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

Component tolerance — where it appears

The fraction by which a part may depart from its marked value, and the quantity that decides what a design can be built to rather than drawn to. It is a distribution rather than a bound, so a worst case computed by adding tolerances describes a circuit almost nobody will build.

Named by 43 essays across 8 fields — each of them below, with the objects they name alongside it.

The closed-loop poles at a gain of 3.05. The locus of the two poles as the amplifier's gain runs from 2.7 to 3.3. It crosses the imaginary axis at a gain of 3.000000 — bisected on the netlist, not quoted — and at 3.05 the real part is 2.500e+2 radians a second, which is an envelope multiplying by 1.17015 every cycle. The crosses are the closed form ω₀(k−3)/2 and they sit on the measured circles.

The gain that is exactly one

An oscillator is designed by making the loop gain one at the frequency where the phase is zero. The gain at which this circuit's poles reach the imaginary axis is 3.000000000000, bisected on the netlist — an equality, not a range. A gain three per cent high multiplies the envelope by 1.0987 every cycle and reaches the rails in 61 milliseconds; three per cent low divides it by the same factor. A one per cent resistor cannot hold the condition, and neither can any other component.

applied · Oscillator The same millivolts, subtracted — and what four resistors leave behind. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz. The second curve is the same measurement made differentially: a unity-gain difference amplifier across the sensor, its four resistors carrying the worst-case skew 0.1% allows. It divides the error by 500.5 — 53.99 dB against a closed-form (1 + G)/(4·tolerance) of 53.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 998.9 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.

The rejection four resistors decide

Measuring a ten-millivolt sensor against a ground that somebody else's hundred milliamps is also using puts five hundred microvolts of their current into the reading. Subtracting the two ends of that conductor with a difference amplifier removes it — by 53.99 decibels with one-tenth-per-cent resistors, against a closed form of 53.98, which is a factor of five hundred and not a removal. The amplifier has nothing to do with it: the number is one plus the gain over four times the resistor tolerance.

instruments · Common-mode rejection What a mistuned arm leaves at 1300 Hz. computed by solving, not by drawing. Neither curve is the null's depth — the null is still bottomless, it has simply moved — but the depth at the frequency the notch was designed for, which is the number a filter is bought for. One per cent components leave -32.9 dB if both errors go the same way and -78.8 dB if they oppose, a factor of 197 from the same tolerance on the same two parts. The slopes are 20.0 and 40.0 decibels per decade: first order in the error on the product LC, second order in the error on the impedance level.

What actually fills a null

A ten per cent error in the two components of a notch's arm leaves the null three hundred decibels deep — it moves it rather than filling it. What fills it is loss, at twenty decibels per decade of arm resistance exactly. And the depth at the frequency the notch was designed for splits into two orders depending on which way the two errors go: one per cent parts leave 79 decibels one way and 33 the other.

filters · Null depth 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 first stage of 100 buys 40.0 dB of rejection, and gives it back above 10.0 kHz. computed by solving, not by drawing. The rejection of a three-amplifier instrumentation amplifier against frequency, beside the one-amplifier difference stage it is built around. At low frequency the two differ by 39.99 dB against 20 log 100 = 40.00 dB, and the reason is that the input stage passes a common-mode voltage at exactly unity: the common-mode gain of the whole instrument is 1.998 mV/V, which is the difference stage's own. So the four resistors around the last amplifier decide the rejection and the two that set the gain do not — ten per cent between them moves it by less than a hundredth of a decibel. What ends it is bandwidth: above 10.0 kHz, which is the amplifier's gain–bandwidth divided by the gain that bought the rejection, the differential gain falls and the rejection falls with it at twenty decibels a decade.

The four resistors that decide, and the two that do not

A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.

instruments · Common-mode rejection Fourteen decades of imbalance, fourteen digits gone, and a matrix in perfect health. computed by solving, not by drawing. A Wheatstone bridge walked towards balance, with the relative error of the solved output against a closed form that cannot lose digits. The condition number of the nodal matrix is 505.0 at every imbalance and the smallest pivot is 2.0e-3 of the matrix norm — orders above the 1e-12 at which this solver refuses to answer at all. Neither number moves, and the answer still loses one digit per decade of imbalance, reaching 33% at δ = 1e-15. The bound drawn over it is the round-off divided by the imbalance, which the measurement stays under at every point. The third curve is the same closed form written as ½ − 1/(2+δ) — algebraically identical, and it loses its digits at the same rate, which is where the loss lives: in the subtraction of two nearly equal numbers, not in the matrix.

The matrix that is ill, and the answer that is not

Every other essay here treats the solve as exact, and it is not. A bridge walked towards balance loses one digit per decade of imbalance and has none left at a part in 10¹⁵ — on a matrix whose condition number never moves and whose smallest pivot stays four orders above the threshold this solver refuses at. A feedback amplifier does the opposite: the solver declines to answer at a gain of 10⁹, one decade after it returned an answer that was exact to the last bit.

networks · Conditioning A difference quotient is best at a step of 1e-5, and is 1e+6 times worse at 10⁻¹¹. computed by solving, not by drawing. The worst disagreement between the adjoint network's derivatives and a central difference quotient of the same quantities, against the fractional step the quotient is taken with, on a 7-element LC ladder filter at 1000 Hz. The curve has a minimum because two errors pull opposite ways: the curvature the quotient neglects falls as the square of the step, and the digits its subtraction destroys rise as one over the step. The best it reaches is 1.4e-9, against the 2.3e-11 that the two-thirds power of the machine epsilon predicts. The exact route costs 2 solves against 15, and has neither error term.

