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

Dynamic range — where it appears

The span between the smallest signal a circuit resolves above its own noise and the largest it carries before distorting. Both ends move with the measurement bandwidth, so a range quoted without one is a number about a circuit that has not been specified.

Named by 11 essays across 4 fields — each of them below, with the objects they name alongside it.

The usable range of one stage, in a 10 kHz measurement. computed by solving, not by drawing. The floor is the Johnson noise of a 1 kΩ source in the measurement's own noise bandwidth — 501.6 nV, using 15.7 kHz rather than the 10 kHz corner. The ceiling is the drive at which an exponential's distortion reaches one per cent, 1.03 mV. Between them is 66.3 dB, and nothing a designer does moves either number without changing the circuit.

A floor and a ceiling

Every other boundary on this site is a ceiling. This one puts a floor underneath and measures the distance between them — 4.00 nanovolts per root hertz at the bottom, one per cent of harmonic distortion at 1.03 millivolts at the top, and 66.3 decibels of range in a ten-kilohertz measurement. Both ends are computed, neither is on a datasheet, and the arrangement of the stages decides which one moves.

noise · Dynamic range Where a converter stops measuring the signal and starts measuring the resistor. computed by solving, not by drawing. The quantisation floor is q/√12 and halves with every bit; the Johnson floor of a 1 kΩ source in 100 kHz is 1.266 µV and does not move. They cross at 18.80 bits. Below that the converter is the limit; above it the resistor is, and a further bit buys a more precise measurement of thermal noise. A resolution quoted without a source impedance and a bandwidth is not a resolution — which is the same sentence the instruments field makes about a probe.

The floor a converter sets

A converter's resolution is quoted as a number of bits, which is a property of the converter. What it can actually resolve is a property of the circuit in front of it, and the two cross: measured against the Johnson noise of a 1 kΩ source in 100 kHz of bandwidth, the quantiser is the limit up to 18.80 bits and the resistor is the limit above it. Past that crossing every further bit buys a more precise measurement of thermal noise.

digital · Quantisation 6.02 bits + 1.76 dB, measured — and the overload that is blamed on it. computed by solving, not by drawing. A coherent sinusoid is quantised and the ratio is read off the error sequence, never from the formula. One step inside full scale the two agree to 0.15 dB at every bit count from 8 to 16. Driven to exactly full scale the same converter measures 0.83 dB worse at 8 bits when driven to exactly full scale, and the reason is in the count beside it: 327 of 8192 samples hit the top code. A mid-tread converter's largest code is one step below full scale, so a sinusoid that reaches full scale is already overloading — half a step of overload, at the peak, where the error correlates with the signal. The loss halves with every bit for the same reason the clipped count does — 0.83, 0.47, 0.21, 0.06, -0.13 dB — and by sixteen bits it is inside the measurement's own scatter. The formula was never the approximation.

Six decibels a bit, and the half step blamed on it

Every converter data sheet quotes 6.02N + 1.76 dB and every bench measurement comes up short of it, which is usually explained by calling the formula an approximation. Measured one step inside full scale it agrees to 0.15 dB at every bit count from eight to sixteen. Driven to exactly full scale the same converter loses 0.83 dB at eight bits — because 327 of 8192 samples hit the top code, and a mid-tread converter's largest code is one step below full scale.

digital · Quantisation kT/C, on a circuit whose resistance is a clock and moves by 1e+4. computed by solving, not by drawing. The noise a switched-capacitor low-pass leaves on its holding capacitor, against the ratio of the two capacitors, measured by marching a seeded sequence through the recursion the circuit obeys. It is 6.328 µV at every ratio drawn — kT/Ch, with the holding capacitor and nothing else in it — while the resistance the arrangement behaves as moves from 1.00 kΩ to 10.0 MΩ across the same axis. The two lines below it are the two sampling events that make it: what the input switch leaves on the switched capacitor, and what the sharing switch leaves behind when it opens. Neither is the answer and their sum is, exactly, because the pole's own bandwidth factor is the same expression.

The noise a clock does not make

A hundred-megohm resistor has a noise density of 1.27 microvolts per root hertz. A switched capacitor that behaves as a hundred megohms has none of it: the noise on the capacitor it charges is kT/C, with the holding capacitor in it and nothing else — not the clock, not the switched capacitor, not the on-resistance. And it is exact rather than asymptotic, at every capacitor ratio from a thousandth to ten.

noise · Kt over c What 10 ps of aperture jitter is worth, in bits. computed by solving, not by drawing. Samples are taken at instants displaced by a seeded Gaussian of 10 ps and the error is measured against the same sinusoid sampled exactly. The line is −20 log(2π f × jitter), which the measurement matches to 0.12 dB across three decades. The penalty is exactly twenty decibels a decade of input frequency, because the error is the signal's slope times the timing error and nothing else — so a converter holds 16 bits only up to 199 kHz and 12 bits up to 3.18 MHz. An aperture figure quoted without an input frequency states no resolution at all.

