Circuit Response
every model has an edge, and the figure carries it
Power, and the part that does no work

The degrees a thermocouple cannot see

Every thermal answer in this collection has been one temperature, and a core makes its heat in its volume and loses it from a surface, so it has two. Solved as a conduction problem, a twenty-millimetre core in still air is 2.59 kelvin hotter in the middle than on the outside — 2.9 per cent of a ninety-kelvin rise, which is why the lumped answer has been good enough. Cool the same core on a plate and the gradient does not shrink; it grows to 3.37 kelvin and becomes 78 per cent of what is left.
15 min read 8 figures Every model has an edgeMeasured, not tabulatedTwo routes to a number

Assumes: The loss that depends on what it causes · The area a curve cannot have

Every thermal statement this collection has made is about one temperature. A thermal resistance carries watts from a junction to an ambient; a fixed point solves T = Tₐ + Rₜₕ·P(T); a ladder of three stages carries a pulse. All of them treat the part as isothermal — including the pulse the heatsink does not feel, whose three stages are three lumps rather than a continuum — and all of them were right to, for a reason that is worth measuring rather than assuming.

A core is not isothermal. It makes its heat in its volume — every cubic millimetre of ferrite carrying flux dissipates — and it loses that heat from its surface. So there is a gradient inside, the middle is hotter than the outside, and the two temperatures answer different questions: the surface is what a thermocouple reads, and the middle is what decides whether the material has run out of flux.

The boundary that is a starting point named this as the reason its ignition temperature is a slight overestimate. Here is the slight.

What is solved

Steady conduction with volumetric generation, through the thickness of the core, with a surface film at each face:

−k T″ = q(T), with −k T′ = h(T − Tₐ) at the faces.

For a uniform q that has a closed form: the surface sits q·d/2h above ambient and the middle sits a further q·d²/8k above the surface. Divide the two and the ratio is h·d/4k, so the internal gradient’s share of the whole rise is Bi/(Bi + 2) with the Biot number Bi = h·d/2k. One dimensionless number decides how much of a part’s temperature rise is inside it, and neither the loss nor the size appears in it alone.

The generation is not uniform, which is what makes this worth solving rather than evaluating. A ferrite’s loss falls with temperature — the exponent nobody put in is where that dependence is measured rather than fitted — so the hot middle of the core makes less heat than its cool faces, and the gradient comes out below the closed form. Forty-one cells, a tridiagonal solve, the loss density evaluated at each cell’s own temperature, and the whole thing iterated to a fixed point.

The temperature through a 20 mm core that makes its own heatcomputed by solving, not by drawing. Conduction with volumetric generation, solved on forty-one cells with a surface film at each face and the loss density evaluated at each cell's own temperature — so the middle of the core makes less heat than its faces do, which a closed form for uniform generation cannot express. The peak is 115.05 °C and the surface 112.45: a gradient of 2.59 kelvin, which is 2.9 per cent of the 90.0 kelvin rise. That share is Bi/(Bi + 2) — 3.0 per cent at a Biot number of 0.063 — so it is decided by how well the surface is cooled and not by how much heat is made.255075100-10-50510position through the core, millimetrestemperature, degrees Celsiuspeak 115.0 °Csurface 112.5 °Cthickness20 mmsurface coefficient25 W/m²KBiot number0.0625peak115.05 °Csurface112.45 °Cinternal gradient2.593 K…share of the rise2.9%Bi/(Bi+2)3.0%solved, then checked — generation in the volume, removal at the face2.9% of the rise is inside the core
Fig. 1 The profile through a twenty-millimetre core in still air. The peak is 115.05 degrees and the surface 112.45 — a gradient of 2.59 kelvin on a rise of ninety. The closed form for uniform generation gives 2.73, and the measured value sits five per cent below it because the middle is making less heat than the faces. Drag the surface coefficient.

The five per cent shortfall is the check. A solver that returned qd²/8k exactly would be one that had not noticed the material was in it, so the agreement being close and the direction being consistent is what says the coupling is real and small — the same shape of evidence as the calibration in the wire that is not a foil, where a substitution had to be exact at one geometry before its error elsewhere meant anything.

