Every essay — page 4
The floor, which bounds from below
Every other boundary here is an upper one. This is the other end, and gain does not help because gain amplifies it too. The only figures on this site whose content is a sample — so every number is run across seeds and quoted with its spread — and the one place a bandwidth is not the −3 dB point but π/2 times it.
The window the square-root law has
Averaging N readings divides the noise power by N only if the readings are independent, and samples of a filtered noise are independent only over a band. Measured on the exact sampled autocorrelation, the rule that a reading's noise is the density over twice the window is right at one front-end bandwidth and wrong on both sides — a factor of three wide at six hundred samples, asking the sample rate to exceed the front end's noise bandwidth by about ten rather than by two. Above it the folded noise arrives at full amplitude; below it the front end has removed noise the rule is still charging for.
The order of the null, not the number of them
A mean rejects the mains exactly and rejects it one per cent off frequency by forty decibels, whatever the window length. Two means in cascade of the same total length settle in the same time, cost a third more noise bandwidth and hold that forty decibels over ten times the band — the factor being ten raised to minus the depth in decibels over forty, because a first-order null rises linearly through a target and a second-order one quadratically. Splitting the two lengths to straddle the frequency buys a further factor of exactly root two, for no noise at all, and puts the design two parts in ten thousand from a cliff.
Power, and the part that does no work
A nodal solve computes real power as a check on its own answer and then throws it away. This field reads it out — and then the imaginary half beside it, which sizes the cable, heats the transformer and is billed for. A correction capacitor is exact at the load it was computed for and at no other; a power factor is cos φ only while the current is a sinusoid, and a rectifier's is not.
The current that does no work
A solved network has been reporting its own real power on every page of this collection, as the second of the two checks each answer passes before it is drawn. What that check discards is the imaginary half — the power that flows out to a reactance and back again, does nothing, and is still carried by the cable, still heats the transformer, and is still on the bill.
The capacitor that was right once
Cancelling a load's reactive power needs one division and no iteration, and the answer is exact. It is exact for the load it was computed from, at the frequency it was computed at, and the figure here is about what it does everywhere else — which includes making the installation worse than it was before anything was fitted.
Three phases, and the wire that carries nothing
Three sources a third of a cycle apart, feeding three equal loads, return a current of 5×10⁻¹⁵ amperes down the wire between the star points. That is zero, and the whole of three-phase distribution rests on it. What is worth measuring is how fast it stops being zero, and the answer is that an eleven per cent imbalance in one phase puts a tenth of a line current down a conductor often sized on the assumption that it carries none.
The neutral that carries more than a line
Three balanced loads draw currents summing to 5.3 × 10⁻¹⁵ amperes in the neutral. That is a theorem about sinusoids, and it uses nothing except that each current is a single frequency. A harmonic of order three is shifted by 360° between phases, which is no shift at all — so the third harmonics add, and for any conduction angle narrow enough that the pulse trains stay disjoint the neutral carries exactly √3 times a line current.
The load that takes the most
A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.
What a meter multiplies by
An average-responding meter rectifies, averages and multiplies by 1.1107, which makes it exactly right for a sinusoid and wrong for everything else by the ratio of two form factors — 11.07 per cent high on a square wave and 35.9 per cent low on a rectifier drawing its current in sixty degrees. It is also exactly right at one other waveform, a 145.90 degree conduction angle, which is nobody's sinusoid. Beside it a true-RMS meter that reaches nine harmonics is two per cent low on a square wave and never within one per cent of anything narrower.
The far end that rises
Voltage regulation is quoted as a percentage: how far the voltage at the end of a line falls when the load is applied. The percentage carries neither of the two things that decide it. Past a computable angle of leading load the voltage at the far end goes above the source's — 28.35 degrees for a line of fifty ohms and a hundred of reactance — and with no reactance in the line there is no such angle at all, because the rise is a partial resonance and needs both halves.
