. The open circles are the same quantity in closed form — the overshoot is V(1 − (1 − β)e^(−d/τ)) and the next half starts from it — which the march reproduces to parts in ten thousand.

The delay that is two delays

Modelling one comparator delay applied to both transitions makes the two half cycles equal by construction, and the fix is a change of one line. Made, the duty cycle moves by 0.28 per cent for twenty nanoseconds of skew rather than the 0.40 the obvious expression gives, more hysteresis improves the duty cycle and worsens the volt-seconds at the same time, and ten nanoseconds of skew saturates a hundred-turn core in 231 cycles.

applied · Hysteresis A 4.0:1 flux slope ratio and a sinusoid enclose the same loop to 0.00%. computed by solving, not by drawing. A core driven in flux rather than in field — the way a winding drives it, by integrating a rectangular voltage — around a triangle of ±100 millitesla at a duty cycle of 0.2, whose two slopes differ by 4.00 to one. The loop it traces encloses 2.1006 joules per cubic metre, against 2.1007 for a symmetric triangle and 2.1006 for a sinusoid of the same peak: the same number to 0.001 per cent. That is not an approximation, it is a theorem about the model — a rate-independent locus depends on where the flux went and not on how fast — and it is the prediction that real cores disagree with by tens of per cent. The disagreement is the measurement of what the model has left out.

The duty cycle that costs nothing

A converter drives its core with a rectangular voltage, so the flux is a triangle whose two slopes differ by nineteen to one at a five per cent duty. The play-operator core charges exactly the same for all of them — 2.1006 joules per cubic metre at every duty and for a sinusoid of the same peak, to three parts in ten thousand — because a rate-independent locus depends on where the flux went and not on how fast. Real cores charge tens of per cent more, and the standard correction hides its entire waveform dependence in α − 1, which is the one term a rate-independent model has none of.

magnetics · Magnetic loss

Named alongside it

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

Volt-secondsModel rangeSaturationDuty cycleMagnetic pathB h loopHysteresisLimit cycleMagnetic lossMarchingPropagation delaySteinmetz equation

All concepts