Why the isothermal assumption has been safe

For a ferrite in still air the numbers are small.

A twenty-millimetre core, thermal conductivity around four watts per metre per kelvin, natural convection at ten to thirty watts per square metre per kelvin: the Biot number is between 0.025 and 0.08, and the internal gradient is one to three per cent of the rise. Ninety kelvin of rise, two and a half kelvin of it inside. Everything this collection has computed with one temperature has been within three per cent of the right one.

The temperature through a 20 mm core that makes its own heat. computed by solving, not by drawing. Conduction with volumetric generation, solved on forty-one cells with a surface film at each face and the loss density evaluated at each cell's own temperature — so the middle of the core makes less heat than its faces do, which a closed form for uniform generation cannot express. The peak is 183.50 °C and the surface 181.64: a gradient of 1.86 kelvin, which is 1.2 per cent of the 158.5 kelvin rise. That share is Bi/(Bi + 2) — 1.2 per cent at a Biot number of 0.025 — so it is decided by how well the surface is cooled and not by how much heat is made.
Fig. 2 The same core cooled by nothing but still air at ten watts per square metre per kelvin. The rise is enormous — the peak is 183.5 degrees — and the internal gradient is 1.2 per cent of it. A part in serious thermal trouble is a part that is very nearly isothermal.

That is the first counterintuitive thing here, and it is worth stating plainly: the worse the cooling, the better the isothermal assumption. A poorly cooled part has a huge surface rise and almost all of its temperature difference is between the surface and the air.

And why it stops being safe exactly when the design improves

Now cool the surface properly.

The temperature through a 20 mm core that makes its own heat. computed by solving, not by drawing. Conduction with volumetric generation, solved on forty-one cells with a surface film at each face and the loss density evaluated at each cell's own temperature — so the middle of the core makes less heat than its faces do, which a closed form for uniform generation cannot express. The peak is 35.45 °C and the surface 32.14: a gradient of 3.31 kelvin, which is 31.7 per cent of the 10.4 kelvin rise. That share is Bi/(Bi + 2) — 33.3 per cent at a Biot number of 1.000 — so it is decided by how well the surface is cooled and not by how much heat is made.
Fig. 3 The same core, same loss, on a cold plate at four hundred watts per square metre per kelvin. The rise has collapsed to ten kelvin, which is the design working — and 31.7 per cent of that ten is inside the core.
When the inside of a core is worth solving for. computed by solving, not by drawing. The gradient between the middle of a core and its surface, as a share of the whole rise above ambient, against the Biot number. The dashed line is Bi/(Bi + 2), which is what the two closed forms give when they are divided — qd²/8k over qd²/8k + qd/2h — and the measurement follows it across two decades even though the generation is not uniform. At the poorly cooled end a ferrite core is at 2.08 per cent and the isothermal assumption is worth its name; at the well cooled end it is 78 per cent — and the gradient in kelvin has grown as well, from 2.38 to 3.37, because this material's loss falls with temperature and a cooler core dissipates more. Better cooling does not make a part more nearly isothermal; it makes the part's own gradient the thing that is left.
Fig. 4 The whole range, against the one number that decides it. The dashed line is Bi/(Bi + 2) and the measurement follows it across two decades of surface coefficient, sitting a few per cent below it throughout because of the temperature dependence. At a Biot number of two the internal gradient is half of everything left.

Better cooling does not make a part more nearly isothermal. It removes the surface’s share of the rise and leaves the core’s own, so the internal gradient becomes a larger and larger fraction of a smaller and smaller number — 2.08 per cent at a poor surface, 78 per cent at a good one.

And there is a second effect on top of it, which is a property of this material rather than of conduction. The gradient in kelvin grows too, from 2.38 to 3.37 across the same range, because a ferrite’s loss falls with temperature: cooling the core makes it dissipate more, and more generation in the same conductivity is a steeper profile.

So the two things move the same way and the share moves faster than either. A designer who improves the cooling and keeps computing with one temperature is getting worse at it as the design gets better.

The size of a core, and the thing that scales wrong

The Biot number carries the thickness linearly, so a thicker core is further from isothermal at the same cooling — but the interesting scaling is not that one.

