The floor that is only a floor while nothing flows

The floor a resistor sets established what makes Johnson noise a floor rather than a specification: it depends on a resistance and a temperature and on nothing else. Not on the material, not on the geometry, not on what the resistor is doing. A wirewound and a carbon composition of the same value at the same temperature produce the same 4.00 nanovolts per root hertz, and a resistor sitting in a drawer produces it exactly as a resistor in a circuit does.

Every one of those clauses is true and one of them is doing more work than it looks.

A resistor in a circuit. A resistor in a circuit usually has a current in it. The moment it does, a second generator appears that has none of Johnson’s properties: it depends on how the part was made, it is proportional to the voltage across the part, and its spectrum is $1/f$ rather than flat. It is called excess noise, it is on every resistor data sheet as a noise index in microvolts per volt per decade, and it is not in the expression the floor is computed from.

So the floor is a floor for a resistor doing nothing, and most resistors are doing something.

What the second generator is, and how it is specified #

Excess noise is a modulation of the resistance by the microscopic motion of whatever carries the current — grain boundaries in a carbon composition, the spiral cut in a metal film, the contact between a film and its end cap. Because it is a modulation of a resistance, what appears at the terminals is that fractional fluctuation multiplied by the voltage across the part, and a part with no voltage across it produces nothing at all.

It is specified as a noise index: the root-mean-square voltage produced in one decade of bandwidth, per volt of direct voltage across the part, in microvolts. A good metal film is 0.1 µV/V/decade or better; a thick-film chip resistor is one to ten; a carbon composition is tens. That is a range of a hundred between parts of identical value, which is the clearest possible sign that the quantity is not thermal.

Written as a density, one decade holding $(NI \cdot V)^2$ of mean square with a $1/f$ spectrum gives

$$e_n(f) = NI \cdot V \sqrt{\frac{1}{f \ln 10}}$$

so a 100 kΩ metal film with ten volts across it produces 660 nV per root hertz at one hertz, against a Johnson floor of 40.0. The thermal noise is a sixteenth of the total at that frequency, and it is the only part of the total that anybody computes.

The crossover, and why it moves as the square #

The two generators are equal at a frequency that follows from setting the densities equal:

$$f_c = \frac{(NI \cdot V)^2}{\ln 10 \cdot 4kTR}$$

and the shape of that expression is worth reading before its value. The numerator has the voltage squared in it and the denominator has the resistance. So doubling the voltage across a resistor moves the crossover up by four, and the crossover is the frequency below which the part is not on the floor anybody computed.

Half a volt across a metal film puts the crossover at two thirds of a hertz, which is below almost any measurement band and is why the effect has a reputation for not mattering. Thirty volts puts it at 2.44 kHz, which is inside every measurement band there is. A factor of sixty in voltage is a factor of 3600 in crossover frequency, and the range from half a volt to thirty volts is the ordinary range of a bias network.

That is the property that makes excess noise awkward to budget for. It is not a fixed contribution that can be added once; it is a contribution whose bandwidth depends on the operating point, so a circuit that is quiet at one bias is not merely noisier at another, it is noisy over a wider band.

The resistance moves it too, and in whichever direction the question is asked #

The crossover expression has a resistance in its denominator as well as a voltage squared in its numerator, and which of the two dominates depends entirely on what is being held fixed — so “is a larger resistor better or worse for excess noise” has two correct and opposite answers.

At a fixed voltage across the part, $f_c \propto 1/R$. Ten volts across a megohm puts the crossover at 27.1 Hz against 271 Hz for a hundred kilohms and 2.71 kHz for ten. A larger resistor is better, by a full decade of crossover per decade of resistance, because the thermal floor it raises is going up as $\sqrt R$ while the excess it produces is not going up at all.

At a fixed current through it, $V = IR$ and the expression becomes $f_c \propto I^2 R$. A hundred microamps through ten kilohms puts the crossover at 27.1 Hz; through a megohm, at 2.71 kHz. A larger resistor is now worse, by a decade per decade, because the voltage across it has gone up in proportion and the voltage enters squared.

