The duty cycle that costs nothing

Both rungs below this one drove the core with a sinusoid, which is what a loss measurement is made with and what a catalogue coefficient is fitted on. Almost nothing drives a core with a sinusoid.

A converter applies a rectangular voltage. The flux is its integral, so what the core sees is a triangle — up during the on time at one rate, down during the off time at another — and the two rates are in the ratio of the two intervals. At a duty of five per cent that is nineteen to one.

The question this rung asks is what that costs, and the answer the model gives is nothing at all.

That answer is worth taking seriously in both directions. It is a strong prediction, it is exactly computed rather than approximately, and it is wrong about real cores by an amount that is itself the most useful number in the argument — because the rate-independent part has been subtracted off by construction rather than estimated, so what is left over is the rate-dependent part and nothing else.

Driving the core the way a winding does #

Everything in the two rungs below was driven in field: a stated H(t), and the material returned B. That is backwards for this question. A winding applies a voltage, the voltage integrates to a flux linkage, and the material then decides what field is needed to hold it. The flux is the independent variable and the field is the answer.

So the core is inverted at every step. That is a bisection rather than a Newton iteration, and the reason is the branch structure: the slope on a branch is μ₀ over most of a saturated excursion and a thousand times larger just after a reversal, so a Newton step started from the wrong side crosses branches and does not come back. Bisection on a monotone function cannot. The bracket is widened geometrically from the last accepted field, which is where the answer is, so the usual cost is four doublings and forty halvings.

The check that the inversion is right is that the loop reaches the flux it was asked for. It does, to one part in ten thousand at every duty on the slider — which matters, because the whole comparison is between loops that must be the same excursion and not three different ones.

One locus, five waveforms #

The measured areas, at ±100 millitesla:

Three parts in ten thousand across a nineteen-to-one asymmetry, and the residue is the marching grid rather than the model: a triangle with a five per cent leg gets forty-five of nine hundred samples on that leg, and the trapezoid on it is correspondingly coarser.

This is not an approximation and it is not a numerical coincidence. It is a theorem about the model. A play operator’s output is a function of the ordered sequence of fields it has been shown; two drives that visit the same fields in the same order produce the same operator states at every point, therefore the same B at every H, therefore the same locus, therefore the same area. Time does not appear in the model, so nothing in it can distinguish a fast leg from a slow one.

The same argument covers the sinusoid, which is why it is in the table. A sinusoid, a symmetric triangle and a nineteen-to-one triangle of the same peak flux all trace the identical major loop.

What the model does respond to #

The indifference is to the rate, not to the path, and the distinction matters because it is what separates this negative result from a claim that the waveform never matters.

A drive that reverses part-way and comes back traces a minor loop, and a minor loop has an area of its own that is added to the cycle’s. The area a curve cannot have measured one: five excursions of sixty amperes per metre at a bias of a hundred enclose 0.2371 joules per cubic metre each, against 12.481 for the major loop around them.

So a converter running in discontinuous conduction, or one whose switching edge rings, is drawing a different path and the model prices it. What it will not do is charge more for the same path traversed unevenly, and a duty cycle changes only the traversal.

Which is wrong, and the disagreement is the measurement #

Real cores are not indifferent to the waveform. A square-wave drive at a given peak flux dissipates appreciably more than a sinusoid at the same peak, and a strongly asymmetric one more again — tens of per cent, routinely, at the frequencies converters run at.

So the model is wrong here, and the interesting thing is how it is wrong. It is not out by a factor that could be absorbed into a coefficient, and it is not out in a way that varies with the material. It is out by exactly the amount that the loss depends on dB/dt, and it has no term in dB/dt at all.

That is a useful kind of wrong. The disagreement between this model and a measurement is a direct reading of how rate-dependent a particular core is, with the rate-independent part subtracted off exactly rather than estimated. Nothing about it is a fitting residual.

It also identifies the two mechanisms responsible, and both are already in this collection’s vocabulary. The eddy current of the exponent nobody put in is proportional to dB/dt and dissipates as its square, so a fast leg costs more than a slow one at the same excursion. And the excess loss that the same rung named — domain walls moving in bursts rather than smoothly — is rate-dependent for a different reason and by a different power. Neither can be a property of a locus.