The reactance cancelled, and the resonance it buys
Putting a capacitor in series with a feeder cancels part of its reactance and the far end falls less: five per cent of regulation becomes four and a quarter at a quarter compensation and 2.7 per cent at seventy per cent. What comes with it is a series resonance that was not there before, at the line frequency times the root of the fraction cancelled — so a quarter compensation resonates at exactly half the line frequency and a ninth at exactly a third. The line's own inductance is not in that answer and neither is the voltage.
The first cycle, which no steady state contains
Every number this field computes about a rectifier — the ripple, the crest factor, the conduction angle, the power factor — is read from the settled state, and the march that produces them starts from an empty capacitor and throws the first cycle away. That first conduction carries 32.4 amperes against a repetitive peak of 1.23, it is 26 times larger than anything the circuit ever does again, and how large it is depends on when somebody's hand closed the switch.
The inductance that limits, and lifts
Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.
The pair that is worse than either
A bulk capacitor and a ceramic are fitted together because each is good where the other is not, and between them is a frequency at which the pair presents six times the impedance either one does alone. The peak is a parallel resonance between one part's inductance and the other's capacitance, its height is one over the series resistance every data sheet asks to be minimised, and at it the two capacitors exchange 5.87 amps for every amp the load draws.
The capacitor that is not where the load is
The rung below this one connects two capacitors to a load through nothing, and says so. Put three nanohenries of ordinary copper in — one of mounting loop per part and two of plane between the bank and the load — and the anti-resonance moves down to 5.63 megahertz and up to 1.29 ohms, a probe touching the ceramic reads a twelfth of what the load sees at 16.7 megahertz and three and a half times too much at 8.35, and the twentieth capacitor is worse than the second.
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.
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.
The floor and the ceiling move apart
A decoupling bank is judged by two numbers — the lowest impedance it reaches and the highest frequency at which it still meets its target — and with no copper between the parts and the load both improve together as capacitors are added, 5.000 milliohms down to 1.500 and 19.8 megahertz up to 162. Three nanohenries of ordinary board separate them. The bank's own floor still falls 3.33 times while the load's falls 1.49, and the ceiling read at the parts climbs to 82.4 megahertz while the load's peaks at 8.06 and falls to 5.05. At twenty parts the two nodes disagree by a factor of 94 about the same solve.
The loop gain one temperature understates
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.
The energy a unity power factor doubles
Reactive power was computed three ways on this site and the agreement was called a verification. Two of the three are one theorem written twice and cannot disagree about anything; only the third is independent, and what it computes is a difference. Correcting a 20 Ω, 50 mH load to a power factor of 1.000000 takes its reactive power from 1,285 var to nothing and takes the energy stored in the installation from 2,044.9 mJ to 4,089.7 mJ — exactly twice, at every load and every frequency.
The half the neutral does not carry
A star load unbalanced in one phase produces two things, not one. The neutral carries three times the zero-sequence current, which is the half this collection has read; the other half is a negative-sequence set of exactly the same size, rotating backwards, that the neutral never sees. With 0.5 Ω of line in front of 20 Ω loads, losing a phase entirely puts 50.00 per cent negative sequence in the current and 0.8265 per cent in the voltage a switchboard meter reads.
The average a square root pulls low
A true-RMS converter squares, averages and takes the root, and the root of a quantity that ripples averages below the root of its mean. With a hundred-millisecond averager a sine is read one per cent low below 1.86 hertz, where the ripple is still ±20 per cent — and a second filter that steadies the display takes the ripple away and leaves the reading exactly as low as it was. A square wave is read exactly at any averaging time; a rectifier current conducting for twenty degrees needs 2.41 times the averaging a sine does. The implicit converter is the explicit one at half the time constant, and a reading falls 1.38 times slower than it rises.
The load that has two voltages or none
A load that takes a fixed power takes more current as its voltage falls, and on a line with impedance in it every power below a limit is delivered at two voltages and every power above it at none. The limit sits at the load the maximum-power theorem describes, half the source voltage on a resistive line. On fifty ohms and a hundred of reactance a load leading by sixty degrees reaches that limit with its far end at 1.055 of the source, and at nine tenths of it reads 1.172 — so a far end that reads high is not a far end with margin. On a direct-current bus the lower of the two voltages is not a state at all: one per cent below it the bus runs down to nothing in 3.48 milliseconds.
