Power, and the part that does no work

The protection that is gone by the second time

An inrush thermistor is ten ohms cold and holds the first cycle down; then the load current warms it and it settles at 83 degrees and 1.94 ohms, which is 19 per cent of what was bought. That is the design working. It is also a part that takes 198 seconds to recover half its cold resistance, against a reservoir capacitor that empties in tens of milliseconds — so a mains dip in that window hands the rectifier an unlimited inrush into an empty capacitor, which is the exact event the part is on the bill of materials for.

Assumes: The loss that depends on what it causes · The first cycle, which no steady state contains · The direct voltage that is a sawtooth

The first cycle, which no steady state contains measured what happens when a rectifier is switched on: a reservoir capacitor at zero volts across a transformer’s winding resistance, drawing a current that no steady-state analysis of the same circuit contains, because there is no steady state in which the capacitor is empty. The inductance that limits, and lifts added a choke and priced what it does — and what it does at the same time, which is raise the rail.

There is a third way, it costs about thirty pence, and it works by getting hot. A negative- temperature-coefficient thermistor in series with the supply is ten or twenty ohms when cold, which is enough to hold the first cycle down to something the rectifier survives. Then the load current warms it, its resistance collapses, and it stops costing what it cost.

That collapse is a thermal fixed point, and it is the same equation the loss that depends on what it causes solves for a ferrite:

T = Tₐ + Rₜₕ · P(T),

with P(T) = I²R(T) and R falling as T rises. What is different is the sign of everything downstream of that, and what is different about the failure.

The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance.
Fig. 1 The event being limited, from the rung below. The peak depends on which point of the mains cycle the switch closes at, and the worst case is not the average case — which is why the limiting element has to be sized for a first cycle rather than for a mean.

A fit, not a property

The resistance law is the two-point β form a catalogue prints:

R(T) = R₂₅ · exp( β·(1/T − 1/T₂₅) ), with T in kelvin.

β is a fitted constant over a stated range, not a material property, and this collection has attached that caveat twice before to two other exponents — to a Steinmetz exponent in the exponent nobody put in, and to a diode’s ideality factor. It is the same caveat and it bites the same way: a β extracted between 25 and 85 degrees does not predict the resistance at 150, and the useful part of this device’s life is spent above 85.

Everything below is computed from one β of three thousand kelvin and a cold resistance of ten ohms, and the conclusions that matter are about the shape rather than about the third digit.

Stable at every current, and that is the unusual part

Differentiate the map. The loop gain is Rₜₕ·I²·dR/dT, and dR/dT is negative, so the loop gain is negative at every temperature and every current.

There is no thermal runaway here and there cannot be one. The core in the rung below has a negative loop gain over a hundred and fifty kelvin and then a wall, because its flux ceiling closes; this part has no ceiling to close. Push more current through it and it gets hotter, gets less resistive, dissipates less per amp, and settles somewhere new. The equation has exactly one root at every current on the slider.

An inrush limiter's steady state, and how little of it is still a limitercomputed by solving, not by drawing. A negative-temperature-coefficient thermistor in series with a supply, at 1 ampere of load current. The falling curve is what it dissipates at a temperature — I²R with R following the two-point β fit a catalogue prints — and the rising line is what its mounting removes. They cross once, at 83.1 degrees, and the loop gain there is -1.374: negative, so the part is stable at every current rather than below a boundary. What is left of its cold 10 ohms at that temperature is 1.937 — 19.4 per cent. The slider moves the load current, and more current leaves less resistance.051050100150200thermistor temperature, degrees Celsiuswatts83 °C, 1.94 Ωwhat the thermistor dissipateswhat its mounting removesload current1 Acold resistance10 Ωsettles at83.1 °Cresistance left1.937 Ωof the cold value19.4%dissipating1.94 Wloop gain-1.374solved, then checked — stable, and no longer there19% of 10 Ω left
Fig. 2 One crossing rather than two. The dissipation curve falls and the removal line rises, so they meet once and the loop gain there is minus one point three seven — negative, which is a device that cannot run away and also a device that cannot be relied on to stay where it was measured.

At one ampere the crossing is at eighty-three point one degrees, dissipating one point nine four watts, with one point nine three seven ohms left of the ten that were bought. Nineteen point four per cent. At two amperes it is a hundred and twenty-three degrees and eight per cent; at three, a hundred and fifty-four degrees and under five.

The direction is worth stating baldly, because it is the wrong way round. The heavier the load — the larger the reservoir capacitor, the bigger the transformer, the worse the inrush — the less resistance the limiter has by the time it matters. A part sized for the light case protects the light case.