A picosecond, read as bits

Every other boundary in this collection has a frequency, an amplitude or a size on its axis. This one has a duration. A converter that samples at t + δ instead of t gets a value wrong by the slope times δ, so the damage is proportional to input frequency and to nothing else about the part: ten picoseconds holds sixteen bits up to 199 kHz and twelve bits up to 3.18 MHz, falling at exactly twenty decibels a decade. An aperture figure quoted without an input frequency states no resolution at all.

digital · Aperture jitter 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 a cascaded modulator is worth, against how well its two paths match. computed by solving, not by drawing. Two first-order loops marched sample by sample at an oversampling ratio of 64, with the first stage's quantisation error taken as the difference between what its comparator said and what was presented to it — nothing here reads a state a real converter could not. Perfectly matched, the cascade gives 72.5 decibels against the first stage's own 47.5: second-order shaping out of two first-order loops, neither of which can be unstable. The gain with which the first error reaches the second stage is then given an error, and the flat left-hand half of the curve is the arrangement working. It costs three decibels at 5.29 per cent, which is a capacitor ratio — achievable, and not free, and not something the digital side can measure or correct.

Two loops, and the mismatch between them

The rung below marched single loops of second, third and fourth order and found the amplitude at which each stops working, falling with the order — which is why nobody builds a fourth-order single loop. The standard answer is two first-order loops with the first one's error fed to the second and differentiated back out, giving second-order shaping out of parts that cannot be unstable. The cancellation is between an analogue path and a digital one, and it is worth 25 decibels until the two differ by five per cent.

digital · Noise shaping Same 2 ps of jitter: a floor at -121 dB, or a line at -91. computed by solving, not by drawing. A 3.337 MHz sinusoid sampled at 10 MHz with two clocks of identical root-mean-square jitter. One clock's timing error is independent gaussians and the other's is a sinusoid at 130 kHz. The total error power is the same — -87.43 and -87.55 decibels below the signal, against the closed form's -87.55 — and the pictures are not. The random clock spreads it over 2048 bins, -120.5 dB each; the modulated one puts it into two lines at 3.337 ± 0.130 MHz, -91.5 dB down. A specification in picoseconds does not choose between them.

A floor, or a line

Two clocks with identical two-picosecond jitter sample the same sinusoid, and the total error power is the same to a tenth of a decibel — the closed form the rung below computed, right for both. One of them puts that error across two thousand bins at −120 dB each; the other puts it into two lines at −91.5. The gap is the processing gain, it grows with the record length because a density falls and a line does not, and nothing in a specification quoted in picoseconds root-mean-square distinguishes the two cases.

digital · Aperture jitter What each factor of attenuation buys on a 2.0 kΩ source. computed by solving, not by drawing at 12 probe ratios: the one-per-cent frequency bisected on the node with and without the probe, against the frequency a tip capacitance alone would predict. A one-to-one probe reaches 6.79 kHz and a hundred-to-one 692 kHz. The first step, from 1× to 2×, multiplies the bandwidth by 2.03 for a factor of two in signal; the two routes differ by at most 1.8% across the sweep, and they differ at all only because the probe's 1.0 MΩ is already 0.20% of the reading before any frequency is applied.

The probe that takes a tenth

A ten-to-one probe buys an order of bandwidth for a tenth of the signal, and on a two-kilohm source the bandwidth is exact: 6.79 kHz becomes 69.2 kHz. The tenth of the signal is not a tenth of the signal-to-noise ratio. Solved resistor by resistor, the noise referred to the tip goes from 1.782 µV to 55.78 µV — a factor of 31.3 — because the divider that does the attenuating is nine megohms and a megohm, and √(n(n−1)kT/C) on the cable's own capacitance has no source resistance in it at all.

instruments · Probe loading Twenty-four orderings, and 11 of them are choices. computed by solving, not by drawing. Every ordering of the four sections of an eighth-order 0.5 dB Chebyshev, at 16 bits, drawn against the two things an ordering decides. The horizontal axis is the round-off floor the arrangement adds — 24.2 least significant bits at best and 99.8 at worst. The vertical axis is the largest value any section's output reaches, which is what decides whether a word overflows: 0.088 of full scale at best and 0.699 at worst, a range of 18.0 decibels. 13 of the twenty-four are beaten on both counts by another ordering and are simply mistakes; the 11 on the lower-left frontier are the actual choices, and no one of them is best.

Which section goes first

A cascade of four biquads can be assembled in twenty-four orders, all of which realise exactly the same transfer function. They do not cost the same: the round-off floor runs from 24 to 100 least significant bits and the largest value any section reaches runs over eighteen decibels — and the two go opposite ways, so eleven of the twenty-four are genuine choices and thirteen are beaten on both counts by another arrangement.

digital · Digital realisation What the accuracy costs: dynamic range against the resistance scale. computed by solving, not by drawing. The rung below 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

Named alongside it

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

Design tradeoffJohnson noiseQuantisationEffective bitsNoise floorRealisationAperture jitterHeadroomModel rangeNoise bandwidthNoise figureSignal-to-noise ratio

All concepts