What the standard correction actually asserts #

The usual way a catalogue’s sinusoidal coefficients are used on a rectangular drive is the improved generalised Steinmetz equation, which says that the instantaneous loss depends on the local rate of change of flux:

$$P = k_i\,\Delta B^{\,\beta-\alpha}\,\frac{1}{T}\oint \left|\frac{dB}{dt}\right|^{\alpha} dt$$

with $k_i$ chosen so that the expression reproduces the sinusoidal fit when the waveform is a sinusoid. It is a sensible and widely used thing, and it has a property that is easy to miss.

With α exactly one, it charges nothing. The integral of |dB/dt| around a cycle is twice the excursion whatever shape the path has — a triangle’s two legs cover ΔB each, however fast — so every duty cycle gives the same answer, and the equation agrees with the play model exactly.

The whole of its waveform dependence therefore lives in α − 1. At α = 1.2 a five per cent duty costs 1.232 times the symmetric case; at 1.4 it is 1.643; at 1.6 it is 2.331. Doubling the excess roughly doubles the charge.

That is worth stating in the form it deserves. Every asymmetric-waveform penalty in use is an assertion that the loss is rate-dependent, made with coefficients fitted on sinusoids under the assumption that it is not. The equation is not inconsistent — it is an interpolation, and it is a good one — but the quantity carrying its entire prediction is the frequency exponent, which the rung below showed to be a mixture ratio between two mechanisms rather than a material constant.

So the waveform correction inherits the range problem. An α fitted over a decade where the eddy term is negligible is close to one, and the correction it produces is close to nothing; an α fitted where the eddy term dominates is close to two, and the correction is close to a factor. Same material, same equation, two answers, and the fitting window decides.

A converter, with the numbers in it #

The abstraction is worth grounding, because the size of the disagreement decides whether any of this is a design question.

Take the six-cubic-centimetre core the rungs below use, at a hundred kilohertz, with the flux swung to ±100 millitesla. The loop area is 2.1006 joules per cubic metre, so the core dissipates 2.1006 × 10⁵ × 6.0 × 10⁻⁶ = 1.26 watts, at every duty cycle, according to this model.

Now apply the standard correction at an α of 1.4, which is a value a catalogue would print for a power ferrite. At a fifty per cent duty it changes nothing. At twenty per cent it adds 13.6 per cent, or 0.17 watts. At five per cent it adds 64.3 per cent, or 0.81 watts — which on a part running at 1.26 is the difference between a component that is warm and one that needs air moving over it.

The correction is not a rounding. And it is the entire waveform dependence of the design, carried by one exponent, which the rung below measured to be somewhere between 1.00 and 1.99 depending on which decade of frequency the fit was taken over.

A design that is sensitive to a factor of 1.64 and whose factor is carried by a quantity with that much range is a design whose loss figure should be quoted as a range. It usually is not.

What the volt-seconds still fix #

One thing does not move, and it is the same thing that did not move in the inductance the current decides.

The flux excursion is set by the applied volt-seconds and by nothing else. A boundary in volt-seconds established that λ = ∫v dt is the quantity a core’s limit is stated in, and the argument survives the material becoming hysteretic intact: the integral of the voltage across a winding is the change in its flux linkage, which is a statement about Faraday’s law and not about the material.

So a design controls ΔB exactly, whatever the core does. What it does not control, and what changes completely between the duty cycles in the table, is the current: the peak field is 50.22 amperes per metre at every duty above, because the flux excursion is the same, but the shape of the current waveform follows the loop rather than the flux, and a nineteen-to-one triangle in flux is not a nineteen-to-one triangle in current.

That distinction is what makes the flux-driven march necessary rather than convenient. Marching the current through a triangle would be marching the wrong quantity, and it would produce a flux excursion that varied with the duty cycle — which is exactly the confound the exponent nobody put in had to avoid on its amplitude axis and which this rung’s whole comparison would have been ruined by.

The experiment this suggests #

There is a measurement implied by all of it, and it is a cheap one.

