The resistor the noise comes from

The floor a resistor sets established that a warm resistance produces 4kTR volts squared per hertz and that nothing about the resistor except its value and its temperature appears in the expression. The bandwidth noise sees established what that density has to be multiplied by, and that using the −3 dB point instead is twenty-one per cent low for a single pole.

Between them those two essays answer the question completely, for a circuit with one resistor in it.

Every circuit in this collection has more. A third-order active filter has two, a seventh-order one has four, an instrumentation amplifier has six that matter and several that do not, and the question the two essays above answer is not the question any of them poses. How much noise a resistor makes is 4kTR and has been settled since the first essay in this field. How much of it arrives is a different transfer function for every resistor in the circuit, and nothing in the collection so far has computed one.

What is being solved #

A noisy resistor is a noiseless resistance with a voltage source in series with it. That sentence is usually offered as an equivalence to be reasoned with; here it is written down literally, as a netlist, and solved.

For each resistor in turn its node is split — the resistance stays between the original node and a new one, and a source of √(4kTR) volts per root hertz is inserted between the new node and where the resistor used to end. The signal source is grounded rather than removed, which for a voltage source is the same thing and is worth saying because it is not the same thing for a current source. The network is then solved at every frequency on a logarithmic grid, and what comes back is that resistor’s own gain to the output.

The output density is the quadrature sum of those, one term per resistor, and the total is that density integrated. Two things follow that the arithmetic makes exact rather than approximate. The shares of the noise power add to one — not nearly, but to twelve figures — because the total is the integral of a sum of squares and integration is linear. And no resistor’s own contribution ever equals the total, because independent sources add in quadrature and not by addition.

Why the terms may be added at all #

Adding the terms in quadrature is a step, not a notation, and it rests on a claim about the sources rather than about the network: that they are independent of one another. For thermal noise in distinct resistors that claim is as solid as anything in this field. The fluctuation in one resistor is the motion of its own carriers against its own lattice, and two resistors on one die at one temperature share the temperature and nothing else — the temperature sets each variance and says nothing about the sign of either at any instant.

Where the claim would fail is worth naming, because it is a real arrangement and not a hypothetical one. A single physical resistor modelled as two in series is one source written twice, and adding those two in quadrature under-counts by √2. So does any construction in which one noise generator reaches the output by two paths — which is not this, since each path here belongs to a different source, but is exactly what happens to an amplifier’s input current noise when it flows out through two branches at once.

The check that the addition was done correctly is the one already made: the shares sum to one, and they sum to one because the total is computed from the summed spectrum rather than from the parts. If a source had been counted twice or lost, the two routes to the total would disagree, and the figure asserts that they do not to twelve figures rather than to a tolerance.

The construction is cheap to write and it is the kind of machinery that returns a plausible ranking for a mis-stated problem. Every number in it is a solve of a network that has been altered, and nothing about a plausible bar chart says whether the alteration was the right one. So it is calibrated before it is read.

That is the calibration the rest of this page is quoted against, and it is a strong one. A factor of two in the relation between a one-sided density and a sample’s variance — the standard way of getting this wrong — would show as a constant offset of √2 rather than as a ranking that merely looks odd. The same move the assumption that is a geometry makes in a different field: find the case where the old answer is exactly right, agree there, and only then quote a disagreement.

The R–C is also the one network where the rule of thumb is exactly right, and the reason is worth stating because everything below is about where it stops being. There is one resistor, its transfer function to the output is the output’s, so giving it the output’s is not an approximation at all.

The departure, which needs only two resistors #

A third-order Butterworth realised as a cascade has two resistors and they are the same value. Both are 1.59 kΩ, to fifteen figures, because the section’s resistance is √(L/C)/Q and the numbers happen to land there.

Nothing about the resistors distinguishes them. What distinguishes them is where they sit: a section’s own noise passes through every section after it and through none before it, so the last resistor in a chain contributes its noise to the output whole and the first contributes what survives the rest of the filter.

That is the whole mechanism, and it has a consequence that is easy to state and impossible to get from a schematic. The noise of a cascade is not a property of its component values. Two circuits with an identical parts list, an identical response and an identical bill of materials can differ in their noise, and later on this page two of them differ by a factor of 1.499.

