Magnetics for Switching Converters

The energy to be stored, which is what really sets the size:

A useful sanity figure. Inductor size scales with stored energy, so a designer with any experience can guess the core from this number alone — and the area product formalises the guess.

The area product pairs the flux requirement with the window needed for the copper:

Read the two factors: \(LI_{pk}\) is the flux linkage the core must support, and \(I_{rms}\) divided by the current density is the copper cross-section the window must hold. A core is adequate only if it satisfies both, which is why they are multiplied.

Choose a core with \(A_e = 100\ \text{mm}^2\) and \(A_w = 200\ \text{mm}^2\):

Twenty per cent of margin, which is about right — the constants \(K_w\) and \(J\) are rules of thumb, not measurements.

The turns come from the flux limit at peak current:

Why \(B_{max} = 0.3\) T? Ferrite saturates around 0.4 T at room temperature and closer to 0.3 T when hot, so 0.3 leaves margin for the temperature rise found in Problem 2. Powdered iron would allow more; ferrite would not.

Now the air gap, which is what makes an inductor an inductor:

The core's own permeability does not appear, because the gap dominates the reluctance completely. That is deliberate: it makes the inductance depend on geometry rather than on a ferrite property that varies by \(\pm25\%\) and drifts with temperature.

The wire, and here the frequency intervenes. At 8 A and \(4\ \text{A/mm}^2\), 2 mm² of copper is needed — roughly AWG 14, 1.6 mm in diameter. But:

A solid conductor 1.6 mm across conducts only in a 0.2 mm skin, so most of the copper is dead weight and the AC resistance is several times the DC value. The wire has to be subdivided.

Litz wire. Each strand must be no thicker than about \(2\delta\) so that the field penetrates it fully:

And the strands must be transposed, not merely bundled, so that each occupies every position along the length — otherwise the outer strands still carry most of the current and nothing has been gained. That transposition is what distinguishes Litz wire from a handful of parallel wires.

The AC flux swing is not the peak flux. Core loss responds to the excursion, and the inductor's DC bias contributes nothing:

Thirty millitesla against a peak of 303 — a tenth. Using the peak in a Steinmetz calculation would overstate the core loss by a factor of \(10^{2.68} \approx 480\), which is the single most common error in magnetics work.

Scale the loss density from the quoted reference point:

Twenty-five times less than the reference, from a threefold reduction in flux. That steep exponent is why keeping the ripple small is worth so much in core loss — and why a converter with 30% ripple has far less core loss than one with 60%, out of proportion to the ripple itself.

The core loss in watts, for an effective volume of \(6\ \text{cm}^3\):

Twenty-four milliwatts. Essentially nothing.

The copper loss, from the winding resistance:

The comparison:

A ratio of thirty-seven to one, which is typical for a DC-biased filter inductor: the core carries a large steady flux but only swings a little, so it barely notices, while the winding carries the full DC current continuously. Effort spent on a better ferrite here would be entirely wasted.

The temperature rise:

Comfortable — and it validates the \(B_{max} = 0.3\) T choice of Problem 1, since a 19 K rise leaves the ferrite well below the temperature where its saturation flux collapses. Note that this rise is what \(J = 4\ \text{A/mm}^2\) was implicitly targeting; the current density rule exists to produce roughly this answer.

When the balance reverses. The copper-dominates result is not universal:

The general rule: if the AC swing is a small fraction of the peak, copper dominates; if the component swings through its full range every cycle, core loss does. Optimising the wrong one is the most common way to waste effort in magnetics.

Confirm the power from the stored energy, since in DCM the flyback is the energy pump of Set 20:

The specification is self-consistent. This is the check to do first: a flyback's power is fixed by its inductance, peak current and frequency, and if those three do not multiply to the required output, nothing downstream will fix it.

The primary turns, trying \(A_e = 80\ \text{mm}^2\):

But the turns ratio must be exactly 8:1, so \(N_1\) must be a multiple of 8. Rounding down to 40 with 5 secondary turns raises the flux:

Slightly over the 0.3 T target. Move to a core with \(A_e = 90\ \text{mm}^2\):

The gap:

Smaller than Problem 1's 1.37 mm because the stored energy is a fifth as much. A gap of a third of a millimetre is easily produced by grinding the centre limb or by a spacer in all three limbs — though a spacer gaps the outer limbs too, which increases the fringing field near the windings and worsens the proximity effect of Problem 4.

Why a flyback is gapped and a forward is not. The two use the core for opposite purposes:

A gap reduces the inductance, which lets the winding carry far more current before the core saturates — and energy storage is exactly what a flyback needs. A forward converter wants the opposite: the largest possible magnetising inductance, so that the magnetising current stays negligible and the reset burden of Set 22 stays small.

Where the energy actually sits:

Twelve to one, for a typical ferrite permeability and magnetic path length. So the gap holds 92% of the stored energy, in a third of a millimetre of air — and the ferrite, occupying a hundred times the volume, holds the rest. That is what "storing energy in the gap" means quantitatively.