Every derivative, and the one that is zero

How much does this response move if that capacitor is one per cent out? A difference quotient answers it one component at a time, in two solves each, and its best possible accuracy is four parts in a hundred million. Transposing the matrix and solving once more answers it for every component at once, exactly. Pointed at a claim the essays on ladder filters have made from the start and never tested directly — that a doubly terminated ladder filter's response is stationary in every element at its passband maxima — it returns two parts in ten billion, where the cascade realising the identical response returns 0.72.

networks · Sensitivity A ±15% envelope on what a bridge reads permits 29 points of working capacitance. computed by solving, not by drawing. The charge-average capacitance between zero and the rated voltage — the number a reservoir or a hold capacitor obeys — against how the data sheet's stated temperature change is divided between the model's two parameters. Every point honours the envelope exactly: the measured value at zero bias with a one-volt test signal is 15 per cent from nominal at each point on each curve, by construction. The working capacitance is not. At the cold end it is anywhere from -42.9 to -13.9 per cent — 29.0 points of ambiguity at a temperature where the measured value is pinned exactly — and at the hot end from 13.9 to 36.2. Across the whole envelope that is 79.0 points against the 30 the specification bounds, a factor of 2.63. The left-hand end of the upper curve is missing because it is impossible: with the amplitude fixed, no characteristic voltage makes the measured value exceed the zero-bias capacitance.

The coefficient that is about one reading

A class II ceramic's temperature coefficient is a third printed number, and it is a coefficient of the one capacitance a data sheet reports: what a bridge sees at zero bias with a one-volt test. The model behind the part has two parameters, one number does not determine two, and every way of dividing a ±15 per cent envelope between them honours the envelope exactly while putting the working capacitance anywhere across twenty-nine points — and two parts a bridge cannot tell apart differ by 1.80 at the voltage they are used at.

frequency · Real capacitor Matched parts cost nothing; a 2 dB difference between them sets a 112 dB ceiling. computed by solving, not by drawing. The common-mode rejection of a three-amplifier instrumentation amplifier against the gain of its input stage, with each amplifier's own rejection in the netlist as an input-referred error of the common-mode voltage over the rejection. The architecture's own figure rises decibel for decibel with the gain, because the difference stage sees a larger differential signal beside the same common-mode one. The parts' contribution does not rise with anything, and the part of it that matters is not their rejection but the difference between their rejections: two amplifiers of 98 dB that are identical cost 0.000 dB, while 100 dB against 98 dB leaves a ceiling of 111.7 dB with no gain in it. The two mechanisms cross: below a gain of 1903 the four resistors decide everything, and above it more gain buys no more rejection at all — 111.9 dB at a gain of 100000, where the arrangement alone would have been worth 148. The one place the instrument beats its own floor is a gain of 1000, where the two errors cancel; that is a coincidence of signs and not something a design can hold.

The rejection the parts have

Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.

instruments · Common-mode rejection 95 dB of instrument, 290 Hz corner — and the corner belongs to the source. computed by solving, not by drawing. The common-mode rejection of the same three-amplifier instrument measured earlier measured, with 1 kΩ of imbalance between the two source resistances and 10 pF at each input. The instrument's own curve is drawn beside it. Below 290 Hz the two agree; above it the measurement falls at twenty decibels a decade while the instrument does not, reaching 84.0 dB at a kilohertz against the instrument's 95.0. What converts common mode into differential is the difference of the two input time constants — 10.0 ns here — and once it is differential no rejection repairs it.

The corner the instrument has no part in

Three earlier essays measured a three-amplifier instrumentation amplifier's rejection at direct current and found 95 dB, of which the resistors' matching decides one part and the amplifiers' own mismatch another. Connect it to a source with a kilohm of imbalance and ten picofarads at each input and the rejection has a corner at 290 Hz and falls twenty decibels a decade after it — reaching 84 dB at a kilohertz on an instrument that is still doing 95. What converts common mode to differential is the difference of two time constants, and the cure is a capacitor on the quiet input.

instruments · Common-mode rejection A bridge's reading turns over at 1.885 V, where the test level stops mattering. computed by solving, not by drawing. What a bridge reads is the fundamental of the charge waveform over the fundamental of the voltage, so it is a function of the bias AND of the amplitude, and a data sheet names one point of it: zero bias, one volt. The four curves are four test levels on one part. At zero bias they run from 9.999 µF down to 8.312 µF — the harder the drive the lower the reading, because the capacitance is at its maximum there and a sinusoid spends most of its time off the peak. At the rated 5 volts they run the other way, 2.000 µF up to 2.426 µF, because the curve is convex. Between them is one bias where the two effects cancel: at 1.8848 V a tenfold change of test level moves the reading by nothing at all, and that voltage is 0.6645 of the polarisation's own characteristic voltage.

The reading a data sheet does not take

A class II ceramic's temperature envelope leaves its working capacitance 29 points wide at one end and 79 across, because one printed number cannot pin a two-parameter model. One further bridge reading recovers almost all of it — and where the reading is taken decides everything. Turning the test level down to a fiftieth separates five parts a data sheet cannot tell apart by 17.76 per cent; moving the bias to half the rated voltage separates them by 176.02, and pins the working capacitance to ±0.512 per cent from a reading known to one.

frequency · Real capacitor Where one pole goes when each component is 5% high. computed by solving, not by drawing. A series R–L–C, its poles recovered by rooting the determinant, and the derivative of the upper one with respect to each element taken exactly from the two null vectors at the pole. The dashed lines are the first-order prediction for a 5 per cent change; the filled circles are where the root actually goes when the element is changed and the determinant re-rooted. The three directions are the argument: the resistance moves the pole along a circle of constant radius, because the natural frequency does not contain it — its normalised sensitivity has a real part of 3.1e-16. The inductance and the capacitance each carry exactly −½ of the radius, and imaginary parts that are exact negatives. At 5 per cent the prediction is out by 0.122 per cent of the pole's own magnitude.

