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

Headroom — where it appears

How far a node inside a circuit can swing before something clips it, measured against the swing the signal actually asks of that node. It is not a property of the output: in a cascaded filter the binding node is usually an internal one, and a design can lose ten or twenty decibels of range at a node whose voltage nothing reports.

Named by 12 essays across 5 fields — each of them below, with the objects they name alongside it.

Every node of a Chebyshev 5, and the one that clips first. computed by solving, not by drawing. The largest signal each node of the realised network ever carries, over the whole frequency sweep, relative to the input. The output reaches 1.000 times the input and F0b reaches 4.537, so on a ±15 V supply the input can be driven to 3.31 V rather than 15.00 before something clips — and the thing that clips is not the output. With the floor at 249.7 nV that is 142.44 dB of range against the 155.57 dB an instrument on the output would report, a difference of 13.14 dB that no measurement at the output can see.

The ceiling is not at the output

A fifth-order Chebyshev's noisiest node is its output and its largest signal is not. F0b carries 4.537 times the input, so on ±15 V the range is 142.44 decibels and not the 155.57 an instrument on the output reports. Across fifteen realised filters the output reading spans 3.97 dB and the range they actually have spans 22.34 — and one family of the three has no internal peaking at all, at any order.

noise · Dynamic range What the accuracy costs: dynamic range against the resistance scale. computed by solving, not by drawing. The earlier essay found the active realisation's response converging on the passive one's as the resistance scale rises. This is the price. The floor rises as the square root of the scale — fitted at 0.500 — because the resistors are the noise. The largest internal swing rises as the scale itself — fitted at 1.012 — because each gyrator forces the inductor's own current through its own resistors, so an amplifier inside it carries that current times R. Dynamic range on a ±15 V supply therefore falls as the three-halves power: 78 dB at 100 kΩ and 18 dB at 10 MΩ. The passive ladder realising the same response has 141 dB, and its worst internal node carries 1.10 times the input.

Eight amplifiers, and what they add

The rung below realised a Chebyshev ladder out of floating gyrators and found the response converging on the passive one's as the resistance scale rises — twenty-two decibels out at ten kilohms, a twentieth of a decibel at ten megohms. It closed by naming two quantities it had not measured. They are the same quantity: the resistors that buy the accuracy are the noise, and the amplifiers inside the gyrators carry the inductor's own current through them, so the floor rises as the square root of the scale and the ceiling falls as the scale.

filters · Gyrator What flatness costs, in the two places it can be bought. computed by solving, not by drawing. The hold's sinc across a band ending at 0.40 of the sample rate, and the same sinc with a one-zero one-pole shelf fitted to its reciprocal over that band. The droop to be removed is 2.420 dB. Corrected digitally the band comes flat exactly and the flat level sits 2.420 dB below what the uncorrected converter gave at direct current, because nothing may exceed full scale — the price is the disease. Corrected in analogue the band comes flat to 0.2308 dB and the shelf is still rising where the images are, so the worst image at 0.60 fs comes up by 3.799 dB, which is 1.57 times the droop it removed — and that boost is between 3.4 and 4.7 dB at every band on the slider, while the droop it cures runs from 0.58 to 3.75. Neither correction changes the signal-to-noise ratio, because the droop never cost any.

Flatness, and the two currencies it is bought in

The hold's droop takes the signal and everything arriving with it down together, so it costs no signal-to-noise ratio at all — a fact that is never stated and settles what correcting it can possibly be worth. Corrected digitally the price is headroom and is exactly the disease: 2.42 dB of flatness for 2.42 dB of output level. Corrected by an analogue shelf the price is image rejection and is nearly a constant: between 3.4 and 4.7 decibels whatever the band, so it is eight times the droop at a fifth of the clock and nine tenths of it at forty-nine hundredths.

digital · Reconstruction Held one drop and 300 mV up, the cascode's floor falls from 0.832 V to 0.403 V. computed by solving, not by drawing. The output current of two cascode mirrors against their output voltage near the bottom of the supply, each divided by its own value at 5 V, on a model whose transistors saturate. The stacked mirror holds its upper device's base two drops up and stays within one per cent down to 0.832 V; the wide-swing mirror holds it one drop and 300 mV up, which leaves the lower device 0.302 V of collector-emitter voltage, and stays within one per cent down to 0.403 V. Its output resistance at 5 V is 13.2 MΩ against 7.39 MΩ, and its copy ratio 0.9758 against 0.9747.