The ranking, and the ranking it is not #

With more sections the shares spread out, and the spread is the useful part: a budget whose entries were all alike would say only that the noise is what it is.

On that filter the largest resistor is also the largest contributor, which makes the budget look like a listing of the resistances. It is not, and the family that shows it is not is the one with the gentlest skirt.

A Bessel is the interesting case for the same reason it is interesting everywhere else in this collection: flat delay, bought with more delay is a family whose skirt is gentle by construction, and a gentle skirt is a poor filter of the noise of the sections in front of it. Its early resistors therefore contribute nearly what they make. A Chebyshev’s do not, because a Chebyshev’s skirt is the steepest per order available and it falls on its own sections’ noise as readily as on a signal.

What decides a share #

Two things do, and separating them is what makes a budget readable rather than merely true.

The first is position, which is the mechanism the two-resistor case isolates: everything downstream of a resistor filters its noise, and everything upstream does not. On a five-section chain the last section’s resistor is at the output with no filtering at all and the first section’s has four sections’ worth in front of it, so position alone spans most of a decade in contribution.

The second is the section’s own quality factor. A section’s resistance is √(L/C)/Q for a given capacitance, so a high-Q section has the smallest resistance in the chain and therefore the smallest density — and its noise arrives at the output through a response that peaks at the section’s own resonance, which multiplies the narrow band around that peak by Q. The two effects pull opposite ways and the peak wins, which is why the Chebyshev’s 364 Ω resistor is not the quiet one and its 4.65 kΩ resistor is the loud one only because it sits last.

That is also why the ranking cannot be read off the schematic even when it happens to agree with the resistances. On the Chebyshev the two orderings coincide; on the Bessel they do not; and nothing visible in either netlist says which case is in front of the reader. The site’s standing complaint about a schematic — that it is a label rather than the object — arrives here in an unusually literal form, since the symbol on the page carries the one number that turns out not to decide the answer.

The rule of thumb, and why it cannot be repaired #

The sentence almost everybody uses is that a circuit’s noise is its resistance in its own noise bandwidth. It is not a bad sentence — it is exactly the arithmetic the first essay in this field earned honestly on a network with one resistor — and it survives because the two numbers in it are both easy to get. The resistances are on the schematic. The noise bandwidth is a property of the response, and the bandwidth noise sees measures it.

Applying the rule to a cascade means adding the resistances and giving every one of them the output’s transfer function. Applying the budget means giving each one its own. The two calculations differ in nothing else, so their ratio is a measurement of exactly that substitution.

The direction is set by how steeply the filter falls between one resistor and the next, which is a property of the family rather than of the design. A steep filter over-counts, because the resistances being added up include ones whose noise the rest of the filter removes. A gentle one under-counts, because its own noise bandwidth is measured on a response that is still open where its early sections’ noise arrives.

Under-counting is the direction that matters. A design whose predicted floor is 0.791 of its real one has a fifth of its noise budget missing, and it is the family chosen when phase matters — which is to say, whenever a filter has to sit in front of a converter or behind a transducer.

What a budget is for #

A ranking that cannot be acted on is a diagnosis without a treatment, and this one can be acted on precisely, because the entries are sections and a section can be paid for on its own.

Every section in these realisations sits behind an ideal follower. Dividing one section’s resistance by s while multiplying its capacitance by s leaves that section’s transfer function identical — the impedance scale cancels out of it — and lowers its own Johnson noise as √s. No other section can tell that it happened. It is the same move that the same filter a thousand times larger applies to a whole network, applied to one section of one.

The shape of that curve is the argument. Spending buys √s while the section paid for is still the largest contributor and buys almost nothing afterwards, because the total is then held up by the sections that were not paid for. Four times the impedance is worth 28 per cent of the noise; the remaining twenty-five-fold is worth another 28 per cent between them.

There is a price on the other side and it is the ordinary one. Sixteen times the impedance means sixteen times the capacitance in that section, and a follower that has to drive it — the Q the amplifier decides is the essay about what happens when the amplifier in an active section stops being ideal, and a heavily scaled section is exactly the arrangement that gets there first.