Sweep the thickness at a fixed cooling regime rather than a fixed h. A five-millimetre core in still air is 0.24 per cent internal with a gradient of 0.17 kelvin; a thirty-five millimetre core is 6.09 per cent internal with a gradient of 7.2 kelvin. The gradient has grown forty-two times while the thickness grew seven, which is the d² in qd²/8k.

That is the argument for the shape of large magnetics, and it is not usually stated thermally. Beyond some size a core cannot be cooled through its surface at all, because the heat has too far to travel inside it, and the answer is to stop making it thicker: several smaller cores in parallel, a distributed gap, or a shape with more surface per unit volume. A design that doubles a core’s linear dimensions gets eight times the volume, four times the surface and four times the internal gradient — so the middle gets hotter even though the loss density has not changed.

And the loss density is what the whole thing is proportional to. Everything in this essay scales linearly with q, so a design running at half the flux density has half the gradient as well as a quarter of the loss. The Biot number does not move, because it has no loss in it at all — which is the whole reason it is the parameter worth quoting.

What the gradient is actually worth

Three consequences, in ascending order of how much they matter.

A thermocouple reads low. A probe on the outside of a core reads the surface. In still air that is 2.6 kelvin below the peak, which is inside anybody’s measurement uncertainty. On a cold plate it is 3.4 kelvin below a rise of four, which is most of the answer.

A loss measurement made from a temperature rise is out by the same amount, and in the direction that flatters the part: a core whose surface rose by ninety kelvin is assumed to be at ninety, its loss is inferred from a thermal resistance, and the material was actually at ninety-two and dissipating slightly less than the inference says. That is a measurement condition of the kind the edges that move with the room collects, attached to a quantity everybody quotes without one.

And the flux ceiling is decided by the peak. This is the one that changes an answer rather than a measurement — and the ceiling is the material’s own, which the area a curve cannot have computes from a marched loop. A ferrite’s saturation flux density falls with temperature, so the part of the core that runs out of flux first is the hottest part — and every ignition temperature, every runaway boundary and every survivable overload duration in the boundary that is a starting point was computed for a part at one temperature. The middle reaches the ceiling before the average does, so those boundaries are optimistic by the gradient: about three kelvin in still air, and rather more once the part is properly cooled.

Where a wound part's heat comes from, against its own temperature. computed by solving, not by drawing. Two mechanisms on one axis over a hundred and eighty kelvin. The core's own hysteresis loss falls the whole way, from 1.722 watts at room temperature to 0.289 at 203 degrees, because the material softens and the loop shrinks in the field direction. The winding's copper loss does the opposite and does it violently: the saturation flux density falls as the temperature rises, the field needed for a fixed flux swing climbs, and the magnetising current climbs with it — from 40.6 milliamperes to 23.3 amperes. So the total has a minimum at 187 degrees and a wall above it, and the wall is in the copper rather than in the ferrite.
Fig. 5 Why a few kelvin at the top of the range is not a few per cent of anything. The copper loss goes from 40.6 milliamperes’ worth to 23.3 amperes’ worth over the last twenty kelvin, because the saturation flux is falling past the swing the volt-seconds demand. Near that wall, three kelvin is a great deal.
The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance.
Fig. 6 And the two fixed points those three kelvin move. The ignition temperature at 191.1 degrees is where the loss curve crosses the removal line going up; a hot spot three kelvin above the average reaches it three kelvin of average temperature earlier.

Two temperatures, and which model wants which

The reason to keep both numbers rather than settling on one is that the questions a design asks are not all asked of the same temperature, and the site has now met three of them in three places.

The insulation’s temperature index wants the hottest copper. That is not in this essay at all — it is the turns nearest the gap’s worst turn, twenty-seven times its neighbours’ loss, in the middle of a winding with the longest thermal path out.

The material’s flux ceiling wants the hottest ferrite, which is the peak measured here.

And the loss budget wants the average, because the total dissipation is what the thermal path has to carry and an average is exactly what an integral over a volume produces.

