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

Spectral density — where it appears

How much of a fluctuating quantity's mean square sits in each hertz of bandwidth, quoted per hertz or per root hertz. Integrated over a response it gives a total, and the width of the brick wall that would pass the same total is the equivalent noise bandwidth.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed 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.

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.

noise · Johnson noise Two floors on one axis, and the 50.0 mV between them. computed by solving, not by drawing. The flat line is the Johnson current noise of 1 kΩ, √(4kT/R), which has no current in it. The rising line is shot noise, √(2qI), which has no resistance in it. They cross at 50 µA — and the direct voltage across the resistance there is 49.981 mV, which is 2kT/q and contains neither quantity. The slider moves the resistance over six decades; the crossing moves with it and the voltage at the crossing does not move at all, to the last bit of a double.

The floor a current sets

A resistor's noise contains no current and a current's noise contains no resistance, and the two are equal when the direct voltage across the thing carrying the current is 2kT/q — 50.0 millivolts at 290 kelvin, whatever the resistance and whatever the current. It is the only boundary in this collection whose axis is a direct voltage across an element. And a forward-biased junction, which has the same dynamic resistance as some resistor, produces exactly half its noise power at every current.

noise · Shot noise Five resistances, five corner frequencies, and one total on the capacitor. computed by solving, not by drawing. The noise density at a 1 pF capacitor charged through 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ and 1 MΩ, at 290 K. The densities are 100 times apart and the noise bandwidths 1.0e+4 times apart, in opposite directions, so the area under every curve is the same: 63.2762 µV against 63.2762 µV, and √(kT/C) is 63.2762 µV. The resistance has cancelled out of the answer, and the reason is that ½C⟨v²⟩ is the ½kT a degree of freedom in contact with a bath holds — which no arrangement of resistors can change. The claim is about the whole frequency axis and nothing less: inside a 15.9 MHz band the same five networks give 5.05 µV to 63.07 µV, a factor of 12.5.

The total that has no resistor in it

A larger resistor is noisier and makes a narrower filter, and the two dependences are exactly reciprocal: the density goes as the square root of the resistance and the noise bandwidth as its inverse. Five decades of resistance charging one picofarad therefore give five decades of corner frequency, two and a half decades of density, and one total — 63.2762 microvolts at every one of them, which is the square root of kT over C and contains no resistance at all.

noise · Kt over c The switches set the floor below 60.2 MHz and the amplifier sets it above. computed by solving, not by drawing. The two contributions to a switched-capacitor stage's noise floor, against the clock frequency. √(kT/C) is 63.3 µV on a 1 pF hold capacitor and does not move with the clock at all. The amplifier's own 4 nV/√Hz is white and is sampled, so all of it folds into the band: settling to 12 bits in half a clock period needs 8.32 time constants, the closed loop's noise bandwidth is then 8.32 clocks over two, and the number of folds is 8.32 — the same number, exactly. The marched route is a seeded white sequence put through the settling exponential and sampled once per clock, and it agrees with the closed form to 0.05% of variance. The two floors are equal at 60.2 MHz, above which a larger capacitor buys nothing.

The amplifier inside the sample

kT/C is exactly independent of the clock, of the capacitor ratio and of the switch resistance — two essays measured that and found it identically true rather than nearly so. The amplifier in the same loop behaves in the opposite way in every respect: its noise is white, it is sampled, and the number of times it folds into the band is exactly the number of time constants the settling needs. So the switches set the floor below sixty megahertz and the amplifier sets it above, and asking for two more bits of settling costs fifteen per cent more noise before anything else has changed.

noise · Kt over c Against its own shaping the error is 17.4× tonal, not 51×. computed by solving, not by drawing. The share of an order-1 loop's in-band error sitting in its five largest lines, from 0.02 to 0.9 of full scale, with both nulls drawn. The lower level is 1.953 per cent — five lines of a FLAT error over 256 — and it is what this measurement has always been quoted against. The upper level is 5.74 per cent, which is what five lines of the loop's OWN shaping hold with no tone anywhere in them. Measured against the first the error is 51 times tonal at 0.02 of full scale and 26 at 0.9; against the second, 17.4 and 8.9. The direction survives the correction and the size does not.