2200 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 16.00 V with 644 mV of ripple, against the 727 mV the expression I/2fC gives — 11.4% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 21.8° of each half cycle and carries 2.98 A at the peak, which is 18.6 times the 160 mA the load draws.
Fig. 3 What the load current actually is, from the direct voltage that is a sawtooth: not a steady draw but a train of conduction pulses whose root-mean-square value is well above the mean. The thermistor is heated by the square, so it is the pulse shape and not the average that sets where it settles.

The iteration diverges while the device does not

The map’s slope at the operating point is minus one point three seven, and that number has a consequence the rung below flagged and did not have an example of.

A fixed point is stable when Rₜₕ·P′ < 1 — one-sided, because the physics is C·dT/dt = P(T) − (T − Tₐ)/Rₜₕ and its equilibrium is decided by the sign of P′ − 1/Rₜₕ. The plain iteration T ← Tₐ + Rₜₕ·P(T) converges when |Rₜₕ·P′| < 1 — two-sided, because it is a map and not a differential equation.

At one ampere this part sits squarely in the gap. It is unconditionally stable and its iteration diverges. Started from the ambient, the first step lands at three hundred and twenty-five degrees, which is outside the range the fit is stated over, and the model refuses on step two — while the part itself sits on a bench at eighty-three degrees all afternoon.

Under-relaxing the step by a factor of about three converges in thirteen iterations to eighty-three point one one five degrees, which is the number the sign-change scan bisects to. Nothing about the device changed; only the arithmetic used to ask it a question.

The iteration a fixed point is usually found by, drawn as the cobweb it is. computed by solving, not by drawing. The curve is the map T → 25 + 45·P(T): guess a temperature, ask the model what it dissipates there, and read off the temperature that much power reaches. The diagonal is where the two agree. Starting from the ambient and stepping between the two draws a staircase, and where it lands is the operating point — 88.76 degrees, against 88.76 from the iteration, which is two routes to one number. The slider moves the thermal resistance; the map's slope at the crossing is the loop gain, and the shape of the staircase is what the sign of that gain looks like.
Fig. 4 The same staircase drawn for the core in the rung below, where the slope is small and negative and the iteration walks straight in. Steepen that slope past minus one and the staircase spirals outward instead, on a fixed point that has not moved.

The map is not the physics. It is a convenience for finding a root, and the conditions under which a convenience works are not conditions on the thing it is being used to study.

Where the heat has to go before any of this is true

There is a quantity in the equation that has not been earned and it is the thermal resistance. Thirty kelvin per watt is a disc thermistor with leads, in still air, not touching anything — and the number moves by a factor of three depending on whether it is standing off the board, lying on it, or sandwiched between two electrolytics that are themselves warm.

That matters more here than it does for the core, because of the sign. A better mounting gives a lower thermal resistance, a lower temperature, a higher resistance left in circuit and therefore more dissipation — the opposite of the usual direction, in which cooling something better reduces what it costs. Cooling this part better makes it a better limiter and a worse conductor, and the design has to choose which it wanted.

The choice is a real one and it is usually made by accident. A thermistor mounted flat against a board near an inlet gets the airflow, ends up cool, keeps four ohms instead of two, and dissipates four watts instead of two — a difference that shows up as a warm patch on a board and never as a design decision.

An inrush limiter's steady state, and how little of it is still a limiter. computed by solving, not by drawing. A negative-temperature-coefficient thermistor in series with a supply, at 3 ampere of load current. The falling curve is what it dissipates at a temperature — I²R with R following the two-point β fit a catalogue prints — and the rising line is what its mounting removes. They cross once, at 154.1 degrees, and the loop gain there is -2.122: negative, so the part is stable at every current rather than below a boundary. What is left of its cold 10 ohms at that temperature is 0.478 — 4.8 per cent. The slider moves the load current, and more current leaves less resistance.
Fig. 5 Three amps through the limiter. It settles at 154.1 °C with 0.478 Ω left of its cold ten — 4.8% — and a gain of −2.12. Where the heat has to go before any of this is true is into the part’s own mass: every number here is a steady state, and the transient that reaches it is the subject of the rung below.

What is left, and for how long

The design question nobody asks about this part is not what it does when it is warm. It is what it does when it has been warm.

The thermistor’s own thermal time constant is its heat capacity times its mounting resistance — for a disc of this size, about two minutes. Its resistance follows the reciprocal of the temperature through an exponential, so the return is slower than the cooling: from a hundred and forty degrees it takes a hundred and ninety-eight seconds to recover half its cold resistance and four hundred and thirty-two to recover nine tenths.