Using one of them for all three is the ordinary mistake, and the direction it errs in is different in each case: the average understates the insulation’s temperature by a lot, understates the material’s by a few kelvin, and is exactly right for the budget. A design that computes one temperature and compares it against three limits is passing two of them on a technicality.

The three lengths in the problem, and which one is the core’s

It is worth separating the quantities, because “thermal resistance” is used for all of them and they do not behave alike.

The film, 1/hA, which is the surface to the air. It is what a heatsink or a fan changes, it is usually the largest term, and it is the only one a designer has much control over.

The conduction inside the part, d/kA scaled by whatever fraction of the path the heat travels. For distributed generation that is d/8kA for the middle-to-surface difference, and it is fixed by the material and the size.

And the spreading resistance where the core meets whatever it is mounted on, which is neither of the above and is not modelled here at all.

The Biot number is exactly the ratio of the first two, and the reason it is the useful parameter is that it says which of them the design is currently limited by without needing either in absolute terms.

One cycle at 1.00 Hz, 2% duty, 5 W average. computed by solving, not by drawing. The same average power delivered as a pulse of 250 watts for 2 per cent of each cycle. The thermal path is a three-stage ladder marched from its own steady state — the heatsink's time constant is eight minutes and the period is 1000.0 ms, so marching to the steady state would be fifty million steps to answer a question about one cycle. The junction reaches 349.0 K above ambient against a steady-state answer of 48.5 K: an excess of 620 per cent. The case, one stage down, swings 11.495 K against the junction's 308.9 — a factor of 27 — so a case-temperature measurement understates the junction's excursion by that much and reports something close to the steady state instead.
Fig. 7 The same separation in time rather than in space, from the transients field. A three-stage ladder carrying a two per cent duty pulse puts the junction 349.0 kelvin above ambient against a steady-state 48.5, while the case one stage down swings 11.5 against the junction’s 308.9 — a factor of twenty-seven. A case measurement says nothing about a junction for the same reason a surface measurement says nothing about a middle.
Where the heat is, turn by turn, 0.4 mm from a 1 mm gap. computed by solving, not by drawing. Every turn carries the same current and every turn is the same wire, so a winding's loss is usually quoted as one number. It is not one number here. Turn 4 dissipates 37.5 times its direct-current loss and the turns at the ends dissipate 1.41 times — a spread of 26.6 across a winding whose data sheet has one resistance in it. The dashed profile is the same winding moved to four millimetres from the wall, where the spread is 1.10. A hot-spot temperature computed from an average loss is computed from a quantity no turn has.
Fig. 8 And the other hot spot in the same window, which is not in the core at all. Four tenths of a millimetre from a gap, one turn of eight dissipates 37.5 times its direct-current loss and the end turns 1.41 — a spread of 26.6 in a winding whose data sheet has one resistance. The copper’s non-uniformity is an order of magnitude worse than the core’s.

What a bench measurement can and cannot resolve

The gradient measured here is between two and ten kelvin, and the honest question is whether an instrument could tell.

A thermocouple bonded to a core surface is good for about a kelvin against the surface it is bonded to, and considerably worse against the material, because the bond has its own thermal resistance and the probe conducts heat away along its own wires. Two and a half kelvin is at the edge of what that arrangement resolves.

Drilling a core to put a probe in the middle changes the object: a hole is a place where no flux flows and no heat is generated, and it interrupts the conduction path the measurement is about. An infra-red camera sees the surface only, which is the temperature that was never in question.

So the internal gradient belongs to the class of quantities this collection keeps meeting — a real effect, of a known size, that no ordinary instrument reads directly and that therefore has to be computed if it is to be known at all. The two-routes discipline applies with more force rather than less: the closed form for uniform generation and the coupled solve agree to five per cent, and the five per cent is itself explained by the material’s own coefficient.

Where the model stops

One dimension. A core is solved here as a slab of stated thickness, which is the right model for the centre limb of an E-core between two flat faces and the wrong one for a toroid or a pot core, where the heat leaves in two directions and the effective d is smaller.

