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

Signal-to-noise ratio — where it appears

The ratio of a signal's power to the power of the noise accompanying it, in the same bandwidth. It is what decides whether a measurement exists at all rather than how accurate it is, and gain cannot improve it because gain multiplies both terms.

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 these essays measure 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 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 One capacitor moves three quantities, and the expression's own answer peaks by 1.18 dB. computed by solving, not by drawing. The bandwidth, the peaking and the total output noise of a 1.0 MΩ transimpedance stage against its feedback capacitor, swept from 0.30 to 4.20 times what the classical expression asks for. Over that factor of fourteen the bandwidth falls from 334 to 55.4 kHz, the total noise from 230 to 65 µV, and the peaking from 10.0 dB to nothing. The expression's own answer sits at 1.18 dB of peaking, 0.82× is where the loop reaches forty-five degrees, and √2× is where the response is flat — so the choice usually quoted as maximally flat is neither of the two conditions it is quoted for. The faint families are the same three quantities at 3 pF and 300 pF of diode: normalised this way they are one curve, so the diode sets the scale and the multiple sets the shape.

The factor the expression leaves out

The classical compensation for a photodiode amplifier is quoted both as the forty-five degree choice and as the maximally flat one, and it is neither: it leaves 1.18 dB of peaking and 52.4° of margin. Flat is at exactly √2 times it — fitted at 1.4186 against 1.4142, at every detector from 3 pF to 1 nF. And the third quantity the capacitor is supposed to trade, the noise, does not move at all inside the signal band: three compensations spanning a factor of fourteen give 8.37 against 8.36 µV in a 4.36 kHz measurement and 230 against 65 µV over the whole plane.

feedback · Transimpedance 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 The direct-current error a bigger feedback resistor does not fix. computed by solving, not by drawing, at direct current with the amplifier in the netlist. A 1.0 nA photocurrent into 100 MΩ gives -0.100 V. The bias current flows in the same resistor, so its contribution is the ratio of the two currents — 0.1000% at 25 °C, and the same percentage at 1 MΩ and at 1 GΩ, which is checked by changing the resistor rather than by reading an expression. The amplifier's 100 µV of offset behaves the other way: it is multiplied by one plus the resistor over the diode's own leakage, so it grows with the resistor and with temperature. Both terms double every ten kelvin, for two different reasons, and at 85 °C they are 6.40% and 0.740% against 0.100% and 0.110% at 25.

The error a bigger resistor cannot help

Every other quantity in a photodiode amplifier improves with a larger feedback resistor: the signal grows as R and the resistor's own noise as √R, so the ratio goes as √R. The bias current does not behave like either — it flows in the same resistor the signal does, so its contribution is the ratio of the two currents with the resistance cancelled, 0.1 per cent at 25 °C and the same 0.1 per cent at a megohm and at a gigohm. The offset behaves the other way, growing as one plus the resistor over the diode's own leakage, and at 85 °C the two are 6.40 and 0.74 per cent against 0.10 and 0.11 at room temperature.

feedback · Transimpedance What a feedback tee buys, and the single thing it charges. computed by solving, not by drawing. Two 50 kΩ resistors with a 6.250 kΩ tap give 500.0 kΩ of transimpedance, which is R₁(1 + R₂/R₃) + R₂ solved on the netlist. That expression has no term for the noise gain, and the tap sets it to 1 + R₁/R₃ = ×9.00 — against exactly one for a single resistor of any value, because at direct current the source is a capacitor. One quantity then does all three things at once. The signal and R₁'s own noise are multiplied together, so the tee buys no signal-to-noise ratio at all: 1846 against 5582 in a hertz at a nanoamp for one resistor of the same transimpedance, a factor of 3.02 which is the square root of the noise gain. The amplifier's own voltage noise and offset are multiplied where they were not before. And the phase margin RISES, from 1.8° to 15.9°, because starting the noise gain high shortens its climb to the peak — so the tee is a compensation that charges for itself in noise, and a capacitor across the feedback is the same compensation for nothing. With no tap the three effects are absent rather than small.

