Circuit Response
every model has an edge, and the figure carries it
Series

Shot noise — the series

7 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Two floors on one axis, and the 50.0 mV between them. computed by solving, not by drawing. The flat line is the Johnson current noise of 1 kΩ, √(4kT/R), which has no current in it. The rising line is shot noise, √(2qI), which has no resistance in it. They cross at 50 µA — and the direct voltage across the resistance there is 49.981 mV, which is 2kT/q and contains neither quantity. The slider moves the resistance over six decades; the crossing moves with it and the voltage at the crossing does not move at all, to the last bit of a double.

    The floor a current sets

    A resistor's noise contains no current and a current's noise contains no resistance, and the two are equal when the direct voltage across the thing carrying the current is 2kT/q — 50.0 millivolts at 290 kelvin, whatever the resistance and whatever the current. It is the only boundary in this collection whose axis is a direct voltage across an element. And a forward-biased junction, which has the same dynamic resistance as some resistor, produces exactly half its noise power at every current.

    part 1 · noise
  2. A junction's noise against its own resistance's: exactly one at zero volts, and a half only far from it. computed by solving, not by drawing. A junction carries two currents at once, Is·e^(V/nVt) forwards and Is backwards, and each has its own shot noise. Their noise over the Johnson noise of the junction's own conductance is n(1 + e^−u)/2. At n = 1 it is 1.000000 at zero volts, 0.5676 at 50 mV, within one per cent of 0.50 only above 115.1 mV — where the forward current is ninety-nine saturation currents — and 2.978 at −40.00 mV of reverse bias. The half the forward-biased junction is known for is the limit of this curve, not its value.

    The junction that is a resistor at zero volts

    A forward-biased junction makes half the noise power of a resistor of its own dynamic resistance, and that half is a limit rather than a value. Kept with the saturation current that flows backwards across it, the ratio is one exactly at zero volts, 0.5676 at 50 millivolts, and within one per cent of the half only above 115.1 — at ninety-nine saturation currents, which is a picoampere on a small silicon diode and a microampere on a leaky one. A photodiode held at zero volts has the Johnson noise of its shunt resistance and nothing else, and it becomes shot-noise-limited at 49.981 millivolts of photocurrent drop.

    part 2 · noise
  3. A junction and its resistor in one loop: equal shares at 12.50 mV, and quietest against both at 49.98 mV. computed by solving, not by drawing. A junction carrying 25 µA in series with a resistor, the loop closed into a short and solved as a netlist with each noise current injected across its own element. Against the drop across the resistor: the bare junction's 2qI, the bare resistor's 4kT/R, each one's share of what reaches the outside, and the total. The shares are equal at 12.50 mV (500 Ω), not at the 49.98 mV where the bare floors cross; there the resistor supplies 80.0 per cent and the total is 0.5556 of either floor. The total is below both floors at every drop.

    The resistor in the same loop

    A resistor's noise and a junction's are equal as bare densities at 49.98 millivolts of drop, and a junction in series with the resistor that carries its current is the arrangement every current source is built from. In one loop each noise current has to cross the other element, so the two supply equal shares at 12.50 millivolts, a quarter of the crossing; at the crossing itself the resistor supplies 80 per cent and the loop is 2.553 decibels below both floors, which is further than it gets anywhere else. The same resistor multiplies the stage's input-referred noise by five.

    part 3 · noise
  4. A transistor's two noise generators are one current: their product is 0.8008 nV·pA/Hz at every bias. computed by solving, not by drawing. The input voltage noise of a bipolar stage, √(2kT·rₑ), is its collector current's shot noise referred through gₘ, and falls as the current rises; its input current noise, √(2qI_C/β), is its base current's, and rises. At β = 100 their product is 2kT/√β = 0.8008 nV·pA/Hz at every collector current from a microampere to ten milliamps, so the best noise figure, 0.4139 dB, does not depend on the bias. What does is where it is: the optimum source resistance times the current is √β·Vt = 249.9 mV, and the two lines cross where that resistance is a kilohm, at 250 µA.

