The junction that is a resistor at zero volts

The floor a current sets ended on an identity. A forward-biased junction carrying a current I has a dynamic resistance of kT/qI; its shot noise, multiplied by that resistance, is a voltage density; and the current cancels out of the comparison with a resistor of the same value. The junction makes exactly half the noise power, at every current.

That essay also gave the reason, and the reason is where this one begins. 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 held away from thermal equilibrium by whatever supplies its current, so the relation does not bind it, and the junction is free to be quieter.

Stated that way the result has an edge the identity does not show. Equilibrium is not somewhere far away from an operating point. It is at zero volts, and every junction passes through it on the way to being biased. A junction with no voltage across it is a passive element at the temperature of its surroundings, and a passive element at that temperature has the Johnson noise of its conductance whatever it is made of. So the ratio that is a half at a milliamp has to be one somewhere, and the question is how it gets from one to the other and how far from zero the half actually is.

The current a meter reads, and the two a junction carries #

The identity took the junction’s shot noise on its net current, 2qI. That is the step to look at.

Shockley’s law is a difference. The current through a junction is $I = I_s\left(e^{V/V_T} - 1\right)$, and the two terms are two physical currents rather than one current and a correction. The first is carriers crossing the barrier forwards, a stream that grows exponentially with the voltage. The second is carriers crossing it backwards, a stream of size Iₛ that the applied voltage does not touch. What a meter reads is their difference.

The noise does not take the difference. Each crossing, in either direction, is an independent event, so each stream is a Poisson process with its own shot noise, and the noise powers of independent processes add whatever directions their currents run in:

$$S = 2q\left(I_s e^{V/V_T} + I_s\right)$$

The conductance is the slope of the net current, and only the forward term has a slope, $g = I_s e^{V/V_T}/V_T$. Dividing the one by 4kT times the other, with u = V/Vₜ,

$$\frac{S}{4kT\,g} = \frac{1 + e^{-u}}{2}$$

Far into forward bias e⁻ᵘ is nothing and the ratio is the half. At zero volts it is one. In reverse bias it grows without limit, because the conductance goes to nothing and the backwards stream does not.

One at zero volts, by necessity rather than by fit #

The value at zero volts deserves more than a reading off a curve, because it is the one number on the page that did not have to come out of this particular arithmetic.

At zero volts the two streams are equal, the net current is nothing, and the noise is 4qIₛ. The junction’s resistance there is Vₜ/Iₛ, which gives a Johnson current noise of 4kT·Iₛ/Vₜ, and because kT is qVₜ that is 4qIₛ as well. The ratio is exactly one, and it had to be: an element with no power flowing into it, at the temperature of everything around it, has the noise the floor a resistor sets measures and no other. A model that returned anything else at zero volts would be describing a junction that passes net noise power to a resistor at its own temperature, which is heat flowing between two bodies at one temperature with nothing driving it.

So the two-stream picture passes a test it was not built to pass. Shockley’s law is a statement about currents and says nothing about equilibrium; the shot-noise formula is a statement about counting and says nothing about temperature. Put together, at the one voltage where thermodynamics has something to say, they agree with it to the last digit a double carries.

The figure also shows what the ratio conceals by being a ratio. At zero volts the noise is 0.08 femtoamps per root hertz and the resistance is two and a half teraohms, which is to say that a small silicon junction at zero bias is an enormous resistor making a tiny noise — the right amount of it for its size, and irrelevant beside almost anything connected to it. The ratio says the junction is honest; the densities say whether anyone would notice.

Where the half begins, in volts and in amperes #

The excess of the ratio over its forward limit is e⁻ᵘ of that limit, exactly. One per cent of excess therefore needs u = ln 100, which is 115.1 millivolts at 290 kelvin, and one tenth of a per cent needs ln 1000, 172.6 millivolts. Those voltages contain nothing about the junction. Every junction with an ideality factor of one reaches the half at the same voltage.

The current does not share that property, and the current is what a designer reads. At u = ln 100 the forward stream is 100 Iₛ and the net current is 99 Iₛ, so the half is within one per cent of right from ninety-nine saturation currents upwards — and saturation currents differ between junctions by many decades.

The three-quarters point is worth having as a landmark because it is exact and easy to remember: a junction carrying its own saturation current forward has a voltage of Vₜ ln 2 across it, 17.3 millivolts, and makes three quarters of the Johnson noise of its conductance. A decade of current either side of that, the ratio is 0.9545 and 0.5455 — each within a tenth of its limit, so the whole passage from one value to the other happens across about two decades of current centred on the saturation current.

