The two generators that are one current

The floor a circuit has describes an amplifier’s noise with two generators: a voltage noise eₙ in series with its input and a current noise iₙ across it, 4 nV/√Hz and 0.6 pA/√Hz for the part it draws. Their ratio is the source resistance at which the amplifier is quietest relative to the source, 6.67 kilohms, and their product fixes how quiet that is, a noise figure of 1.138 dB. The bowl, and the bottom of it takes the same two numbers further and finds that the bowl the noise figure makes against source resistance is one shape, scaled by its depth.

Both essays treat the two generators as properties of a part, like a gain or an offset, and the bowl essay says so directly: they come from different mechanisms inside the device, nothing requires them to be related, and that independence is what makes the product and the ratio two separate design variables. Three parts with the same product and ratios sixteen apart are, on that account, three different components for three different sources.

For one very common input stage that is not what the device does. A bipolar transistor has one current flowing in at its emitter. Most of it leaves by the collector and a fraction 1/β leaves by the base, and both are a current crossing a junction — each has the shot noise the floor a current sets puts at 2qI. The voltage generator is the collector’s shot noise referred to the input; the current generator is the base’s. They are one current divided two ways, and dividing it fixes how the two are related.

Where each of the two comes from #

The voltage generator is the collector’s shot noise. The collector current I꜀ carries 2qI꜀ of current noise. A voltage at the base produces a collector current through the transconductance gₘ = I꜀/Vₜ — the small-signal transconductance how small is small signal linearises — so a voltage generator at the input that would produce the same noise current has a density of 2qI꜀/gₘ², which is 2kT·rₑ with rₑ = Vₜ/I꜀. It falls as the current rises, because the transconductance rises faster than the noise does.

The current generator is the base’s shot noise. The base current I꜀/β is a current crossing a junction too, and its noise 2qI꜀/β flows in whatever resistance the source presents. It rises as the current rises.

The product of the two densities is then

$$e_n^2\,i_n^2 = 2kT\,\frac{V_T}{I_C}\cdot\frac{2qI_C}{\beta} = \frac{4(kT)^2}{\beta}, \qquad e_n i_n = \frac{2kT}{\sqrt\beta}$$

and the collector current has cancelled out of it. The minimum noise factor the device-noise essays write as 1 + 2eₙiₙ/4kT becomes

$$F_{\min} = 1 + \frac{1}{\sqrt\beta}$$

with no current, no area and no temperature in it. At a current gain of a hundred it is 1.1, which is 0.4139 dB, and in the figure above the product stays constant to twelve figures over four decades of collector current at every current gain its slider offers.

The half that was already known #

The voltage generator deserves a second look, because it has appeared before under a different name.

$2kT\,r_e$ is exactly half the Johnson noise of a resistor of value $r_e$, and that is the identity a bare junction satisfies when it is forward biased. A transistor’s base–emitter junction is a junction, and its collector current is that junction’s forward stream carried across to another terminal. So the input voltage noise of a bipolar stage is the junction noise of its emitter, and the check that the two are the same number to twelve figures at a microampere, a hundred microamps and ten milliamps is a check that the transistor arithmetic and the junction arithmetic are one arithmetic.

It also says which of the two generators the emitter-resistor loop was about. The resistor in the same loop solved a junction with its emitter resistor and found the input-referred noise to be 2kT·rᵈ plus the resistor’s 4kTR; the first term is eₙ² here. That loop left the base current out, correctly for a current source. For an amplifier the base current is the other generator, and leaving it out is leaving out half the noise figure.

The bowl, drawn at a transistor’s own numbers #

At a hundred microamps and a current gain of a hundred, the two generators are 1.415 nV/√Hz and 0.566 pA/√Hz. Those are a part’s numbers in exactly the form the device-noise essays use, so the same picture can be drawn for them.

Nothing about the bowl’s shape has changed from the bowl essay, and it should not have: the essay’s closed form contains only eₙ and iₙ. What has changed is the meaning of the two inputs. For the part in that essay, 4 nV and 0.6 pA were two numbers on a data sheet. Here they are one collector current and one current gain, and moving the current moves both numbers together in opposite directions.

