The floor a current sets

The noise field opens with a floor that belongs to a resistance. Four kTR volts squared per hertz, in which nothing about the resistor appears except its value and its temperature — not its material, not its power rating, not what it is made of. A different resistor of the same value at the same temperature makes the same noise.

There is a second floor, of the same kind and with a different constant in it, and it belongs to a current. Twice the elementary charge times that current, amperes squared per hertz, in which nothing about the device appears at all — not its resistance, not its area, not its temperature. A current of one amp is one over q carriers a second however it was produced, and the granularity of that count is the noise.

Two floors, one containing only a resistance and one containing only a current. They cross, and where they cross is what this essay is for.

The condition, and what is not in it #

Setting the two densities equal is a line of algebra:

$$2qI = \frac{4kT}{R} \quad\Longleftrightarrow\quad IR = \frac{2kT}{q}$$

and the left-hand side of the second form is the direct voltage across the thing carrying the current. So the crossing is at a voltage, and the voltage is a constant of nature and a temperature — 49.98 mV at 290 K, 51.70 mV at 300 K — with no resistance and no current in it.

That is unusual enough in this collection to be worth dwelling on. Every other boundary here is a frequency, an amplitude, a size or a duration, and every one of them is computed from the components of the circuit it bounds. This one is computed from Boltzmann’s constant and the charge on an electron, and it is the same number in every circuit ever built at that temperature.

The right-hand column is the whole point and the figure asserts it as such: seven resistances spanning six decades, seven crossing currents spanning six decades, and one voltage identical to the last bit of a double at all of them.

A designer’s version of the same statement. If the direct drop across a device is more than about fifty millivolts, its shot noise is larger than the Johnson noise of its own resistance; if it is less, the other way round. Nothing has to be looked up.

Why the constant is 2q and not q #

The factor of two is a convention rather than physics and it is worth pinning down, because half the disagreements about noise arithmetic are about it.

A Poisson process of rate $\lambda$ has a variance equal to its mean, and the two-sided spectral density of the resulting current is $qI$. Every quantity in this collection is a one-sided density — defined over positive frequencies only, so that integrating it over a bandwidth gives the mean square directly — and folding the negative frequencies onto the positive ones doubles it. Hence $2qI$, and hence $4kTR$ rather than $2kTR$ beside it.

The two conventions differ by exactly a factor of two in every expression, so a crossing computed consistently in either gives the same answer; a crossing computed in a mixture of the two is out by two, which is a decibel and a half and looks like a real effect. Both floors on this site are one-sided, both integrate directly against a noise bandwidth, and the noise-bandwidth essay’s π/2 is the other place that same care is needed.

What a density is worth in a real bandwidth #

A density is per root hertz and no instrument measures a hertz. What matters is the density times the square root of the noise bandwidth — not the −3 dB point, which for a single pole understates it by 21%, and which the noise-bandwidth essay measures.

A milliamp of shot noise is 17.9 pA/√Hz. In ten kilohertz of noise bandwidth that is 1.79 nA rms, which is 1.8 parts per million of the milliamp — so shot noise is a fractional fluctuation of about two parts per million on a milliamp in an audio bandwidth, and of two parts in ten thousand on a nanoamp in the same bandwidth. The fraction goes as $1/\sqrt{I}$, which is the reason a small current is a noisy current and a large one is not.

The same arithmetic on the junction: a milliamp junction is 0.4473 nV/√Hz, which in ten kilohertz is 44.7 nV rms across its own 25 ohms. That is the floor under a bipolar input stage, and it is the number a low-noise amplifier’s several-nanovolt specification is built out of by putting several such junctions in parallel.

The condition on the mechanism, which is not small print #

Both densities cannot belong to the same object, and understanding why is what keeps the crossing from being nonsense.

A resistor carrying a direct current has no shot noise. Its carriers are not independent: each one scatters off the lattice many times in transit, and the scattering correlates them, so the current’s granularity is smoothed away and what is left is the thermal agitation the resistance already has. Passing a hundred milliamps through a resistor does not add noise to it.

Shot noise needs a barrier — a junction, a vacuum gap, a tunnelling contact — where each carrier crosses independently and the crossing is a Poisson event. So the two lines in the figure describe two different objects, and the crossing is a comparison between them rather than a transition in one.

That distinction is not a caveat added afterwards; it is what makes the essay’s second result possible.

A resistance that is not a resistor, and is quieter #

A forward-biased junction carrying a current has a small-signal resistance of $kT/qI$ — 25 Ω at a milliamp, 25 kΩ at a microamp. It is a resistance in every sense a small-signal analysis cares about: it appears in the netlist, it sets gains, it forms corners with capacitances.