The drop the lower device did not need

A cascode mirror's floor is a whole base-emitter drop plus the upper device's own minimum, because the stack that biases the upper base holds it two drops above ground. The lower device needs only a few hundred millivolts. Hold the upper base one drop and 300 mV up instead, and on a model whose transistors saturate the one-per-cent floor falls from 0.832 V to 0.403 V. The output resistance does not fall with it: it rises from 7.39 to 13.2 MΩ, because the upper device's base current no longer comes out of the reference. The cost has moved into the margin itself. Below about 150 mV the lower device saturates, and at 50 mV the resistance is 0.51 MΩ and the copy ratio 0.7665.

semiconductors · Cascode A mirror, cascoded or not, delivers two transistors' shot noise at low frequency — and cascoding it does not add a third. computed by solving, not by drawing. The output noise current of a 1 mA NPN mirror into a short, in units of one device's collector shot noise 2qI, against frequency, for a plain two-transistor mirror, the stacked cascode mirror and the wide-swing cascode with its upper base held by a separate diode at 100 µA. At 1 kHz they deliver 1.986, 1.960 and 1.965: the reference transistor's shot noise, copied, and the output transistor's own. At 100 MHz the mirror's own pole and the capacitance at the output transistor's collector have begun to take noise away, and the three read 1.681, 1.196 and 1.512.

The base current the mirror counts twice

A current mirror delivers two transistors' shot noise — the reference device's, copied, and the output device's own — and cascoding it does not add a third: 1.960 and 1.965 of one device's noise for the stacked and wide-swing cascodes against 1.986 for the plain mirror. What the cascode adds is its upper output transistor's base noise, and there the two arrangements differ by a factor of four. In the wide-swing mirror the base current goes to a separate bias and counts once, one part in β. In the stacked mirror it goes into the reference branch, which mirrors it back into the node it left, and it counts 3.82 times at β = 150 — the same base current that took 43 per cent of that mirror's output resistance. The reference resistor's noise, meanwhile, adds almost nothing even at 1.8 V, because it takes its place in the diode's node rather than adding to it.

semiconductors · Cascode An unregulated supply's output against its load: 3.38 Ω of slope with perfect parts, 3.64 with 0.3 Ω in the winding, 3.32 with it in the capacitor. computed by solving, not by drawing, marched with the diodes in the netlist: a 17 V peak centre-tapped rectifier into 1000 µF, loaded from 200 Ω down to 25 Ω. The mean output with perfect parts falls from 15.99 V at 80 mA to 14.33 V at 573 mA, an average slope of 3.380 Ω against the textbook's 1/(4fC) = 5 Ω, drawn dashed without its diode drop. With 0.3 Ω in each half-winding the slope is 3.642 Ω; with the same 0.3 Ω in the capacitor it is 3.317 Ω. The slider is the resistance.

The resistance a supply's regulation counts

An unregulated supply's output falls as its load rises, and the slope is its output resistance: 4.17 Ω at 80 mA and 2.80 at 575 for a thousand microfarads, below the textbook's 1/(4fC) = 5 Ω and falling, because the diodes conduct for longer as the load grows. A winding resistance adds to that slope at about its own value, 0.94 to 1.18 of it; a capacitor's adds −0.14 to −0.25. The guess that each should count by the share of the cycle it carries current in — a sixth and five sixths — is wrong in both places, and the right weighting is the mean current each carries: all of the load's through the winding, none through the capacitor.

applied · Unregulated supply With a leak of a hundred periods the flux settles to a spread of 1.22% of the swing at β = 0.5 — 0.759 of what independent duty errors would give. Seeded: three runs of sixty thousand periods of the event map, 5 mV of threshold noise, β = 0.5, driving a magnetising branch whose flux leaks with τ = L/R. The stationary spread of the period-end flux in volt-nanoseconds against τ in periods, quoted here as a percentage of one half cycle's swing V·T/2 = 5493 V·ns: 1 → 0.146% (closed form 0.145%), 3 → 0.233% (closed form 0.232%), 10 → 0.401% (closed form 0.397%), 30 → 0.685% (closed form 0.673%), 100 → 1.25% (closed form 1.22%), 300 → 2.16% (closed form 2.10%). The closed form linearises one period, keeping the leak inside each half and the covariance −AB between consecutive periods' imbalances; dropping that covariance gives the upper curve, 1.61% at a hundred periods. For a slow leak the spread is V·σ·|B − A|·√(τ/T), the drawn asymptote.