Drive a core at a fixed peak flux and a fixed frequency, and vary only the duty cycle. Everything rate-independent is held constant by construction — same path, same extrema, same locus — so whatever the loss does is the rate-dependent part, isolated, with no fitting and no subtraction of a modelled term.

That is a better instrument than the usual one. The standard way to separate hysteresis loss from eddy loss is to measure P/f against f and read the intercept as the hysteresis term and the slope as the eddy term, which assumes the split is exactly linear-plus-quadratic and inherits every error in that assumption. A duty-cycle sweep assumes only that the loss has a rate-independent part, which is what a loop is.

The prediction is specific enough to fail. This model says the duty-cycle sweep is flat to three parts in ten thousand. A real core will not be, and the shape of the departure — whether it follows the $\oint |dB/dt|^{\alpha}$ form the standard correction assumes, or something else — is a statement about the mechanism that nothing in a frequency sweep can make, because a frequency sweep changes the rate and the number of cycles together.

Where the model is worth trusting #

The negative result is strong enough to say clearly, and it is not the same as “this model is no good”.

Three sentences, in the order they should be believed.

For the loop’s shape, its area at a stated flux, and the amplitude law, the model is doing the work: it is those quantities that gave the coercivity, the remanence, the wiping-out property and the exponent curve, and none of them needs a rate.

For a waveform correction, it has nothing to say and says nothing, which is better than a model that produces a plausible number. A model that returned a small duty-cycle penalty here would be inventing a rate dependence out of arithmetic, and the invented number would be indistinguishable from a measured one.

That refusal-by-construction is the same discipline as an exact answer to a different question’s: an answer that is exactly right about the wrong quantity is more dangerous than no answer, and knowing which quantity a model is exactly right about is most of what using it consists of.

The same shape appears in a band rather than an edge, where a switch modelled as ideal is right about the steady states and silent about the transition, and in how small is small signal, where a linearisation is exact about a derivative and says nothing about an amplitude. In all three the model’s silence is the honest part, and the failure mode is a reader who does not notice which question was answered.

And the boundary is a frequency, which is where this field usually ends up. Below the frequency at which the eddy term matters, this model is the whole of the core loss and the waveform genuinely does not matter. The rung below drew that crossing: ten megahertz for a ferrite at five ohm-metres, six hundred and thirty hertz for a silicon-steel lamination. A ferrite converter at a hundred kilohertz is four decades inside the range where the loop is the answer; a mains transformer is a decade outside it.

Which is a satisfying place for this ladder’s third rung to end, because it is the shape the whole collection is built around. The model is not approximately right everywhere. It is exactly right below a frequency, exactly silent above it, and the frequency is computable from the material’s own resistivity and the thickness it was cut into — two numbers on the same data sheet as the loss coefficients it is being used instead of.

The current inside the iron is where that frequency is measured, and it is lower than the phrase “eddy currents at high frequency” suggests: 465 hertz for a 0.35 mm sheet, above which the classical $d^2f^2B^2/6\rho$ is 1.74 times the truth at a kilohertz and forty-seven times at ten. So the duty-cycle independence measured here holds for a converter running at a few hundred hertz and does not for one running at a hundred kilohertz, which is every converter anybody builds.

That does not make the result useless; it makes it a reference. What a real core charges above the crossing is this essay’s rate-independent locus plus a rate-dependent term, and the correction the industry applies — hiding the whole waveform dependence in α − 1 — is a fit to the second one wearing the first one’s clothes. The exponent nobody put in is the argument for separating them: the frequency exponent of the rate-independent part is exactly one, a theorem rather than a fit, so any measured exponent above one is entirely the eddy term and the two can be told apart by their own scaling rather than by fitting a third parameter.

The result on this page is therefore best read as a statement about what a converter’s waveform can and cannot cost. Below the crossing, nothing: a triangle whose two slopes differ nineteen to one charges what a sinusoid of the same peak does, to three parts in ten thousand. Above it, everything the eddy term charges, which is a function of $dB/dt$ and so is very much a function of the duty. One mechanism is blind to the waveform and the other sees nothing else.