The part in the device-noise essays is worth reading back through this arithmetic, because it gives a check on how far the account goes. A bipolar stage with an input voltage noise of 4 nV/√Hz would run at 12.5 µA; to have a current noise of 0.6 pA/√Hz at that current it would need a current gain of about eleven. Equivalently, a noise figure of 1.138 dB is 1 + 1/√β at β ≈ 11. No small-signal transistor chosen for low noise has a current gain that low, so that part is not a bare bipolar input with nothing else in it — its current noise is larger than its collector current explains, which is the ordinary state of an integrated amplifier whose input current is shaped by more than one device. The two generators of a real part can be independent; the claim of this essay is only that in a bipolar input stage they begin as one current.

One bowl, slid by the bias #

The ratio of the two generators is the optimum source resistance, and it is the other half of the arithmetic:

$$R_{\mathrm{opt}} = \frac{e_n}{i_n} = \frac{\sqrt\beta\,V_T}{I_C}$$

The collector current is in it, inversely. So the bias current does to the bowl exactly what the bowl essay’s product family did with three different parts: it keeps the floor and moves the optimum.

That product is a voltage, $\sqrt\beta\,V_T$, and it is the second direct voltage to come out as the answer to a noise question — the first was the 2kT/q at which a resistor’s floor and a current’s cross. It says that a bipolar stage is at its noise optimum when the source resistance would drop 249.9 millivolts at the collector current, or equivalently when the source resistance drops $\sqrt\beta\,V_T/\beta$ = 2.499 millivolts at the base current. Nothing has to be looked up but the current gain.

The practical instruction is the reverse of the one the bowl essay’s width measurement was approaching. That essay asked how far a source can miss a part’s fixed optimum before a decibel is lost, and answered in factors — thirty on its quietest part, two on its noisiest. For a bipolar stage the optimum is not fixed: it is wherever $\sqrt\beta\,V_T/I_C$ puts it, and the collector current can be chosen. So the question of matching the source to the amplifier becomes the question of biasing the amplifier to the source, and the answer is a current rather than a transformer.

The equal-product family is one part #

The bowl essay’s sharpest result was three amplifiers with the same product and optima a factor of four apart — equally good, and good for different sources, and indistinguishable by the single noise figure a data sheet quotes. Drawn at a bipolar stage’s product, the family is the same construction.

The three “parts” are one transistor at three collector currents a factor of four apart, and the claim that they are indistinguishable by noise figure and different in which source they suit is, for a bipolar stage, a claim about its bias. The bowl essay’s warning survives exactly: a noise figure quoted without the source resistance it was measured at says nothing about which circuit the part is for. What changes is the remedy. For three different amplifiers the remedy is to pick the right one; for one transistor it is to set its current.

The densities at the optima keep the bowl essay’s other result too. At its own optimum the stage sits on 4kTRₒₚₜ multiplied by the same Fₘᵢₙ, so the absolute noise in volts falls as the optimum resistance falls — a factor of four here for sixteen in resistance — and a bipolar stage biased for a small source is quieter in volts than the same stage biased for a large one, by the square root of the ratio of the two sources.

The resistance that is not shot noise #

Every number so far has come from two shot noises and nothing else, and the result that the floor contains no current depends on that. A real transistor has a resistance in series with its base — the spreading resistance of the base region between the contact and the active junction — and that resistance has thermal noise of its own, 4kT·rᵦ, which adds to the voltage generator and contains no current at all — the same kind of addition an emitter resistor makes, which what a resistor in the emitter buys prices in linearity and the resistor in the same loop prices in noise.

With rᵦ included the product is no longer constant, but the calculation stays short. For a given source resistance Rₛ, the collector current that minimises the noise figure is still $\sqrt\beta\,V_T/R_s$ — the base resistance’s noise does not depend on the current, so it cannot move the minimum — and the noise factor there is

$$F = 1 + \frac{r_b}{R_s} + \frac{1}{\sqrt\beta}$$

The two terms after the one are the base resistance’s share and the shot noises’ share, and they are equal at Rₛ = rᵦ√β.

The marks are the second route. A golden-section search over the logarithm of the collector current, which is given only the two generators and never sees the expression above, lands on the same current to a part in a hundred thousand and the same noise factor to a part in a thousand million, at every marked source and at every base resistance the slider offers. The expression says where to look; the search says what is there, and they agree.

What the figure says is that the transistor has two regimes and the source picks between them. Above rᵦ√β the floor is set by shot noise and is the same 1 + 1/√β that contains no current; biasing correctly delivers it, and the only way to improve it is a larger current gain. Below rᵦ√β the floor is set by the base resistance, and no choice of bias helps: into 50 ohms the best this transistor can do is 3.222 dB, eight times the shot-noise floor in excess noise, at a collector current of five milliamps.

