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

Thermal resistance — where it appears

Kelvin of temperature rise per watt dissipated, along a path from a device to its surroundings. It behaves exactly as an electrical resistance does, so a die, a case and a heatsink form a ladder that a network solver handles unchanged, with temperature for voltage and power for current.

Named by 12 essays across 7 fields — each of them below, with the objects they name alongside it.

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.

transients · Reverse recovery A fit to the held curve reads the series resistance falling to nothing at 532 K/W. computed by solving, not by drawing. Each point is a three-parameter fit — a constant, an ideality factor and a series resistance — to the held forward curve between 10 and 100 mA, for a junction built with 0.6 Ω and no temperature coefficient on it, mounted at the thermal resistance on the axis. With no thermal resistance the fit returns 0.580 Ω with a residual of 43.9 µV. At 350 K/W it returns 0.184 Ω, an ideality of 1.086 and a residual of 34.0 µV. The resistance it reports reaches zero at 531.7 K/W and is negative beyond.

The resistance a slow curve cannot see

A diode's series resistance is read off the top of its forward curve, and a bench curve is a slow one: each point is held until the junction has warmed to it. Through 350 kelvin per watt the held curve sits 39.7 millivolts below the pulsed one at 100 milliamps, and the three-parameter fit that reads 0.580 ohms from the pulsed curve reads 0.184 from the held one — with a smaller residual. The fitted resistance reaches zero at 531.7 kelvin per watt, and the resistance it hid is what keeps the junction from folding back: with 0.05 ohms instead of 0.6 the held curve turns over at 87.6 milliamps.

limits · Diode model 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 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.

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.

power · Thermal feedback 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 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.

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.

power · Thermal feedback The thermal network's step response, and the sum of its own stages — 0.95 per cent apart at worst. computed by solving, not by drawing. A step of power into a three-stage thermal ladder, and the junction's rise divided by it. The solid curve is exact: the impedance is a continued fraction in s, its denominator has 3 real negative roots, and the partial-fraction expansion of Z(s)/s is a sum of that many ordinary exponentials — no march, no step size. The dashed curve is the sum every account of a thermal path writes, each stage's own resistance times 1 − exp(−t/RC) with its own local time constant, and it is an approximation because the stages load each other. What that costs is 0.950 per cent, once, at 12.9 ms — between the fastest stage's 2.4 ms and the next one's 200 ms, which is the only place two stages are moving together. It is one-sided: the sum never reads low.

Two ladders the terminals cannot tell apart

A thermal path drawn as a ladder and the same path drawn as a sum of exponentials are called different models of one object, and the difference between them has never been priced because pricing it needs an exact answer. Solved in closed form, the sum is 0.950 per cent high at worst and never low; the marched netlist is right to a part in 21,169; and the largest disagreement in the picture was 2.919 per cent that has nothing to do with heat at all, which reading the curve one sample differently removes.

transients · Reverse recovery 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.

magnetics · Thermal feedback A diode thermometer measuring its own sense current, at 600 K/W. computed by solving, not by drawing. The same fixed point as the core and the thermistor, on a junction: the dissipation is I·V and V falls with temperature, so the loop gain is negative and the equation has one root at every current. What it costs is two errors. The junction sits above ambient by 0.29 millikelvin at a microamp and 4.26 kelvin at ten milliamperes, which a calibration removes; and a kelvin of ambient produces less than a kelvin of junction, by 1/(1 − R_th·dP/dT), which it does not. After a calibration at 25 degrees the reading at 85 is out by -583 millikelvin at ten milliamperes and -0.09 at a microamp. The coefficient itself moves too — -2.403 against -1.613 millivolts a kelvin — so a quoted tempco carries a sense current as well as a junction.

The sensor inside its own answer

A junction driven from a current source cannot run away, because its forward voltage falls with temperature and its loop gain is therefore negative. What that costs is a thermometer that is warmer than what it is measuring by 4.26 kelvin at ten milliamperes, and — the part a calibration cannot remove — under-reports every change in ambient by 9,584 parts per million, because the sense current's own dissipation falls as the reading rises. Calibrated at 25 degrees, it is out by 583 millikelvin at 85.

semiconductors · Thermal feedback The shunt's resistance as a function of what it is measuring. computed by solving, not by drawing, as a fixed point: the shunt dissipates I²R, its temperature rises by 20 K per watt, and at 50 ppm/K its resistance rises with its temperature — so the resistance the reading is divided by depends on the reading. Iterated to convergence it agrees with the closed form R₀/(1 − αθI²R₀) to 2.2e-16. Along the burden-voltage optimum, where R = u⁄I, the dissipation is I·u rather than I²R, so the temperature rise is 155 mK per ampere and the error is the FIRST power of the current — fitted exponent 1.0007 over five decades. That is the only one of the shunt's errors with the current in it, and it puts a term in I² into the reading, which is a curvature no single-current calibration removes. The upper curve is a shunt of fixed resistance, where the error is quadratic. The fixed point stops existing at 129 kA and never at a current a shunt will see.