Its noise is not a resistor’s. Multiply its shot-noise current by its own resistance to get a voltage density:

$$v_n^2 = 2qI \left(\frac{kT}{qI}\right)^2 = \frac{2(kT)^2}{qI} = \frac{1}{2}\cdot 4kT r_d$$

and the current cancels out of the ratio entirely. A junction produces exactly half the noise power of a resistor of its own resistance, at every current. With an ideality factor $n$ the same arithmetic gives $n/2$ — the $n^2$ in the voltage density is divided by the $n$ in the resistance — so it is half only for an ideal junction. “Every current” means every forward current well above the saturation current, and whether a junction with $n \neq 1$ really follows the arithmetic is a separate question: the junction that is a resistor at zero volts keeps the current this page leaves out and takes up both.

A factor of $\sqrt{2}$ in voltage and exactly two in power, at every row. The figure asserts the ratio to a part in $10^{12}$ because it is an identity rather than a measurement.

Two consequences follow. A junction is the only two-terminal element in ordinary use that beats $4kTR$ — everything else either equals it or is worse — which is why the input stage of a low-noise amplifier is built out of them and not out of resistors. And the reason it beats it is not that something has been improved: it is that the junction is not in thermal equilibrium. A resistor’s noise and its resistance are two faces of one fluctuation-dissipation relation and neither can be had without the other; a biased junction is being held away from equilibrium by the supply, and the relation does not apply.

Two more settings walk the resistance up the slider, and the point of walking it rather than quoting the closed form is that the crossing current moves by four decades while the voltage across the resistance at that crossing does not move at all.

Where fifty millivolts turns up #

A bias resistor. A megohm feeding a hundred nanoamps into a base has fifty millivolts of drop only at fifty nanoamps, so at any ordinary bias current the base current’s own shot noise exceeds the resistor’s thermal noise. That is why an amplifier’s input current noise is quoted as a separate generator: it is shot noise on the bias current, $\sqrt{2qI_B}$, and for a hundred nanoamps that is 0.18 pA/√Hz.

A photodiode. The whole design problem is that the signal is a current and the noise floor is that current’s own shot noise, so the signal-to-noise ratio improves only as the square root of the light. No amount of feedback resistance changes it — a larger resistance raises signal and shot noise together and only helps against the amplifier’s contribution, which is the transimpedance essay’s subject and not this one.

A current source’s output. An undegenerated transistor’s output current has $\sqrt{2qI}$ of noise on it whatever its output resistance is, and this is the number that limits how quiet a biasing network can be. Degenerating it with an emitter resistor lowers that at every resistance, and not only past fifty millivolts: the device’s noise current and the resistor’s each have to push through the other element to reach the output, the total becomes $2qI(1+2x)/(1+x)^2$ with $x = IR/V_T$, and that is below both $2qI$ and $4kT/R$ at every drop — furthest below the lower of them, by 2.55 dB, at exactly the fifty millivolts of this page’s crossing. A one-to-one mirror doubles every term and changes none of the ratios. The resistor in the same loop measures it, and it is the same crossing arriving as a design rule, though not in the form it first suggests.

The top of the slider is worth reaching separately, because a ten-megohm resistance carrying five nanoamps is not a contrived arrangement — it is the input of a photodiode amplifier, and the crossing it sits at decides which of the two mechanisms its designer should be reading about.

How it would be measured #

Nothing in this essay is hard to check on a bench, and the check is worth describing because it is the experiment that separates the two mechanisms rather than assuming them.

Take a resistor and a junction with the same small-signal resistance — a kilohm, and a junction at 25 µA. Amplify each with the same amplifier into the same bandwidth and read the output noise. The junction should read a factor of $\sqrt2$ below the resistor, and the ratio should not change when the current is moved provided the junction’s resistance is re-matched each time.

Then pass a direct current through the resistor and repeat. Nothing should change, at any current, until self-heating raises T. That is the experiment that shows shot noise needs a barrier: the same current through two things of the same resistance produces noise in one of them and not in the other.

The third measurement is the crossing itself, and setting it up shows something the algebra above hides. A junction cannot be biased to its own crossing: its dynamic resistance is $kT/qI$, so the product $I r_d$ is $kT/q$ — exactly half the crossing voltage, at every current there is. That is the factor-of-two result seen from the other end, and it says a junction is permanently on the quiet side of its own line and cannot be moved off it.

To build the crossing, put a resistor in series with the junction and choose it so that its drop is fifty millivolts at the operating current — two kilohms at 25 µA. The resistor’s bare thermal noise and the junction’s bare shot noise are then equal as densities, which is what the crossing says. What they are not is equal contributions to the current noise of the pair. In series, each noise current has to push through the other element to reach the outside: the junction’s is divided by $(1+x)^2$ and the resistor’s by $(1+1/x)^2$, with $x = IR/V_T$, which is two here. So the resistor supplies four fifths of the pair’s noise and the junction one fifth, and the total is five ninths of either bare density rather than twice it. The two contributions are equal at a quarter of the crossing voltage, 12.50 millivolts — five hundred ohms at 25 µA — which is the degeneration rule from the current-source paragraph arriving as a bench setup.