The same filter, twice, with the sections written in the other order #

The strongest statement the budget makes needs no scaling at all.

Ideal followers separate the sections, so the filter’s response is the product of the section responses, and multiplication commutes. Writing the sections in a different order therefore produces the same filter. It does not produce the same noise, because a section’s noise is filtered by everything after it, and reordering changes what is after what.

Fifty per cent of the noise, from a decision that changes nothing a network analyser can see. The mechanism is the one from the two-resistor case, applied to the loudest section rather than to a pair of equal ones: the high-Q section is where most of the noise is made, and putting it first puts the whole of the rest of the filter between that noise and the output.

The last clause of the caption is the reason this is not simply free, and it is the subject of the rung above this one: the arrangement that buries the noise is the arrangement that puts a resonant rise at the front of the chain, where the signal has not yet been attenuated. The floor and the ceiling move in opposite directions, and choosing between two orderings on the floor alone chooses the worse one.

What it does not say #

It does not say the rule of thumb should be abandoned. Between 0.79 and 1.55 is a factor of two on a quantity that a design usually wants to a factor of two, and the rule costs a multiplication where the budget costs one network solve per resistor per frequency — some tens of thousands of factorisations for a seventh-order filter on a decent grid. For a first pass it is the right instrument.

It says the rule has a range, and that the range is a property of the family rather than of the frequency or the amplitude. That is an unusual shape of boundary for this collection, whose edges are almost always a hertz or a volt, and it makes the rule harder rather than easier to use safely: there is no number to stay below, only a question about how steeply the thing falls between its own resistors.

It also says nothing about the amplifiers. Every follower in these realisations is a nullor, and a nullor is silent — it has no voltage noise, no current noise and no bandwidth. Real ones have all three, and the floor a circuit has measures what a real one adds: a part with 4 nV/√Hz in series with its input and 0.6 pA/√Hz across it. On the filters above, three or four such amplifiers sit exactly where the resistors do, and eight amplifiers, and what they add is the essay that puts them into a ladder and reads the floor again. Every total on this page is therefore a lower bound, and it is stated as one.

The grid is the last of the assumptions. The budget integrates on a logarithmic grid with the trapezoidal rule in linear frequency, which is the pairing the bandwidth noise sees explains at length, and it is under-resolved wherever the response has a feature narrower than a grid interval. The filters here have no such feature; a high-Q resonator would, and the honest response to one is a grid with points where the circuit has them rather than a wider tolerance.

Where else a budget is owed #

The machinery takes a netlist, an output node and a frequency grid, so every circuit in this collection that is written as a netlist can have one, and several of them are asking for it.

The instrumentation amplifier is the clearest case. The four resistors that decide, and the two that do not already establishes that its rejection is set by two ratios and is untouched by the other two — a statement about the signal path that says nothing about which of the six resistors the noise comes from. Those are different questions with different answers, and the second has never been asked here.

The passive ladder is the case where the answer is likely to be flattest and is worth having for that reason. A ladder is not a cascade is about a realisation with no internal buffering at all, so every element’s noise reaches the output through the whole network rather than through a tail of it, and the two resistors a ladder was designed between names the only two resistances in it. A ladder ought therefore to have a budget with two entries and no ordering to get wrong, which would make it the counter-example to everything above rather than another instance of it.

And the divider is the smallest circuit with a non-trivial budget in it. The divider, and the thing it does not know about computes what a load does to a ratio; the two resistors in it also have unequal shares of the output noise, for the reason this page is about, and the arithmetic is short enough to be done by hand and checked against the solve.

The number worth carrying #

Two identical resistors, 60.0 and 40.0 per cent. That is the whole finding in the smallest circuit that can hold it, and everything above is the same statement in larger networks: 64.1 per cent from one of three, 1.551 times the truth from the rule that adds them up, and 1.499 times the noise from writing the same sections in the other order.

The habit that goes with it is a question rather than a number. A schematic gives every resistor equal standing — it is a symbol with a value beside it, and the values are what a bill of materials holds. The circuit gives them nothing of the kind: each one has a position, the position is a transfer function, and the transfer function is most of the answer. Asking which resistor before asking how much resistance is the whole of the method, and one solve per resistor is what it costs.