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

Current noise — where it appears

A generator in parallel with an input, describing the fluctuating current a device draws whatever is connected in front of it. Flowing in the source impedance it becomes a voltage, so the source resistance that minimises a circuit's total noise is the ratio of the two densities.

Named by 15 essays across 3 fields — each of them below, with the objects they name alongside it.

The floor an amplifier adds, against the source it is given. computed by solving, not by drawing. A part with 4.00 nV/√Hz of voltage noise and 0.60 pA/√Hz of current noise is quietest into 6.67 kΩ, where its noise figure is 1.138 dB. That resistance is the ratio of the two generators and the floor there depends only on their product. Matching the same part for maximum power into its own 1 MΩ input instead — a resistance 150 times larger — costs 12.57 dB.

The floor a circuit has

A resistor's noise is 4kTR and there is nothing to choose about it. An amplifier adds two generators that belong to the device — 4 nV/√Hz in series with its input and 0.6 pA/√Hz across it — and because one matters most into a small source and the other into a large one, there is a source resistance at which their sum is least. It is 6.67 kΩ, it is the ratio of the two, and it is not the resistance that transfers maximum power.

noise · Device noise 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.

noise · Shot noise Two curves that only rise, and the gap between them that has a minimum. computed by solving, not by drawing. The source's own Johnson density, √(4kTR), and the amplifier's total input-referred density, √(4kTR + eₙ² + (iₙR)²), for a part with 4 nV/√Hz and 0.6 pA/√Hz. Neither curve has a minimum: the total is 4 nV/√Hz at a source of nothing, is 4.196 at 100 Ω, and rises without limit. What has a minimum is the ratio, at 6.67 kΩ, where the noise figure is 1.138 dB and the total density is 11.780 nV/√Hz — 2.81 times noisier in volts than at 100 Ω, where the noise figure reads 10.41 dB. The two statements are about different questions and the figure is what stops them being confused.

The bowl, and the bottom of it

An amplifier's noise figure has a minimum against source resistance and its input-referred noise has none: the 4 nV/√Hz part reads 1.138 dB into 6.67 kΩ and 10.41 dB into 100 Ω, and is 2.81 times noisier in volts at the first. The bowl is one shape scaled by its own depth, so the quieter the part the flatter it is — ±30.1 times for a decibel on the best of four, ±2.16 on the worst — and three parts of equal eₙiₙ share a floor of 0.3138 dB at optima 16 times apart.

noise · Device noise 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.

noise · Shot noise 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.

noise · Shot noise The resistors own the floor between 2.91k Ω and 85.9k Ω, and the part owns it outside. computed by solving, not by drawing. The noise at the load of the two-path compensation against the impedance of its own feedback network, with the resistors scaled together and the compensation capacitor taken down in proportion so that Rf·Cf — the handover between the two feedback paths — does not move. Four contributions are integrated over 10 Hz to 100 MHz: the amplifier's 4 nV/√Hz, fitted as the 0.005 power of the impedance and so flat; the two resistors' √(4kTR), the 0.501 power; and the amplifier's 0.60 pA/√Hz flowing in the feedback resistor, the 0.997 power. Two different powers of one quantity cross twice. The resistors carry more than half the power only between 2.91k Ω and 85.9k Ω; outside that window, in both directions, the part does. The part's share is least at 15.8k Ω, which is not eₙ/iₙ — it is that ratio multiplied by the noise gain of 2.000 and again by 1.187, the square root of the ratio of the bandwidths the two generators actually see; there it carries 26.26 per cent. The model stops where the amplifier's output current does: at 100 Ω the feedback resistor alone draws 10 mA a volt.

The window the resistors own

Eight essays have priced one compensated stage, and the fourth of them left two of the amplifier's own generators named and uncounted. With the current generator put in the netlist the floor at ten kilohms goes from 28.88 microvolts to 30.13, and the resistors carry more than half the noise power only between 2.91 kΩ and 85.9 kΩ — outside that window, in both directions, the part does. The flicker corner turns out to be worth 1.00009 in this stage's own band, and 3.474 one band away.

feedback · Capacitive load 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.

noise · Shot noise The cascode device's collector noise is cancelled below 7.22 MHz and its base noise never is. computed by solving, not by drawing. The output noise current of a cascode stage at 1 mA and β = 150, driven from 1 kΩ, into a short, source by source, with 20 pF at the node between the two devices. At low frequency the upper device's base shot noise contributes 0.667% of the lower device's collector noise power, 1/β, and its collector shot noise 4.88 × 10⁻⁵, about 1/β². The collector's contribution overtakes the base's at 7.22 MHz, where the admittance reaches gₘ/√β — 7.21 MHz from the lower half's solved admittance, against 25.1 MHz from the node capacitance alone, because the lower device's Cμ, seen from its collector, is multiplied by its own gain; it passes the lower device's own collector noise near 316 MHz.