That is also why emitter degeneration is a noise technique and not only a linearity one, for the current a stage delivers. The device brings shot noise that cannot be reduced; the resistor brings thermal noise that falls as the resistance rises; and in one loop each also shields the output from the other. Past 12.5 millivolts of degeneration the resistor supplies most of what is left, and a few hundred millivolts is where the device’s contribution has become a detail — 4.8 per cent of it at 250 mV. Read at the stage’s input instead, the same resistor adds its thermal noise in full, which is the other half of the account and belongs to the essay about the loop rather than to this one.

What this essay does not claim #

That shot noise is white for ever. It is white while the transit time is short compared with a period. Above the reciprocal transit time the carriers’ arrivals stop being independent and the density falls; for a silicon junction that is in the gigahertz, well above anything in this collection, and it is a boundary rather than an absence of one.

That the crossing is a transition. It is not, and the section above says why. A resistor does not acquire shot noise as its drop passes fifty millivolts. What the crossing says is which of two different components is quieter at a given operating point, and that is the question a design actually asks.

That the factor of two makes a junction a good resistor. It makes it a quiet one. It is also strongly temperature-dependent, non-linear at any drive worth having — a millivolt is four per cent of the thermal voltage — and has a capacitance across it. The half applies to the small-signal resistance and the small-signal resistance is valid over a few millivolts.

That the total is the larger of the two. Where they are comparable, the powers add — for the reason the networks field’s superposition essay gives, that the two mechanisms are uncorrelated and their cross term averages away. At the crossing the total is $\sqrt{2}$ times either one.

The four floors this field measures #

Shot noise is the second of four floors, and the field’s whole business is which of them is in charge. The floor a resistor sets is the first, measured two independent ways. The floor a circuit has is the amplifier’s own, with an optimum source resistance. The total that has no resistor in it is the sampled floor, which contains neither a resistance nor a bandwidth. A floor and a ceiling is where all four are put under a ceiling and become a range. And The one current a constant is right at is where the fifty millivolts this page’s crossing sits at turns up as a device model’s own boundary.

The gate #

The crossing voltage is asserted across the whole slider, not at the setting drawn: seven resistances from ten ohms to ten megohms, and the drop at the crossing agreeing to a part in $10^{14}$. That the crossing exists is arithmetic; that it is the same voltage every time is the claim.

The junction’s half is asserted as an identity, to $10^{-12}$, rather than as a measurement with a tolerance. It is a ratio of two closed forms and the site’s rule is that a residual with a closed form is a measurement while a residual without one is a disappointment — this one has no residual at all.

And the two densities are asserted equal at the crossing current, computed from the crossing and then fed back into both expressions, so that the figure’s edge mark is checked against the curves it is drawn on rather than against the algebra that produced it.

The one boundary whose axis is a voltage across a component #

Fifty millivolts is an unusual number for this collection to end an essay on, because it is not a frequency, an amplitude of a signal, a size or a duration. It is a direct voltage across the element carrying the current, and it is the only such axis here.

What makes it that rather than a current is worth restating: a resistor’s noise contains no current and a current’s noise contains no resistance, so the question “which dominates” cannot be answered in either variable alone. Their ratio is $2kT/q$ divided by the voltage across the thing, which is a pure number, so the boundary lives on the one axis that both mechanisms share. Whatever the resistance and whatever the current, a component with fifty millivolts across it is at the crossing.

That has a direct consequence for two circuits measured elsewhere. The ammeter that is a resistor optimises a burden voltage and gets 7.75 millivolts, which is a factor of six below this crossing — so a shunt designed for accuracy is in the regime where its own thermal noise dominates the shot noise of the current through it, and the noise budget can be written with the resistor alone. Where the trouble is at the input is the other way round: a photodiode’s shot noise is proportional to its photocurrent and there is no deliberate voltage across the junction at all, so the crossing is not reached and the resistor’s noise is the one that has to be argued with — which is what a megohm feedback resistor’s 127 nV/√Hz is doing in that essay’s budget.

The junction’s factor of two #

The second result on this page — that a forward-biased junction produces exactly half the noise power of a resistor with the same dynamic resistance, at every current — is the one that decides whether a device can be used where a resistor was.

It matters most where a bias network is in the signal path. A bias point is a solution, not a choice establishes that a junction’s dynamic resistance is a slope of a solved operating point rather than a component value, so a designer choosing between a diode and a resistor for a given impedance is choosing between two parts with the same small-signal resistance and a factor of two in noise power — three decibels, free, in the direction that favours the junction.

What the junction charges instead is everything else about being a junction: a temperature coefficient of 1.828 millivolts per kelvin, a resistance that moves with the current through it, and a nonlinearity whose amplitude boundary how small is small signal puts at 7.3 millivolts. A resistor has none of those and is three decibels noisier. Which of the two trades better is a question about the circuit, and the point of this essay is that the noise half of it has an exact answer rather than a rule of thumb.