The load that may be complex
Freed of the constraint that it be a resistance, the best load is the source's conjugate — found here by a two-dimensional search on the solved network rather than assumed — and it takes the available power at exactly fifty per cent efficiency whatever the source's reactance. A load that may only be a resistance takes 2/(1 + √(1+x²)) of that, and at a source reactance of twice its resistance that is exactly the golden ratio less one, 0.618034. The resistor-only load is also the MORE efficient of the two, rising towards one while the conjugate sits at a half for ever.
The headroom that is the line's own charge
A line of 50 + j100 ohms delivers at most 0.6180 of a matched resistive line's power to a unity-power-factor load, with 0.5878 of the source voltage left. Give the line its own shunt capacitance — a π, half at each end, B|Z| = 0.4 in all — and the nose moves to 0.7025 at 0.6895: 13.7 per cent more power and 17.3 per cent more voltage. It is headroom, but not the headroom the far end advertises, which with no load rises 21.1 per cent. The capacitance turns the source and line into a Thevenin equivalent with more voltage behind more impedance, and it moves a voltage threshold's meaning in opposite directions depending on whether it is referred to the source or to the unloaded far end.
The regulator that pushes past the nose
A regulator that raises a ratio whenever its load's voltage is low is a stabiliser only while raising the ratio raises the voltage, and on a line that stops being true at exactly the nose: the load's voltage, n·V₀ times the load resistance over n²R plus that resistance, peaks at a turns ratio of the square root of the load over the line and falls beyond it. Ask the load for 90 per cent of the nose power and the regulator settles at n = 1.519 — unless it starts above n = 2.925, where the same setpoint is met on the wrong side of the peak, and then it collapses the voltage. Ask for 101 per cent and the voltage climbs to 9.950 volts, the most the line allows, and is below half its setpoint 256 seconds later. Near the nose the collapse takes a time that grows as the inverse square root of the excess: 2,521 seconds at a hundredth of a per cent.
The noise a true-RMS meter reads low
A true-RMS converter reads a sine low by an amount that falls as the square of its frequency, and for anything but a slow sine that amount vanishes: 3.96 parts per million at ten times the averager's corner. Noise is not a sine. Its square fluctuates at every frequency down to zero, and the averager passes a share of that set by its own bandwidth against the noise's, so the reading is low by 1/(16Bτ) — the first power, not the second. Measured on seeded noise at Bτ = 10 it is 0.600 per cent low against a predicted 0.625; at 100, 672 parts per million. One reading scatters by eighteen times that, so no single reading shows the bias and the mean of a few hundred is nothing but bias.
The cycle a converter has to know
Summing a waveform's square over a whole number of periods reads its root-mean-square exactly: no averager, no ripple, no bias. It needs the period, and a period known one per cent long puts a hundredth of a period too much into the window. Wherever that extra piece falls, the reading moves — on a sine by up to 0.50 per cent over one period, and by 0.46 per cent over ten, because the extra piece grows with the window as fast as the window does. The worst error is δ(CF² − 1)/2, set by the crest factor and the period error and not by how many periods are summed, until the excess reaches half a period. A square wave is read exactly from any window, and a 60° rectifier current twice as badly as a sine.
Devices, and the amplitude they stop being linear at
An operating point is where a transcendental equation and a linear network agree, and finding it is Newton's method on the whole netlist. Past that, the thing a linear model cannot express at all: distortion. It arrives seven times sooner than gain error does, its harmonics are Bessel functions of the drive, and a differential pair removes every even one of them exactly.
A bias point is a solution, not a choice
The phrase "the diode drops 0.7 volts" is a constant standing in for the root of a transcendental equation. Solved properly, from a five-volt supply through a kilohm, it drops 0.692544 V — and from forty-eight volts through the same kilohm it drops 0.754459 V, because the drop moves about sixty millivolts for every decade of current through it.