How long an inrush limiter stays useless after it has done its job. computed by solving, not by drawing. The part cools on its own thermal time constant of 120 seconds, and its resistance follows the reciprocal of the temperature through an exponential, so the return is slower than the cooling. From 140 degrees it takes 198 seconds to recover half its cold 10 ohms and 432 to recover nine tenths. A mains dip long enough to empty the reservoir capacitor and short enough to leave the thermistor hot presents the rectifier with an unlimited inrush into an empty capacitor — which is precisely the event the part is on the bill of materials for. The slider moves the temperature it is recovering from.
Fig. 6 The cool-down, in seconds. The straight comparison is with the reservoir capacitor, which empties in tens of milliseconds — four orders of magnitude faster — so there is a wide window in which the capacitor is discharged and the limiter is not there.

Now put the two time scales beside each other. A supply’s hold-up is set by its reservoir capacitor and its load, and it is tens of milliseconds by design; that is the whole point of the capacitor. The limiter’s recovery is hundreds of seconds.

So there is a window — from about fifty milliseconds after a mains interruption to about five minutes after it — in which the reservoir is empty and the thermistor is hot. A dip in that window presents the rectifier with the full inrush into an empty capacitor with no series resistance at all, which is worse than the cold start the part was bought to prevent, because a cold start at least had ten ohms in it.

An inrush limiter's steady state, and how little of it is still a limiter. computed by solving, not by drawing. A negative-temperature-coefficient thermistor in series with a supply, at 0.3 ampere of load current. The falling curve is what it dissipates at a temperature — I²R with R following the two-point β fit a catalogue prints — and the rising line is what its mounting removes. They cross once, at 41.1 degrees, and the loop gain there is -0.490: negative, so the part is stable at every current rather than below a boundary. What is left of its cold 10 ohms at that temperature is 5.969 — 59.7 per cent. The slider moves the load current, and more current leaves less resistance.
Fig. 7 Three tenths of an amp: 41.1 °C, 5.969 Ω left of ten — 59.7% — and a gain of −0.49. What is left, and for how long, is the whole of the protection: at a tenth of the current the part still has six of its ten ohms, and at three amps it has half an ohm.

Why it is not usually a disaster, and when it is

Three things usually save it, and all three are conditions rather than guarantees.

A brief interruption often does not fully discharge the reservoir, so the second inrush charges from part-way rather than from zero and the peak is smaller. The transformer’s own winding resistance is still there and still limits — it is what the first cycle, which no steady state contains measured before any limiter was added, and on a small supply it dominates anyway. And a rectifier’s non-repetitive surge rating is generous, so surviving one extra event is likely.

None of those is available on a supply with a large reservoir behind a stiff source — a switched-mode front end with a bulk capacitor and no transformer between it and the mains — which is exactly where an inrush limiter is not optional. That is why relays exist: a thermistor with a relay across it, closed once the supply is up, keeps the thermistor cold and gives it back for the next event.

How long an inrush limiter stays useless after it has done its job. computed by solving, not by drawing. The part cools on its own thermal time constant of 120 seconds, and its resistance follows the reciprocal of the temperature through an exponential, so the return is slower than the cooling. From 100 degrees it takes 147 seconds to recover half its cold 10 ohms and 380 to recover nine tenths. A mains dip long enough to empty the reservoir capacitor and short enough to leave the thermistor hot presents the rectifier with an unlimited inrush into an empty capacitor — which is precisely the event the part is on the bill of materials for. The slider moves the temperature it is recovering from.
Fig. 8 And the recovery from a hundred degrees: 147 seconds to half the cold resistance and 380 to nine tenths. Why it is not usually a disaster is that a second switch-on within minutes is unusual; when it is a disaster is a supply that cycles on a fault, where every retry sees a limiter that has not cooled and passes a larger inrush than the one before.

The efficiency the part costs by existing

Two watts at one ampere is not a footnote on a hundred-watt supply, and it scales the wrong way. At three amperes the settled resistance is under half an ohm but the dissipation is four point three watts, because the current has gone up faster than the resistance has come down.

That is not an accident of these numbers. At the fixed point the dissipation is (T − Tₐ)/Rₜₕ, so whatever the resistance does, the power is fixed by the temperature the part reaches — and the temperature rises with current. The thermistor dissipates whatever it takes to sit where the falling curve meets the rising line, and the only way to make that less is to cool it better, which lowers the temperature and raises the resistance and makes the loss worse again.

What this rung adds to the one below

The core in the rung below and the thermistor here are the same equation with the sign of dP/dT reversed and one qualitative difference: the core has a wall and the thermistor does not.