The tee that charges for its own compensation

A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.

feedback · Transimpedance A band from zero to its top frequency takes 4.77 dB less from 10 ps of jitter than a tone at the top: it is the rms frequency that counts, not the highest. computed by solving, not by drawing. Forty-eight equal tones at random phases filling a band from 0 to 1 of its top frequency, sampled 16,384 times at instants displaced by a seeded 10 ps Gaussian, at seven top frequencies from 100 kHz to 100 MHz. The measured signal-to-error ratio (dots) is 88.64 dB at 1 MHz, against −20 log(2π × rms frequency × jitter) = 88.76 dB from the band's power-weighted rms frequency, 580 kHz — and against 84.04 dB for a tone at 1 MHz itself. The gap is 4.77 dB at every frequency, 10 log(3/(1 + r + r²)) for a flat band from r to 1 of the top: 4.77 dB (a factor of √3 in frequency) for a band from zero, nothing for a narrow one. At 1 MHz the band holds 14.43 bits where the tone holds 13.67.

The frequency a band moves at

The jitter rule is written for a sinusoid: ten picoseconds holds a tone at 1 MHz to 84.04 dB. A signal that fills the band up to 1 MHz holds 88.64 dB against the same clock, because each sample's error is its own slope times its own timing error, and a band's slope is set by its rms frequency — the top frequency over √3 for a flat band, 4.77 dB better. Pink content up to 1 MHz is 11 dB better than the rule and content falling as 1/f² is 29 dB better. So the same ten picoseconds hold sixteen bits to 199 kHz for a tone, 344 kHz for a flat band and 734 kHz for a pink one: an aperture figure needs a signal as well as a frequency.

digital · Aperture jitter The same 0.1 ps of period jitter: 107.0 dB at any record length if each edge is independent, 80.3 dB over 4096 samples if the edges walk. computed by solving, not by drawing, averaged over 8 seeded clocks at each of 5 record lengths. A 10.01 MHz tone sampled at 160 MHz by a clock whose every period has a spread of 0.1 ps, in two ways: each edge independently displaced (white), and each edge displaced from where the last one landed (a free-running oscillator's walk). The sampled tone is fitted by least squares for amplitude, delay and a small frequency offset, and what is left is the error. White jitter gives 107.2, 107.0, 107.0, 107.1, 107.0 dB — the closed form 107.0 dB, at every length. The walk gives 92.6, 86.4, 80.3, 74.5, 68.2 dB for 256, 1024, 4096, 16384, 65536 samples, falling 10.1 dB a decade, as the period spread times √(n/15): a jitter with no value until the length of the look is named.

A jitter that grows with the look

Two clocks with the same 0.1 ps spread in every period sample the same 10 MHz tone. If each edge is displaced independently, the error is 107.0 dB down at any record length. If each edge is displaced from where the last one landed — a free-running oscillator — the edges walk, and the error after fitting the best delay and frequency is 92.6 dB down over 256 samples, 80.3 over 4,096 and 68.2 over 65,536: ten decibels a decade of record, as the period spread times √(n/15). The one oscillator honestly has jitter figures from 0.10 to 4.7 ps depending on the length named. A locking loop is what gives the jitter a value: with a time constant of a hundred periods it stops growing at 87.0 dB.

digital · Aperture jitter With a 2 ps reference and a 0.1 ps-a-period oscillator the best loop has a time constant of 20.0 periods, where the clock's error is 0.453 ps — 90.9 dB on a 10 MHz tone. computed by solving, not by drawing, with seeded locked clocks as dots. A free-running oscillator whose period has a spread of 0.1 ps, locked by a first-order loop to a reference whose timing error, as the phase detector sees it each period, is 2 ps rms and white. The loop passes the reference's error below its corner and the oscillator's walk above it, so the stationary error is σₒ²/(1 − a²) + σᵣ²(1 − a)/(1 + a) with a = exp(−1/τ). The oscillator's part rises as √(τ/2)·σₒ and the reference's falls as σᵣ/√(2τ); they are equal, and their sum least, at τ = 20.00 periods, the ratio σᵣ/σₒ, where the clock's error is 0.4528 ps — √(σᵣ·σₒ). Seeded clocks of 65,536 periods: τ = 2, 1.000 ps; τ = 5, 0.656 ps; τ = 20, 0.452 ps; τ = 80, 0.649 ps; τ = 300, 1.187 ps; τ = 1000, 2.105 ps.