The distortion a linear model cannot have
A small-signal model's output is a scaled copy of its input by construction, so it has no second harmonic and asking it for one is not a hard question but a meaningless one. Measured on the curve itself, an exponential produces one per cent of harmonic distortion at 1.03 mV of drive — seven times sooner than the 7.30 mV at which its gain is one per cent wrong.
What a pair cancels, and what it only halves
A differential pair's transfer characteristic is an odd function, and an odd function driven symmetrically produces no even harmonics at all. Measured, the second harmonic comes out at 10⁻¹⁶ of the fundamental — the arithmetic's own floor, not a small physical residue. The third harmonic is a different story, and it comes out at exactly half the single stage's, which is a reduction and not a cancellation.
The frequency a device sets for itself
A common-emitter stage's bandwidth is decided by two picofarads between its collector and its base. The Miller approximation says how — lump it at the input, multiplied by one plus the gain — and predicts 643 kHz where the solved network gives 504 kHz. Twenty-two per cent optimistic, and it has no room at all for the second pole or for the zero in the right half-plane that the network also has.
Two millivolts a kelvin, and the wrong sign
Every number in the semiconductor field was computed at 300 K, and the model had no temperature in it at all. Putting it in moves a diode's drop by 1.828 mV/K — downwards, although the thermal voltage in the exponent is rising, because the saturation current rises by nine orders of magnitude across the same range. Two temperature dependences of one device, of opposite sign, from one solve.
What matching does about temperature
A pair cancels the 1.8 mV/K that broke the previous essay, and what it leaves behind is exact: a saturation-current mismatch of m shows up as an offset of the thermal voltage times ln(m), which is proportional to absolute temperature and therefore drifts in proportion to itself. 3 333 parts per million per kelvin, at every mismatch on the slider, because the ratio is 1/T and contains nothing about the device at all.
The copy, and its two errors
Every account of a current mirror leads with the base currents: two are stolen from the reference, so the copy is beta over beta plus two, which is 1.32 per cent at beta of 150 and is what a third transistor is spent on. The Early effect is a footnote and is nine times larger over any useful swing — 11.2 per cent between one volt and ten. Moving beta from 20 to 1000 changes the first by a factor of forty-five and the second by nothing at all.
The buffer that is not a buffer
An emitter follower is reached for when something has to be driven without being loaded: unity gain in, high impedance seen, low impedance presented. The last of those is a number with a range, and the range is narrow. At a kilohm of source the emitter presents 19.08 ohms at low frequency and 67 ohms at 29 megahertz, because the current gain that made it small is falling — and the peak is worst in the middle of the slider, so it cannot be avoided by making the source stiffer or softer.
What a resistor in the emitter buys
Degeneration is described as a trade: give up gain, get linearity. Measured on the transfer curve, the two sides of that trade are not the same size. Dividing the gain by six moves the amplitude at which distortion reaches one per cent by thirty-five times — the square of the factor — and the amplitude at which the gain is one per cent out by only twelve. So the two edges close up, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known.
The input that pushes back
An emitter follower with a capacitor on its emitter has a negative resistance looking into its base — 1182 ohms of it at 3.4 megahertz for a nanofarad, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it: with ten ohms of source the loop goes unstable above 74.9 nanohenries, which is seven centimetres of wire. The repair is the opposite of the instinct — a hundred ohms of source raises the threshold to 913 nanohenries, so the fix for a follower that oscillates is to make the thing driving it worse.
Where the two exponents come from
What a resistor in the emitter buys measured two exponents and could explain neither: the distortion edge moves as the square of the degeneration factor and the gain edge as its 0.950 power, and at a factor of exactly three halves the gain error vanished. The degenerated transfer curve has no closed form forwards and an exact one backwards, and reverting that series gives all three. The distortion exponent is exactly two; the gain edge is proportional to D squared over the root of the absolute value of three minus twice D, which is infinite at three halves and tends to a three-halves power; and the two edges are 7.07 apart on the bare device and 1.62 at a factor of eleven, against 7.06 and 1.61 measured.