The spread a winding resistance holds

A relaxation oscillator's threshold noise walks an ideal core's flux without limit. Put the winding's resistance back and the walk stops at a spread: 1.22% of a half cycle's swing when L/R is a hundred periods, growing as the square root of L/R, while a comparator skew's offset grows as L/R itself. The anticorrelation between consecutive duty errors is worth 24% of that spread at β = 0.5 and 62% at β = 0.2 — but only in a core slow enough to remember several periods. And the resistance that bounds the walk is a boundary in its own right: with a millihenry and a fifth of the swing to spare, an ohm tolerates 12.9 mV of noise, and a hundred picoseconds of skew needs at least 0.14 Ω.

applied · Hysteresis With a 300 ps skew and 5 mV of noise a core with 100 ns of headroom saturates after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. Seeded: 2,000 drives of an ideal core by the event map at β = 0.5, 5 mV of threshold noise and a comparator skew of 300 ps, each run until the summed high-minus-low time leaves ±100 ns. A histogram of the counts on logarithmic bins, against the closed-form first-passage distribution, between two barriers, of a walk with drift 0.200 ns a period and spread 1.89 ns per root period. Mean 501.9 against 499.9; quantiles 1% 184, 10% 275, 50% 463; the empirical and closed-form cumulative distributions differ by at most 0.014.

The count a core survives

A relaxation oscillator with a 300 ps comparator skew and 5 mV of threshold noise saturates an ideal core with 100 ns of volt-second headroom after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. The mean does not see the noise at all until the noise is large enough to saturate the core the other way first; the early tail sees nothing else. Every skew and noise level falls on one curve in one ratio, the skew's count over the count at which its ramp overtakes the walk. Read as a burst, a thousand periods with one failure in a thousand allows 150 ps of skew with no noise, 93.5 ps at a millivolt, and none at all above 2.41 mV.

applied · Hysteresis The Chebyshev 5's sections, each node's peak before and after equal-maxima scaling. computed by solving, not by drawing. Unscaled (lighter), section by section the largest signal anywhere inside reaches 4.537×, 3.109×, 1.000× the input, so the ceiling is 4.537× at F0b. With a gain of 0.220 at the input, 1.000 and 1.602 between sections and 2.833 at the output, every section peaks at 1× (darker) and the response is unchanged. The floor rises from 249.7 to 883.8 nV, so the range gains 2.16 dB of the 13.14 dB the internal peak was costing.

The ceiling a gain gives back

A fifth-order Chebyshev loses 13.14 dB of its range because its first section's capacitor node carries 4.537 times the input. The textbook repair is to scale: divide the signal in front of the loud section, restore it later, so that every node peaks at the same level. Done exactly, with silent gains of 0.220, 1.000, 1.602 and 2.833 and the response unchanged to the last digit, it lowers the ceiling to the input and raises the floor from 249.7 to 883.8 nV — and gives back 2.16 dB of the 13.14. Every section between the division and the restoration has its noise multiplied by what the signal was divided by. On the even-order Chebyshevs scaling loses half a decibel; on the seventh-order it recovers 3.77 of 22.45. Reordering the sections and scaling them together gives 4.75 dB. The rest of the ceiling belongs to the high-Q section and no gain moves it.

noise · Dynamic range The supply's mean output against its load, with no leakage inductance, 0.3 mH and 1 mH in the winding. computed by solving, not by drawing, marched with the diodes in the netlist: 17 V peak into 1000 µF, 50 mΩ of winding throughout. Without inductance the output falls 1.69 V from 200 Ω to 25 Ω, an average slope of 3.43 Ω. With 0.3 mH it falls 0.62 V, 1.14 Ω, from a higher start; with 1 mH, 0.42 V, 0.77 Ω. An element that carries the load's current and drops no mean voltage makes the regulation better, not worse.

