The noise a clock does not make

The rung below this one established a result that reads at first like a trick and turns out to be the whole of the subject: the noise voltage stored on a capacitor charged through a resistance is $\sqrt{kT/C}$, and the resistance is not in it. A large resistor makes a large noise density over a small bandwidth and a small resistor makes a small density over a large one, and the product is a constant with a Boltzmann constant, a temperature and a capacitance in it.

That result was derived on a resistor and a capacitor, and its reasoning goes through a resistance before cancelling it. So it is fair to ask what happens when the resistance is not a resistance at all — when it is a capacitor being shuttled by a clock, behaving as a hundred megohms because it moves a picofarad’s worth of charge ten thousand times a second.

The naive reading predicts disaster. A hundred-megohm resistor has a noise density of $\sqrt{4kTR} = 1.27$ microvolts per root hertz. Ten of them in a filter would put the floor somewhere around a millivolt, and a technique that fakes large resistances would be unusable for anything with dynamic range.

The measured answer is that the arrangement’s noise is $kT/C_h$, with the holding capacitor in it, and that no clock frequency, no switched capacitance and no on-resistance appears anywhere. It is not approximately that. It is identically that.

Two sampling events, and neither is the answer #

The circuit does one thing per clock period and it does it in two halves, so it takes two noise samples and they are independent.

In the first half the switched capacitor is connected to the input through a switch. It charges, and when the switch opens it holds the input voltage plus whatever noise the switch’s own resistance left on it — variance $kT/C_s$, by the rung below this one, with the on-resistance already cancelled out.

In the second half it is connected to the holding capacitor and the two share charge. When that switch opens, the two capacitors are in series around the loop the switch was in, so the charge it randomises has variance $kT\,C_{\text{ser}}$ with $C_{\text{ser}} = C_sC_h/(C_s + C_h)$ — and that charge moves the holding capacitor’s voltage by $q/C_h$.

Writing $a = C_s/C_h$ for the ratio and $p = 1/(1+a)$ for the pole, one clock period is

$$v[n+1] = p\,v[n] + (1-p)\left(v_{\text{in}} + n_1\right) + \Delta$$

with $\operatorname{var}(n_1) = kT/C_s$ arriving divided by $1+a$, and $\operatorname{var}(\Delta) = kT\,a/\left[(1+a)C_h\right]$.

Summing the stationary variance of that recursion:

$$\sigma^2(1 - p^2) = \frac{kT\,a}{(1+a)^2C_h} + \frac{kT\,a}{(1+a)C_h} = \frac{kT\,a(2+a)}{(1+a)^2C_h}$$

and $1 - p^2 = a(2+a)/(1+a)^2$ is the same expression to the last symbol. Every $a$ cancels.

$$\sigma^2 = \frac{kT}{C_h}$$

Not to first order in the ratio. Identically, for every ratio, at every clock.

What the cancellation is doing #

It is worth pausing on that, because a cancellation that complete usually means one of two things and here it means the interesting one.

Each of the two contributions on its own is a smaller number than the answer. At a ratio of a fiftieth the input sample contributes 4.452 microvolts on a hundred picofarads and the sharing event contributes 4.496; the total is 6.328, which is the two added in quadrature. Neither is the answer and their sum is.

The share each takes is a closed form and it moves with the ratio while the total does not: the input sample contributes $1/(2+a)$ of the variance and the sharing event $(1+a)/(2+a)$. At a small ratio they split it almost exactly in half — 0.49505 against 0.50495 at $a = 0.02$, measured, against 0.49505 and 0.50495 predicted. At $a = 1$ it is a third and two thirds. At $a = 10$, one twelfth and eleven twelfths.

So the arrangement does not have one noise source that happens to give $kT/C$. It has two, their proportions change by two orders across the axis, and their sum is constant — which is a stronger statement than the rung below this one makes and a more surprising one.

Measured, by a route that shares no algebra with it #

An identity that falls out of an algebraic cancellation is exactly the kind of result that is wrong when a sign is wrong, so it is measured rather than believed.

Two independent seeded sequences are drawn, scaled to the two variances above, run through the recursion the circuit obeys, and the variance of what comes out is counted. Across ratios from 0.001 to 10 — a factor of ten thousand, over which the equivalent resistance moves from a kilohm to ten megohms — the measured variance agrees with $kT/C_h$ to within 1.3 standard errors of the estimate.