The derivative of a root

An earlier essay turned one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder filter stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder filter's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.

networks · Sensitivity At a stationary point a tolerance has a mean, not a spread. computed by solving, not by drawing. Six hundred ladder filters with every reactance drawn independently from ±1 per cent, measured at the ripple peak and at a frequency between the peaks. Away from the peak the distribution is centred on nominal and 325 of 600 are above it. At the peak none is: the whole distribution lies below, with a mean of -0.0030 per cent and a worst case of -0.0140. A yield calculation that assumes a symmetric spread at a frequency that has a stationary point is wrong in both directions at once — it allows parts above a limit that cannot exist, and it misses that the whole batch has moved.

The tolerance that can only take away

At a doubly terminated ladder filter's ripple peak the first derivative of the magnitude with respect to every reactance is zero to ten digits, which an earlier essay measured and which says nothing about how much the response moves. This says how: every second derivative is negative, so of six hundred ladder filters built from one per cent components not one is above nominal, the mean has shifted rather than the spread having grown, and doubling the tolerance quadruples the damage instead of doubling it.

networks · Sensitivity Five tolerances, and the response moves in two directions. computed by solving, not by drawing. The eigenvalues of the relative second-derivative matrix of a fifth-order 0.5 dB Chebyshev ladder's magnitude at its lower ripple peak, over its five reactances. Two are of order one — -0.9473 and -0.8051, both negative — and the other three are 3.9e-9, which is zero at the precision the arithmetic has. So the quadratic form is negative semi-definite of rank two, and there is a three-dimensional subspace of component errors that the peak cannot see. The open circles are the same matrix computed by four re-solves per pair, sharing no adjoint arithmetic with the filled ones: they agree to parts in ten thousand on the two that are there and place the three zeros about two decades higher, which is the price of differencing a difference.

The three tolerances that do nothing

An earlier essay computed every second derivative of a ladder filter's magnitude at a ripple peak, found them all negative, and built six hundred ladder filters to argue that no combination of tolerances could raise the response. The whole matrix says so outright — and says something six hundred samples could not have found, because a sample of a five-dimensional box never lands on a three-dimensional subspace: two of the five eigenvalues are of order one and the other three are nine decades down.

networks · Sensitivity One component of a gyrator moves the response by 0.50; the passive ladder's inductor moves it by 3e-8. computed by solving, not by drawing. The magnitude sensitivity at the passband maximum of a gyrator ladder, against the resistance scale the gyrators are built at, for a single component and for the combination of both halves that actually changes the synthesised inductance. The passive ladder realising the identical response is at 3.2e-8. A single component sits at about a half whatever the resistance scale is, and the combination falls as one over it — the 0.97 power over 3 decades. So the stationarity has not been destroyed, it has become a statement about a combination of components rather than about a component, and independent parts do not come in combinations.

One inductor, and ten components

The rung below built an inductor out of an amplifier and ended with a boundary that is not a frequency: one end of it is soldered to ground. A ladder's series inductors are floating, so making one takes four amplifiers rather than two — and the four-amplifier version is exactly floating, to five figures, and reproduces the ladder's response to a hundredth of a decibel. It does not reproduce its stationarity. Each of a gyrator's ten components moves the response by exactly a half, where the inductor it replaced moved it by ten to the minus eight.

filters · Gyrator The most sensitive direction at one ripple peak is not the one at the next. computed by solving, not by drawing. The eigenvector of the largest eigenvalue of the relative curvature matrix, at each of the 2 ripple peaks of a 5th-order 0.5 dB Chebyshev ladder, drawn as its components over the reactances in order along the ladder filter. Every one is symmetric under the filter's own reversal and no two of them are the same direction: the largest overlap between any pair is 0.0603, which is 86.5 degrees apart. The eigenvalue that belongs to them grows from -0.9473 at 554.9 Hz to -12.005 at 897.9 Hz, a factor of 12.7, so the peak nearest the band edge is both the most sensitive place in the passband and sensitive to a different combination of parts. Each direction is confirmed by a second computation of the whole matrix — four re-solves per pair, sharing no adjoint arithmetic — which returns the same direction to 7.6e-7 radians.

The direction a response is most sensitive to

An earlier essay diagonalised a ladder filter's curvature at one ripple peak and read only the eigenvalues: rank two of five, three directions of nothing. The eigenvectors say what the two directions are, and doing it at every peak rather than one changes the conclusion. The most sensitive combination at 554.9 Hz and the one at 897.9 Hz are 86.5 degrees apart, the curvature that belongs to them grows from −0.556 to −75.4 across a ninth-order passband, and exactly one combination survives the whole band at every order — which turns out to be the ripple depth.

networks · Sensitivity The sensitivity of a pole against the room it has. computed by solving, not by drawing. A series R–L–C whose damping is walked from 0.3 to 0.999999, which slides its two poles together along a straight line and changes nothing else. The exact derivative of a pole with respect to the capacitor climbs from 0.5241 to 353.6 as the gap between them falls from 19078 to 28.28 radians a second. The fitted exponent over the closest four is -1.0000, and the product of the two is the natural frequency itself — 9999.6894 against 9999.6894, at every damping drawn and not merely in the limit, which a closed form gives and this computation never sees. The resistor's curve runs at 2ζ times the capacitor's — below it at 0.3 and at twice it by the time the poles have met — and the inductor's lies exactly under the capacitor's throughout.

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.

transients · Poles The reading is a count of decades: 1.0288 parts per thousand of them. computed by solving, not by drawing. 17 marched tests, three families of absolute time — a tenth of a second, one second and ten seconds of short — plotted against the number of decades between the short and the reading. The families lie on one another, which is the finding: the answer is not a property of the part alone and not a property of either duration, it is a count of the decades of relaxation time the test leaves in. The line is a least-squares fit through the origin at 1.0288e-3 per decade; the model's own capacitance per decade of relaxation time is α = 1.0343e-3, which nothing in the fit was told — the fit sits 0.53 per cent under it, because the charge that comes back is shared with the slow branches it came off. The worst residual is 4.80 per cent, at the narrowest ratio drawn, and 1.23 per cent over the 8 tests that are two decades wide and read before the slowest relaxation the model has; the 3 read after it fall away to 4.49 per cent, which is where the law ends. Families a hundred times apart in absolute time differ by at most 0.84 per cent, which is the whole of the collapse.