The exponent that is a square
Every device in this collection so far has been an exponential, and the whole of its arithmetic — 59.5 millivolts a decade, a distortion edge at 1.03 millivolts, 3,333 parts per million a kelvin — comes out of that one law. A field-effect device obeys a different one, and the interesting part is that it obeys both: an exponential below threshold and a square above it. The square law is within one per cent over a factor of 1.06 in overdrive at half a micron, and the two errors that bound it point opposite ways.
The point the device is never at
A device's linearity is specified by one number, and that number is a place on no curve. The third-order intercept is where two straight lines would cross if both went on being straight, and neither does. For a differential pair the crossing sits π²/4 — 7.84 decibels — above the largest output the device can produce at any drive whatever, and the arithmetic that says so contains no tail current and no temperature.
What the fourth order says about the third
A closed form derived by neglecting the fourth-order term is a claim with an error, and the error is a quantity the fourth-order term can be asked about. Carried one order further, the degenerated exponential's reversion predicts 7.33 per cent where the leading expression was measured to be 6.89 per cent optimistic — and takes the residue to 0.362 per cent. The correction has its own edge, in the same quantity, and it is measured too.
The order that stops helping
Three essays in this field have derived expressions for the amplitude at which a degenerated stage's distortion reaches a target, each one order longer than the last, and each one nearer the measurement. This is where that stops. The error of an expression truncated at order m grows as the m-th power of the drive — 2.00, 4.00 and 5.88 measured — so every added order buys a range that ends sooner than the last one bought, and above 8.9 thermal voltages the six-term expression is further from the device than the four-term one.
The device that never sees the swing
A second transistor standing between the first and the load does two things that every reference gives one expression each for, and one of the two expressions has no ceiling in it. The output resistance is not gₘrₒ² — that is 248 megohms here, and the measurement is 11.5 — it is βrₒ, because the upper device's base draws current and shunts the very node the feedback works through. The bandwidth really is fourteen times better, and what it costs is two volts of a five-volt supply.
The source that holds to the supply
Putting a second transistor on each branch of a current mirror is always described as buying output resistance and costing headroom, and both halves of that are measured here rather than repeated. The resistance goes from 82 kΩ to 7.39 MΩ, the floor rises by 0.71 volts — and the range over which the current is actually what it was set to goes from 1.70 volts to 9.09, because a plain mirror's current never stops climbing. Forty-three per cent of the resistance that should be there is missing, and it is in the reference branch.
The error that is a distribution
The rung below solved a mirror and separated two errors — one that falls with beta and one that does not. Neither is what limits a real mirror. Two transistors on the same die differ, a fractional difference in saturation current is a fractional difference in collector current with nothing dividing it, and the honest object is a spread rather than a number: mean 3.84 per cent, standard deviation 1.96, worst of three hundred 8.43. Degeneration divides it by one plus gm·R and stops at the resistors' own tolerance, and where it stops is a voltage — a hundred millivolts, containing nothing but the ratio of two tolerances.
The resistance that is below zero
An emitter follower's output resistance is 5.5 Ω at direct current and −21.9 Ω at 257 MHz, and the sign is not the transistor's: with an ideal source at the base there is no negative band at all, and a hundred nanohenries of wire between the source and the base produces one from 110 to 301 MHz. A capacitance resonating inside that band is a resonator with loss of the wrong sign, so 4.7 to 100 pF on the emitter oscillates while 1 pF and 470 pF do not — a band of load capacitance with quiet ground on both sides of it.
The refusal, and what it was protecting
A current mirror's model has declined to answer below two hundred millivolts of collector-emitter voltage since it was written, because it has no base-collector junction and would return a forward-active current for a saturated transistor. Put the junction in and the refusal turns out to have been placed where the model it protects is still right to seven parts in ten thousand — and the mirror's real failure is somewhere else entirely: a saturated output takes six and a half per cent off an output that is sitting at five volts.