Standard errors, rather than a percentage, and the reason is worth stating because it is the same trap the field’s averaging argument fell into. A first-order recursion whose pole sits at $1 - 10^{-3}$ has an effective sample count three orders below its length: the variance of a sample variance is $2\sigma^4/n$ times $(1+p^2)/(1-p^2)$, so at the small-ratio end of this axis a run of three hundred thousand samples has a spread of several per cent however honestly it is taken. A fixed tolerance would be slack at one end and chosen-to-pass at the other. The spread is computable from the pole and the sample count alone, so the assertion is made against it.

The folding, which is where the microvolts went #

The algebra above is complete and it is not an explanation. It says the answer is $kT/C_h$; it does not say what became of the equivalent resistance’s very real noise density. The account that does is about sampling, and it is the same account the field about turning signals into numbers gives for a different purpose.

The switch’s on-resistance produces noise over a bandwidth set by the on-resistance and the switched capacitor — several megahertz, typically, and far wider than anything the filter passes. When the switch opens, that whole wideband process is sampled, so everything above half the clock folds down into the band below it. The variance does not change when it folds: sampling a wideband process rearranges where its power sits in frequency and does not destroy any of it, which is exactly what the aliasing argument says in the other direction.

So the total is conserved through the fold, and the total is $kT/C_s$ — a stored variance with the resistance already cancelled out of it. That is why the on-resistance never appears: it decides the bandwidth being folded and the density being folded, in inverse proportion, and the fold conserves their product.

A reader who wants one sentence for it: the equivalent resistance’s noise is all still there, it has been folded into the band, and the amount that arrives is set by the capacitor that held it while it was being folded.

The one place a faster clock does help #

Everything above says the clock does not appear. There is a case where it does, and it is worth separating from the main result rather than left as a caveat, because it is a different mechanism entirely.

Nothing in this essay’s recursion has any noise from anything but the switches. A real stage has an amplifier in it, and an amplifier has a voltage noise density that is a property of the part rather than of the sampling. That density is folded by the sampling too, and how much of it folds in depends on the amplifier’s bandwidth against the clock — so a stage whose amplifier is much faster than its clock folds a great deal of amplifier noise into the band, and one whose amplifier is only just fast enough to settle folds much less.

Which gives the design rule that looks paradoxical beside everything above: make the amplifier exactly as fast as it needs to be to settle in a half period, and no faster. The switches’ contribution is $kT/C_h$ whatever happens; the amplifier’s contribution is proportional to the excess bandwidth it was given for nothing.

That is a real trade and this essay does not measure it, because it needs an amplifier noise model inside the sampled loop and the machinery here has a switch and two capacitors. It is named because leaving it out silently would make the result above sound like a complete account of a real stage’s floor, and it is not — it is a complete account of the part that cannot be designed away.

What it means for a design #

The consequences are the reason the technique is usable at all, and they are not the ones the equivalent resistance suggests.

Raising the clock does not help. It lowers the equivalent resistance and it lowers the corner frequency in exactly the same proportion, so the noise bandwidth and the noise density move together and the product does not move. This is the resistor result again with a different quantity doing the cancelling.

Making the switched capacitor smaller does not hurt. It raises the equivalent resistance, which is usually the whole point, and the floor is unchanged.

Only the holding capacitor matters, and it matters as $1/\sqrt{C_h}$: 63.3 microvolts on a picofarad, 20.0 on ten, 6.33 on a hundred, 2.00 on a nanofarad. So the capacitor sizes in a switched-capacitor design are chosen by the noise budget and not by the response, which is why the areas in such a design look nothing like the schematic suggests.

The temperature is in it and nothing else is. A floor that contains $k$, $T$ and a capacitance is a floor that can only be lowered by cooling or by area, and both of those are expensive in ways a circuit designer cannot argue with. It is the same shape as the thermal floor the field opens with, arrived at by a circuit that contains no resistor and is not in thermal equilibrium with anything in the obvious sense.

And stages accumulate. Each stage is an independent sampler, so the variances add and the noise grows as $\sqrt{N}$ while the signal does not: on a hundred picofarads a one-volt signal has a signal-to-noise ratio of 101.0 decibels through one stage, 97.9 through two, 94.9 through four and 91.9 through eight. A sixth-order filter is three stages and costs 4.8 decibels against one.