The junctions on which it matters #

Stated plainly, since the figure could be read as a warning about every diode: for a small silicon junction the departure is a sub-picoampere matter. With a saturation current of 10⁻¹⁴ A the half is within one per cent of right from 990 femtoamps and within two per cent from 490. No bias network, input stage or current mirror in an ordinary circuit runs a junction that gently, and the forward identity, half the Johnson noise of the junction’s conductance, stands for all of them without qualification.

The weight of this essay rests on two other kinds of junction.

Junctions with a large saturation current. The saturation current scales with area and falls by orders of magnitude as the barrier height rises, so a large-area junction or a low-barrier one can have an Iₛ many decades above the small-signal diode’s, and it rises steeply with temperature on top of that — the reason two millivolts a kelvin comes out with the sign it does. The middle curve of the figure reaches the half at a nanoamp, the rightmost at a microampere. On such a part the half-noise argument for using a junction as a quiet resistance is true at operating currents and false at bias currents, and which one a given circuit sits at is a question the circuit has to answer. The constant that is a window is the reminder that the saturation current is itself the output of a fit over a range of currents rather than a constant of the part, so the “ninety-nine Iₛ” of any particular junction inherits that fit’s window.

Photodiodes held at zero volts. This is the case where the departure is not a small correction at the edge of a range but the whole of the answer, and it takes the rest of the essay.

A photodiode held at zero volts #

A photodiode read by a transimpedance amplifier with its anode and cathode at the same potential — the arrangement where the trouble is at the input analyses — sits at exactly the bias where the ratio is one. Its dark junction is carrying no net current; its two streams are equal; its noise is the Johnson noise of its zero-bias resistance, which a photodiode’s data sheet calls its shunt resistance and which that essay puts at hundreds of megohms.

That settles a question a noise budget otherwise has to guess at: what the diode itself contributes in the dark. Not the shot noise of a dark current, because at zero volts there is no net dark current to take a shot noise of. Not zero, because there are two opposing streams. Exactly 4kT divided by the shunt resistance — and, because that value is fixed by equilibrium rather than by the mechanism, it is the right value whether the shunt is the junction’s own saturation current or surface leakage in parallel with it. At zero volts nothing about how the conductance arises can change its noise.

Light adds a third stream, the photocurrent, flowing in the reverse direction and independent of the other two, with its own 2qI. The floor at the amplifier’s input is then a resistance’s noise plus a current’s noise, and the floor a current sets has already said where two such floors cross.

The crossing voltage 2kT/q was found by comparing a resistor with a current flowing through something else, and there the two floors belonged explicitly to two different objects. Here they belong to one. The dark photodiode is its own resistor, the illuminated photodiode is its own current source, and the component becomes shot-noise-limited when its photocurrent would drop 49.981 millivolts across its own shunt resistance. It is the same boundary, arriving without a resistor anywhere in the circuit.

The feedback resistor is the same kind of noise #

The amplifier holding the photodiode at zero volts needs a feedback resistor, and that resistor puts its own 4kT/R𝒻 into the same summing node. Every noise current at that node is now either a resistance’s or the photocurrent’s, and resistances in parallel have the Johnson noise of their parallel combination, so the crossing condition keeps its form with the combination in place of the shunt:

$$I_{ph}\,\left(R_0 \parallel R_f\right) = \frac{2kT}{q}$$

That turns into a rule a designer can apply without a calculator. The stage’s output signal is the photocurrent times the feedback resistance, and at the crossing that is $\frac{2kT}{q}(1 + R_f/R_0)$ — 50.48 millivolts for the stage in the figure, and within a per cent of fifty for any stage whose shunt resistance is a hundred times its feedback resistance. A transimpedance stage whose output signal is well above fifty millivolts is limited by the light’s own statistics; one whose output is well below it is limited by resistors, whatever its gain and whatever its bandwidth, since all three noise currents are white and scale with bandwidth together.

No amount of feedback resistance changes a photodiode amplifier’s shot-noise floor, and that stays true here: raising R𝒻 raises the signal and the shot noise’s output voltage in the same proportion. What it changes is where the other floor is. A larger feedback resistor makes less current noise, moves the crossing to a smaller photocurrent, and so extends downwards the range of light over which the detector is as good as its photons allow. That is the noise half of the reason the factor the expression leaves out keeps finding megohms in the feedback path.