The inductance that flattens the supply

An unregulated supply's output resistance counts each resistance by the mean current it carries, and an inductance in the winding carries the whole mean current and drops no mean voltage — so by that rule it should count for nothing. Marched, it counts for less than nothing: 0.3 mH of leakage takes the output resistance at 100 Ω from 3.88 Ω to 0.67, and the mean output falls 0.62 V from 200 Ω to 25 Ω instead of 1.69. The effect has an interior optimum — 0.03 mH makes it slightly worse, 0.3 mH best, 3 mH back to 2.86 Ω — because the inductance first rings the reservoir up and then turns the rectifier toward choke input.

applied · Unregulated supply Forty thousand periods at β = 0.5: each period's error against the same period's duty error, correlated by 0.008. Seeded: forty thousand periods of the event map, β = 0.5, 5 mV of threshold noise; one in twenty is drawn, each scaled by its own standard deviation (3.4 ns for the period, 2.49 ns for the high-minus-low time). A period is three draws with coefficients A, A + B, B and its duty error the same three with A, B − A, −B, so their covariance is σ²[A² + (A + B)(B − A) − B²], which is zero at every β. Measured, the correlation is 0.0080, inside the ±0.0100 that forty thousand samples resolve. With normal draws, uncorrelated is independent: the cloud is round.

Every draw shared, and nothing in common

A relaxation oscillator's period error and its volt-second error are made of the same threshold draws, and they are exactly uncorrelated. A period weighs its three draws A, A + B and B; the high half less the low weighs the same three A, B − A and −B; the products sum to A² + B² − A² − B², which is zero at every hysteresis. Over forty thousand periods the measured correlation is 0.008 against a resolution of 0.010. What the two do share is one draw with the period on either side, ±0.157 at β = 0.5, and over any whole burst those two lags cancel term for term. Four thousand bursts judged against twenty-five pairs of limits fail both ways exactly as often as the product of the two rates says: the margins specified separately double-count nothing.

applied · Hysteresis Scaling a Chebyshev 5 with real 4 nV/√Hz amplifiers everywhere: 118.18 dB at equal maxima and 118.32 dB with no gain stages at all, so the gains cost 0.14 dB, where silent ones promised 2.16 dB. computed by solving, not by drawing: the dynamic range on ±15 V of a fifth-order Chebyshev with gains d^α in front of each section and after the last, against α. Every amplifier has a voltage noise of 4 nV/√Hz, the sections' own and the stages', and each stage carries a kilohm in its divider or lower feedback arm (solid). With every amplifier and gain silent (dashed), equal maxima recovers 2.16 dB over no scaling, 142.44 dB to 144.60 dB. With real ones: 118.32 dB with no stages, and with stages 115.75 dB at α = 0.2, 116.27 dB at α = 0.4, 116.86 dB at α = 0.6, 117.50 dB at α = 0.8, 118.18 dB at α = 1. Merely having the stages is a step down at the smallest α, since each is a real amplifier in the path, and the curve climbs from there to equal maxima. The response is unchanged to a part in a billion at every α.

The floor the gains bring

Equal-maxima scaling recovers 2.16 decibels of a fifth-order Chebyshev's range when its gains make no noise and its amplifiers make none either. Give every amplifier in the circuit a voltage noise of 4 nV/√Hz, the sections' own and the stages', and the unscaled filter reads 118.32 dB over a 1 MHz band and the scaled one 118.18: the last gain has nothing after it to filter its noise, so it multiplies 4 µV of wide-band noise by 2.83, and the lower ceiling is spent paying for it. At 1 nV/√Hz scaling costs 6.61 dB, and at 16 nV/√Hz it gains 0.99, so a noisier amplifier makes it look better. A narrower measuring band does not rescue it. On a Chebyshev 7, with a node 13.26 times the input, the same arrangement gains 6.15 dB.

noise · Dynamic range

Named alongside it

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

Design tradeoffDynamic rangeOutput impedanceRealisationJohnson noiseNoise floorRandom walkRelaxation oscillatorSeeded generatorVolt-secondsCascodeConduction angle

All concepts