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.
The length that is a voltage
The square law's band is closed from above by a parameter that is not a bias, a current or a temperature: it is the channel length, wearing a voltage's units. Swept, the band goes from a factor of 4.74 on a thirty-micron device to 1.021 at fifty nanometres — and the overdrive a stage actually biases itself to barely moves at all, so the two cross at 7.93 microns and every shorter device is biased outside the band. The band was also measured in the wrong quantity: the square law is exact in the current somewhere at every length, and its error in the transconductance is never below 42.5 per cent at fifty nanometres and reaches one per cent only above 6.502 microns.
What the cure at the base costs
The rung below bisected the smallest base resistor that stops an emitter follower oscillating and got 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.
The product that is not the third
Every distortion result in this field is odd-order, and the two-tone machinery has computed the second-order product on every call since the day it was written and thrown it away. On a bare exponential it is the drive over twice the thermal voltage — 1.934 × 10⁻² of the fundamental at a millivolt, against 1.870 × 10⁻⁴ for the third-order product, a ratio of 4Vₜ/a and a hundred and three to one. A differential pair puts it at 6.2 × 10⁻¹⁶. And degeneration removes it as D⁻² where it removes the third order as D⁴/|3 − 2D|, so a stage linearised until its third-order product is negligible is a stage whose distortion is almost entirely the one nobody measured.
The mismatch that cancels itself
A current mirror's copy error is spread by three things the two transistors can differ in, and the population that measures it has always drawn all three at once. Turned on one at a time, a two per cent spread of saturation currents gives 1.910 per cent of copy error and a two per cent spread of current gains gives 0.00022 — because the gains enter only as a sum of reciprocals, which has no first derivative where they are equal. Then the standard cure un-cancels it, by a factor of 86.
Before the steady state
Everything on the frequency axis assumes a settled circuit. These are the figures about getting there — and about the one limit no transfer function contains, where a step becomes large enough that the response stops being a scaled copy of a smaller one.
One step, computed twice
A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.
Where the behaviour is written down
Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.
The step that is too big
A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.
The cliff before the fastest settling
Settling time against damping is not a smooth curve with a minimum. It falls by a third in one step of a sweep of five thousandths, and the fastest damping sits on the edge of that step — so a design a hundredth of a damping ratio to the left of the optimum settles forty-eight per cent slower, with a waveform that looks no different.
The half that never arrives
Charging a capacitor from a step loses exactly as much energy as it stores, and the resistance it is lost in does not appear in the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms, to nine figures. Drive the same network with a ramp instead and the loss falls as two time constants over the ramp, with no floor beneath it at all.
A sum that is exact, and the estimate that is not
Add each capacitor's value times the resistance seen at its own terminals with the others removed, and the total is the ratio of the first two coefficients of the denominator polynomial — a theorem, holding to a part in a billion at every spread tested. Divide one by two pi times it and you have a bandwidth estimate that is 14 per cent low with three equal capacitors and never once optimistic. Two settings of the slider have the same three time constants and bandwidths two per cent apart, which is why the sum can never be more than an estimate.
Two numbers without solving for the waveform
Where a step response starts and where it ends are two limits of the transfer function, and neither needs the waveform. Both are exact here — 1.000000000 volts at the end and the whole step at the first instant — and one of them is a lie waiting to happen: take the damping to zero and the final-value theorem still returns 1.000000 for a response that swings between 0 and 2 for ever. Its condition is not on the transfer function but on where the poles are, and the practical condition is narrower still: at five ohms the poles are safely in the left half-plane and sixty cycles is not enough time.
The cancellation that leaves a tail
A pole and a zero placed on top of each other disappear from the response. Miss by one per cent and the magnitude changes by 0.078 decibels, which no measurement would report as a fault, while the time to settle to a thousandth goes from 6.9 time constants to 245 — thirty-six times longer. The settling time has a closed form containing neither the fast circuit nor the doublet's separation as such, and its consequence is blunt: settling to a part in ten thousand needs a cancellation good to a part in ten thousand, however fast the amplifier in front of it is.
