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

Voltage divider — where it appears

Two impedances in series across a source, whose ratio sets the fraction of it appearing at their junction only while nothing draws current there. What it does not know about is the load, and the error that ignorance costs is the whole content of the first essay in this collection's networks field.

Named by 14 essays across 5 fields — each of them below, with the objects they name alongside it.

A 10 kΩ + 10 kΩ divider, solved with its load. The unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does.

The divider, and the thing it does not know about

A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

networks · Divider Where a switch is a switch: a band, and the 6.43 MHz at which it closes. computed by solving, not by drawing. A switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1.0% of being ideal only for loads between 49.5 Ω and 1.01 MΩ — 4.31 decades, and both edges are the same part. The upper edge is a frequency as well as a resistance, because the off-capacitance shunts the open switch: it falls a decade per decade above 318 Hz and meets the lower edge at 6.43 MHz, where the band closes and no load at all will do. Checked by scanning every load at 1.3 times that frequency and finding the best possible error to be 1.17%.

A band rather than an edge

Every other boundary in this collection is one-sided: a model is true below a frequency, or below an amplitude. A switch is a switch only for loads between 49.5 ohms and 1.01 megohms — bounded at both ends by the same part — and the upper end is a frequency as well as a resistance, so the band narrows as the frequency rises and shuts completely at 6.43 megahertz, above which no load resistance at all will do.

limits · Ideal switch A divider of 4 equal 1.0% resistors, solved 3000 times. computed by solving, not by drawing. Every resistor drawn from its tolerance band and the divider solved, 3000 times. The worst case is ±1.000% — the part tolerance itself, and it does not improve when the divider is built from more parts — while the measured spread is 0.2944% and the worst of 3000 draws reached 82% of the bound. The root-sum-square, offered as though it were a standard deviation, is 1.70 of one here: for uniformly distributed parts it is √3 σ, a coverage of about 92%.

The tolerance that is not on any part

Four one per cent resistors in a divider give an answer whose worst case is one per cent, whose measured spread is 0.29 per cent, and whose root-sum-square bound — offered everywhere as though it were a standard deviation — is 1.70 of one. Adding parts does not move the worst case at all and shrinks the spread as one over their root, so the gap between the promise and the fact widens with every resistor. And the same arithmetic draws a boundary in tolerance rather than in frequency: an R–2R ladder is a twelve-bit converter only while its resistors are inside 0.14 per cent.

networks · Component tolerance A threshold crossed once, in 20k samples of noise. With no hysteresis the comparator changes its mind 22.7 times on average and as many as 29, on a signal that crosses the threshold once. The vertical bars are the range over twelve seeds. The expected number of extra transitions falls below a tenth at 3.57 standard deviations — so the hysteresis a threshold needs is set by the noise under it and not by the signal over it.

Two thresholds because there is a floor

A comparator with one threshold, watching a slow signal cross it once, changes its mind 22.7 times on average and as many as 29 — because there is noise under the signal and no threshold is ever crossed once. Hysteresis fixes it, and how much is needed is a multiple of the noise's own standard deviation rather than a voltage: 3.57 of them here. The multiple grows with how long the threshold is watched, and slowly — a hundredfold longer record needs three times the hysteresis, not a hundred times.

applied · Hysteresis A 350 Ω bridge, with ideal leads. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 0.498% at 1.0% — so the tangent is worth one per cent only up to 2.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 4.040%.

The bridge that is linear near one point

The expression for a Wheatstone bridge's output is V delta over four, and the solve says it is V delta over two times two plus delta — low by half the fractional change, exactly, at every change tested. So the tangent is worth one per cent only up to a two per cent change and a tenth of a per cent only up to two tenths. Driving the bridge from a current source instead halves the departure at every point and doubles both edges, which is a change of one component and no change at all to the four resistors being measured.

applied · Bridge The capacitor across the upper resistor: 90.9° of margin at 836.5 pF, and less ripple past it. The regulator's phase margin against a capacitor across the upper divider resistor, with the output ripple the 1000 µF reservoir leaves beside it. With no capacitor the margin is 46.47° at a crossover of 9.73 kHz, the output impedance at 10 kHz is 1.95 Ω, the worst rail rejection is 5.79 dB and the ripple 60.62 mV. The margin is greatest, 90.929°, at 836.5 pF — a zero at 6.34 kHz and a pole at 25.4 kHz around a crossover moved to 16.9 kHz — where the output impedance at 10 kHz is 828 mΩ, the worst rail rejection -1.50 dB and the ripple 52.33 mV. At 10 nF the margin has fallen back to 66.34° and the ripple is 35.40 mV.