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.

transients · Dielectric absorption Two of the four errors are divided by the gain; the best the instrument gets is 110.1 dB, at ×776. computed by solving, not by drawing. The four mechanisms that limit a three-amplifier instrumentation amplifier's common-mode rejection, each measured alone against the gain of its input stage and then all four together, at 50.0 Hz with 1 kΩ of imbalance between the source resistances and 10 pF at each input. The difference stage's four resistors and the difference amplifier's own rejection are injected after the gain, so their common-mode gain is a constant — 1.998 mV/V and 0.0100 mV/V — and the rejection they allow rises decibel for decibel with the gain. The input pair's mismatch and the source's time-constant gap are injected before it, so they are amplified by exactly the gain the signal is and the rejection they allow is flat. The four add as complex numbers: the two largest are real and of opposite sign, they cancel at a gain of 776, and what is left there is the source's 3.142 µV/V, which is purely imaginary because it is ωΔτ. The instrument's best is 110.07 dB against the source's own 110.06, and above that gain more of it buys nothing.

The errors that arrive before the gain

Four earlier measurements each found a different owner of one instrument's common-mode rejection and each measured it alone. Solved together, the four are complex numbers that add — to 1.25 parts in ten thousand — and two of them carry a factor of the first stage's gain while two do not. The two that do cancel the two that do not at a gain of 776, and what is left there is 110.066 decibels, which is exactly the number the cable sets. Better-matched amplifiers move that gain from 93 to 3392 and do not move the ceiling by a hundredth of a decibel.

instruments · Common-mode rejection The fastest damping is a surface, and the band is worth 5 times the third pole. computed by solving, not by drawing. Each point is the last settling cliff, bisected — the damping at which the first overshoot's peak lands exactly on the band's edge, which is where the fastest settling is. Across the five bands the optimum moves by 0.231 of damping ratio; across a third pole from 1.5 times the natural frequency out to a second-order response it moves by 0.047. The two axes are worth 5.0 to one, and the expensive one is the specification rather than the parasitic. The classic 0.78 for fastest two per cent settling is the second-order curve's value at ±2%, 0.7797; at ±1% the same response wants 0.8261.

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%.

transients · Damping 300 mirrors built to one design, with 2% device mismatch. computed by solving, not by drawing. Every pair in the population is a full Newton solve of the same netlist with two saturation currents drawn from a normal distribution, the Early conductances iterated to self-consistency for each. The mean is 3.836 per cent, which is the systematic error the rung below computed with identical devices (3.937 per cent) — the mismatch does not move it. The spread about it is 1.955 per cent, which is the device mismatch arriving with nothing dividing it, and the worst pair of the 300 is 8.43 per cent out. A design whose specification is the mean has specified the one mirror nobody has.

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.

semiconductors · Current mirror At Q = 2 and a hundred times the corner, one arrangement is 1.97% out and the other 2.83%. computed by solving, not by drawing. The same second-order section designed twice — a follower with a capacitance ratio, and equal passives with the Q supplied by a closed-loop gain of 2.5000 — built with the same one-pole amplifier and swept over its gain-bandwidth, with both pole pairs recovered by rooting the determinant. The unity-gain arrangement's error is Q over the ratio: 2.00 per cent at a hundred times the corner. The equal-component arrangement's is K²/2 over the ratio, 3.13 per cent, because its amplifier is a gain-of-K stage and therefore has K times less bandwidth to spend. Below about thirty times the corner both laws fail, and the second one changes sign.

Where the Q comes from

Two second-order sections with no component value in common have the same transfer function to 1.8×10⁻¹⁰ of a decibel, and are not the same circuit. Against a slow amplifier the follower's Q error is Q over the gain-bandwidth ratio and the gain stage's is K²/2 over it — so the second is worse below Q = 3.08 and better above it. Against component tolerance the follower's worst element carries ½ and the gain stage's carries 2Q − ½, nineteen times as much at Q = 5. Neither oscillates at any gain-bandwidth at all.

filters · Q enhancement An order-8 Butterworth: the whole sketch is 3.0103 dB out at the corner, and its 4 sections −5.85 to +8.17 dB. computed by solving, not by drawing. The error of the straight-line sketch against the solved response — for each buffered section of an order-8 Butterworth lowpass at 1.00 kHz, and for the whole cascade. At the corner the sections are out by −5.852 dB (Q = 0.5098), −4.418 dB (Q = 0.6013), −0.915 dB (Q = 0.9000), +8.175 dB (Q = 2.5629), which add to −3.0103 dB: the whole filter's error, the same 10 log 2 as a single pole. The whole sketch is never further out than that anywhere; the section with the highest quality factor is +8.343 dB out at 1.04 kHz. The quality factors multiply to 1/√2.

The corner error a filter hides in its sections

The straight lines of an eighth-order Butterworth filter are 3.0103 dB out at the corner and nowhere worse — the same as one pole, at every order. The lines of the four sections it is built from are out by −5.85, −4.42, −0.92 and +8.17 dB there, which must add to the whole because the sections multiply. The whole sketch never gets worse with order and the worst section's error grows as 20 log(n/π). And the section the sketch misrepresents most is the one whose frequency error moves the filter most: 0.312 dB for one per cent, against the 0.173 dB any section's stopband shift gives.

frequency · Asymptotic approximation Two 1000 µF parts of 100 mΩ each against one 2000 µF part of 100 mΩ: 709 mV against 728 mV of ripple, and a crest factor of 14.41 against 12.61. computed by solving, not by drawing, marched with each capacitor behind its own series resistance. Two identical 500 µF, 60 mΩ parts give 1.330303 V of ripple and one 1000 µF, 30 mΩ part 1.330303 V — the same network. Two 1000 µF parts of a stated resistance each, against one 2000 µF part of the same resistance: at 10 mΩ, 704 mV against 704 mV of ripple and a crest factor of 17.40 against 16.93; at 30 mΩ, 704 mV against 705 mV of ripple and a crest factor of 16.51 against 15.48; at 100 mΩ, 709 mV against 728 mV of ripple and a crest factor of 14.41 against 12.61; at 300 mΩ, 758 mV against 867 mV of ripple and a crest factor of 11.45 against 9.490; at 1000 mΩ, 1.013 V against 1.334 V of ripple and a crest factor of 8.170 against 6.620. The pair's ripple is lower only where one part's resistance is past the ripple's minimum, and its peak current is higher at every resistance.