The resistance that depends on the reading

Three of a shunt's errors are free of the current being measured, which is the whole content of the burden-voltage optimum. The fourth is not: the shunt dissipates, warms, and its resistance rises — so the divisor the reading uses is a function of the reading. Solved as a fixed point it agrees with R₀/(1 − αθI²R₀) to 2×10⁻¹⁶, and along the optimum, where the dissipation is I·u* rather than I²R, the error is the FIRST power of the current: 7.75 ppm at an ampere, 775 at a hundred, fitted exponent 1.0007.

instruments · Current sensing 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.

magnetics · Thermal feedback A 1 ms pulse of 100 A warms the element 98.7 mK: 4.94 ppm at its end, 2.86 ppm averaged, against 775 ppm held. computed by solving, not by drawing, from the exact step response of the shunt's thermal network — element, terminations and board, 0.2, 4 and 15.8 K/W with 5 mJ/K, 0.1 J/K and 2 J/K — expanded into its Foster modes. A 100 A pulse of 1 ms through the 77.5 µΩ shunt the burden optimum picks, dissipating I·u* = 0.775 W: the element rises 98.7 mK by the pulse's end and cools after it. At 50 ppm/K that is 4.94 ppm of error at the end of the pulse and 2.86 ppm averaged over it, where the same current held long enough for the board to settle would read 775 ppm high.

The pulse that ends before the heat

A shunt's self-heating error is a fixed point in the steady state — 775 ppm at 100 A for the shunt the burden optimum picks, through 20 K/W. A pulse never reaches it. Heat leaves the element through a ladder network — the element's own 5 mJ/K, its terminations, the board — and a pulse short against the element's millisecond heats it adiabatically, by P·t over its heat capacity alone: 0.737 ppm for 100 µs at 100 A, a thousandth of the held figure, 10.7 ppm for 10 ms. Along the burden optimum the error is linear in the current, so the current a pulse may carry for 100 ppm is exact: 12.9 A held, 936 A for 10 ms, 13.6 kA for 100 µs. A reading averaged over the pulse carries half the error of one taken at its end while the heating is adiabatic, and the section-by-section thermal model a data sheet quotes is 10% wrong at 7 ms.

instruments · Current sensing At θ = 4 K/W the transistor holds at 311.7 K; the second crossing at 333 K runs away. computed by solving, not by drawing. The power a transistor dissipates against its junction temperature (solid), each point an electrical operating point solved at that temperature with the saturation current and thermal voltage it has there, for a transistor biased from a stiff 1 Ω Thévenin source to carry 100 mA from 15 V at 300 K with no emitter resistor; and the heat a sink of 4 K/W removes (dashed). Operating points are the crossings: 311.7 K with a thermal loop gain of 0.64 and 332.5 K with a thermal loop gain of 1.43. Above θ* = 4.43 K/W the two meet and vanish.

The bias point that heats itself

A transistor driven from a stiff base voltage to carry 100 mA from 15 V dissipates 1.5 W, and every kelvin that heats it raises the current it carries. On a 4 K/W heat sink the heat balance has two solutions — 311.7 K, which holds, and 332.5 K, which runs away — and above 4.43 K/W it has none. A tenth of an ohm in the emitter moves that edge to 7.69 K/W, and from a quarter of an ohm there is no edge below 520 K at all. The designer's estimate from the loop gain at room temperature puts the edge at 11.04 K/W, two and a half times too high, because the loop gain grows as the part warms.

semiconductors · Operating point Two 1% parts sharing 90% of their error: the divider's output spreads 0.129% at a half, 0.236% at ten to one and 0.258% at a hundred to one. computed by solving, not by drawing, over 20,000 seeded builds at each of seven ratios, every ratio computed exactly: a divider of two 1% parts, each with a uniform part's 0.577% spread and sharing a fraction ρ = 0.9 of it, with the top resistor N times the bottom. The output ratio's standard deviation (dots) against N, beside N/(N + 1)·√2·σ·√(1 − ρ) (line) and the same law for the other shared fractions (faint). N 1: 0.1286%; N 2: 0.1712%; N 5: 0.2145%; N 10: 0.2356%; N 20: 0.2477%; N 50: 0.2536%; N 100: 0.2581%. What is shared cancels at every ratio; what is not reaches the output with a leverage of N/(N + 1), which is a half for a halving divider and nearly one for a large ratio — so the same two parts make a ten-to-one divider 1.83 times worse.

The small leg of a large ratio

A divider of two 1% parts sharing ninety per cent of their error spreads 0.129 per cent at a half and 0.236 at ten to one — the same two parts, nearly twice as bad, because what they do not share reaches the output with a leverage of N/(N + 1) rather than one half. Correlation cancels at every ratio; the leverage does not. Built from equal units of one array, a divider of N to one spreads as σᵤ·√(N/M), so at any one area a ten-to-one ratio costs √10 against a half and needs ten times the units to match it, and the units it needs are the bottom leg's. And a ratio far from a half heats its own parts unequally: at 50 V across 11 kΩ the top of a two-part ten-to-one divider runs 46 K hotter than the bottom and the ratio moves 1,055 ppm with perfectly matched coefficients, where a chain of eleven equal units moves at most 29.

networks · Component tolerance

Named alongside it

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

Fixed pointModel rangeLoop gainThermal feedbackTemperature coefficientMagnetic lossMarchingMeasurement conditionSelf-heatingStabilityThermal runawayDesign tradeoff

All concepts