The capacitor across the upper resistor

The rejection essay said a regulator reproduces its reference times its divider's four, and that a capacitor across the lower divider resistor brings that down to one at high frequency. Measured, the loop peaks the reference's gain to 5.68 near its crossover before any capacitor is added; a nanofarad across the lower resistor raises the peak to 11.2; and the capacitor that brings it down belongs across the upper resistor, where a nanofarad keeps the gain from ever exceeding four. The same capacitor is a lead pair in the loop: 836.5 picofarads takes the phase margin from 46.5 to 90.9 degrees, and ten nanofarads, past that optimum, still holds 66 while cutting the output ripple from 60.6 millivolts to 35.4.

applied · Regulator Three dividers in a row, and where the error actually is. computed by solving, not by drawing. Three two-resistor dividers cascaded with nothing between them. The product of their ratios is 0.1250 and the solved output is 0.076923, 38.5% low at a staircase of ×1. Decomposed stage by stage with the rest of the chain in place — and the product of those three is the answer exactly — they are 0.3846, 0.4000, 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is at the front, which is the opposite of where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion, fitted exponent -0.993, so a decade a stage is within 4.88% and two decades within 0.50% — which is why a chain that has to be right is built as a staircase and not out of one value repeated.

The stage that is wrong is the far one

Three identical ten-kilohm dividers in a row give 0.076923 rather than the product of their ratios, 0.125 — 38.5 per cent low. Decomposed stage by stage with the rest of the chain in place, and the product of those three is the answer exactly, they are 0.3846, 0.4000 and 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is entirely at the front, which is the opposite end from where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion — fitted exponent −0.993.

networks · Divider The current divider, and the resistance that is not in the branch. computed by solving, not by drawing. A current source into two parallel branches, the metered one 10.0 kΩ and the other 1.00 kΩ. The metered branch takes 0.090909 of the current, which is the OTHER branch's resistance over the sum; writing the subscripts the way a voltage divider writes them gives 0.90909, a different number at every ratio but one. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with the roles exchanged: one per cent of error at 111.1 Ω, which is that resistance over ninety-nine to 1.7e-12%. And the headline of the loaded divider holds in the dual too — two current dividers of identical ratio read 0.04762 and 0.09090 into one hundred-ohm meter — while a perfect ammeter reads them identically.

The branch the other resistance decides

A voltage divider's output is set by the resistance the output is taken across; a current divider's is set by the resistance the current does not go through. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with one word changed: one per cent at a meter resistance of R/99 where R is what the meter looks back into — and removing an ideal current source means OPENING it, so that R is the two branches in series, 11 kΩ here, not the 909 Ω of their parallel combination — a factor of twelve in the same construction on the same network.

networks · Divider The R–2R ladder is the chain designed for the rest of itself: every node halves exactly, in place and only in place. computed by solving, not by drawing. A current source into an 8-bit R–2R ladder, each bit's 2R returned to a virtual earth. With the rest of the ladder in place every node splits the current arriving at it exactly in half, so bit k carries 2 to the −k of the source and the step at the major carry is exactly one least significant bit. A stage alone, into a short, would send a third on rather than a half; it is right because of what is connected to it. The same ladder with every resistor equal — the three-divider chain's habit — gives 0.61803, 0.23607 and then falls away by 2.6182 a stage, which is the square of the golden ratio, 2.6180, to the fourth figure.

The R–2R ladder, built for the rest of itself

Three equal voltage dividers in a row deliver 0.076923 of their input instead of 0.125. Their error sits in the front stage, at 0.3846, 0.4000 and 0.5000 per stage. The dual chain, three current dividers fed by a current source, gives the same three numbers to the last digit, and its repair runs the other way: each stage's resistance must fall, not rise. The R–2R ladder is the chain whose stages are wrong alone and right in place, halving exactly at every node. It still keeps its error at the front: a one per cent error in its first resistor makes an eight-bit ladder step backwards at the major carry, and the same error in its last shunt moves that step by 0.0034 LSB.

networks · Divider Four 1% parts spread evenly across their band: the divider's error has a spread of 0.289%, and 8.6% of builds are beyond ±0.5%. computed by solving, not by drawing: a divider of four equal 10 kΩ resistors tapped at the middle, built 6000 times from seeded parts uniform across ±1%, each copy's output error taken from the ratio in closed form and the first four hundred also solved on the netlist, agreeing to 7e-16. The inset is the parts' own distribution; the histogram is the divider's error. Its standard deviation is 0.2894%, half the parts' 0.5780%, because each part moves the output by a quarter of its own error and four quarters add in quadrature to a half. 8.60% of builds fall beyond ±0.5%, which is half the worst case the printed tolerance allows.