Two capacitors that are one

Two identical reservoir capacitors in parallel are not a new circuit to be marched: 500 µF at 60 mΩ twice gives 1.330303 V of ripple and so does 1000 µF at 30 mΩ once, to the sixth decimal. So a pair against one part of the same capacitance is one curve read at two resistances, and the pair always sits at the lower one — which lowers the ripple only past the curve's minimum, 709 mV against 728 at 100 mΩ a part, and raises the peak current at every value, a crest factor of 14.41 against 12.61. Mismatch the pair and the current still divides by capacitance, so a part with twice the resistance carries 2% less current and 1.93 times the heat.

applied · Unregulated supply A twin-T's null: everything fills it, and at one order. computed by solving, not by drawing. A twin-T of 10 kΩ and 10 nF nulls at 1591.5 Hz, where a low-pass T of two resistors and a high-pass T of two capacitors deliver equal and opposite contributions. With exact components the null is -Infinity dB, which is the arithmetic's floor. A tolerance fills it at 20.0 decibels per decade — first order, where an LC notch splits into twenty and forty depending on which way two errors go, because there a product can be left alone and here there is no product. Loss fills it at 20.0 decibels a decade too, and barely moves it — 1591.55 to 1595.52 Hz across four decades of series resistance — because a resistance of r in a capacitor branch is a fractional error of r/R and nothing else: one ohm gives -88.5 dB against a hundredth of a per cent of tolerance's -92.0. So a notch's two mechanisms are not exchanged here — they are one mechanism, because a twin-T's null is stationary in nothing and an LC notch's frequency is stationary in the product of its two components.

The null that is stationary in nothing

An LC notch has two mechanisms and they do different things: a tolerance moves its null rather than filling it, because the resonance depends on a product two errors can leave alone, and loss fills it at twenty decibels per decade of arm resistance. A twin-T has no resonance and no arm. Measured, everything fills it at twenty decibels a decade and nothing moves it — one ohm in series with its capacitors gives 88 decibels where a hundredth of a per cent of tolerance gives 92, because a series resistance in a capacitor branch is a fractional error of r over R and nothing else.

filters · Null depth Inside a loop of gain 1000, a twin-T error of +0.10% raises the peak's Q from 250 to 333, and the same error the other way lowers it to 200. computed by solving, not by drawing. An amplifier of gain 1000 with a 10 kΩ, 10 nF twin-T as its feedback network, whose closed-loop gain A/(1 + A·T) peaks at the null, 1591.5 Hz. With exact components the peak is 1000 and its quality factor 250.2, a quarter of one more than the gain. With the twin-T's shunt elements −0.10% out the other way the peak falls to 800 and the Q to 200.2; +0.10% raises them to 1333 and 333.5, and +0.30% to 3973 and 994. At +0.40% the loop oscillates.

The sign the null leaves behind

Put a twin-T in the feedback path of an amplifier of gain A and its null becomes a peak with a quality factor of (A + 1)/4. A filled null is not a small phasor of any phase. It is a real number, minus a quarter of the error in the shunt elements, and once the loop multiplies it by A its sign decides everything. At a gain of a thousand a 0.1 per cent error lowers the Q from 250 to 200 one way and raises it to 334 the other, and at 0.4 per cent the second direction oscillates. Every gain falls on one curve, the exact-null Q divided by 1 − A·t/4. So a Q is set to ten per cent only by a null deeper than 20 log(44·Q): one per cent parts allow a Q of 9.6.

filters · Null depth 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 On a 300 kHz amplifier the stage lags 27.9° at the null: the Q stays at 25.4 and the peak moves 1.02% below it. computed by solving, not by drawing. The 10 kΩ, 10 nF twin-T, exact, as the feedback network of a stage of gain 100 whose own bandwidth is the amplifier's gain–bandwidth over 100, so that at the null, 1591.5 Hz, it lags 27.95° (solid); dashed is an ideal stage. The quality factor is 25.440 against 25.247 — hardly moved — and so is the peak's height, 99.7 against 100.0. What moves is where the peak is: 1.024% below the null, where 2·f₀/GBW is 1.061%. At the null itself the gain is 88.3, 0.886 of the peak. With the shunt elements 1% out, the Q is 33.72 against 33.50.

The peak the amplifier moves

A twin-T around a stage of gain A makes a peak of quality factor (A + 1)/4, and a real stage has a bandwidth of its own, so its gain at the null is complex. The expectation is that the lag turns the loop's gain and the Q rises by 1/cos φ, doubling at 60°. It does not: a single pole that turns the gain by φ also shrinks it by cos φ, and the two cancel in the width of the peak — the Q holds to about one part in the gain up to 60°, and the loop still oscillates at the same shunt error, 4.1% at a gain of 100, to three per cent. What the stage's bandwidth moves is the peak itself: 2·f₀/GBW below the null, with the gain nowhere in it. At 45° of lag the frequency the circuit was built to select sits on its own −3 dB edge.

filters · Null depth A bridged-T with R₁/R₂ = 100 bottoms out at 34.2 dB, and in a loop of gain 1000 it makes a Q of 4.76. computed by solving, not by drawing. A bridged-T of two 10 nF capacitors, R₁ = 100 kΩ bridging them and R₂ = 1.00 kΩ from their junction to ground, which puts its minimum at the twin-T's 1591.5 Hz: its own transfer (dashed), bottoming at 0.0196, 34.2 dB, rather than a null; and the loop it makes as the feedback network of a gain of 1000 (solid), a peak of 48.5 with a Q of 4.760. Dotted, for scale, is the exact twin-T in a loop of gain 100: a Q of 25.2.

