Circuit Response
every model has an edge, and the figure carries it
The floor, which bounds from below

The floor a resistor sets

A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.
17 min read 7 figures Two routes to a numberEvery model has an edgeRefused, not extrapolated

Every boundary this collection has measured so far is an upper one. A model is true up to a frequency, up to an amplitude, up to a size, and the figures are about where the ceiling is.

This field is the other end. Below some level a circuit’s output stops being a scaled copy of its input and becomes a sample of something else, and no amount of gain moves the level, because gain amplifies the floor along with the signal.

The noise of a 1.6 kΩ resistor through a 10.0 kHz filtercomputed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz.±619.3 nV, the root mean square1200 of 192001 stepssampled, this seed621.2 nVsampled, 6 seeds619.3 nV ± 1.79%integrated from |H|²620.6 nV…ignoring the hold626.3 nVnoise bandwidth15.03 kHzthe −3 dB point10.00 kHzsolved, then checked — a sample against an integralagreeing to 0.21%
Fig. 1 The noise of a 1.6 kΩ resistor through a 10 kHz filter, computed two ways. The trace is a seeded white sequence marched through the network by the trapezoidal rule; the table beside it compares the root-mean-square of that trace, the mean across six seeds with its spread, and the same quantity integrated from the solved |H(f)|² with no random number anywhere. The slider is the filter’s corner.

What a warm resistor does

A resistor in thermal equilibrium produces a voltage across its own terminals whose one-sided spectral density is 4kTR volts squared per hertz. That is the whole of Johnson–Nyquist noise, and two things about it are remarkable enough to state separately.

It depends on nothing about the resistor except its resistance and its temperature. Not the material, not the geometry, not the manufacturing process. Two resistors of the same value at the same temperature produce the same noise, which is why this is a floor rather than a component specification.

It is flat with frequency. There is no corner, no roll-off, no shape — the density is the same at one hertz and at one gigahertz, which is what “white” means and what makes the quantity a density rather than a voltage.

At room temperature the numbers work out to 0.895 nV/√Hz for fifty ohms, 4.00 for a kilohm, 12.7 for ten kilohms and 127 for a megohm. The scaling is with √R, since the density is proportional to R and the voltage to its square root, so a hundredfold change in resistance is a tenfold change in noise.

The one number worth memorising is the middle one: a kilohm is four nanovolts per root hertz. Everything else follows from √R.

Why a density needs a bandwidth

A density is not a voltage, and turning one into the other requires a bandwidth — which is where almost every mistake in this area is made.

The noise voltage out of a system is √(density × bandwidth), and the bandwidth in that expression is not the −3 dB point. It is the equivalent noise bandwidth: the width of the rectangular filter that would pass the same noise power. For a single pole that is π/2 = 1.571 times the −3 dB point, so using the corner frequency instead understates the noise by 21%.

That factor is the subject of the bandwidth noise sees and is measured there rather than quoted. Here it is enough to say that the figures on this page all use it, and that the table beside the trace prints both numbers so the difference is visible: for the 10 kHz filter drawn, the noise bandwidth is 15.5 kHz.

A single pole, and the brick wall that passes the same noisecomputed by solving, not by drawing and integrated over 267 frequencies. The equivalent noise bandwidth is 1.5706 times the −3 dB point, and π/2 is 1.5708. A noise voltage computed with the corner frequency instead is 20.2% low. The area under the curve and the area of the rectangle are the same number.00.500101234frequency ÷ the −3 dB point|H|², the power that gets throughhalf the power: the −3 dB pointthe brick wall: 1.5706× the corner1.571×solved, then checked — the area integrated, not tabulated1.571× the corner, not 1×
Fig. 2 A single pole’s squared magnitude and the brick wall that passes the same noise power. The two enclose the same area, and the rectangle is π/2 times as wide as the −3 dB point — measured by integrating the solved response, and agreeing with π/2 to four decimal places.

Two routes, as far apart as this collection gets

The comparison on this page is between a simulation and an integral, and they could hardly share less.

The sampled route generates a sequence of independent normal deviates from a seeded stream, scales them to the required density, holds each one for a sample interval, and marches the network forward in time with the trapezoidal rule. It produces a voltage that varies, and its root-mean-square is the answer.

The analytic route integrates the spectral density against |H(f)|² over the solved response and never draws a random number at all.