A floor, or five tones

The field's sharpest statement about a one-bit loop is that three quarters of its in-band error sits in five lines, against the 1.953 per cent a white error would put in any five — a factor of thirty-eight. The comparison is to a white error, and a shaping loop exists to make its error anything but white. Measured against the loop's own transfer function, which the loop reproduces line by line to a part in five hundred when it is dithered, the same error is 17.4 times tonal at a fiftieth of full scale and 8.9 times at nine tenths. And 99.5 per cent of it at the quiet end is two harmonics of the input.

digital · Noise shaping Three averagers passing the same noise, and three different half-power points. Integrated by eight-point quadrature on every lobe, with the tail past the last lobe in closed form. A mean over 20 ms has the response sin(πfT)/(πfT), and the area under its square is 25.000000 Hz against the 25 Hz of 1/(2T) — the brick wall drawn shaded. A mean over the window passes half its power at 22.147 Hz, so the noise bandwidth is 1.12880 times that frequency; A one-pole averager passes half its power at 15.915 Hz, so the noise bandwidth is 1.57080 times that frequency; Two means in cascade pass half their power at 23.919 Hz, so the noise bandwidth is 1.04521 times that frequency. Every curve drawn encloses the same area; they differ in where they spend it.

The filter an average is

A mean taken over a window is a filter, and the area under its squared response is exactly one over twice the window — 25 hertz of noise bandwidth for twenty milliseconds, passing half its power at 22.15. Built to the same noise, a one-pole averager passes half its power at 15.92 hertz and takes 2.33 times as long to settle to one per cent, and two means in cascade pass half at 23.92 and take 1.25 times as long. Between its nulls a mean rejects the mains no better than the one-pole does, and one per cent off a null it rejects it by forty decibels however many cycles the window holds.

noise · Noise bandwidth What is warm in a capacitor, by its two loss models. computed by solving, not by drawing. One 100 nF capacitor of loss tangent 0.02, written as a 3.183 Ω resistance in series with it and as a 7.958 kΩ resistance across it — the pair that converts exactly at 10.0 kHz and nowhere else. The noise at the terminals is 4kT times the real part of the impedance, so the two models give the same density at 10.0 kHz and are 33.0 dB apart at 100 Hz and 40.0 dB apart at a megahertz. The dots are the same quantity computed the other way — the resistor split out of the netlist, a source put in its place and the network re-solved — agreeing to 3.3e-16. The reactance itself contributes nothing at either end: a lossless capacitor has no real part and is not warm.

Only the real part is warm

Johnson's 4kTR is the special case of a statement about impedances: the noise across any passive two-terminal in equilibrium is 4kT·Re{Z}, so a reactance contributes nothing however large it is. That turns a modelling convenience into a noise figure. A 100 nF capacitor of loss tangent 0.02 written as 3.183 Ω in series and as 7.958 kΩ across it — the pair that converts exactly at 10 kHz — gives 0.226 and 10.10 nV/√Hz at 100 Hz, 33 dB apart, and 40 dB apart the other way at a megahertz.

noise · Johnson noise The floor a biased resistor is not standing on. computed by solving, not by drawing. A 100 kΩ resistor with 10 V across it. Its Johnson noise is 40.02 nV/√Hz and does not depend on the voltage; its excess noise is a 1/f density proportional to that voltage, 0.1 µV per volt per decade here, and the two are equal at 271 Hz. Below that the resistor is dominated by a generator that is a property of how it was made rather than of its resistance, and the crossover moves with the SQUARE of the voltage. Splitting the same total resistance into eight parts in series leaves the Johnson noise exactly where it was and divides the excess by √8 = 2.828, taking the total at 1 Hz from 660.2 to 236.4 nV/√Hz. With no voltage across it the second generator is absent rather than small.

The floor that is only a floor while nothing flows

Johnson noise depends on a resistance and a temperature and on nothing else, which is what makes it a floor. A real resistor has a second generator that depends on how it was made and on the voltage across it: 0.1 µV per volt per decade for a metal film, rising as 1/√f. On 100 kΩ with 10 V across it the two are equal at 271 Hz, and the crossover moves with the SQUARE of the voltage — 0.678 Hz at half a volt, 2.44 kHz at thirty. Splitting the same total resistance into eight parts in series divides the excess by exactly √8 and leaves the Johnson noise where it was.

noise · Johnson noise

Named alongside it

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

Johnson noiseEquivalent noise bandwidthModel rangeDesign tradeoffKt over cMonte carloNoise floorSeeded generatorSettling timeVerificationAliasingAveraging

All concepts