The total that has no resistor in it

A resistor’s noise density is $\sqrt{4kTR}$. Ten times the resistance is $\sqrt{10}$ times the noise, which is one of the few pieces of design advice in this subject that is genuinely simple: if the circuit is too noisy, make the resistors smaller.

Put a capacitor across the resistor and the advice stops working, completely, in a way that is exact rather than approximate.

Two dependences that cancel #

The resistor decides two things about the output noise and they are reciprocal.

The density, which is $\sqrt{4kTR}$ and therefore proportional to $\sqrt{R}$. Over the five decades of resistance drawn, that is a factor of 100.

The noise bandwidth, which for a single pole is $\pi/2$ times the corner frequency, and the corner is $1/2\pi RC$. So the noise bandwidth is $1/4RC$, proportional to $1/R$, and over the same five decades it is a factor of $10^4$.

Multiply them and the mean square is

$$\overline{v^2} = 4kTR \cdot \frac{1}{4RC} = \frac{kT}{C}$$

with the resistance cancelled out of it identically. Measured by integrating the solved response of each of the five networks over its whole axis, the five totals agree to better than a part in a hundred thousand, and the middle one is 63.2762 µV against a closed form of 63.2762 µV.

That number — 63.3 microvolts on a picofarad at room temperature — is worth committing to memory, because it is the floor under a very large number of circuits and there is nothing to be done about it.

Why no arrangement of components can avoid it #

The cancellation looks like an algebraic coincidence and it is not. It is a thermodynamic statement, and the thermodynamic version says why no cleverness will get round it.

A capacitor in contact with a bath at temperature $T$ is a degree of freedom, and a degree of freedom in thermal equilibrium holds $\tfrac12 kT$ of energy. The energy stored on a capacitor is $\tfrac12 C\overline{v^2}$. Setting those equal gives $\overline{v^2} = kT/C$ immediately, with no resistor mentioned anywhere in the derivation — which is exactly why no resistor appears in the answer.

The figure asserts that form directly: the energy computed from the integrated noise voltage is $\tfrac12 kT$ to four parts in ten million.

The consequences follow at once. Two resistors are no better than one — a divider onto the same capacitor produces four noise terms and they still sum to $kT/C$. A filter in front changes nothing, provided the capacitor is still in equilibrium with something warm. And an inductor obeys the same law in current: $\tfrac12 L\overline{i^2} = \tfrac12 kT$ gives 2.0 nA on a millihenry, by the identical argument.

The third route, which is samples #

The two routes above are both integrals, which is a weaker pairing than this collection usually insists on: one integrates a closed form and the other integrates a solved response, and they share the idea of integrating.

So the figure adds a third that shares nothing with either. A seeded white sequence of the right density is marched through the middle network in the time domain and its root mean square taken over six seeds. It gives 62.73 µV against the integral’s 63.28, with a seed-to-seed spread of 5.9 per cent.

The comparison has to account for one thing before it means anything: a sampled sequence has no power above its own Nyquist frequency, and the total being predicted is over the whole axis. At two hundred samples per corner the tail beyond Nyquist carries 0.64 per cent of the variance, so the marched route is expected low by 0.32 per cent of the amplitude — and the assertion is made against the seed-to-seed spread rather than against a tolerance chosen afterwards.

The edge, which is the whole frequency axis #

Every claim above is about the total over all frequency. In a finite measurement band the resistance is back, and the figure measures how far back.

Taking a band equal to the middle network’s own corner frequency — 15.9 MHz for a picofarad through ten kilohms — the five networks give 5.05 µV to 63.07 µV: a factor of 12.5, and the same factor at every capacitance on the slider, because the band is defined in terms of the circuit rather than in hertz.

That is the sense in which the resistance still matters. A high resistance puts all of its noise inside a narrow band, so a measurement that only looks at part of the axis sees nearly all of it; a low resistance spreads the same total over a much wider band, so the same measurement sees a small fraction. The totals are equal and the shapes are not, and almost every real measurement is a shape question.

The practical reading: if the thing downstream integrates the whole axis — a sample-and-hold, a switched-capacitor stage, anything that takes a snapshot — the resistance is irrelevant and $kT/C$ is the answer. If it is band-limited, the resistance is back and smaller is better.

What the axis has to reach #

The integral is over all frequency, and a figure has to stop somewhere, so it is worth saying how far “all” has to go before the answer stops moving.