The residue that has only one sign

A bridged-T puts its minimum where a twin-T puts its null, with one fewer part, and its residue there is exactly 2R₂/(2R₂ + R₁): real, like the twin-T's, but positive for any values the parts can take. In a loop of gain A the two networks follow one law — the Q falls by 1 + A·ρ — and they sit on opposite sides of it. The twin-T's residue is zero with exact parts and either sign with real ones, so its loop's Q can be anything, up to oscillation. The bridged-T's loop is always damped: it rises with the gain and stops at a ceiling near √(R₁/R₂)/2, 5.00 for a ratio of 100, with no gain up to a million able to make it oscillate. What the ceiling buys is steadiness: at a Q of five, ±1% parts move the twin-T's loop by 9.1% and the bridged-T's by 2.0%.

filters · Null depth Two coupling capacitors 1% apart reject common mode by only 46.0 dB at their corner, 0.159 Hz — and perfectly at direct current and far above it. computed by solving, not by drawing at 281 frequencies: one common-mode source driving two coupling networks, each 1 µF into 1 MΩ, with the capacitors 1% apart. The difference between the two outputs, per volt of common mode, agrees with the difference of the two first-order high-passes to 6.9e-14. It is zero at direct current, where both block, and zero far above the corner, where both pass; between, it peaks at the corner, 0.159 Hz, at 5.000e-3 — half the mismatch — which is 46.0 dB of rejection. Behind a 90 dB amplifier (grey), the coupling network decides the rejection from 0.000503 Hz to 50.3 Hz.

The rejection two capacitors lose at their corner

AC-couple both inputs of a differential instrument and the two coupling networks become part of its common-mode path. Two 1 µF capacitors one per cent apart turn common mode into difference at exactly half their mismatch at their shared corner — 46.0 dB of rejection at 0.159 Hz, from a pair that is perfect at direct current and far above. Behind a 90 dB amplifier they decide the instrument's rejection over five decades around that corner. A 10 V common-mode step arrives as a 36.8 mV differential pulse one time constant later. Four hundred pairs of 20 per cent electrolytics reject a median 25 dB. Coupling the difference instead of each side removes the dip and hands the common mode to the amplifier.

instruments · AC coupling 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 One per cent more C moves the Q = 5 pair straight down, parallel to the axis, and the response by up to 3.20% at 1.093 ω₀. computed by solving, not by drawing. The upper pole of the Q = 5 series RLC low-pass before and after a 1% increase in C, solved from both netlists (upper panel: the displacement, magnified to a fixed length, beside the tangent to the circle |s| = ω₀ and the direction parallel to the axis), and the fractional change in the response's magnitude against frequency (lower panel): solved as the ratio of the two responses (solid) and predicted as the displacement projected onto the two legs (dashed, on top of it), agreeing to 3.9% of the largest change. The pole moves 0.00 × 10⁻⁴ ω₀ towards the axis and 49.88 × 10⁻⁴ ω₀ down. The largest change in the response is −3.198% at 1.093 ω₀: beside the peak, where the short leg is at forty-five degrees.

A tolerance is an arrow

Change one part of a series RLC low-pass by one per cent and its pole pair moves by a few ten-thousandths of ω₀ — and the response moves by the displacement projected onto the two distances that make it, to within four per cent of the solve. The direction is set by the part. A resistor moves the pair round its circle, away from the axis, and is felt at the peak: 0.99 per cent at Q = 5. A capacitor moves it straight down, parallel to the axis, and is felt beside the peak: +1.78 and −3.20 per cent at the half-power points and almost nothing at the peak itself. Because the short leg is ω₀/2Q long, the capacitor's effect grows by half a per cent for every unit of Q, to 10.1 per cent at Q = 20. Two hundred builds from 1% parts spread by ±2.5 per cent at the band edge.

frequency · Frequency response 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 The dual bridged-T is the same function: at a ratio of 100 both floor at 0.0196 and both loops make a Q of 4.76. computed by solving, not by drawing. The transfer of an RC bridged-T — two 10 nF capacitors bridged by 100 kΩ with 1 kΩ from their junction to ground — and of its dual — two 10 kΩ resistors bridged by 1 nF with 100 nF from their junction to ground — over two decades about 1591.5 Hz. The two curves are one to 5.7e-8 of each other, which is the solver's own conditioning: the dual's capacitor ratio plays the resistor ratio's part exactly, the residue is 2/(2 + n) for both, and in a loop of gain 1000 both make a Q of 4.760. What differs is which components carry the ratio.

The dual that is the same function

Swap a bridged-T's resistors for capacitors and its capacitors for resistors and the transfer does not change: two 10 kΩ resistors bridged by 1 nF with 100 nF to ground are the RC network's 100 kΩ and 1 kΩ around two 10 nF capacitors, to 5.7×10⁻⁸, which is the solver's own conditioning. The residue is 2/(2 + n) for both and the loop's ceiling √n/2 for both. So the choice between them is not about the response; it is about which components carry the ratio. Near the ceiling the Q moves by half the ratio's error, and a ratio of two capacitors at ±5% spreads the Q by 10.0% where two resistors at ±1% spread it by 2.1% — and drifts 2.0% over a hundred kelvin where the resistors drift 0.25.

filters · Null depth With the capacitor tracking the inductor, one per cent of drift moves the Q = 5 response by at most 1.00%, at the peak — against −2.73% beside it with the capacitor held. computed by solving, not by drawing, from two netlists. The Q = 5 series RLC low-pass with its inductor raised by 1% and its capacitor changed by (1.01)^(−t) − 1, so that at t = 1 their product is unchanged; here t = 1. The fractional change in the response against frequency (solid), against the inductor alone with the capacitor held (dashed). The upper pole moves 9.90 × 10⁻⁴ ω₀ towards the axis and 0.99 × 10⁻⁴ ω₀ up it, and ω₀ by 0.00%. The largest change is 1.00% at 1.000 ω₀; with the capacitor held it is −2.73% at 1.115 ω₀. The slider is the tracking.

