The average an integrator can hold

The average that is not N times quieter found that averaging repeated samples of a switched capacitor buys the textbook $1/\sqrt N$ only when each sample has had an RC or more to forget the last, and that an average over a stated time can never gather more than one independent sample per $2RC$ of tracking. It ended by naming the other way to average: add the samples’ charges into a feedback capacitor, cycle after cycle, which is what a switched-capacitor integrator does.

The integrator changes both terms of the trade. Its samples are exactly independent, because each cycle ends with the sampling capacitor emptied into the feedback capacitor and begins with a fresh closure. And it pays for each sample twice, because each cycle has two switch phases. What it cannot do is hold the sum for ever, and the reason it cannot is the one component the averaging arithmetic usually leaves out: the amplifier.

The integrator as a sum #

In each cycle the sampling capacitor $C_s$ charges to the input, then its charge is transferred into the feedback capacitor $C_f$ around an amplifier whose inverting input is held at virtual earth. The output steps by $(C_s/C_f)$ times the input. After $N$ cycles of a steady input $x$ it has stepped $N$ times, and dividing the output by $N\,C_s/C_f$ returns the input.

Each switch phase leaves $kT/C_s$ of noise on the sampling capacitor — the noise a clock does not make found that number independent of every resistance and clock — and the two phases are independent, so each cycle adds $2kT/C_s$ referred to the input: 89.5 µV rms for a picofarad at room temperature. With an ideal amplifier the $N$ contributions add in power and the input-referred noise of the estimate is $\sqrt{2kT/(C_sN)}$. That is an average of $N$ independent samples, each twice as noisy as a single kT/C.

So far the integrator is a slightly worse averager than a sampler with long tracking, trading a factor of two in variance per sample for perfect independence and no tracking time. The trade is interesting only because of what happens next.

Why a cycle costs two draws #

The factor of two deserves its own paragraph, because it is the whole of the integrator’s disadvantage per sample and it comes from the switches rather than from anything the integrator does.

In the sampling phase the capacitor is connected between the input and ground through two switches, and when they open it holds the input plus a kT/C draw, exactly as a sample-and-hold does. In the transfer phase it is connected between ground and the amplifier’s virtual earth through two other switches, and it discharges into the feedback capacitor; when those open, the charge it failed to deliver is another kT/C draw, independent of the first because the switches and the circuit around them are different. The two draws reach the feedback capacitor with the same weight, so each cycle carries $2kT/C_s$.

That is the price of an integrator whose transfer does not depend on the sampling capacitor’s parasitic capacitance to ground — the property the four-switch arrangement exists for. A simpler integrator with one switch phase would pay once, and would carry its sampling capacitor’s bottom-plate capacitance straight into its gain. The noise a clock does not make met the same accounting in a switched filter, where the sharing switch’s draw and the input switch’s draw had to be added to get kT/C exactly; here they add to twice it, because nothing shares.

The leak a finite gain makes #

An amplifier with open-loop gain $A$ does not hold its inverting input exactly at earth: it sits at $-v_{out}/A$. Some charge that should have been transferred into $C_f$ is instead left on $C_s$ and on the input node, and on the next cycle it is thrown away. The integrator forgets a fraction of its own output every cycle. To first order the retained fraction is

$$p = 1 - \frac{C_s}{A\,C_f}$$

per cycle — 0.99975 for a gain of a thousand with a quarter the capacitance sampling. The output is then a leaky sum, $y_N = (C_s/C_f)\sum_k p^{N-k}(x + n_k)$, and the estimate divides it by the signal gain the same sum gives, $\sum_k p^k$, so that the steady input is still returned exactly.

The noise on that estimate is the per-cycle variance times $\sum p^{2k}/(\sum p^k)^2$. For $p = 1$ that is $1/N$. For any $p < 1$ it tends, as $N$ grows, to $(1-p)/(1+p)$, and substituting the leak gives

$$\sigma^2_\infty = \frac{2kT}{C_s}\cdot\frac{1-p}{1+p} \;\approx\; \frac{kT}{A\,C_f}$$

The integration stops improving. Its best achievable noise is $\sqrt{kT/(AC_f)}$ — 1.00 µV for a gain of a thousand and 4 pF — reached after about $2AC_f/C_s$ cycles, 8,000 here, which is the effective number of samples a leaky sum can hold.

Checked twice #

The closed form is checked against a seeded march of the leaky sum itself: the recursion $y \leftarrow p\,y + n$ run for 16 and for 256 cycles, three thousand times each, at the drawn gain and at infinite gain. The march’s variance is required to match the formula within six of its own standard errors, which at three thousand trials is a test to about fifteen per cent in variance and a tighter one in its agreement across four cases.