A single pole’s tail falls as $1/f^2$ in power, so the variance beyond a frequency $f$ is proportional to $1/f$ — which is a slow enough convergence to be worth checking rather than assuming. Integrating from a millionth of the corner to a million times it captures everything but a part in a million, and that is what the figure does: six decades either side of each network’s own corner, which for the hundred-ohm case means reaching past a hundred gigahertz.

That is far outside where the model is true, and saying so is the honest part. A hundred-ohm resistor is not a resistor at a hundred gigahertz, and a real circuit’s noise is cut off long before that by its own parasitics. So the exact $kT/C$ is the answer for an idealised single pole, and a real capacitor with an inductance in series with it collects slightly less — which is one of the few directions in which reality is on the designer’s side.

The axis the figure draws is narrower than the axis it integrates, and that is deliberate: the drawn band is framed on the five corner frequencies so that every knee is visible at every setting of the slider, while the integral runs past both ends of it.

Where this decides a design #

The result’s importance is out of all proportion to how simple it is, because it sets the floor for every circuit that samples.

A sample-and-hold. The switch is a resistance and the hold capacitor is a capacitance, and when the switch opens, the noise voltage on the capacitor at that instant is frozen onto it. Its root mean square is $\sqrt{kT/C}$, whatever the switch is made of. A one-picofarad hold capacitor has 63 µV of noise on every sample, and a lower on-resistance switch does not help.

A converter’s front end. Put those 63 µV against a one-volt full scale and compare with a quantiser’s own noise of one least significant bit over $\sqrt{12}$: the two are equal at 12.16 bits. So a one-picofarad sampling capacitor puts a twelve-bit floor under a converter regardless of how many bits the converter has, and reaching sixteen requires 26 pF or more — which costs acquisition time, because the same capacitor has to be charged through the same switch.

A switched-capacitor filter. Every switching event deposits $kT/C$ onto its capacitor, and a filter with many stages accumulates them. The capacitor sizes in such a design are set by noise rather than by the response, which is why the response equations look free of it and the areas do not.

The acquisition time, which is the price of the repair #

The only lever on $kT/C$ is $C$, and making $C$ larger costs time, so the two constraints meet and the meeting is where a sampling design actually lives.

Charging a capacitor through a switch of resistance $R_{\mathrm{on}}$ to within a fraction $\epsilon$ of the final value takes $R_{\mathrm{on}}C\ln(1/\epsilon)$. Settling to half a least significant bit of a sixteen-bit converter is $\ln(2^{17})$ — about 11.8 time constants.

So doubling the capacitance to buy half a bit of noise doubles the acquisition time, and the only way to keep the time is to halve the switch resistance, which costs area in an integrated design and charge injection in any design. That is the trade a converter’s front end is built around, and it is why the sampling capacitor is one of the numbers that gets argued about.

Working it through for the twelve-bit floor above: a one-picofarad capacitor through a hundred-ohm switch acquires to sixteen bits in 1.2 nanoseconds and has 63 µV of noise. Twenty-six picofarads gets the noise to a sixteen-bit floor and takes 31 nanoseconds through the same switch, which is a thirty-megahertz sample rate before anything else is accounted for. The two numbers are the same design read from its two ends.

The other quantity that does not contain the resistor #

This collection has met the same shape of result once before, in a different field and about a different quantity, and the pair is worth holding together.

Charging a capacitor through a resistor from a voltage source dissipates $\tfrac12 CV^2$ in the resistor — exactly half the energy the source delivers — whatever the resistance is. Ten ohms and a hundred kilohms lose the same energy, to nine figures.

The two results are the same kind of statement about the same kind of network: a resistance sets a time scale and not a total. What the resistance actually decides in the charging case is how long the loss takes, and in the noise case how wide a band the noise occupies. In neither case does it decide the answer.

The difference is what happens when an attempt is made to escape. The charging loss can be reduced without limit by charging slowly — a ramp instead of a step, with the loss falling as $2\tau/T$ and no floor under it. The noise cannot be reduced at all, because it is not a consequence of how the capacitor was charged but of the fact that it is warm.

What the temperature does, and what it does not #

The result contains $T$, and it is the only thing besides the capacitance that it contains.