The two cases are not academic. A resistor across a supply rail is the first; a resistor in the collector of a transistor, or anywhere in a bias chain that sets a current, is the second. Both are everywhere, and the sign of the design rule flips between them.

What holds in both cases is the one thing the splitting rule uses: the total is fixed by the circuit and the number of parts is not. Splitting does not change the resistance, the current or the voltage across the string, so it is the only variable in the expression a designer can move without changing the circuit — which is why it is the repair, and why the two contradictory rules above are mostly a distraction from it.

At a megohm and ten volts the thermal floor is 126.6 nV per root hertz, four times higher than at a hundred kilohms, and the excess density at a hertz is unchanged at 660. So the same excess noise has become a smaller fraction of a larger total — which improves the crossover and makes the circuit noisier. That is the trap in reading a crossover frequency as a figure of merit: it fell because the floor rose.

Splitting the resistor, which is the one repair that works #

The excess noise of a part is proportional to the voltage across that part. So a resistance built from $n$ equal resistors in series, carrying the same current, has $V/n$ across each of them.

Each contributes $NI \cdot V/n$ in a decade. They are independent — different pieces of film, different grain boundaries — so they add in quadrature, and the total over $n$ of them is

$$\sqrt{n}\cdot\frac{NI\cdot V}{n} = \frac{NI \cdot V}{\sqrt{n}}$$

Excess noise falls as the square root of the number of parts. And the Johnson noise does not move at all: $n$ resistors of $R/n$ in series have $4kT(R/n)$ each, summed over $n$ of them, which is $4kTR$ exactly. The figure checks that part by part rather than claiming it, because the whole claim is that one quantity moves and the other does not.

Eight parts in series take the total at one hertz from 660 nV per root hertz to 236, which is a factor of 2.8 — $\sqrt8$ acting on the excess term alone while the 40 nV of Johnson noise stays where it is. Sixteen parts would give a factor of four on the excess, and the improvement stops being worth anything once the excess has fallen to the thermal floor, which for this part and this voltage happens at about $n = 270$.

This is the only repair in these essays that lowers a noise floor by adding components. Every other route — a lower resistance, a lower temperature, a narrower band — trades something. This one trades board area and part count for a term that had no business being there, and the reason it works is that the offending generator is proportional to a voltage that can be divided while the resistance is not.

It is also why a precision divider for a high voltage is built from a string of resistors rather than from one, and why the usual explanation for that — voltage rating, power dissipation, temperature coefficient matching — is incomplete. Those are all real, and the noise is a fourth reason nobody lists.

Take the voltage down far enough and the second generator leaves the measurement band entirely.

Where the floor stops being a floor #

The distinction this essay adds is between two kinds of lower bound.

Johnson noise is a bound on what a resistance of that value at that temperature can do, and it cannot be evaded by any choice of part. That is what the floor a resistor sets established and it is still true.

Excess noise is a bound on what this part can do, and a different part of the same value does better. A wirewound has essentially none — its conduction is through bulk metal with no granular structure, and the difference is in the manufacturing rather than in anything the resistor that is only a resistor would measure — so a wirewound 100 kΩ with ten volts across it sits on the thermal floor at every frequency. A carbon composition of the same value might be a hundred times worse than the metal film measured here, which would put its crossover at 2.7 MHz.

So the honest statement of a resistor’s floor has two parts and only one of them is physics. The thermal part is a constant of nature and the excess part is a purchase order. A design that needs to be on the floor below a hertz is not a design that needs a low-noise amplifier; it is a design that needs the right kind of resistor, and the amplifier is the wrong place to have looked.

That pattern — a fundamental bound with a manufacturing term sitting on top of it in the same units — is not unique to resistors. It is exactly the shape of the junction that is a resistor at zero volts, where a diode’s shot noise is fundamental and its excess is not, and of the noise a clock does not make, where an oscillator’s thermal floor is computable and its close-in phase noise is a property of the device.

For comparison, the quantity all of this is measured against, drawn as it was first established.

