Concept

Noise figure — where it appears

How much worse a stage's signal-to-noise ratio is at its output than at its input, in decibels, for a stated source resistance. It is minimised at a source resistance equal to the ratio of the amplifier's voltage and current noise densities, which is not the resistance that transfers the most power.

Named by 18 essays across 4 fields — each of them below, with the objects they name alongside it.

The usable range of one stage, in a 10 kHz measurement. computed by solving, not by drawing. The floor is the Johnson noise of a 1 kΩ source in the measurement's own noise bandwidth — 501.6 nV, using 15.7 kHz rather than the 10 kHz corner. The ceiling is the drive at which an exponential's distortion reaches one per cent, 1.03 mV. Between them is 66.3 dB, and nothing a designer does moves either number without changing the circuit.

A floor and a ceiling

Every other boundary these essays measure is a ceiling. This one puts a floor underneath and measures the distance between them — 4.00 nanovolts per root hertz at the bottom, one per cent of harmonic distortion at 1.03 millivolts at the top, and 66.3 decibels of range in a ten-kilohertz measurement. Both ends are computed, neither is on a datasheet, and the arrangement of the stages decides which one moves.

noise · Dynamic range
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
Where a converter stops measuring the signal and starts measuring the resistor. computed by solving, not by drawing. The quantisation floor is q/√12 and halves with every bit; the Johnson floor of a 1 kΩ source in 100 kHz is 1.266 µV and does not move. They cross at 18.80 bits. Below that the converter is the limit; above it the resistor is, and a further bit buys a more precise measurement of thermal noise. A resolution quoted without a source impedance and a bandwidth is not a resolution — which is the same sentence the instruments field makes about a probe.

The floor a converter sets

A converter's resolution is quoted as a number of bits, which is a property of the converter. What it can actually resolve is a property of the circuit in front of it, and the two cross: measured against the Johnson noise of a 1 kΩ source in 100 kHz of bandwidth, the quantiser is the limit up to 18.80 bits and the resistor is the limit above it. Past that crossing every further bit buys a more precise measurement of thermal noise.

digital · Quantisation
Averaging a white sequence, and averaging a pink one. computed by solving, not by drawing. Both sequences are the same seeded white stream, one of them put through the 1/f network. Averaged in non-overlapping blocks, the white one's spread falls as n to the -0.510 ± 0.006 across five seeds — the √N law — and the pink one's as n to the -0.087 ± 0.013, which is very nearly not at all. A thousand-sample average buys a factor of 36.9 on the first and 2.1 on the second.

The corner where averaging stops working

Average N samples and the noise falls by √N. That is a statement about independent samples, and flicker noise's samples are not independent — its correlation extends over every time scale, which is what a spectrum with no bottom means. Measured on the same seeded stream filtered and not: the white sequence falls as the −0.510 power of the block length and the pink one as the −0.087 power, so a thousand-sample average buys a factor of 36.9 on one and 2.1 on the other.

noise · Device noise
The same three stages, in two orders. computed by solving, not by drawing by Friis's cascade. With the low-noise amplifier first the chain's noise figure is 3.31 dB; with the mixer first it is 12.01 dB. The gain is identical either way — 48.0 dB — so the 8.70 dB is bought with nothing but an ordering. Everything after the first stage contributes 26.0% of the total. The 2.0 dB of cable in front has gain below one, so it multiplies every later stage's contribution rather than dividing it.

The loss in front, counted twice

A decibel of cable before an amplifier costs a decibel of signal, which everybody expects, and a decibel of noise figure, which is a separate decibel arriving from a separate place. Measured across the slider it is exact: 1.31 dB of chain noise figure becomes 2.31 with one decibel of cable and 9.31 with eight, every time, while the 8.70 dB that stage ordering is worth does not move at all. A lossless reactive network in the same position costs nothing, because only the real part of an impedance is warm.

noise · Johnson 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
Twenty metres buys 20 dB of apparent match and costs 10 dB of noise figure. computed by solving, not by drawing. The same cable and the same 200 Ω load as the reading, with the amplifier that is actually behind the instrument. The rising trace is the return loss the instrument reads, which is the load's own 4.44 dB plus twice the one-way loss. The lower pair is the chain's noise figure: a 2 dB amplifier with the cable in front of it, counted as a matched attenuator whose noise factor is its loss, and counted honestly from the available gain of a lossy line driven by a source that reflects 0.60. The first says the exchange rate is exactly two decibels of match per decibel of noise figure, at every length here. The second is higher everywhere — by (1−Γ²u²)/(1−Γ²), which is 0.028 dB at five centimetres and 1.938 dB, the load's own mismatch loss, once the cable is long enough to have absorbed the reflection. At twenty metres the instrument reads 24.4 dB and the chain costs 13.92 dB against the amplifier's own 2.