That is the reason low-noise bipolar parts for low-impedance sources are built with very small base resistance, and why a common way to get it is to put several transistors in parallel: N identical devices sharing a collector current have N base resistances in parallel, and the same total current, so the shot-noise terms are unchanged and the base-resistance term falls by N. The figure’s two settings are a factor of ten apart in rᵦ, which is ten devices, and they move the corner a decade.

Current gain as the only lever, and why it is a square root #

With the base resistance out of the way, one number is left in the floor. The current gain enters as a square root because it divides only one of the two generators: β sets the base current’s noise and has nothing to do with the collector’s, so the product goes as β to the minus a half.

Quadrupling the current gain halves the excess noise factor, from 0.1 to 0.05 — 0.4139 dB to 0.2119 — and doubles every optimum source resistance. That is a slow lever: a transistor with β of 30 has a floor of 0.728 dB and one with β of 400 a floor of 0.212, so thirteen times the current gain buys about half a decibel. It is also a lever with an edge, because the current gain the arithmetic needs is the gain at the frequency being measured, and a transistor’s current gain falls at high frequency. Where it has fallen the floor rises with it, which is one of the reasons a bipolar stage’s noise figure climbs toward its transition frequency.

Correlation, flicker, and the inputs this does not describe #

The account above is the low-frequency, small-signal limit, and three things lie outside it.

The two shot noises are treated as uncorrelated. At low frequency the collector and base currents are carried by different events — carriers that cross to the collector and carriers that recombine in the base — and treating their noises as independent is standard. At frequencies approaching the transistor’s transition frequency the two become partly correlated, and the minimum noise factor then depends on a correlation term that this page does not compute.

Both generators are white here. The base current in particular has a flicker component whose density rises as 1/f at low frequency, and the corner where averaging stops working is the collection’s account of what such a component does. With it included the current generator rises at low frequency and the voltage generator does not, so the product is no longer constant across frequency and the optimum source resistance falls toward low frequency. The bias-independence of the floor holds in the white region and is not claimed below the corner.

A field-effect input is not described at all. A field-effect transistor’s gate current is a leakage current rather than a fraction of its channel current, so its current noise is not tied to the current that sets its voltage noise, and the two generators there really are independent in the sense the bowl essay assumed. The arithmetic of this page is specific to an input whose current noise is a fixed fraction of its output current, which is a bipolar transistor’s defining property — and the same property the current the instrument draws meets as a bias current whose own shot noise overtakes a source resistor’s thermal noise at 2kT/qI.

The temperature is 290 kelvin throughout, which is the reference a noise figure is defined against, and the one the loss in front, counted twice uses to turn a cable’s loss into the same kind of decibel.

Still open: the count, and the input with no current #

The count under the density. Every essay on shot noise here has used it as a density, 2qI, and the density is exact at every current, because it is a statement about the variance of a count and the count of independent carriers is Poisson whatever its size. What is not exact at every current is reading that variance as a Gaussian floor. In a window of length T a current I delivers N = IT/q carriers; with T set by a noise bandwidth B as 1/(2B), N = I/(2qB), and the shot-noise-limited signal-to-noise ratio is exactly √N. At a picoampere in ten kilohertz that is 312 carriers and a ratio of 17.67; the skewness of the count, 1/√N, reaches a tenth at about a third of a picoampere in the same bandwidth; and at a femtoamp nearly three windows in four contain no carrier at all. The base current of a bipolar stage biased at a microampere with a current gain of a thousand is ten times that picoampere. Marching a seeded Poisson arrival train through a filter and measuring its distribution against a Gaussian of the same variance would put a current on the edge of the Gaussian floor — the amplitude at which the white-noise model stops being true while its density stays right.

The field-effect input, whose current noise has no current behind it. The section above excludes field-effect inputs because their two generators are not one current. That makes them the natural contrast: an input whose voltage noise follows its channel current and whose current noise is gate leakage, so that the product does move with bias and the bowl’s floor is a design variable rather than a constant. Measured against the bipolar floor at the same source resistance, it would say from which source resistance upwards the leakage-limited input wins, and that is a boundary in ohms this collection has not drawn.

The paralleled input. The base-resistance section gives the reason for putting N devices in parallel and stops at the base resistance. Paralleling also changes the collector current per device, and so the optimum source resistance of the combination, and it multiplies the capacitance at the input. Finding the number of devices that minimises the noise figure into a given low-impedance source, with the capacitance’s bandwidth cost beside it, would turn the one-sentence rule into a measured one.