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

Device noise — the series

7 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The floor an amplifier adds, against the source it is given. computed by solving, not by drawing. A part with 4.00 nV/√Hz of voltage noise and 0.60 pA/√Hz of current noise is quietest into 6.67 kΩ, where its noise figure is 1.138 dB. That resistance is the ratio of the two generators and the floor there depends only on their product. Matching the same part for maximum power into its own 1 MΩ input instead — a resistance 150 times larger — costs 12.57 dB.

    The floor a circuit has

    A resistor's noise is 4kTR and there is nothing to choose about it. An amplifier adds two generators that belong to the device — 4 nV/√Hz in series with its input and 0.6 pA/√Hz across it — and because one matters most into a small source and the other into a large one, there is a source resistance at which their sum is least. It is 6.67 kΩ, it is the ratio of the two, and it is not the resistance that transfers maximum power.

    part 1 · noise
  2. Averaging a white sequence, and averaging a pink one. computed by solving, not by drawing. Both sequences are the same seeded white stream, one of them put through the 1/f network. Averaged in non-overlapping blocks, the white one's spread falls as n to the -0.510 ± 0.006 across five seeds — the √N law — and the pink one's as n to the -0.087 ± 0.013, which is very nearly not at all. A thousand-sample average buys a factor of 36.9 on the first and 2.1 on the second.

    The corner where averaging stops working

    Average N samples and the noise falls by √N. That is a statement about independent samples, and flicker noise's samples are not independent — its correlation extends over every time scale, which is what a spectrum with no bottom means. Measured on the same seeded stream filtered and not: the white sequence falls as the −0.510 power of the block length and the pink one as the −0.087 power, so a thousand-sample average buys a factor of 36.9 on one and 2.1 on the other.

    part 2 · noise
  3. Two curves that only rise, and the gap between them that has a minimum. computed by solving, not by drawing. The source's own Johnson density, √(4kTR), and the amplifier's total input-referred density, √(4kTR + eₙ² + (iₙR)²), for a part with 4 nV/√Hz and 0.6 pA/√Hz. Neither curve has a minimum: the total is 4 nV/√Hz at a source of nothing, is 4.196 at 100 Ω, and rises without limit. What has a minimum is the ratio, at 6.67 kΩ, where the noise figure is 1.138 dB and the total density is 11.780 nV/√Hz — 2.81 times noisier in volts than at 100 Ω, where the noise figure reads 10.41 dB. The two statements are about different questions and the figure is what stops them being confused.

    The bowl, and the bottom of it

    An amplifier's noise figure has a minimum against source resistance and its input-referred noise has none: the 4 nV/√Hz part reads 1.138 dB into 6.67 kΩ and 10.41 dB into 100 Ω, and is 2.81 times noisier in volts at the first. The bowl is one shape scaled by its own depth, so the quieter the part the flatter it is — ±30.1 times for a decibel on the best of four, ±2.16 on the worst — and three parts of equal eₙiₙ share a floor of 0.3138 dB at optima 16 times apart.

    part 3 · noise
  4. At 100 Hz the bowl's bottom is at 2.11 kΩ and 3.10 dB, against 6.67 kΩ and 1.14 dB in the white region. computed by solving, not by drawing: the spot noise figure against source resistance of an amplifier with 4 nV/√Hz and 0.6 pA/√Hz of white noise, whose voltage noise has a flicker corner at 10 Hz and whose current noise has one at 1 kHz. At 100 Hz the spot densities are 4.20 nV/√Hz and 1.99 pA/√Hz, so the optimum is 2.11 kΩ — found by search on the drawn curve at 2.11 kΩ — with a floor of 3.102 dB. Dashed: the white-region bowl, 6.67 kΩ and 1.138 dB. A source left at the white optimum reads 4.492 dB at this frequency. The slider is the frequency.

    The source resistance the flicker moves

    An amplifier's optimum source resistance is the ratio of its voltage noise to its current noise, and both densities rise below their flicker corners — which are rarely the same frequency. With the voltage noise's corner at 10 Hz and the current noise's at 1 kHz, a part whose optimum is 6.67 kΩ at high frequency wants 2.11 kΩ at 100 Hz and 699 Ω at 1 Hz, tending to the white value times the square root of the ratio of the corners. Over the audio band the drift costs 0.012 dB; over a band from 1 to 100 Hz, choosing the white optimum costs 3.77 dB, and the spot optimum at the band's centre recovers all but a tenth of a decibel.

    part 4 · noise
  5. With a correlation of −0.3 between the two generators the bowl's bottom stays at 6.67 kΩ and its floor moves from 1.14 dB to 0.83. computed by solving, not by drawing, with the optimum found twice — in closed form and by a direct search of the noise factor, which agree to a part in a million. The 4 nV/√Hz, 0.6 pA/√Hz amplifier of the earlier essays, its two generators given a real correlation coefficient γ = −0.3. The noise figure against source resistance: the cross term 2γ·e·i·R adds the same fraction of the source's own noise at every resistance, so the bowl moves vertically and its bottom stays at e/i = 6.67 kΩ. The floor's excess over one scales as 1 + γ: 0.827 dB against 1.138 for independent generators. The slider is the correlation.