No winding. The copper’s own loss is generated somewhere else and removed by a different path, so it is left out of the conduction problem entirely and only the core’s share is used — the share the core the solver has to remember separates from the copper’s inside a marched netlist. That is conservative for the core’s gradient and it means the numbers here are not a whole part’s thermal model.

And the surface coefficient is a number. Natural convection depends on the temperature difference it is driving, roughly to the power of a quarter — the same shape of dependence the loss that depends on what it causes has to resolve on the generation side — so h is not constant across the range swept — which matters for the absolute temperatures and not for the Biot relationship, since that is evaluated at each h separately.

The number worth carrying

Bi/(Bi + 2), and a Biot number between 0.02 and 0.08 for a ferrite in air.

The habit that goes with it is the one about when to stop simplifying. The isothermal assumption is not merely adequate for a ferrite core in still air; it is adequate by a factor of thirty, and solving a profile to find that out is the right use of a model rather than a waste of one. What the same model then says is where the assumption expires — and it expires in the direction of better engineering, which is the direction nobody re-checks an assumption in.

What the gradient does to the loops that were closed on one temperature

A lumped temperature is not merely a simplification here; it is the state variable of every thermal feedback result in this collection, and a gradient means those loops were closed on the wrong number.

The loss that depends on what it causes closes the loop: a ferrite’s saturation flux falls with temperature and its permeability rises, both move the loss, and the temperature becomes a fixed point rather than a product — with a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists. Every one of those numbers is a solution of an equation in one temperature.

The boundary that is a starting point then marches the same equation rather than solving it, and finds trajectories three kelvin apart on either side of 189.6 degrees going opposite ways. Three kelvin is the same size as the gradient measured on this page — 2.59 kelvin in still air and 3.37 on a cold plate — so the initial condition that decides whether a core runs away is comparable with the difference between the temperature at its centre and the one at its surface.

Which is the sharpest consequence of the result here and is worth stating as a question rather than as a correction. A runaway starts where the material is hottest, and a thermal model with one temperature in it does not have a hottest place. Whether the ignition threshold should be compared against the centre temperature or the mean is a modelling decision that none of those essays had to make, and the gradient measured here is how much it is worth.

The direction of the second result is what makes this worth an essay rather than a footnote. Cooling the core on a plate does not shrink the gradient; it grows it, to 3.37 kelvin, and takes it from 2.9 per cent of the rise to 78 per cent of what is left. So the lumped model is best where the cooling is worst, and the case in which somebody has taken trouble over the thermal design is the case in which the model they are checking it with has stopped being adequate.

That is the same inversion the sensor inside its own answer finds on a much smaller object: a junction used as a thermometer under-reports every change in ambient by 9 584 parts per million, because its own dissipation falls as the reading rises — an error that a calibration at one temperature cannot remove and that gets proportionally worse as the mounting gets better. In both cases the quantity being neglected is a difference between where the heat is made and where it is measured, and in both cases improving the heat path changes the difference rather than removing it.

Part 5 on thermal feedback

One argument about Thermal feedback, and one of 6 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

The temperature a part cannot come back from, and how long it takes to leave. computed by solving, not by drawing. The same fixed-point equation as the rung below, marched in time with a thermal capacitance rather than solved for its steady states: C dT/dt = P(T) − (T − T_a)/R_th, stepped adaptively on the temperature change because dT/dt goes through zero at each fixed point. Every trajectory starting below 191.1 °C returns to 88.8, however far above the operating point it began; every one starting above it leaves the material's range entirely, the closest in 0.2 minutes. The two nearest starts are 3.0 kelvin apart. The ignition temperature is a boundary in the STARTING CONDITION, and no steady-state analysis contains one. The boundary that is a starting point Part 4 — A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes. What one temperature costs the loop gain of a part that has a gradient. computed by solving, not by drawing. The thermal loop gain of a 30 mm core, solved as a body with its own internal temperature profile and again as a single lump at that profile's mean, against the Biot number. Both are negative, so the core is a stabilising feedback either way — but the body's loop is the more negative of the two at every point, by 3.0 per cent at a Biot number of 0.108 and 38 per cent at 10.8. A lumped calculation therefore understates how stable a wound part is, and the amount it understates by is not a property of the material but of how well the surface is cooled relative to how well the inside conducts. Below a Biot number of about a tenth it is worth under two per cent and the lump is the right model; at the cooled end the part has 7 kelvin inside it and half the feedback is invisible to a single temperature. The loop gain one temperature understates Part 6 — Every thermal loop gain this collection has computed was computed at a single temperature, because a lumped fixed point has only one — and the essay that measured the gradient inside a core recorded, without measuring it, that this makes each of those numbers a lower bound. It is a lower bound by three per cent where a ferrite usually sits and by thirty-eight per cent at the well-cooled end, always in the direction that makes the part safer than the calculation said. The obvious candidate for what decides it is refused: three geometries at one Biot number are 3.3 times apart.