    The two generators that are one current

    An amplifier's noise is two generators, a voltage in series with its input and a current across it, and the essays on its noise figure treat them as independent numbers. In a bipolar input stage they are one current's shot noise divided two ways, by the collector and the base, and their product is 2kT/√β at every collector current: 0.8008 nV·pA per hertz at a current gain of a hundred, from a microampere to ten milliamps. The best noise figure, 0.4139 dB, does not depend on the bias. The bias decides only where the best source is, and 50 ohms of base resistance decides what the best actually is below 500.

    part 4 · noise
  5. 1 pA in 10.0 kHz: 312 carriers a window, and a Gaussian of the same variance that looks right and is not in its tails. computed by solving, not by drawing. A direct current of 1 pA counted in windows of 1/(2B) for a noise bandwidth of 10.0 kHz: the mean count is N = I/(2qB) = 312.1, and the count is Poisson, drawn as bars, against a Gaussian of the same mean and variance, drawn as a curve. The variance is N exactly, so 2qI is right; the signal-to-noise ratio is √N = 17.67 and the skewness 1/√N = 0.0566. The count exceeds its mean by five standard deviations with probability 7.84 × 10⁻⁷, against the Gaussian's 2.87 × 10⁻⁷. A seeded arrival train counted in 20,000 windows gives a mean of 312.200 and a variance of 314.461.

    The count under the density

    Shot noise's density, 2qI, is exact at every current, because it is the variance of a count of independent carriers, and that count is Poisson at any size. Reading that variance as a Gaussian floor is not exact. A picoampere in a 10 kHz noise bandwidth is 312 carriers a window, and it passes five standard deviations above its mean 2.74 times as often as a Gaussian of the same variance says. At 10 femtoamperes it is 359 times as often, and 4.4 per cent of the windows are empty. A single pole reads a third more skewness than a count does, (4/3)/√N against 1/√N. The density is right and the shape is wrong, and the shape is what a threshold uses.

    part 5 · noise
  6. A field-effect input beats a bipolar one biased for each source from 2.83 kΩ to 5 GΩ. computed by solving, not by drawing. The noise figure a field-effect input shows a source — drain current 1 mA, voltage noise 2.31 nV/√Hz from its channel, current noise 0.566 fA/√Hz from 1 pA of gate leakage — against a bipolar input of β = 100 and 50 Ω of base resistance with its collector current set for each source, whose best is 1 + rb/Rₛ + 1/√β. The field-effect input is quieter between 2.83 kΩ and 5 GΩ; the closed forms √β·(Rₙ − rb) and 2Vₜ/(√β·Iₗ) give 2.83 kΩ and 5 GΩ. Its own best, 0.0007 dB, is at 4.08 MΩ; the bipolar floor never goes below 0.414 dB.

    The window a field-effect input wins in

    A bipolar input's two noise generators are one current's shot noise split two ways, so their product, and with it the best noise figure, is fixed at 1 + 1/√β, 0.414 dB at a gain of a hundred, whatever the bias. A field-effect input's are two unrelated things. Its voltage noise is its channel's and its current noise is the shot noise of its gate leakage, so their product is a design variable: 0.0007 dB at a milliampere of drain and a picoampere of leakage. Against a bipolar input biased for each source, the field-effect input wins in a window of source resistance with two ends in closed form, from 2.83 kΩ, set by the two inputs' noise resistances, up to 5 GΩ, set by the gate leakage. A nanoampere of leakage brings the top down to 5 MΩ, and 438 nA closes the window.

    part 6 · noise
  7. A field-effect input against a bipolar one biased for each source, at 25 °C. computed by solving, not by drawing. The field-effect input (dark) has 1 mA of drain current and a gate leakage that is 1 pA at 25 °C, from the junction's own law; the bipolar input (pale) is at its best current for each source. It wins between 2.83 kΩ and 5.14 GΩ. At 25 °C the ceiling was 5.14 GΩ; it falls in proportion to the leakage, which the slider's temperature multiplies.

    The window that closes from the top

    A field-effect input beats a bipolar one biased for each source between 2.83 kΩ and 5.14 GΩ at 25 °C, and the top of that window is twice the thermal voltage over √β times the transistor's gate leakage: it falls as fast as the input junction leaks. Carried through the junction's own law, the ceiling is 39.5 GΩ at 0 °C, 913 MΩ at 50 and 5.79 MΩ at 150 — while the floor does not move from 2.83 kΩ until 200 °C. A 1 GΩ sensor leaves the window at 48.6 °C, a 10 GΩ one at 16.4, about 33 kelvin a decade. The leakage's doubling interval is not ten kelvin but 7.5 at 0 °C and 15.0 at 150, so the rule of thumb is right at one temperature only.

    part 7 · noise

All series