The noise the cascode device does add

A cascode's upper device is said to add no noise, because it is a common-base stage fed from a high impedance and its own noise current has nowhere to go but round itself. That is true of its collector noise, which reaches the output at 4.88 × 10⁻⁵ of the lower device's, about one part in β². It is false of its base noise. The partition of the emitter current into collector and base cannot cancel, so it arrives whole: one part in β, 0.667 per cent at a current gain of 150. The cancellation also needs the node between the devices to stay high-impedance. The lower device's own base-collector capacitance, seen from its collector, is multiplied by its gain, so the upper device's collector noise comes through above 7.2 MHz with 20 pF at that node, and at 100 MHz the cascode stage is up to 0.55 dB noisier than the plain one.

semiconductors · Cascode 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.

noise · Shot noise A 1 kΩ bias resistance at the upper base is attenuated to 7.97 × 10⁻⁶ at low frequency and passes the lower device's noise at 15.2 MHz. computed by solving, not by drawing. The output noise current of the cascode stage at 1 mA and β = 150, driven from 1 kΩ, source by source, with the upper device's base fed through 1 kΩ instead of held at signal ground. At low frequency the resistance's noise reaches the output as 7.97 × 10⁻⁶ of the lower collector's, about 2R/(gₘrₒ²): its voltage is followed to the node between the devices and becomes a current only through the lower device's rₒ. The upper base's own shot noise, flowing in the same resistance, is 0.683%. The resistance's noise passes the lower device's at 15.2 MHz, and at 100 MHz the stage's noise figure is 6.07 dB worse than with the base held.

The resistor the upper base is held through

A cascode's upper base is drawn at signal ground, and on a board it is held by a bias network with a resistance of its own. At low frequency that resistance's noise is harmless for the reason usually given — the upper device follows its base and turns the voltage into a current only through the lower device's output resistance — so 1 kΩ reaches the output at 8 × 10⁻⁶ of the lower device's noise. But the attenuation belongs to rₒ, and rₒ is shunted by the node's capacitance from 23.5 kHz, three hundred times below the frequency at which the upper collector's own noise stops cancelling. Unbypassed, 1 kΩ costs 6.07 dB at 100 MHz, more than 100 Ω or 100 kΩ does, and 100 pF across it makes any resistance free.

semiconductors · Cascode A mirror, cascoded or not, delivers two transistors' shot noise at low frequency — and cascoding it does not add a third. computed by solving, not by drawing. The output noise current of a 1 mA NPN mirror into a short, in units of one device's collector shot noise 2qI, against frequency, for a plain two-transistor mirror, the stacked cascode mirror and the wide-swing cascode with its upper base held by a separate diode at 100 µA. At 1 kHz they deliver 1.986, 1.960 and 1.965: the reference transistor's shot noise, copied, and the output transistor's own. At 100 MHz the mirror's own pole and the capacitance at the output transistor's collector have begun to take noise away, and the three read 1.681, 1.196 and 1.512.

The base current the mirror counts twice

A current mirror delivers two transistors' shot noise — the reference device's, copied, and the output device's own — and cascoding it does not add a third: 1.960 and 1.965 of one device's noise for the stacked and wide-swing cascodes against 1.986 for the plain mirror. What the cascode adds is its upper output transistor's base noise, and there the two arrangements differ by a factor of four. In the wide-swing mirror the base current goes to a separate bias and counts once, one part in β. In the stacked mirror it goes into the reference branch, which mirrors it back into the node it left, and it counts 3.82 times at β = 150 — the same base current that took 43 per cent of that mirror's output resistance. The reference resistor's noise, meanwhile, adds almost nothing even at 1.8 V, because it takes its place in the diode's node rather than adding to it.

semiconductors · Cascode At 100 Hz the bowl's bottom is at 2.11 kΩ and 3.10 dB, against 6.67 kΩ and 1.14 dB in the white region. computed by solving, not by drawing: the spot noise figure against source resistance of an amplifier with 4 nV/√Hz and 0.6 pA/√Hz of white noise, whose voltage noise has a flicker corner at 10 Hz and whose current noise has one at 1 kHz. At 100 Hz the spot densities are 4.20 nV/√Hz and 1.99 pA/√Hz, so the optimum is 2.11 kΩ — found by search on the drawn curve at 2.11 kΩ — with a floor of 3.102 dB. Dashed: the white-region bowl, 6.67 kΩ and 1.138 dB. A source left at the white optimum reads 4.492 dB at this frequency. The slider is the frequency.