The limit is checked separately: at a million cycles the formula’s variance is required to equal $kT/(AC_f)$ to two per cent, and the effective count to equal $2AC_f/C_s$ to two per cent. Both approximations are first order in $1/A$, so at a gain of a hundred they are the loosest, and they still hold.

A tenfold change in gain, drawn #

The slider moves the gain by factors of ten, and two settings are worth fixing on the page because they are the two ends of what an integrator’s amplifier is likely to be.

At a gain of a hundred the knee is at 800 cycles and the floor at 3.16 µV. That is a floor of 3.16 µV on the input, reached in under a thousand cycles and never passed, however long the conversion is made. A designer who extends the integration to gain resolution, as an integrating converter’s datasheet would suggest, buys a gain error and nothing else.

At ten thousand the knee is at 80,000 cycles and the floor at 0.316 µV. The pattern is the finding: a tenfold gain buys a tenfold longer useful integration and a threefold lower floor, because the floor is an average of $2AC_f/C_s$ samples and the average improves as the root of its count. An amplifier’s gain is ordinarily specified for the accuracy of a closed-loop gain; here it is a noise specification, and it has to be read at the frequencies the integration runs at, which for a slow integrating measurement is close to direct current, where the gain is highest.

The floor the sampling capacitor is not in #

The most useful thing about the floor is what is missing from it. $C_s$ has cancelled: a larger sampling capacitor has less noise per cycle, $2kT/C_s$, but leaks faster, since the leak $C_s/(AC_f)$ grows with it, and it holds fewer cycles in exact compensation.

This is the same kind of cancellation the total that has no resistor in it found for kT/C itself: a quantity that makes the noise larger also makes the bandwidth narrower, in exact proportion, and the total is left with neither. Here the sampling capacitor sets both the noise per cycle and the rate at which cycles are forgotten, and the floor keeps neither.

What the floor does contain is the design’s two levers. The feedback capacitor, as $1/\sqrt{C_f}$: four times the feedback capacitance halves the floor, and costs four times the area and a quarter of the integrator’s gain per cycle. And the amplifier’s gain, as $1/\sqrt A$: ten times the gain buys $\sqrt{10}$, 3.16 µV at a hundred falling to 1.00 at a thousand and 0.316 at ten thousand. The gain is the lever the averaging arithmetic never mentioned, and it is what decides how long an integrating measurement is worth running.

What the leak does to a single integration #

The floor is a statistical statement. One integration shows what it means for one reading.

The two estimates share every noise sample, and for the first few thousand cycles they are nearly identical. Then the leaky one stops converging: by the time it has integrated the late cycles it has forgotten most of the early ones, so it is always an average of about the most recent eight thousand, and its wander settles to the width of an eight-thousand-sample average. The ideal one keeps narrowing. After twenty thousand cycles their standard deviations are 0.633 and 1.007 µV, and the gap widens as the square root of the integration length.

There is a second cost the trace does not show, because the estimate divides by the leaky sum’s own gain. The signal gain of $N$ leaky cycles is $\sum p^k = (1 - p^N)/(1 - p)$, and at 20,000 cycles with a leak of one part in four thousand that is $(1 - e^{-5})/5$ of $N$ — 0.199. An integrator whose output was divided by $N$, as though the amplifier were ideal, would read a fifth of its input. The leak is a gain error long before it is a noise floor, and a design that calibrates the gain has only then reached the question this essay is about.

The feedback capacitor as a budget #

Because the floor goes as $1/\sqrt{C_f}$, the feedback capacitor is the design’s noise budget, and two other things it controls point the same way.

The swing. The output after $N$ cycles is $N(C_s/C_f)x$, and it has to stay inside the amplifier’s output range. A larger feedback capacitor makes each cycle’s step smaller and lets the integration run longer before it clips — so the same change that lowers the floor also lengthens the integration the floor needs. The two move together, which is why an integrating front end is usually drawn with a feedback capacitor several times its sampling capacitor, as the four to one here.

The leak itself. The leak per cycle is $C_s/(AC_f)$, so a larger feedback capacitor forgets more slowly, and the knee moves out in proportion. It is the same statement as the floor’s, read in time rather than in volts.

What pushes the other way is area and the charge the amplifier has to supply each cycle. A capacitor four times larger is four times the silicon, and in a switched-capacitor filter — the filter that samples is one — the capacitor ratios are set by the response, so the feedback capacitor cannot be enlarged without enlarging everything else with it. The integrator used as a measurement is free of that constraint, which is the reason its feedback capacitor can be chosen for the floor alone.