Where this goes next

The temperature a part cannot come back from, and how long it takes to leave. computed by solving, not by drawing. The same fixed-point equation as the rung below, marched in time with a thermal capacitance rather than solved for its steady states: C dT/dt = P(T) − (T − T_a)/R_th, stepped adaptively on the temperature change because dT/dt goes through zero at each fixed point. Every trajectory starting below 191.1 °C returns to 88.8, however far above the operating point it began; every one starting above it leaves the material's range entirely, the closest in 0.2 minutes. The two nearest starts are 3.0 kelvin apart. The ignition temperature is a boundary in the STARTING CONDITION, and no steady-state analysis contains one. The boundary that is a starting point A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes. The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance. The loss that depends on what it causes Every thermal figure in this collection has had the power handed in. A ferrite's has no business being: its saturation flux falls with temperature, its permeability rises, and both move the loss. Closing that loop makes the temperature a fixed point rather than a product — and the fixed point has a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists. The frequency at which a pulse train becomes an average. computed by solving, not by drawing. The same 5 watts of average dissipation at every frequency, delivered 2 per cent at a time. The flat line is the steady-state answer, which does not know about the frequency. The falling curve is the marched peak junction temperature, which does. They meet at 308 Hz, and that frequency is not a property of the converter: it is a fraction of one junction time constant per period — f·τ = 0.738 at this duty, with τ = 2.40 ms, and between 0.78 and 0.56 across the duties on the slider. A hundred-kilohertz converter fits 240 periods inside that time constant, and at the top of the sweep — 10.0 kHz — the steady state is already exact to 0.46 per cent, so the averaged-power fixed point is right and this is the measurement that says why. The march puts 48 steps inside each pulse, which is what the answer is sensitive to: at six it put the boundary 19 per cent too high. The pulse the heatsink does not feel A thermal resistance iterated to a fixed point with a diode or a switch is a statement about a power — so it assumes that a hundred and fifty watts for two per cent of the time is three watts. The die's own heat capacity decides whether that is true, and it decides it at a frequency: above 308 hertz the junction integrates, by a hundred kilohertz the fixed point is exact to five parts in ten thousand, and at one hertz the same average power on the same heatsink puts the junction three hundred kelvin hotter. The fringing field, and the turns standing in it. computed by solving, not by drawing. The slot on the left is the gap, cut through the centre leg to the core's own symmetry plane where the potential is zero. Flux crossing it does not stay in the slot: it bulges into the window and crosses the copper at right angles to the layers, which is the one direction Dowell's expression and every ladder in this collection assumes has no field in it. The turns are shaded by their own loss. The worst is turn 4, level with the gap, at 32.5 times its direct-current dissipation; the best is 1.39 times. Same wire, same current, same winding, and a spread of 23.4 between them. The turns nearest the gap A gapped inductor's flux does not turn a corner into the iron on its way out of the gap; it bulges into the window and crosses the copper at right angles to the layers. Four tenths of a millimetre from a one-millimetre gap, the worst turn of an eight-turn winding dissipates 37.5 times its direct-current loss and the winding as a whole 14.1 times. Move the same winding three millimetres further out and those become 2.9 and 2.4 — and the distance that governs it is 0.60 millimetres, which is not the gap length and does not scale with it.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Biot numberDesign tradeoffHeat conductionHot spotMagnetic lossMeasurement conditionModel rangeThermal feedbackThermal resistance