The cable that hides two things

A length of cable improves a return-loss reading by twice its loss and raises a noise figure by once it, so the rule of thumb is two decibels of apparent match per decibel of floor. Both halves are owned here and neither essay had the other. Put together they say what an acceptance limit costs: making a 4.0:1 load read 1.50:1 spends 6.53 decibels of noise figure, against a mismatch that was itself costing 1.938 — and the exchange rate is not two but 2(1−Γ²)/(1+Γ²), which is 0.94 where a pad is actually short.

lines · Line loss
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 6 dB pad at 290 K turns a cold source into 217.2 K and a 290 K one into 290 K. computed by solving, not by drawing, every resistor's Johnson noise carried to the pad's open output. A matched 50 Ω T-pad of 6 dB (arms 16.61 Ω, shunt 66.93 Ω) at 290 K, fed from a 50 Ω source at the temperature on the horizontal axis. The output's noise temperature (dots) is Tₛ/L + T·(1 − 1/L), with T the pad's own temperature, to nine figures (line): a slope of 1/L = 0.2512 and an intercept of 217.2 K — the pad's own noise, (1 − 1/L)·290 K, seen at its output. The dotted diagonal is no pad at all. The two cross at 290 K, where the source is at the pad's temperature and the output is too.

The noise a pad adds is its temperature

A matched resistive pad of loss L at physical temperature T presents a noise temperature of Tₛ/L + T(1 − 1/L) at its output, for a source at Tₛ,, and summing its four resistors' Johnson noise to the open output confirms it to nine figures. Referred to its input its own noise is (L − 1)·T — 288.6 K at 3 dB, 28,710 K at 20 dB — so its noise figure in decibels is its loss in decibels at 290 K, which Friis states from the definition and the resistors reach by themselves. At its output a pad can never add more than its own temperature: cooled to 4 K, a 20 dB pad adds 3.96 K there, and still 396 K referred to its input. Most of a heavy pad's noise is made in its last resistor — 82.6% at 20 dB. And set between a 400 K source and a 290 K load, a pad stops the net flow of noise power at a temperature that depends on its loss, or, below 1.40 dB, at none.

noise · Johnson noise
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 50 Ω source over 20.0 Hz–20.0 kHz: the best turns ratio is 10.7, giving 1.30 dB; the white optimum's 11.5 gives 1.31, and no transformer at all 13.2. computed by solving, not by drawing, the band integrals checked by quadrature. The drifting part of the essay before — 4 nV/√Hz and 0.6 pA/√Hz white, flicker corners at 10 Hz and 1 kHz — fed from a 50 Ω source through an ideal step-up transformer of ratio n, so it sees n²·50 Ω. The noise figure over 20.0 Hz to 20.0 kHz against n: least at n = √(√(E/I)/Rₛ) = 10.73, where the amplifier sees 5.76 kΩ, at 1.298 dB. The ratio that matches the white optimum, 11.55, gives 1.310 dB; the source connected directly, 13.23 dB. The slider is the band.

One turns ratio for a band

A 50 Ω source and an amplifier whose optimum is 6.67 kΩ want a step-up transformer, and the ratio that matches the white optimum, 11.5, is the one a design note computes. Over the audio band it is nearly right: 1.31 dB against 1.30 at the band's own best ratio, 10.7. Over 1 Hz to 100 Hz it is badly wrong — 9.2 dB where a ratio of 4.8 gives 5.4 — because the band's noise lives in its bottom decade, where the flicker has moved the optimum down. One ratio for a two-decade band costs at most 0.24 dB against a ratio that could follow the optimum. The transformer's winding resistance multiplies the whole noise factor by one plus the winding's resistance over the source's, not just its own share: a tenth of the source costs 0.41 dB where its own noise alone would cost 0.31. And 115 amplifiers in parallel are an ideal 10.7:1 transformer, noise for noise.

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
A 20 dB line from 300 K down to 4 K, fed from a 300 K source: how much of the noise at its cold end each piece of it makes. computed by solving, not by drawing: the line as forty matched pads, each at the temperature of its own place on a linear gradient, each pad's law held to the resistor-by-resistor sum. The noise temperature at the cold end is 67.7 K — the continuous line's closed form gives 67.63 K — of which 3 K is the source's, attenuated, and the rest the line's own. Each piece emits (1 − 1/L)T at its own temperature and is attenuated by every piece after it, so the warm pieces at the start, which emit most, are attenuated most: the warm half of the line supplies 17.99 K, the cold half 46.71 K.

The end a lossy line looks like

A line running from room temperature into a 4 K stage is a chain of pads, each at its own temperature, and its noise at the cold end is each piece's emission attenuated by everything after it. Cut into forty pads along a linear gradient, a 20 dB line fed from a 300 K source delivers 67.7 K to the cold end, where the continuous closed form says 67.63; its cold half supplies 46.7 K of that and its warm half 18.0. Read from the warm end, the same line's own noise is 184 K at 10 dB against 89.7 at the cold: a lossy line looks like the end it is read from. And twenty decibels placed as one attenuator at the 4 K stage deliver 6.96 K, where at room temperature they deliver 300.

noise · Johnson 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.

Current noiseOptimum source resistanceVoltage noiseJohnson noiseDesign tradeoffShot noiseAvailable powerImpedance matchingModel rangeNoise floorFlicker noiseFriis cascade

All concepts