The source resistance the flicker moves

An amplifier's optimum source resistance is the ratio of its voltage noise to its current noise, and both densities rise below their flicker corners — which are rarely the same frequency. With the voltage noise's corner at 10 Hz and the current noise's at 1 kHz, a part whose optimum is 6.67 kΩ at high frequency wants 2.11 kΩ at 100 Hz and 699 Ω at 1 Hz, tending to the white value times the square root of the ratio of the corners. Over the audio band the drift costs 0.012 dB; over a band from 1 to 100 Hz, choosing the white optimum costs 3.77 dB, and the spot optimum at the band's centre recovers all but a tenth of a decibel.

noise · Device noise With a correlation of −0.3 between the two generators the bowl's bottom stays at 6.67 kΩ and its floor moves from 1.14 dB to 0.83. computed by solving, not by drawing, with the optimum found twice — in closed form and by a direct search of the noise factor, which agree to a part in a million. The 4 nV/√Hz, 0.6 pA/√Hz amplifier of the earlier essays, its two generators given a real correlation coefficient γ = −0.3. The noise figure against source resistance: the cross term 2γ·e·i·R adds the same fraction of the source's own noise at every resistance, so the bowl moves vertically and its bottom stays at e/i = 6.67 kΩ. The floor's excess over one scales as 1 + γ: 0.827 dB against 1.138 for independent generators. The slider is the correlation.

Two generators with one cause

The two-generator model of an amplifier assumes its voltage and current noise independent. Give them a correlation γ and the source sees a cross term, 2·Re(γ·e·i·Z*), linear in its resistance and its reactance. A real correlation leaves the quietest source resistance at e/i and moves the floor in proportion to 1 + γ: 0.83 dB instead of 1.14 at γ = −0.3. An imaginary one moves the optimum off the resistance axis — 5.77 kΩ with 3.33 kΩ of capacitive reactance at γ = 0.5j — and the noise figure's contours become circles about it; a resistive source cannot use it at all. If the two flicker components share one cause, the sign of the sharing decides the low-frequency floor: at 1 Hz, 3.9 dB or 18.0 instead of 15.1. And the correlation is measurable with three resistors, best with the middle one next to the optimum.

noise · Device noise 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.

noise · Shot noise Paralleling a 4 nV/√Hz amplifier on a 50 Ω source, summed through 100 Ω from each: the best is 1.19 dB at 140 parts, against 1.14 dB at 134 parts with a perfect summing node. computed by solving, not by drawing: the noise figure from a 50 Ω source against the number of amplifiers in parallel, each 4 nV/√Hz and 0.6 pA/√Hz, with a perfect summing node (dashed) and with their outputs averaged through 100 Ω each into one node (solid), which adds a thermal 4kTR/N at the node. The summing resistors' noise falls as 1/N exactly as the amplifiers' voltage noise does, so their ratio is the same at every N, and a summing resistor comparable with the resistance an amplifier's voltage noise is worth is a constant factor on that term.

The resistors that sum a hundred amplifiers

A hundred amplifiers in parallel across a 50 Ω source are an ideal transformer of ratio ten, and the summing network between them and the output is where the comparison was left. Its resistors turn out to be the least of it. N resistors averaging N outputs make the noise of R/N, falling exactly as the amplifiers' voltage noise does, so the resistor equal to a 4 nV/√Hz amplifier is about a kilohm at every N, and at 133 parts 200 Ω costs only a tenth of a decibel. The stage that was forgotten is whatever reads the sum. Averaging leaves it alone, so a following amplifier as good as one of the parts costs 12.14 dB, and each part needs a gain of 26 to bring that under a tenth. Parallel amplifiers then match a transformer with one ohm of winding.

noise · Device noise

Named alongside it

The objects these essays reach for when they reach for this one.

Noise figureVoltage noiseOptimum source resistanceShot noiseJohnson noiseDesign tradeoffModel rangeNoise floorThermal voltageCascodeDynamic resistanceFlicker noise

All concepts