They share the netlist and nothing else — not a time step, not a frequency, not a draw. And they agree to a fraction of the seed-to-seed spread, which is itself measured rather than assumed: six seeds give a spread of about one per cent, and the two routes differ by a tenth of that.

Getting them to agree took finding a real error

They did not agree at first, and what happened next is the part of this page worth reading.

At one integration step per sample the two routes disagreed by between two and seven per cent. That is many times the seed-to-seed spread, so it was not sampling noise; and it moved with the sample rate rather than shrinking with the run length, which is the signature of a discretisation error rather than of a noisy estimate. Two were found.

The trapezoidal rule was interpolating between noise samples rather than holding them. Stepping once per sample makes the integrator treat the drive as piecewise linear through the sample points, which is a different input from the piecewise constant one whose density was computed. Stepping eight times per sample makes the drive genuinely held, and the disagreement fell to about one per cent.

The held sample has a spectrum of its own. A sequence held constant for h seconds does not have a flat density; it has one shaped by sinc²(πfh), falling to 0.405 of its low-frequency value at the Nyquist frequency. The analytic route was integrating a flat density that the sampled route never had. Putting the sinc² into the integral took the disagreement from 0.8–1.2% to 0.07–0.3%.

The evidence that both diagnoses were right is that the residual shrinks as the sample rate rises in exactly the way each mechanism predicts: with the hold correction in place the two routes agree to 0.30% at thirty samples per corner, 0.11% at fifty, and 0.07% at a hundred.

The alternative was to widen a tolerance until the check passed. It would have taken one line and it would have thrown away both findings.

The noise of a 1.6 kΩ resistor through a 1.00 kHz filtercomputed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 196.0 nV across six seeds, spread 1.84%. Integrating the same density against the solved |H(f)|² gives 196.3 nV — -0.11% apart, well inside the spread. The noise bandwidth is 1.50 kHz against a −3 dB point of 1.00 kHz.±196.0 nV, the root mean square1200 of 192001 stepssampled, this seed196.5 nVsampled, 6 seeds196.0 nV ± 1.84%integrated from |H|²196.3 nV…ignoring the hold198.1 nVnoise bandwidth1.50 kHzthe −3 dB point1.00 kHzsolved, then checked — a sample against an integralagreeing to 0.11%
Fig. 3 The same comparison with the filter ten times narrower. The noise voltage falls by √10, the noise bandwidth falls by ten, and the two routes still agree inside the sampling spread. Nothing about the argument depends on where the corner is, which is what makes the agreement a check on the machinery rather than on a particular case.

The factor of two that decides everything

Between a sequence of samples and a spectral density there is a conversion, and it contains a factor of two that is very easy to lose. Losing it would make every noise voltage on this site wrong by √2 — forty-one per cent — and nothing except the two-route comparison would notice.

A sequence of independent samples with variance σ², taken every h seconds, carries all its power in a band from zero to 1/2h. So its one-sided spectral density is 2hσ², and generating a sequence of stated density requires σ = √(density/2h).

There is no way to check that by inspection. It is one line, both plausible answers differ by two, and a figure drawn with the wrong one looks entirely reasonable — noise is noise, and a trace with forty-one per cent more of it than it should have is not visibly different from a correct one.

What catches it is the comparison. The analytic route computes the output from the density directly and knows nothing about σ; the sampled route computes it from σ and knows nothing about the density. A factor of two in the conversion is a factor of √2 in the disagreement, which is forty-one per cent against a seed-to-seed spread of one per cent — impossible to miss and impossible to explain away.

That is what a second route is for, and it is worth being explicit about the shape: the check is valuable in proportion to how little the two routes share. Here they share the netlist. A version of the analytic route that took σ as its input would share the conversion too, and would confirm a mistake instead of catching it.

What a seed is doing in a figure

This is the only field on this site whose figures are samples, and that changes what they have to carry.

A trace drawn from one seed shows one draw from a distribution of traces. Its root-mean-square is a statistic with a spread, and quoting it as though it were a measurement would be quoting a number that changes if the figure is redrawn. So every number on this page that is claimed about behaviour rather than about one dataset is run across seeds, and the spread is printed beside the middle.