Two parts that drift together

A one per cent change in the inductor of a Q = 5 resonant low-pass moves its pole pair mostly downwards, and the response by 2.73% beside the peak. Let the capacitor fall by the same fraction as the inductor rises and their product is fixed, so ω₀ is: the pair can only go round its circle, sideways near the top, and the response changes by 1.00% at the peak and nowhere else — the change a resistor makes. The difference grows with Q: at Q = 20 one per cent of drift moves the band edge 10.2% with the capacitor held, 20.7% with it drifting the same way, and 1.00% with it tracking. Over −40 to 85 °C an inductor of +150 ppm/K moves a Q = 20 band edge 9.1% with a zero-coefficient capacitor and 0.57% with a −150 ppm/K one. Correlation between two parts' errors is itself a tolerance.

frequency · Frequency response 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 The bootstrap reduced both ways: Thévenin's resistance diverges at A = 2, Norton's conductance crosses zero there. computed by solving, not by drawing at 121 gains, R₁ = R₂ = 10 kΩ. The Norton current is Vs/R₁, 0.500 mA, at every gain to a part in a billion (dashed, at 1), because shorting the node also shorts the bootstrap's input. The Norton conductance times R₁ is a straight line, 2 at A = 0 through zero at A = 2 to −1 at A = 3 (solid). The Thévenin resistance over R₁ is its reciprocal (faint, clipped at ±3): 0.5, 1 at A = 1, infinite at A = 2, −1 at A = 3. One network, two coordinate systems, and only one of them is singular exactly where the network is interesting.

The reduction that stays finite

The bootstrap's Thévenin pair diverges at A = 2: its resistance and its open-circuit voltage both pass through infinity and the solve that finds them refuses. Its Norton pair does not. The short-circuit current is Vs/R₁, 0.5 mA, at every gain to a part in a billion, and the conductance is a straight line through zero. A bench that drives a current and reads a volt loses the network for every gain from 1.992 to 2.008; a bench that drives a volt and reads a current sees it to a nanosiemens throughout. A 0.1% error in the gain moves the conductance by a constant 0.1 µS and the resistance by an amount that diverges at the pole — and four hundred bootstraps built from 0.1% parts to sit on it have conductances within ±0.3 µS of zero and resistances that are a median 10.6 MΩ, 196 of them negative.

networks · Equivalent circuit The gain a Wien oscillator needs, when one of its four parts is off its value. computed by solving, not by drawing: for each part moved alone from −5% to +5%, the gain at which the poles reach the axis, bisected on the netlist, against 1 + Rs/Rp + Cp/Cs (to ten digits at every point). A series resistor or shunt capacitor 1% high raises the gain needed to 3.0100; a shunt resistor or series capacitor 1% high lowers it to 2.9901. Every part moves the frequency the same way, by half its error: 1% high on any one of them, 1.58 kHz against 1.59 kHz. The gain needed moves by the error itself, a third of it relative to three.

The gain a tolerance demands

A Wien oscillator needs a gain of exactly three only when its four network parts are exactly equal. In general it needs 1 + Rs/Rp + Cp/Cs — bisected on the netlist to ten digits — so a series resistor 1% high asks for 3.0100 and a shunt resistor 1% high for 2.9901. Four thousand seeded oscillators from 1% parts set to the textbook's comfortable 3.02 leave a quarter of them silent; 99.9% start only at 3.072, and at that setting the median unit distorts 2.6% — the limiter has to absorb a margin the tolerance forced. The frequency spreads by one part's own spread, 0.58% for 1% parts, as large as the limiter's own 0.771% shift. Parts from one array, tracking to 0.05%, need only 3.0035, because the gain depends on ratios and the frequency on products.

applied · Oscillator The gain a Wien oscillator needs above three, against its amplifier's gain–bandwidth product. computed by solving, not by drawing: the closed-loop gain at which the oscillating pair reaches the axis, bisected on the netlist with a one-pole amplifier of 100 dB direct-current gain, for equal 10 kΩ and 10 nF parts. At 10 times the frequency, 3.15339; at 30 times the frequency, 3.02349; at 100 times the frequency, 3.00257; at 1000 times the frequency, 3.00012. Every point agrees with 3(1 + δ) + 3ε²/(1 + δ + 3ε), where ε = kf₀/GBW and δ = k/a0, to two parts in ten million. The bandwidth's share falls as the square of the ratio; below it sits a floor of 9.00e-5 from the finite direct-current gain, and the two are equal near a ratio of √(3a0) = 548.