Two things are not in this arithmetic and belong in any real budget. The amplifier’s own voltage noise appears at its output multiplied by a noise gain of 1 + R𝒻/R₀, and at low shunt resistance that can exceed everything here; and a photodiode’s capacitance turns that noise gain into a rising function of frequency. Both are the transimpedance essays’ subject. What this page supplies to them is the diode’s own dark term, which is a resistance’s noise, computed rather than assumed.

An ideality factor the arithmetic cannot carry #

The derivation above used an ideality factor of one, and the obvious generalisation is to write the exponent as V/nVₜ and repeat it. The algebra goes through, and gives a ratio of n(1 + e⁻ᵘ)/2 — a forward limit of n/2, which is the factor the forward identity takes at an ideality factor other than one.

It also gives a ratio of n at zero volts.

That cannot be right, and the reason is the one given two sections ago: at zero volts the junction is in equilibrium and its noise is 4kT times its conductance, whatever the ideality factor is. So the picture of independent crossings of a single barrier, applied with n = 2, contradicts thermodynamics at the one bias where thermodynamics can be checked. An ideality factor near two belongs to a different mechanism — carriers recombining inside the depletion region rather than crossing it — and that mechanism’s noise is not two independent Poisson streams of the sizes the exponent suggests.

What the noise of such a junction is between zero volts and far forward bias is not computed anywhere in this collection, and this essay does not claim to know it. Its value at zero volts is known, and is one. Its forward limit of n/2 rests on the same independent-crossing picture that fails at zero volts, so it is unproved for n ≠ 1 as well, and a design leaning on a junction with a measured ideality factor of 1.5 to be three quarters as noisy as a resistor is leaning on arithmetic rather than on a measurement. The figure keeps the curve because the contradiction is informative; its edge note says it is not a model.

The two limits, read together #

The whole of the result fits in one line. A junction’s noise over the Johnson noise of its own conductance is a function of the voltage across it, and at an ideality factor of one it is (1 + e⁻ᵘ)/2. Everything else on the page is a reading of that line.

At zero volts it is one, and the junction is a resistor in the only sense noise cares about. Forward, it is the familiar half, reached at a fixed voltage and at a current that is a fixed multiple of the saturation current, which is why the half is a safe assumption for a small silicon junction at any current a circuit would use and an unsafe one for a large or leaky junction at a small current. Reverse, it grows without bound, which is the everyday statement that a reverse-biased photodiode’s dark current carries full shot noise and has almost no conductance to go with it.

Two routes agree at every bias drawn — the two streams summed and divided by 4kT times a derivative, and the closed form — and a third route, thermodynamics, agrees with both at the one bias where it applies. That last agreement is the evidence that the two-stream picture is the right one for this junction, and the failure of the same agreement at n = 2 is the evidence that it is not the right one for that one. An identity is best checked at the setting where something independent can check it; zero volts is that setting for every junction there is.

Still open: a loop, and a transistor #

Three questions follow from what this essay leaves open, and each needs something it did not have.

A junction in a loop with the resistor carrying its current. Every comparison so far has been between two objects standing side by side: a junction’s noise beside a resistor’s. In a real circuit the resistor that sets a junction’s current is usually in series with it, and then each noise current has to push through the other element to reach the outside. That changes who supplies the noise and by how much, and it changes the voltage at which the two contributions are equal — the question the resistor in the same loop measures.

A transistor whose two input generators are both a current’s noise. The collector current of a bipolar transistor is a junction’s forward stream, and its base current is another. An amplifier’s voltage noise and current noise are usually treated as two independent numbers; built from one transistor they are the shot noise of one current divided two ways, and that fixes their product. The two generators that are one current is that argument.

The junction whose ideality factor is not one, measured rather than extrapolated. The section above leaves a gap with a known value at one end: a junction dominated by recombination has the Johnson noise of its conductance at zero volts, and nothing here says what it has at a milliamp. The honest way to close it is a model of the recombination current’s own noise — carriers generated and captured inside the depletion region, each event partly correlated with the barrier it sits in — required to return one at zero volts before its forward value is believed. A model built that way would say whether the three-quarters a junction with n = 1.5 appears to promise is real, and it is the one place in this argument where the answer is not already implied by the arithmetic.