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 24 essays across 7 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 ladder 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 this collection has made since its ladder essay and never tested directly — that a doubly terminated ladder'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 the rungs below 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 rungs of this argument 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 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

The rung below turns one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder 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'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 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 At a stationary point a tolerance has a mean, not a spread. computed by solving, not by drawing. Six hundred ladders 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's ripple peak the first derivative of the magnitude with respect to every reactance is zero to ten digits, which the rung below measured and which says nothing about how much the response moves. This says how: every second derivative is negative, so of six hundred ladders 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

The rung below computed every second derivative of a ladder's magnitude at a ripple peak, found them all negative, and built six hundred ladders 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. Every one is symmetric under the ladder'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

The rung below diagonalised a ladder'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

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Model rangeDesign tradeoffComponent sensitivityAdjoint networkCommon-mode rejectionPolesVerificationDifference amplifierGain–bandwidth productMeasurement conditionNonlinearityNumerical error

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