Two consequences of the gap:

Which is why the winding should be kept a few gap-lengths clear of the centre limb's gap, and why distributed-gap materials such as powdered iron are sometimes preferred despite their higher core loss. And (ii) the leakage inductance of Set 22 rises with the gap, because flux that leaves the core near the gap may not link both windings — so the gap that makes the flyback work also feeds its snubber.

The skin depth at 65 kHz:

A quarter of a millimetre — about AWG 30. Any conductor thicker than about half a millimetre is already carrying current unevenly at this frequency.

But skin effect is the smaller problem. In a multi-layer winding, each layer sits in the magnetic field produced by every layer inside it:

So the outermost layer of a four-layer winding sits in four times the field of the innermost, and dissipates far more than its share. That is proximity effect, and unlike skin effect it gets worse with every layer added.

The AC resistance factor, from Dowell's analysis in its practical approximate form for \(h/\delta \le 1.5\):

With \(h = \delta\) and four layers:

Nearly three times the DC resistance, and therefore nearly three times the copper loss — from a winding whose wire gauge was chosen correctly for the current. The copper is there; the current is not using it.

Interleaving halves the effective layer count. Splitting the primary and placing the secondary between the halves — a P–S–P arrangement — makes the field return to zero in the middle:

Halved, for no extra copper and no extra core — only a different winding order. The mechanism is that the magnetomotive force builds up from zero to a maximum and back to zero across the window, rather than climbing monotonically, so no layer ever sees the full field.

How the factor grows with layers:

Roughly quadratic in the layer count. An eight-layer winding wastes seven times the copper loss it should — and interleaving it recovers only to the four-layer figure, so tall windings must be attacked with foil, Litz or a different core shape as well.

What interleaving costs. Nothing is free:

Read the second row both ways. Lower leakage shrinks the snubber loss of Set 22 — a substantial second benefit — but the increased interwinding capacitance couples common-mode noise straight across the isolation barrier, worsening conducted EMI. A screen between the windings fixes that and costs more window still.

The naive expectation. At fixed flux density, the area product scales inversely with frequency:

Which is the sales pitch for high-frequency conversion: eight times the frequency, nearly five times less magnetics volume.

Now check the core loss. The loss density rises steeply with frequency:

Applied to a volume that has shrunk by 4.62:

Six times the core loss in a component that is a fifth the size — so thirty times the loss density per unit surface area. It would melt. The naive scaling is not merely optimistic; it is thermally impossible.

Hold the core loss constant instead and solve for the flux density:

The flux density must halve. Ferrite's saturation limit is no longer the constraint at 500 kHz — core loss is, and it binds at half the flux the material could physically support.

Which reduces the size benefit:

A real reduction of 2.75 in volume rather than 4.62 — still worthwhile, but a good deal less than the arithmetic suggests. And this is before accounting for the semiconductor switching loss, which rose linearly with frequency in Set 2 and gets no such relief.

What else changes at 500 kHz:

Skin depth falls to under a tenth of a millimetre, so Litz strands must be finer and more numerous, and foil windings become attractive. Proximity effect worsens for the same winding, since \(h/\delta\) rises — and it enters Problem 4's expression to the fourth power.

The material choice moves too:

Notice the trend: high saturation and high frequency do not coexist. A material with a large \(B_{sat}\) gets there through a mechanism — metallic conduction — that also produces eddy currents, and every high-frequency material sacrifices flux density to suppress them.

The output inductor first. Following Problem 1's procedure:

About a third of Problem 1's requirement, since the stored energy is a third as much. A core with \(A_e = 50\ \text{mm}^2\) and \(A_w = 130\ \text{mm}^2\) gives 0.65 cm&sup4; — adequate.

Its turns and gap:

The transformer carries no DC bias — the magnetising current is a small parasitic — so it is sized by volt-seconds rather than by energy:

Note \(\Delta B = 0.25\) T, not 0.3: a forward transformer's flux swings from near zero up to the peak and back, so the whole excursion must fit below saturation with margin. A bridge transformer would swing \(\pm0.25\) T and need half the turns — the core-utilisation advantage of Set 22, Problem 4, made concrete.

The transformer's core loss is significant, unlike the inductor's, because its flux swings fully:

Scaling from 100 kW/m³ at 100 kHz and 100 mT:

Fifteen times the inductor's core loss, from a smaller core — entirely because \(B_{ac}\) is four times larger and enters to the power 2.68. This is the reversal that Problem 2's table predicted.

The complete magnetics budget:

Against Set 23's transformer allocation of 1.10 W for a 65 W adapter — the same order, which validates the budget. Magnetics typically account for 2 to 3% of throughput in a well-designed converter of this class.

Where to improve, and where not to:

The fourth row is the free one. Problem 4 showed interleaving halves the AC resistance for nothing but winding order, which would take the transformer copper from 0.50 to 0.25 W — a 19% cut in total magnetics loss at no cost in parts. The third row is a genuine optimisation: more turns lower \(\Delta B\) and the core loss with it, but lengthen the winding, so there is an optimum where the two are comparable.