The shape a tolerance does not print

A resistor marked 1% promises only that its value lies within one per cent. How the values fill that band is not printed, and it decides almost everything a design does with the part. Four populations that all honour ±1% — spread evenly, Gaussian with the band at three standard deviations, screened from a wider population, and sorted with the best parts removed — give a four-resistor divider spreads of 0.289, 0.168, 0.267 and 0.378 per cent, a factor of 2.26 under one label. Specify the divider at ±0.5 per cent and the yield is 91.4, 99.8, 93.7 and 85.3 per cent. The shape's signature fades as one over the number of parts, so it bites hardest in the circuits with fewest.

networks · Component tolerance Parts sharing 90% of their error: the divider's ratio spreads 0.091% and the chain's total resistance 0.559% — one from the same four parts, cancelling, and one adding. computed by solving, not by drawing, over 20,000 seeded builds: four 1% resistors whose errors share a fraction ρ = 0.9 of their variance — as parts from one reel, one wafer or one temperature do — built into a divider tapped at the middle. The ratio's error (solid bars) has a standard deviation of 0.0912%, half the parts' 0.577% times √(1 − ρ); the total resistance of the same four in series (outline bars) has 0.5587%, the parts' spread times √(ρ + (1 − ρ)/4). Correlation that makes a ratio better makes a sum worse, by the same arithmetic.

What a ratio does not see

Parts from one reel, one wafer or one temperature share part of their error, and a circuit that uses them as a ratio cancels exactly the part they share. Four 1% resistors sharing ninety per cent of their error make a divider whose ratio spreads 0.091 per cent and a series chain whose total spreads 0.559 — the same four parts, one answer six times better than independent parts would give and one nearly twice as bad. The ratio's spread falls as √(1 − ρ), the total's rises as √(ρ + (1 − ρ)/4). From −40 to 85 °C the chain moves by up to 0.76 per cent and the ratio by 0.135, inside the ±25 ppm/K the coefficients differ by. And a resistor array whose parts share 99.5 per cent of their error puts a divider inside ±0.1 per cent every time that four discrete parts manage it one time in four.

networks · Component tolerance With Rₜₕ = 10 kΩ, a tenfold range of β moves the collector current by 17.3%. computed by solving, not by drawing. The collector current of a transistor biased from a 2 V Thévenin source through 1, 10 and 100 kΩ, with 4.7 kΩ in the collector, 1 kΩ in the emitter and a 12 V supply, against its current gain from 30 to 1000, each point a Newton solve of the netlist with both junctions in the device. Dotted: one resistor from the supply to the base, 876 kΩ, sized for the same current at β = 100; it is proportional to β until it saturates. With Rₜₕ = 10 kΩ the current moves by 17.3 per cent from β = 50 to 500.

The bias the gain does not set

A transistor's current gain varies tenfold across the parts one type number covers, and a bias network is judged by how little of that reaches the collector current. One resistor from the supply to the base passes all of it — 0.61 mA at β = 50, 1.16 at 100, and saturation from 200. The four-resistor network passes 98.5 per cent of a tenfold change with a 100 kΩ divider, 17.3 with 10 kΩ and 3.4 with 1 kΩ, and never less than the 1.78 per cent the ratio β/(β + 1) keeps however stiff the divider. Each tenfold cut in that spread costs a tenfold rise in the divider's standing current, so the design is a point on a hyperbola rather than a rule.

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 A megohm part with one picofarad to its mounting, heavier at one end: 0.433 pF with that end driven and 0.233 pF turned round, 1.56° against 0.84° at 10 kHz. computed by solving, not by drawing: the angle of the impedance of a 1 MΩ resistor as 200 sections, with 1 pF in total from its film to its mounting tapered as c(x) ∝ 1 + a(1 − 2x), a = 0.6, read at the terminal with the far end grounded. Driving the end nearer the mounting (solid) it presents 0.433 pF and 1.56° at 10 kHz; turned round (dashed), 0.233 pF and 0.84°. The first-order weight is ∫c(x)(1 − x)²dx, which is C(1/3 + a/6) and C(1/3 − a/6). The part's total capacitance has not changed, nor has its resistance, and neither of the readings is that total.

The resistor turned round

A resistor's stray capacitance is usually drawn as one capacitor across its ends, and a stray to the part's mounting is not one number. Read at the terminal with the far end grounded it is weighted by the square of the distance from ground, so a uniform stray counts as a third of itself, and turned round a part with a tapered stray reads differently: 0.433 pF one way and 0.233 pF the other from one picofarad in all. In the current through the part the weight is x(1 − x), a sixth, and the current reaching the far end is 3 dB down at ωRC = 9.18 rather than at one. Two readings of one part separate the total from the taper, and a distributed stray's phase never reaches a capacitor's: it stops at 45 degrees.

frequency · Real resistor

Named alongside it

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

Model rangeComponent toleranceDesign tradeoffLoadingMonte carloComponent sensitivityThevenin equivalentOutput impedanceSeeded generatorCurrent dividerDevice matchingDuality

All concepts