What the noise index does not specify #

That the spectrum is exactly $1/f$. It is close to $1/f$ over many decades and is measured as such, and the exponent is not exactly one for any real part — values between 0.9 and 1.2 are ordinary. The noise index specification assumes exactly one decade-independent, which is what makes a single number possible, and the error in that assumption is a few decibels over the range where the excess dominates.

That the noise index is a tight specification. It is a maximum, and parts of the same type vary by a factor of several. Two resistors from one reel can differ by ten decibels in excess noise and be identical in every other respect, which is a property no amount of measurement of one of them reveals about the other, and is the same statistical situation the distribution the bench cannot see describes for component tolerance.

That alternating current across the part does nothing. The specification is written in terms of direct voltage because that is how it is measured, and the mechanism responds to the root-mean-square voltage whatever its frequency. A resistor carrying a large signal has excess noise proportional to the signal, which makes it a multiplicative noise rather than an additive one — a quite different object, and one this figure does not measure.

That splitting the resistor is free. Eight parts have eight times the parasitic capacitance to ground, eight tolerances, eight temperature coefficients and eight solder joints. The noise argument says the excess falls as $\sqrt n$; it does not say the network is better.

The crossover solved, and the root-n rule summed part by part #

The crossover is solved rather than read off, by setting the two densities equal and confirming that they are equal there to floating-point precision.

Its square-law dependence on the voltage is checked as a ratio, between the crossover at a voltage and at half of it, which must be four — a claim about the shape of the expression rather than about either value.

The $\sqrt n$ rule is summed part by part over one, two, four, eight and sixteen resistors and checked against $\sqrt n$ to a part in $10^{12}$, rather than being quoted.

The Johnson noise is summed part by part too, over the same splits, and checked not to move — which is the other half of the claim and the half that would be easy to assume.

And the whole second generator is refused at zero volts. With nothing across the part the excess density is exactly zero at every frequency, not small, so the total returns to the thermal floor identically — which is what makes “the floor is a floor for a resistor doing nothing” a measurement.

A constant of nature with a purchase order on top of it #

The useful way to hold this is that a noise floor computed from physics is a lower bound on a quantity, not a prediction about a component.

$4kTR$ is what a resistance must produce. A resistor is a resistance plus a manufacturing process, and the process contributes a second generator that the physics has nothing to say about. The two arrive in the same units at the same terminals and only one of them is in the expression, so a budget that contains the expression and not the data sheet is a budget with a term missing — and the missing term is the larger one below a frequency that depends on the operating point.

The same sentence with different nouns covers most of this field. Only the real part is warm is the version for a capacitor, where the physics fixes the noise once the loss is known and the loss is a property of the dielectric. The junction version is shot noise plus flicker. The oscillator version is a thermal floor plus close-in phase noise. In every case the fundamental part is computable from two numbers and the other part is not computable at all — it is measured, on that part, by whoever made it.

Which is why a noise budget is not finished when the physics is done, and why the most useful thing a figure like this can produce is not a density but a frequency: the one below which the computable part of the answer stops being the answer.

Still open: the exponent that is not one, the modulation that is not additive, and the metal foil #

Measuring the exponent rather than assuming it. Everything here uses a $1/f$ spectrum because the noise index specification does. Building the same measurement on a seeded pink sequence with a settable exponent — the apparatus the corner where averaging stops working already has — would say how much the crossover moves for an exponent of 0.9 or 1.2, and therefore how much of the number above is the specification’s own assumption.

Excess noise as a multiplication rather than an addition. The mechanism modulates a resistance, so with a signal across the part the noise is proportional to the signal and appears as sidebands rather than as a floor. That is a different object from the one measured here and it is the one that matters in an attenuator carrying a large signal, where it sets a distortion-like limit that no additive budget contains.

And the part where the excess is genuinely absent. A bulk metal foil resistor has no granular conduction and essentially no excess noise, so its floor is thermal at every voltage. Measuring the crossover for a part with a noise index a hundred times lower would put it below a millihertz, at which point the question becomes whether anything else in the circuit is that quiet — and the answer is almost certainly the amplifier, which is where the floor a circuit has starts.