    Two generators with one cause

    The two-generator model of an amplifier assumes its voltage and current noise independent. Give them a correlation γ and the source sees a cross term, 2·Re(γ·e·i·Z*), linear in its resistance and its reactance. A real correlation leaves the quietest source resistance at e/i and moves the floor in proportion to 1 + γ: 0.83 dB instead of 1.14 at γ = −0.3. An imaginary one moves the optimum off the resistance axis — 5.77 kΩ with 3.33 kΩ of capacitive reactance at γ = 0.5j — and the noise figure's contours become circles about it; a resistive source cannot use it at all. If the two flicker components share one cause, the sign of the sharing decides the low-frequency floor: at 1 Hz, 3.9 dB or 18.0 instead of 15.1. And the correlation is measurable with three resistors, best with the middle one next to the optimum.

    part 5 · noise
  6. A 50 Ω source over 20.0 Hz–20.0 kHz: the best turns ratio is 10.7, giving 1.30 dB; the white optimum's 11.5 gives 1.31, and no transformer at all 13.2. computed by solving, not by drawing, the band integrals checked by quadrature. The drifting part of the essay before — 4 nV/√Hz and 0.6 pA/√Hz white, flicker corners at 10 Hz and 1 kHz — fed from a 50 Ω source through an ideal step-up transformer of ratio n, so it sees n²·50 Ω. The noise figure over 20.0 Hz to 20.0 kHz against n: least at n = √(√(E/I)/Rₛ) = 10.73, where the amplifier sees 5.76 kΩ, at 1.298 dB. The ratio that matches the white optimum, 11.55, gives 1.310 dB; the source connected directly, 13.23 dB. The slider is the band.

    One turns ratio for a band

    A 50 Ω source and an amplifier whose optimum is 6.67 kΩ want a step-up transformer, and the ratio that matches the white optimum, 11.5, is the one a design note computes. Over the audio band it is nearly right: 1.31 dB against 1.30 at the band's own best ratio, 10.7. Over 1 Hz to 100 Hz it is badly wrong — 9.2 dB where a ratio of 4.8 gives 5.4 — because the band's noise lives in its bottom decade, where the flicker has moved the optimum down. One ratio for a two-decade band costs at most 0.24 dB against a ratio that could follow the optimum. The transformer's winding resistance multiplies the whole noise factor by one plus the winding's resistance over the source's, not just its own share: a tenth of the source costs 0.41 dB where its own noise alone would cost 0.31. And 115 amplifiers in parallel are an ideal 10.7:1 transformer, noise for noise.

    part 6 · noise
  7. Paralleling a 4 nV/√Hz amplifier on a 50 Ω source, summed through 100 Ω from each: the best is 1.19 dB at 140 parts, against 1.14 dB at 134 parts with a perfect summing node. computed by solving, not by drawing: the noise figure from a 50 Ω source against the number of amplifiers in parallel, each 4 nV/√Hz and 0.6 pA/√Hz, with a perfect summing node (dashed) and with their outputs averaged through 100 Ω each into one node (solid), which adds a thermal 4kTR/N at the node. The summing resistors' noise falls as 1/N exactly as the amplifiers' voltage noise does, so their ratio is the same at every N, and a summing resistor comparable with the resistance an amplifier's voltage noise is worth is a constant factor on that term.

    The resistors that sum a hundred amplifiers

    A hundred amplifiers in parallel across a 50 Ω source are an ideal transformer of ratio ten, and the summing network between them and the output is where the comparison was left. Its resistors turn out to be the least of it. N resistors averaging N outputs make the noise of R/N, falling exactly as the amplifiers' voltage noise does, so the resistor equal to a 4 nV/√Hz amplifier is about a kilohm at every N, and at 133 parts 200 Ω costs only a tenth of a decibel. The stage that was forgotten is whatever reads the sum. Averaging leaves it alone, so a following amplifier as good as one of the parts costs 12.14 dB, and each part needs a gain of 26 to bring that under a tenth. Parallel amplifiers then match a transformer with one ohm of winding.

    part 7 · noise

All series