The diode that conducts backwards
Every diode in this collection is an instantaneous function of its own voltage, which is exact for an operating point and has no time in it at all. A conducting junction holds a charge, and until that charge is gone it cannot block: drive its current down at a hundred amperes a microsecond and it conducts 3.83 amperes backwards for 48 nanoseconds, against the one ampere it was carrying forwards. The expression every reference gives for that peak is 17 per cent high there, and is right to a per cent only above sixteen thousand amperes a microsecond.
The heat a recovery leaves behind
The essay below this one measured how much current a diode conducts backwards and for how long, and stopped there. Both numbers are multiplied by a voltage somewhere, and the surprise is where: while the junction is still conducting it holds almost nothing, so nine tenths of the energy is dissipated in the transistor pulling the current down and not in the diode. Repeat it a hundred thousand times a second and it is 1.5 watts, the lifetime rises with temperature, and above 1.28 megahertz the diode's own loop has no fixed point at all.
Two loops, and one heatsink
The rung below this one found that nine tenths of a reverse recovery's energy is dissipated in the transistor and not in the diode, and then computed the diode's junction temperature with all of that energy in it. Repaired, the diode alone survives to 1.28 megahertz instead of 128 kilohertz — a factor of exactly the ninety per cent. What replaces the number is the arrangement that exists: two devices with two different positive temperature loops on one piece of aluminium, giving out at 135 kilohertz, and it is the switch that goes.
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 capacitor that remembers
Charge a capacitor, short it for ten seconds, open it, and it climbs back to a fifth of a per cent of where it was. Nothing leaked and nothing was gained: some of the dielectric had not finished discharging. The same defect measured as an admittance says the part is 0.593 per cent more capacitance at a tenth of a millihertz than at a kilohertz, and measured in a sample-and-hold it says a millisecond of hold costs a hundred parts per million — thirteen bits, on a part specified at nothing.
Three cliffs, and where they are
The rung below sweeps a second-order step's damping, finds the settling time falling by a third in one step of a five-thousandth sweep, and calls it the cliff. There are three of them between 0.3 and 0.98, one for each excursion that stops leaving the band, and adding a third pole moves all three left and makes all three shallower — so the classic 0.78 for fastest two per cent settling is a second-order number, and at a third pole one and a half times the natural frequency the answer is 0.745 and 0.78 is on the wrong side of the step.
The gap a derivative needs
The derivative of a pole is exact and has no step size in it, and beside the formula sits a sentence nobody had measured: it divides by a quantity that vanishes when two poles meet. Driven together, the sensitivity climbs as the reciprocal of the gap — fitted exponent −1.0000, the product a constant 1.00000 times the natural frequency — while the largest change it still describes falls as the gap *squared*. A one per cent capacitor is outside first order once the poles are 3194 radians a second apart, which is an ordinary critically damped design.
The half a switch keeps
The rung below found that charging a capacitor from a step loses half the delivered energy whatever the resistance, and that a ramp takes the loss down as 2τ/T with no floor. A staircase of N settled risers costs one Nth of the step, exact to four parts in a million, and the law ends at a dwell of 5.272 time constants. A switch is the other half of the same product and buys nothing at all: with the supply held at five volts and the channel conductance ramped over a thousand time constants, the loss is 1.00000000 of ½CV².
Ten seconds, and fifteen minutes
A data sheet's dielectric absorption is quoted as a property of the part. It is not: the same modelled capacitor reads 0.4050 per cent with a tenth-of-a-second short and 0.0305 per cent with a thousand-second one, and 0.0047 against 0.2948 depending on when the reading is taken. Seventeen marched tests collapse onto one line — the recovery is 1.0288 parts per thousand for every decade between the two durations — and four dielectrics the specified test declares identical read a factor of 3.31 apart one decade away from it.
