Circuit Response
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Series

kT/C noise — the series

7 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · noise
  2. kT/C, on a circuit whose resistance is a clock and moves by 1e+4. computed by solving, not by drawing. The noise a switched-capacitor low-pass leaves on its holding capacitor, against the ratio of the two capacitors, measured by marching a seeded sequence through the recursion the circuit obeys. It is 6.328 µV at every ratio drawn — kT/Ch, with the holding capacitor and nothing else in it — while the resistance the arrangement behaves as moves from 1.00 kΩ to 10.0 MΩ across the same axis. The two lines below it are the two sampling events that make it: what the input switch leaves on the switched capacitor, and what the sharing switch leaves behind when it opens. Neither is the answer and their sum is, exactly, because the pole's own bandwidth factor is the same expression.

    The noise a clock does not make

    A hundred-megohm resistor has a noise density of 1.27 microvolts per root hertz. A switched capacitor that behaves as a hundred megohms has none of it: the noise on the capacitor it charges is kT/C, with the holding capacitor in it and nothing else — not the clock, not the switched capacitor, not the on-resistance. And it is exact rather than asymptotic, at every capacitor ratio from a thousandth to ten.

    part 2 · noise
  3. 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.

    part 3 · noise
  4. Subtracting removes kT/C entirely and doubles the amplifier — worth 31× at a megahertz and a loss above 60 MHz. computed by solving, not by drawing. The noise on one sample of a switched-capacitor stage, and on the difference of two samples taken a settled interval apart, against clock frequency. The reset level is the same number in both samples and cancels exactly; the amplifier's own noise is two independent samples and its variance doubles, measured at 2.000 against the 2 the correlation predicts. At a megahertz that is 63.8 µV down to 11.53 — 31 times in power. The two curves cross at 60.2 MHz, which is where the amplifier's own noise equals kT/C, and above it the subtraction costs more than it removes.

    The sample that is subtracted

    Three earlier measurements found floors that no gain moves and no filter reaches, because both arrive as numbers already sampled. One of them can be subtracted: the reset level a capacitor holds is the same number in two consecutive samples and cancels exactly. What it costs is that the amplifier's own noise is not — two samples of it are independent, so its variance doubles. That is thirty times better at a megahertz, a loss above sixty, and the crossing is the one an earlier essay computed for a different question.

    part 4 · noise
  5. With 1 RC of tracking, sixteen samples average like 7.8 independent ones. computed by solving, not by drawing. The rms noise on the average of N consecutive held samples of a 1 pF capacitor, against N, for track times of 0.25, 1, 3 and 10 RC between samples, with the 1/√N law for independent samples dotted. A sample shares exp(−Tₜ/RC) of its value with the one before, so at 1 RC the correlation is 0.368 and sixteen samples are worth 7.81 independent ones; 1,024 are worth 473.6. The closed form is checked against a seeded march of the held sequence.

    The average that is not N times quieter

    Averaging sixteen samples of a 1 pF capacitor is supposed to take its 63.3 µV of kT/C noise down by four. With one RC of tracking between samples it takes it down by 2.8: the switch has not been closed long enough for the capacitor to forget the last sample, each shares 37 per cent of its value with the one before, and sixteen samples are worth 7.8 independent ones. At a quarter of an RC they are worth 2.6. Sampling faster never helps past a ceiling: however quickly the samples come, an average gathers at most one independent sample per 2RC of time — the continuous average's own limit. And a sampler that settles its signal to 8 bits has already waited long enough for its noise.

    part 5 · noise
  6. An amplifier gain of 1000 caps a switched-capacitor integration at 8000 cycles' worth of averaging. computed by solving, not by drawing. The noise, referred to the input, of the estimate a switched-capacitor integrator makes by integrating a steady input for N cycles, with 1 pF sampling and 4 pF feedback capacitors, against N, for amplifier gains of 100, 1,000, 10,000 and infinity. Each cycle adds 2kT/Cs — 89.5 µV rms — and an ideal integrator averages it as 1/√N without limit. A finite gain leaks, and the noise stops falling at √(kT/(A·Cf)) = 1.000 µV for A = 1000, reached after about 8000 cycles. The closed form is checked against a seeded march.

    The average an integrator can hold

    A switched-capacitor integrator is an averager whose samples are exactly independent, because every cycle resets the sampling capacitor — and it pays for that with two switch phases, 2kT/Cs a cycle, 89.5 µV rms with a picofarad. An ideal integrator averages that as 1/√N without end. A real one leaks through its amplifier's finite gain, and the leak caps the average at 2A·Cf/Cs cycles: 8,000 for a gain of a thousand and a 4 pF feedback capacitor, where the noise stops at √(kT/(A·Cf)) = 1.00 µV. That floor contains the feedback capacitor and the gain and not the sampling capacitor at all, which decides only how long the integration takes to reach it.

    part 6 · noise
  7. Averaging an input that drifts 1 µV a cycle through a 1 pF switched-capacitor averager: too short a memory keeps the noise, too long a lag, and the best is 16.5 cycles at 27.2 µV. computed by solving, not by drawing, and seeded. A switched-capacitor exponential averager — 1 pF sampling capacitor, an equal damping capacitor for unity gain, memory set by the feedback capacitor — fed a ramp of 1 µV a cycle. The rms error against the input (solid) is the lag (dashed), (a/(1 − a))·r, and the switch noise (dotted), 2kT(Cs + Cd)/(Cf²(1 − a²)), together; seeded marches of sixty thousand cycles (dots) agree to 5%. The best memory is 16.55 cycles against (kT/(Cs·r²))^(1/3) = 15.88, with 27.2 µV of error: 15.5 µV of lag and 22.3 µV of noise.

    The leak worth building in

    An integrator that never forgets cannot follow an input that moves. Give a switched-capacitor integrator a deliberate leak — a damping capacitor that samples the output each cycle and takes its charge away — and it becomes an exponential average with a memory of Cf/Cd cycles. Fed an input drifting 1 µV a cycle, through a 1 pF sampling capacitor, too short a memory passes the switch noise and too long a memory lags the drift; the best, on the exact closed form and on seeded marches of sixty thousand cycles, is 16.5 cycles, for 27.2 µV of error against 89.5 µV in a single sample. The best memory follows a cube-root law, the cube root of kT/(Cs·r²), ten times the drift for a memory 4.64 times shorter, and at it the noise variance is twice the squared lag. And the leak is not free: the damping capacitor is switched too, and at unity gain it samples half of the averager's noise.

    part 7 · noise

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