The spread is not decoration. It is where the tolerance for the two-route comparison comes from: the figure asserts that the sampled and integrated routes agree to within two and a half times the measured seed-to-seed spread, which is a tolerance taken from the measurement rather than chosen to make it pass.

The generator itself comes from verify-kit rather than from a copy written here — the same counter-based stream, the same seeding, the same summaries that normaldistribution.xyz uses, after that site found two defects in the obvious version. A third copy would have been a third chance to reintroduce them.

Three families at order 3, and how much noise each lets throughcomputed by solving, not by drawing. All three are normalised to the same −3 dB point, so the bars compare the noise each admits for the same nominal corner: butterworth 1.047, chebyshev 0.999, bessel 1.074. Chebyshev's is the lowest and at order five it drops below one — its response is already falling inside the band while its −3 dB point sits beyond the ripple, so it passes less noise than a brick wall at its own corner would.-40-2001001k10kfrequency (hertz)gain (decibels)the −3 dB point they sharenoise bandwidth ÷ the −3 dB pointButterworth1.0472×Chebyshev0.9992×Bessel1.0736×a brick wall at the cornersolved, then checked — three integrals, one cornerChebyshev at 0.999× the corner
Fig. 4 The noise bandwidth of three filter families at the same −3 dB point. Every number on this page needs one of these, and they are not all the same: a three-pole Butterworth passes 1.047 times its corner and a three-pole Chebyshev passes 0.999 — less than a brick wall at its own corner, which is the next essay’s finding.

Where the floor comes from in a real circuit

The resistor is the simplest case and rarely the dominant one. Two other contributions matter and are worth naming, because a floor computed from resistance alone is optimistic.

The amplifier’s own noise. Every active device contributes a voltage noise and a current noise referred to its input, and the current noise flowing in the source resistance produces a voltage that adds to the rest. The three combine in quadrature, and which dominates depends on the source resistance: voltage noise at low source impedance, current noise at high, and the resistor’s own thermal noise in between. The resistance at which the total is minimised is a real design parameter and this site does not compute it, because it needs device noise parameters that no model here carries.

Excess noise. Real components produce more than thermal noise at low frequency — a density rising as 1/f below a corner that can be anywhere from a few hertz to a few megahertz depending on the device. Nothing on this page models it, and for measurements that average over long times it is the dominant term.

Both of those are stated rather than measured, which is a shortfall by this collection’s own standard. The honest description of what is here is: the thermal floor, computed exactly, with everything that sits above it named.

The same three stages, in two orderscomputed by solving, not by drawing by Friis's cascade. With the low-noise amplifier first the chain's noise figure is 1.31 dB; with the mixer first it is 10.01 dB. The gain is identical either way — 50.0 dB — so the 8.70 dB is bought with nothing but an ordering. Everything after the first stage contributes 6.9% of the total.quiet stage first — noise figure 1.31 dBlow-noise amplifier1.2589mixer0.0900intermediate amplifier0.0030noisy stage first — noise figure 10.01 dBmixer10.0000low-noise amplifier0.0259intermediate amplifier0.0030each bar is that stage's contribution to the noise factor, after division by the gain in front of itsolved, then checked — contributions, not totals8.7 dB from an ordering
Fig. 5 What happens to a floor when it passes through a chain of amplifiers, from the field’s third essay. The first stage decides the answer and the rest very nearly do not — the same three stages in two orders give 1.31 dB and 10.01 dB of noise figure — which is the practical consequence of everything on this page.
The usable range of one stage, in a 10 kHz measurementcomputed 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.1e-81e-71e-61e-51e-41m10m100m1amplitude (volts)Johnson noise of 1 kΩ: 501.6 nV1% distortion, exponential: 1.03 mV1% gain error, exponential: 7.30 mVlinear step, unity follower: 79.58 mV66.3 dB of rangesolved, then checked — a floor and a ceiling, both computed66 dB between them
Fig. 6 The floor and the ceiling of one stage on a single amplitude axis, from the field’s third essay. The left-hand mark is the quantity this page computes; the right-hand ones are the amplitudes at which distortion and slewing take over. The distance between them is what “dynamic range” means, and both ends of it come out of measurements rather than out of a specification.

Where the constant comes from

The expression 4kTR has a factor in it that is worth accounting for, since every other number on this page depends on it and it is the kind of constant that gets misremembered.