Where the equivalent resistance would have been right #

There is a way to get the naive answer, and identifying it is the clearest statement of what the cancellation is doing.

Take the same switched capacitor and load it with something that is not a capacitor — a resistor, say, or an amplifier input with a wide bandwidth and no hold. Now there is no $C_h$ to store the charge and integrate the noise, the arrangement’s output bandwidth is set by whatever the load is, and the noise seen is the equivalent resistance’s density times that bandwidth. The microvolts per root hertz are real; they were always real. What made them harmless was that the only thing they could drive was a capacitor, which turned a density into a stored variance and cancelled the resistance out of the answer.

That is the same sentence the rung below this one ends on, and it is worth having twice: $kT/C$ is not a statement about a resistor. It is a statement about what a capacitor is allowed to know, and it holds for any way of charging it — including one with no resistor in it anywhere.

The three floors in one stage #

A switched capacitor has three noise floors and only one of them contains the clock. This page is kT/C, which contains neither the clock nor the switch resistance. The amplifier inside the sample is the amplifier’s own, which folds entirely into the band and therefore does depend on the clock. The sample that is subtracted is the arrangement that removes one of the three. A resistor made of a clock is the circuit all three live in, and The bandwidth noise sees is the factor that turns any density into a voltage. A floor and a ceiling is the range they are the bottom of.

What is checked #

Three assertions, and the first is made against a statistical spread rather than a tolerance.

That the measured variance is $kT/C_h$ at every ratio, from a thousandth to ten, asserted against the standard error the estimate can have given the pole and the sample count — which is a per cent at one end of the axis and five at the other, and would be a number chosen to pass if it were one number.

That the equivalent resistance the arrangement behaves as moves by four orders across those same ratios, which is what makes the first assertion say something: an answer that did not move while nothing else moved either would be no evidence at all.

And that each of the two sampling events contributes less than the total, so that the cancellation is between them rather than one of them being the answer with the other as a correction.

What the technique does and does not escape #

A hundred megohms of behaviour with none of a hundred megohms’ noise is a strong claim, so it is worth setting out precisely which of the switched-capacitor arrangement’s costs this result removes and which it leaves.

It removes the density, and it does so for the reason the total that has no resistor in it gives: a resistor charging a capacitor produces a density going as the square root of its resistance and a noise bandwidth going as its inverse, so five decades of resistance give one total — 63.2762 microvolts, which is the square root of $kT/C$ and contains no resistance at all. The switched arrangement inherits that total and never has the density, because there is no continuous band over which to integrate one.

It does not remove the amplifier, and the amplifier inside the sample is where that is measured: the amplifier’s 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.

And one of the two floors can be taken away again. The sample that is subtracted shows the reset level a capacitor holds being the same number in two consecutive samples and cancelling exactly, at the cost of the amplifier’s noise not cancelling — two samples of it being independent, so its variance doubles. That is thirty times better at a megahertz and a loss above sixty, with the crossing being the same one the rung below computed for a different question.

Three essays, three floors, and one arrangement in which which of them dominates is decided by a clock rate rather than by any component value.

Why this is not a free lunch, stated precisely #

A hundred megohms with no hundred-megohm noise sounds like a violation of something and is not, and the reason is worth stating because it says exactly what has been bought.

A resistor’s noise is large because its bandwidth is large: the density goes as $\sqrt R$ and the noise bandwidth of whatever it charges goes as $1/R$, which is why the total that has no resistor in it finds one total across five decades of resistance. A hundred-megohm resistor charging a capacitor does not deliver more total noise than a kilohm charging the same capacitor; it delivers the same total over a much narrower band. The switched arrangement has no continuous band at all — it delivers a fixed quantity of charge noise per sampling event — so what it escapes is a density that was never the total in the first place.

Which means the honest claim is narrower than “quieter” and more useful. The arrangement is quieter per root hertz and identical in total, and it matters wherever a density is what the design is budgeted in — a spectral floor, an in-band noise figure, a resolution over a stated bandwidth. Where the budget is a total in volts, the resistor and the clock give the same answer, and this essay’s result is a change of units rather than a saving.

Which is exactly the distinction a floor, or a line draws about a converter’s timing error, one field over: two clocks with identical totals putting that total in two entirely different places, twenty-eight decibels apart, with nothing in a specification quoted as a total to distinguish them. A total and a distribution are different objects, and most of what a noise budget is for is the second one.