The gain an amplifier asks for, squared

A Wien oscillator built round a real amplifier needs more than a gain of three, and the excess is easy to overrate: the amplifier's lag moves the frequency in proportion to the lag and the required gain only in proportion to its square. At a gain–bandwidth product a hundred times the oscillation frequency the frequency is 4.22 per cent low and the gain needed is 3.00257 — fifty-one times smaller in proportion. The square is also what makes the bias spread: an amplifier 30 per cent slow asks for 1.94 times the typical excess. Against 1 per cent parts the amplifier matches the network's share at a ratio of 16; against a network matched to 0.05 per cent, not until 86.

applied · Oscillator At a fixed output a 0.5 mV junction mismatch's second harmonic falls as (1 + gₘRₑ)⁻² and a 1% resistor mismatch's as (1 + gₘRₑ)⁻¹: past gₘRₑ = 1.93 the resistors set it. computed by solving, not by drawing, every waveform solved arm by arm in currents. A pair on a 1 mA tail, its output held at what the bare, balanced pair gives for 10 mV of peak drive; the second harmonic over the fundamental from a 0.5 mV VBE mismatch alone (dashed), from a 1% mismatch between the emitter resistors alone (dotted), and from both in the same sense (solid), against one plus the degeneration. Between gₘRₑ = 5 and 20 the junction's falls with slope −2.00 and the resistors' with −0.89, still approaching −1, since the local slope of gₘRₑ/(1 + gₘRₑ)² is 1/gₘRₑ − 1. The resistors' mismatch is an input offset of Vₜ·gₘRₑ·δ, which grows with the degeneration that divides it, so the two cross where gₘRₑ = ΔVBE/(Vₜδ) = 1.93; beyond it each decibel of gain given up buys one decibel of distortion, not two.

The resistor mismatch that is only an offset

Two emitter resistors that differ by δ unbalance a degenerated pair, and with an ideal tail they do it in the simplest way possible: the arm currents always sum to the tail, so the mismatch adds the constant (Iₜ/2)·R·δ = Vₜ·gₘRₑ·δ to the input and nothing else. Solved arm by arm, a 1% mismatch at gₘRₑ = 3 is the balanced pair moved by 0.776 mV to 4 × 10⁻¹⁶ of the tail, and a trim of that size leaves no second harmonic at any drive. What separates it from a junction's offset is that it grows with the degeneration meant to divide it: at a fixed output its second harmonic falls as gₘRₑ/(1 + gₘRₑ)², one decibel per decibel of gain rather than two, and it takes over from 0.5 mV of base–emitter mismatch at gₘRₑ = 1.93. For −80 dB at 10 mV of output the resistors must match to 0.823% at gₘRₑ = 1 even with perfect transistors.

limits · Small-signal A 63.7 Ω resistor in series with a gyrator's capacitor moves its series resistance's sign change from 63 Hz to 754 Hz, and 70.0 Ω removes it below the amplifiers' edge. computed by solving, not by drawing. The one-henry synthetic inductor of four 10 kΩ resistors, a 10 nF capacitor and two 1 MHz amplifiers with a gain of 10⁵, its series resistance against frequency with a resistor r₄ in series with the capacitor of 0 (grey), 0.9, 1 and 1.1 times 4/(ωₜ·C) = 63.66 Ω. Without it the resistance is −0.153 Ω at 100 Hz and −25.2 Ω at 1 kHz, negative from 63.0 Hz. r₄ adds ω²·L·C·r₄, the same frequency law, so at exactly 4/(ωₜ·C) the resistance is 0.099 Ω at 100 Hz and −0.075 Ω at 1 kHz and changes sign only at 754 Hz; at 1.1 times it stays positive throughout: 2.44 Ω at 1 kHz.

The resistor that pays the amplifiers back

A synthetic inductor's negative resistance grows as the square of frequency, and so does the real part a resistor in series with the gyrator's capacitor adds: Zᵢₙ = sL/(1 + sCr₄) ≈ sL + ω²LCr₄. Fitted on solves across six designs, the resistor that cancels the amplifiers' term is 4/(ωₜ·C) to 1.8% — 63.7 Ω for 10 nF and 1 MHz amplifiers, whatever the resistance level or the open-loop gain. Put in, it moves the sign change from 63 Hz to 754 Hz; a tenth more removes it below the amplifiers' edge and gives a quality factor of 2,579 at 1 kHz where the bare inductor had −250. The catch is that the cancellation is as good as the gain–bandwidth is known: amplifiers 10% slower than the resistor was chosen for leave −2.9 Ω at 1 kHz. Chosen a tenth over the slowest amplifier a ±30% spread allows, the inductor's Q at 1 kHz runs from 312 to 1,847 and is never negative.

filters · Gyrator The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections.

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

instruments · Compensation A follower section of Q = 5 on an amplifier 100 times the corner: +4.79% as designed from the formula; designed for that amplifier, exact at it, and +1.64% and −0.966% on parts 30% slow and fast. computed by solving, not by drawing: the realised Q of a unity-gain Sallen–Key section against the actual amplifier's gain–bandwidth, relative to the typical value it was chosen for, from the poles of the netlist. Designed from the textbook formula (solid): +6.51% at 0.7, +4.79% at 1, +3.78% at 1.3. Predistorted, with its capacitance ratio and resistors found by iteration on the netlist so that the typical part is exact (dashed): +1.64% at 0.7, exact at 1, −0.966% at 1.3. Predistortion moves the line down without changing its slope much: it removes the error the amplifier has in common and leaves what differs between amplifiers.

The Q designed for the amplifier

A follower section's Q comes out 4.79 per cent high at Q = 5 with an amplifier a hundred times the corner, and nothing stops a designer asking for 4.771 so that the real circuit lands on 5.000. That removes the error the amplifier has in common and leaves what differs between amplifiers: parts 30 per cent slow and fast read +1.64 and −0.97 per cent. So the exchange rate is the spread itself, and it is large. One per cent of Q across ±30 per cent asks for 725 times the corner from the formula and 196 designed for the part. The design runs out at a definite ratio, 7.2 times at Q = 5, below which no value of its parts reaches the Q wanted. On a whole Chebyshev 5 the passband error falls from 1.43 dB to nothing on a typical part and 0.55 on a slow one.

filters · Q enhancement

Named alongside it

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

Design tradeoffModel rangeComponent sensitivityMonte carloQuality factorTemperature coefficientGain–bandwidth productTransmission zeroCommon-mode rejectionLoop gainThe quality factorAdjoint network

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