That difference is what makes their failures different. A wound part fails by having no solution — a root disappears and the temperature has nowhere to settle. A thermistor never fails that way; it fails by succeeding, arriving at a perfectly stable operating point which happens to be one where the thing it was installed to do is no longer being done.

The second failure is much harder to see. Nothing is hot, nothing is out of specification, and every measurement taken on a bench at room temperature reports ten ohms. The part is doing exactly what it was designed to do and the protection is not there.

There is one more thing the pair share and it is a habit rather than a result. Both parts have a data sheet that describes them at one temperature, and for both the temperature they are described at is not the temperature they operate at. A ferrite’s loss curve is printed at a hundred degrees because that is roughly where a converter’s core lives; a thermistor’s resistance is printed at twenty-five because that is a laboratory. The second convention is the one that misleads, and it misleads in the direction of making the part look like it does its job.

Two millivolts a kelvin, and the wrong sign makes the same observation about a junction, and what matching does about temperature makes it about a pair. In each case the coefficient is printed, small, and correct; what is missing is the operating temperature to multiply it by, and the operating temperature is usually decided by the circuit rather than by the specification.

The number worth carrying is a hundred and ninety-eight seconds. Not because it is precise — it is a disc size and a mounting and a β, and any of those moves it by a factor of two — but because it is three orders of magnitude away from the number it has to be compared with, and a gap of that size does not close when the details change.

The number worth carrying about the method is the one that made the map diverge. A device can be perfectly stable and its most natural solution method can fail on it, and the failure looks like a result: a temperature that runs away on the screen, on a part that does not. Anything solved by iterating a physical relation is exposed to that, and the cheapest defence is the one used here — find the roots by scanning a residual for sign changes and bisecting, which does not care about the slope at all, and use the iteration afterwards to see what the approach looks like rather than to find out where it goes.

The event the part is bought for, and the window it is absent in

The 198 seconds is the number this essay turns on, and what makes it a design fault rather than a datasheet curiosity is what is on the other side of the window.

The first cycle, which no steady state contains measures the event: 32.4 amperes into an empty reservoir against a repetitive peak of 1.23, twenty-six times larger than anything the circuit ever does again, and dependent on where in the supply’s cycle the switch closed. That is what the thermistor is fitted to hold down, and it holds it down once.

The direct voltage that is a sawtooth supplies the other half of the arithmetic: a reservoir capacitor of the size that makes the ripple acceptable empties in tens of milliseconds when its supply is removed. So a mains dip of a hundred milliseconds — which is an ordinary event, not a fault — leaves the capacitor empty and the thermistor still at 1.94 ohms, and the second inrush is very nearly the unlimited one.

The repair everybody uses is to short the thermistor out with a relay once the supply has started, which removes the efficiency cost measured above and does nothing about this: a relay that has closed is a relay that has to be opened again, and deciding when is deciding how long a dip counts as a restart. The inductance that limits, and lifts is the alternative with no memory at all — leakage inductance dividing the peak by seven and the energy by five, dissipating nothing, and doing it identically on every event — and what it charges is an output that sits 29 per cent above the peak of its own supply.

A part whose specification is a state

The general shape of this result is worth naming because it is rare in this collection. Almost every boundary here is a value of a variable — a frequency, an amplitude, a length — above or below which a model stops applying. This one is a state: the same part, at the same ambient, in the same circuit, has two resistances four hundred per cent apart depending on what it has been doing for the last three minutes.

The nearest relative is the core the solver has to remember, where a hysteretic core’s state is not a function of its current at all — half an amp is one flux on the way up and a different flux on the way down — so the march’s state vector gains twenty-four more numbers and the Newton loop is forbidden to touch them. A thermistor’s state is one number rather than twenty-four, and the modelling consequence is the same: a solve that asks for its resistance without saying what happened before is asking a question with no answer.

Which is why the 198 seconds is the specification rather than the ten ohms. A component whose value depends on its history is characterised by a time constant and an initial condition, and neither of those appears on the line of a bill of materials that names the part.

And the time constant is the specification that is hardest to test for, because testing it means switching the equipment off and on again after a stated interval — a test that passes at ten seconds and at ten minutes and fails somewhere in between. A qualification procedure that power-cycles once, or that waits for the equipment to cool, is a procedure that never enters the window this essay is about — and the test that would find it is one nobody writes, because the failure it looks for is absent from the model the part is specified by.

Part 2 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:

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.

Fixed pointInrush currentLoop gainPower law fitRectifierReservoir capacitorStabilityThermal feedbackThermal resistance