The best damping is not the one to build
The fastest settling damping is the right-hand limit at a discontinuity, so two thousandths below it costs 41 per cent and two thousandths above it costs 0.34 — a ratio of 120 in the penalty for the same error. With ±2 per cent on the damping ratio the nominal that minimises the worst case is 0.7927 rather than the optimum's 0.7734, and it guarantees 4.243/ωₙ against 5.943. The band moves the optimum by 0.231 of damping ratio and the third pole by 0.047, and 0.78 is exact at ±2% and 55 per cent slow at ±1%.
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.
Shorted instead of opened, and the error changes sign
The same construction with the other capacitors shorted rather than removed sums the reciprocals of the products, and that sum is the ratio of the denominator's two HIGHEST coefficients — the negated sum of the poles, exact to a part in 10¹². Divided by 2π it estimates the lower corner of a band, and it is 16.6 per cent HIGH with three coupling capacitors and never once low. Two settings of the slider give the same three time constants in a different order, the same sum, and corners two per cent apart.
Where the estimate stops being a bound
The sum of open-circuit time constants is never optimistic on a network with real poles, and the claim is about the network rather than about the theorem. On a second-order section the ratio of the estimate to the truth is Q/√(k + √(k²+1)) with k = 1 − 1/2Q², which is exactly 1/√2 at the Butterworth quality factor — its worst point, 29.29 per cent low — and exactly 1 at a quality factor of √2. Above that the estimate is high, by 6.45 times at a Q of ten, and the crossing bisected on the solved response is 1.414213 against 1.414214.
The ringing that belongs to the rule
The trapezoidal rule is stable for every stable circuit and every step size, and it is not damping. March a one-microsecond pole with twenty-microsecond steps and its node reads 1.818, 0.331, 1.548, 0.552 volts — an oscillation at half the stepping rate, its distance from the final volt multiplied by exactly −0.8182 every step, taking twenty-three steps to fall below one per cent. The slow node that pole drives is right to 1.2 × 10⁻⁵ V at a millisecond. One backward-Euler step at the discontinuity cuts the first swing from 0.818 V to 0.048 and two to 0.0023, because backward Euler multiplies the same error by 1/(1 + h/τ) and the trapezoidal rule by (1 − h/2τ)/(1 + h/2τ), which approaches −1.
The phase the rule loses
An inductor and a capacitor with no resistance ring for ever, and two ways of marching them disagree about how. The trapezoidal rule keeps the amplitude exactly — its factor per step has a magnitude of one — and loses phase instead: 2π − 2N·atan(π/N) a cycle, 2.918° at twenty steps a cycle, so ten cycles later it is 29.2° behind the circuit. Backward Euler keeps 0.3901 of the amplitude a cycle at the same step, and after ten cycles 0.0082 per cent of the ringing is left, in a circuit that has no loss. The two errors fall at different rates: the trapezoidal rule's phase as the square of the steps a cycle, backward Euler's amplitude as the first power. A hundred cycles to within one per cent needs 182 steps a cycle of one and 196,404 of the other.
The start a step takes from infinity
The initial-value theorem reads where a step starts off H at infinite frequency. Apply it again to s·H, s²·H and on, and it reads how the step starts: the first r − 1 derivatives are zero for a network r degrees more poles than zeros, and the r-th is the ratio of the leading coefficients. So a step through r sections begins as a power of time — for sections of τ, τ/2, … τ/r, exactly (t/τ) to the r — and reaches one per cent at 10.1 µs through one section, 105 µs through two and 508 µs through five. Put a zero anywhere, even a thousand times above every pole, and the step starts linearly instead, with a slope of twice the zero's time constant over τ² that is the larger term for the first two of them.
The slowest one decides it
One doublet's settling is its own time constant times the log of the tail over the band, so halving the mismatch buys ln 2 of it and no more. Split the same one-per-cent tail into eight doublets spread over four decades and the settling goes from 0.23 milliseconds to 470 — a factor of two thousand — while sixteen times as many doublets over the same spread changes it by less than two. It is the spread that costs and not the number, because the slowest doublet decides the settling however small its own share, and its share enters only through a logarithm.