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

Equivalent noise bandwidth — where it appears

The width of the brick-wall filter that would pass the same noise power as a given response at its own reference gain. For a single pole it is π/2 times the half-power point, which is where a floor computed from the −3 dB frequency is two decibels light.

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 A single pole, and the brick wall that passes the same noise. computed 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.

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.

noise · Noise bandwidth 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 The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range.

What the second path costs at the floor

The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

feedback · Capacitive load Which resistor the noise of a Chebyshev 5 actually comes from. computed by solving, not by drawing, one solve per resistor. Each bar is that resistor's share of the noise power at the output, found by splitting its node, putting a source of √(4kTR) in series with it and re-solving the whole network — so what is drawn is not how much noise each resistor makes but how much of it arrives. The largest contributor is F2R at 64.1 per cent, the smallest F0R at 9.0, and the shares add to 1.000000000000 because noise powers add. A resistor's share of the noise is not its share of the resistance: the largest departure between the two is 3.9 percentage points.

The resistor the noise comes from

Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.

noise · Johnson noise The noise bandwidth of four families at 8 orders, and the one that has none. computed by solving, not by drawing. Every one of the 32 entries is an integral of the realised network's own squared magnitude, divided by that network's own measured −3 dB point. At order one the four families are the same filter and return 1.5706, which is π/2 — the calibration the rest of the table is quoted against. Only Butterworth then does what the ratio is usually said to do: it falls at every order, to 1.0065 at 8. Bessel is least at order 5 (1.0385) and rises to 1.0441; Chebyshev alternates with parity, 0.9637 at five against 1.0686 at six; and an even-order elliptic has no noise bandwidth at all, because its stopband comes back up to a constant — its magnitude at the top of the range moves by 0.00 decades per decade of frequency, so the integral grows with whatever limit it is stopped at.

The ratio that does not walk to one

A single pole passes π/2 times as much noise as a brick wall at its corner, and every account of it says the ratio falls towards one as the skirt steepens. Over thirty-two realised filters only Butterworth does that. Bessel is least at order five, 1.0385, and rises again; Chebyshev alternates with parity and the two branches separate only above 0.1968 decibels of ripple; and an even-order elliptic has no noise bandwidth at all, its integral returning 380 or 38,005 depending on where it was stopped.

noise · Noise bandwidth 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

Named alongside it

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

Johnson noiseSpectral densityBrick wall filterDesign tradeoffSettling timeFilter familiesIntegrationKt over cMonte carloSeeded generatorVerificationAliasing

All concepts