A last connection is worth making because it is easy to mistake one floor for another. The integrator also accumulates errors that are not noise: the amplifier’s own input offset, integrated every cycle, and the charge each switch leaves behind when it opens — the offset that knows the signal finds that charge depends on the input, so it is not even a constant. A leak bounds both in the same way it bounds the noise, which is why they are easy to confuse, and neither is a floor in the sense here: both are systematic and both can be measured and removed. Removing the amplifier’s offset by sampling it each cycle is the usual way, and if that adds a third draw of $kT/C_s$ to each cycle the floor rises by $\sqrt{3/2}$ for the same gain and capacitors — the arithmetic of the leaky sum with a larger per-cycle variance, and the price of a floor that has no offset under it.

Two ways to spend the same time on one capacitor #

The two averaging essays now bound the choice.

A sampler that re-tracks for several RC between reads averages independent kT/C samples at one per track time; the average is limited by how many track times fit into the measurement, and a long average reaches $kT/C$ divided by that count, with nothing to stop it but the time available and — the corner where averaging stops working — the first flicker in the chain.

An integrator averages independent $2kT/C_s$ samples at one per cycle, which can be much faster than a sampler’s tracking, because nothing has to settle to more than the per-cycle transfer; but its average is limited by the amplifier’s gain to $2AC_f/C_s$ cycles, and its floor is $kT/(AC_f)$ whatever the measurement time.

So the integrator wins for short measurements, when its per-cycle rate buys more samples than tracking can, and loses for long ones, when the sampler’s unlimited count passes the integrator’s leak. The crossover is the measurement time at which the sampler’s count equals $2AC_f/C_s$, and it is a property of the amplifier rather than of the switches.

The reset that has to happen anyway #

Every integrating measurement ends with a reset: the feedback capacitor is shorted and the next measurement begins. That reset leaves $kT/C_f$ on the feedback capacitor, once per measurement, and referred to the input it is divided by the integrator’s gain $N\,C_s/C_f$: $\sqrt{kT\,C_f}/(N\,C_s)$, which after a thousand cycles with these capacitors is 0.13 µV — small against the 2.8 µV the average itself has reached by then. It is a one-off draw, and it can be removed entirely by reading the output just after the reset and subtracting it at the end, which is the sample that is subtracted applied to the integrator’s own starting point.

The reset also bounds the leak’s effect in the other direction. An integration that is reset every $M$ cycles never runs long enough to forget its beginning if $M$ is well below $2AC_f/C_s$, and then the leak’s only effect is the gain error above. The floor matters only to a measurement that tries to integrate past the amplifier’s memory.

Where this sits among the resistors a clock makes #

A switched capacitor clocked at $f$ behaves as a resistance $1/(fC_s)$, which a resistor made of a clock establishes; an integrator built from one is the sampled equivalent of an RC integrator, $R = 1/(fC_s)$ into $C_f$. The continuous version leaks through the amplifier’s finite gain too, with a time constant $A\,RC_f$, and the number of cycles in that time constant is $A\,C_f/C_s$ — half the $2AC_f/C_s$ found here, because the effective count of an exponential average is twice its time constant in samples. The sampled and continuous integrators have the same memory, measured in the only unit that makes them comparable.

A leak that is only the gain, and a steady input #

The leak is the amplifier’s gain and nothing else. A real integrator also leaks through the switches’ off-state current and the feedback capacitor’s own insulation, and at long integrations those can exceed the gain’s leak.

The per-cycle noise is two kT/C draws. That is the standard account of a parasitic-insensitive integrator with an ideal amplifier; the amplifier’s own noise, sampled, adds to it, and the amplifier inside the sample is how much.

And the input is steady. An integrator measuring a changing input is a filter, and its response — a sinc in frequency for the ideal integrator, softened by the leak — is a separate calculation from its noise floor.

Still open: the loop around the integrator, the leak that is useful, and the energy per bit #

The integrator inside a feedback loop. A delta-sigma modulator puts this integrator inside a loop that returns its output to its input every cycle, so nothing accumulates and the leak’s floor does not apply as stated. The integrator’s $2kT/C_s$ is then input noise, white, and the loop’s oversampling ratio is what averages it — a different limit with the same per-cycle cost, and the comparison would say when the loop’s average beats the open integrator’s.

A leak chosen on purpose. A leak with a known time constant turns the integrator into an exponential averager, which tracks a slowly changing input where a pure integrator does not. The best leak for an input that drifts at a stated rate is a trade between the drift the average follows and the noise it removes, and it is a number.

The energy per bit. Each cycle charges $C_s$ and drives $C_f$, so the energy of a measurement is proportional to the cycle count, while the noise falls only until the leak’s floor. There is a cycle count past which further integration spends energy for nothing, and at $2AC_f/C_s$ it is exactly where this essay’s figures put the knee.