Cooling helps as $\sqrt{T}$: liquid nitrogen at 77 K reduces the noise by a factor of 1.94 against room temperature, which is worth having and is a long way from removing it. Cooling the resistor alone helps in proportion to the fraction of the total it contributes, which for a single resistor is all of it — but the switch, the substrate and the wiring are all at their own temperatures, and the capacitor’s noise is set by whatever it is in equilibrium with.

Which is the last of the ways this result differs from the ordinary noise arithmetic. In a normal noise budget the terms come from named components and can be attacked one at a time. Here the capacitor is a single degree of freedom in contact with a bath, and the only two numbers in the answer are how large it is and how warm.

The other floors, and the one that cancels #

A total that contains neither the resistance nor the bandwidth is unusual in this field, and it is worth saying against what. The floor a resistor sets contains both. The floor a circuit has contains a source resistance with an optimum. The bandwidth noise sees is the factor that cancels here and nowhere else. The noise a clock does not make is the same total in the circuit that made it famous, and The half that never arrives is the other quantity in the collection that is independent of the resistance it happens in.

What is checked #

Five assertions, and the first two are the essay.

That the five totals agree to a part in a hundred thousand across five decades of resistance — the cancellation, stated as a measurement rather than as algebra.

That the middle one is $\sqrt{kT/C}$ to the same precision, so that the cancelled quantity is the right one and not merely a constant.

That the two halves of the cancellation are each what they are claimed to be: the density rising as the square root of the resistance, asserted to a part in a million, and the noise bandwidth falling as its reciprocal, to two parts in ten thousand. Asserting only the product would have let a compensating pair of errors through.

That the energy is $\tfrac12 kT$, which is the reason for all of it and is a different statement from the first one rather than a restatement.

And that in a finite band the five are not equal — a factor of 12.5 apart — which is the refusal that keeps the claim from being read wider than it is.

Where a floor with no resistance in it decides a design #

A total that contains only $k$, $T$ and $C$ is the kind of result that sounds like a curiosity and is in fact the binding constraint on a large class of circuits, because it says the only thing a designer can do about it is make the capacitor larger.

The noise a clock does not make is where that becomes surprising: a switched capacitor behaving as a hundred megohms produces none of the 1.27 µV/√Hz that a hundred megohms of resistor would, because what lands on the holding capacitor is $kT/C$ with the holding capacitor in it and nothing else — not the clock, not the switched capacitor, not the on-resistance — exactly rather than asymptotically.

The floor a converter sets is where it changes what a specification means. That essay crosses a quantiser’s floor against a source resistance, which is the right calculation for a continuous front end; for a sampling one the source resistance cancels out entirely and what replaces it is the sampling capacitance, a property of the converter rather than of the circuit in front of it.

And the sample that is subtracted is the one place the floor is beaten rather than accepted: the reset level a capacitor holds is the same number in two consecutive samples and cancels exactly. What that costs is the amplifier’s own noise not cancelling, since two samples of it are independent and its variance doubles — thirty times better at a megahertz and a loss above sixty.

The three say the same thing in three registers. The total measured here is not a property of a component that a better component improves; it is a property of a capacitance and a temperature, and the only ways past it are a bigger capacitor, a colder one, or an arrangement that subtracts the same sample twice.

The first of those three is the one a designer actually uses, and it is worth noticing what it costs in the circuit rather than in the noise budget. A larger sampling capacitor is a larger load on whatever drives it, a longer settling time through the same switch resistance, and — through the amplifier inside the sample — more folded amplifier noise, since the number of folds is the number of settling time constants. So the remedy for the floor measured here raises the floor measured there, and the design sits at whichever capacitance makes the sum least. That is an interior optimum in a quantity nobody chooses for noise reasons, which is the same shape as the floor a circuit has’s optimum source resistance and is arrived at from the opposite direction.

The second remedy is worth one line because it is the one that is occasionally taken. Cooling helps in exact proportion to absolute temperature, so liquid nitrogen is a factor of 3.9 in power and just under two in volts — a bit of resolution, at the cost of a cryostat. That the exchange rate is exactly the temperature ratio, with no material property in it anywhere, is the same fact this essay’s title is about.

And the third — subtracting the same sample twice — is the only one that beats the floor rather than lowering it, which is why it is worth a rung of its own. The other two make $kT/C$ smaller; that one removes it and pays in a different currency.