Nyquist’s argument is a thermodynamic one rather than a circuit one. A resistor connected to a transmission line in thermal equilibrium must exchange power with it at the rate equipartition requires, and working out what voltage source would deliver that power gives a mean square of 4kTRB across a bandwidth B. The factor of four is a consequence of the source being loaded by a matched resistance — the available power is v²/4R, so the open-circuit voltage has to be twice what a naive reading suggests.

That derivation is the reason the result contains no property of the resistor beyond its resistance. Nothing in it refers to carriers, or to a material, or to how the component was made, which is why the same number applies to a carbon film resistor, a wirewound one, and the radiation resistance of an antenna pointed at a warm sky.

The number that makes it real

A kilohm in a ten-kilohertz noise bandwidth produces 400 nanovolts.

That sounds negligible until it is put beside something. A twelve-bit converter with a one-volt range has a least significant bit of 244 µV, so the resistor’s noise is a six-hundredth of a bit and genuinely negligible. A sixteen-bit converter’s bit is 15 µV and it is a fortieth. A twenty-four-bit converter’s bit is 60 nV, and the resistor’s noise is nearly seven bits — which is why a twenty-four-bit converter’s specification talks about effective bits and why the last six or seven of them are a property of the source impedance rather than of the converter.

That is the shape of every argument in this field. The floor is a small number, it is a hard number, and whether it matters is decided entirely by what is being asked of the measurement.

The inverse question is the one a design usually asks, and it has an equally short answer. The resistance at which Johnson noise reaches a stated level in a stated bandwidth is R = V²/(4kTB), so reaching one microvolt of noise in a ten-kilohertz bandwidth takes 6.2 kΩ. Anything below that resistance is quieter than a microvolt in that bandwidth, and anything above is not — which turns “how large may this resistor be” into arithmetic rather than judgement, and is the form in which this number is most often needed.

Where four of this site's models stop being trueIn order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is.101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV
Fig. 7 The frequency boundaries this collection has measured. The floor is not on it, because it is not a frequency — it is a level, and it is the one boundary here that gets worse as everything else is made better, since every technique that raises the amplitude at which a circuit stops being linear does nothing at all for the amplitude below which it stops being able to see.

What can be done about a floor

Three things move it, and the list is short because most of what is usually tried does not.

Lower the resistance. The density goes as √R, so a tenfold reduction buys a factor of 3.16. That is real and it is often available, and it costs whatever the lower resistance costs elsewhere in the circuit — more current, more loading, a different bias point.

Narrow the bandwidth. The voltage goes as the square root of the noise bandwidth, so halving the bandwidth buys a factor of 1.41. This is the technique the third essay in this field measures as averaging, and it costs time in exactly the same proportion.

Cool it. The density goes as √T with T in kelvin, so cooling from 290 K to 77 K — liquid nitrogen — buys a factor of 1.94. From 290 K to 4 K it buys 8.5. That is a large factor and it is why the front ends of radio telescopes are cryogenic, and it is unavailable to almost everything else.

What does not move it is gain. An amplifier after the resistor multiplies the noise and the signal by the same number, so the ratio between them is unchanged — and an amplifier adds noise of its own, so the ratio gets slightly worse. That is the single most important consequence of a floor being a floor, and it is why the third essay’s subject is which amplifier comes first rather than how much gain there is.

Where this goes next

101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV Every model has an edge Four assumptions this collection runs on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does. 00.500101234frequency ÷ the −3 dB point|H|², the power that gets throughhalf the power: the −3 dB pointthe brick wall: 1.5706× the corner1.571×solved, then checked — the area integrated, not tabulated1.571× the corner, not 1× The bandwidth noise sees A single pole passes π/2 times as much noise power as a brick wall at its own corner frequency, so a noise voltage computed with the −3 dB point is twenty-one per cent low. Measured by integrating the solved response rather than taken from the table it usually comes from, the ratio is 1.5706 and π/2 is 1.5708. A five-pole Chebyshev's is 0.964 — less than one. 1e-81e-71e-61e-51e-41m10m100m1amplitude (volts)Johnson noise of 1 kΩ: 501.6 nV1% distortion, exponential: 1.03 mV1% gain error, exponential: 7.30 mVlinear step, unity follower: 79.58 mV66.3 dB of rangesolved, then checked — a floor and a ceiling, both computed66 dB between them 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.