The loop a reference allows

A sampling clock is a free-running oscillator held by a loop to a reference, and the reference has an error of its own. With 0.1 ps of period spread in the oscillator and 2 ps of white error in the reference, a first-order loop of 20 periods leaves the clock 0.453 ps from ideal — 90.9 dB on a 10 MHz tone — and that is the best any first-order loop can do: the oscillator's share rises as the square root of the loop's time constant, the reference's falls as one over it, and they are equal at σᵣ/σₒ. The clock's error there is the geometric mean of the two, so a reference a hundred times worse costs ten. On a spectrum the rule reads as one instruction: put the loop's corner where the free oscillator's skirt meets the reference's floor. And it holds only for long captures. Below about seven loop time constants the free oscillator beats every loop.

digital · Aperture jitter With a 100 kHz flicker corner the 0.1 ps oscillator's ratio stops falling ten decibels a decade of capture and starts falling twenty, past 2,277 samples — 1.42 periods of the corner. computed by solving, not by drawing, from the exact expected residual of a summed process, with seeded clocks as dots (24 at each length to 16,384). A free-running oscillator with a 0.1 ps period spread and a flicker of its frequency whose timing spectrum meets the walk's at 100 kHz, sampling a 10.01 MHz tone at 160 MHz. Signal-to-timing-error after removing each capture's best delay and frequency: the walk alone falls ten decibels a decade, 85.7 dB at 1024, 67.6 dB at 65536; with the flicker, 84.1 and 52.9 dB. The flicker's share equals the walk's at 2,277 samples, 14.2 µs, which is 1.42 periods of the corner frequency; beyond it the ratio falls 19.5 dB a decade. Removing a quadratic as well buys 3.80 dB at 65,536 samples and changes no slope.

A frequency that wanders

Below its flicker corner an oscillator's phase noise falls thirty decibels a decade instead of twenty, because the oscillator's frequency itself wanders. A clock with 0.1 ps of period spread and a 100 kHz corner samples a 10 MHz tone at 84.1 dB over 1,024 samples and 52.9 over 65,536 — 14.7 dB worse than the walk alone — because past 2,277 samples its error grows as the capture rather than its square root. That crossing is 1.42 periods of the corner frequency for every corner. Fitting a parabola as well as a line buys 3.88 dB at every length and changes no slope; each further term buys a fixed fraction. And a first-order locking loop cannot stop it: the locked clock's variance grows with the logarithm of the record, 1.92 dB lost between 4,096 and 262,144 samples, until a loop with an integrator holds it at 87.5.

digital · Aperture jitter The quantisation error's power as a function of where the input sits, for four dithers. computed by solving, not by drawing; the error's mean square given the input, integrated over each dither's density through the converter's own quantiser, and checked against seeded draws to 6 × 10⁻³. Undithered it is the square of a sawtooth, x² up to half a step and (1 − x)² beyond. One uniform step makes it x(1 − x): zero at a code, a quarter of a step² between codes — the error's power is carried by the signal. Two steps (triangular) hold it at ¼ everywhere, and one step subtracted after the converter at 1/12 everywhere. Averaged over the input the prices are 3.01, 4.77 and 0.00 dB against the undithered 1/12.

One step of dither per moment

A whole step of uniform dither makes a quantiser's error average to zero wherever the input sits, and leaves its power at x(1 − x) steps²: zero at a code and a quarter between codes, so the noise rises and falls with the signal. Two steps — a triangular dither — hold the power at a quarter everywhere and leave the third moment swinging by 0.096 steps³; three steps fix that too. Each added step fixes one more moment of the error and costs a twelfth of a step² of power: 3.01 dB for one, 4.77 for two. Subtracted after the converter, one step fixes the error's whole distribution for 0.00 dB.

digital · Quantisation

Named alongside it

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

Design tradeoffAperture jitterSeeded generatorMeasurement conditionModel rangePhase noiseTransimpedanceClosed formCorner frequencyDynamic rangeEffective bitsJohnson noise

All concepts