Johnson noise — where it appears
Named by 24 essays across 5 fields — each of them below, with the objects they name alongside it.
The floor a resistor sets
A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.
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.
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.
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.
The total that has no resistor in it
A larger resistor is noisier and makes a narrower filter, and the two dependences are exactly reciprocal: the density goes as the square root of the resistance and the noise bandwidth as its inverse. Five decades of resistance charging one picofarad therefore give five decades of corner frequency, two and a half decades of density, and one total — 63.2762 microvolts at every one of them, which is the square root of kT over C and contains no resistance at all.
The noise a clock does not make
A hundred-megohm resistor has a noise density of 1.27 microvolts per root hertz. A switched capacitor that behaves as a hundred megohms has none of it: the noise on the capacitor it charges is kT/C, with the holding capacitor in it and nothing else — not the clock, not the switched capacitor, not the on-resistance. And it is exact rather than asymptotic, at every capacitor ratio from a thousandth to ten.
The same filter a thousand times larger
Multiply every resistance by a thousand and divide every capacitance by a thousand and the response does not change — not approximately, but to a part in ten to the fifteenth, which is the last bits of a double. So a designer has a free parameter that the design says nothing about, and what decides it is the two quantities that refuse to scale: fifty ohms of amplifier output resistance at one end and two picofarads of stray at the other. Between them the realisation survives over three decades of impedance level and nowhere else.
The sample that is subtracted
Three rungs of this argument have measured floors that no gain moves and no filter reaches, because both arrive as numbers already sampled. One of them can be subtracted: the reset level a capacitor holds is the same number in two consecutive samples and cancels exactly. What it costs is that the amplifier's own noise is not — two samples of it are independent, so its variance doubles. That is thirty times better at a megahertz, a loss above sixty, and the crossing is the one the rung below computed for a different question.
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.
What the second path costs at the floor
The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.
The resistor the noise comes from
Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.
The current the instrument draws
Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get.
Where the trouble is at the input
Every other arrangement in this field has its difficulty at the output — a capacitive load, an isolation resistor, a load that draws current. A photodiode amplifier has it at the input, and the capacitance causing it is not a parasitic: it is the diode's junction, which is the price of its area, and area is what a photodiode is bought for. The feedback resistor's own noise is 127 nV/√Hz against the amplifier's 4, and the amplifier is still ninety-five per cent of the noise at a large diode.
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.
The probe that takes a tenth
A ten-to-one probe buys an order of bandwidth for a tenth of the signal, and on a two-kilohm source the bandwidth is exact: 6.79 kHz becomes 69.2 kHz. The tenth of the signal is not a tenth of the signal-to-noise ratio. Solved resistor by resistor, the noise referred to the tip goes from 1.782 µV to 55.78 µV — a factor of 31.3 — because the divider that does the attenuating is nine megohms and a megohm, and √(n(n−1)kT/C) on the cable's own capacitance has no source resistance in it at all.
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.
Eight amplifiers, and what they add
The rung below realised a Chebyshev ladder out of floating gyrators and found the response converging on the passive one's as the resistance scale rises — twenty-two decibels out at ten kilohms, a twentieth of a decibel at ten megohms. It closed by naming two quantities it had not measured. They are the same quantity: the resistors that buy the accuracy are the noise, and the amplifiers inside the gyrators carry the inductor's own current through them, so the floor rises as the square root of the scale and the ceiling falls as the scale.
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.
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.
The cure that becomes a different circuit
The classical cure for an amplifier's input current is to make the two resistances its inputs look back into equal, and it reads as one instruction. Solved, it is two circuits meeting at 909 ohms — the feedback network's bottom resistor times (G−1)/G — and above that knee there is no resistor to add: the network is scaled to the source, which at a megohm means 11 megohms of feedback and at a gain of 1001 means 1001. Above the knee three different networks become one instrument to twelve figures, the noise penalty settles at 1.41420 against a √2 of 1.41421, and the benefit at 10.49 against two currents whose ratio is ten.
Only the real part is warm
Johnson's 4kTR is the special case of a statement about impedances: the noise across any passive two-terminal in equilibrium is 4kT·Re{Z}, so a reactance contributes nothing however large it is. That turns a modelling convenience into a noise figure. A 100 nF capacitor of loss tangent 0.02 written as 3.183 Ω in series and as 7.958 kΩ across it — the pair that converts exactly at 10 kHz — gives 0.226 and 10.10 nV/√Hz at 100 Hz, 33 dB apart, and 40 dB apart the other way at a megahertz.
The floor that is only a floor while nothing flows
Johnson noise depends on a resistance and a temperature and on nothing else, which is what makes it a floor. A real resistor has a second generator that depends on how it was made and on the voltage across it: 0.1 µV per volt per decade for a metal film, rising as 1/√f. On 100 kΩ with 10 V across it the two are equal at 271 Hz, and the crossover moves with the SQUARE of the voltage — 0.678 Hz at half a volt, 2.44 kHz at thirty. Splitting the same total resistance into eight parts in series divides the excess by exactly √8 and leaves the Johnson noise where it was.
The tee that charges for its own compensation
A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.
Which way the noise goes
Two warm resistors joined together each drive the other, and the net flow is 4kΔT·R₁R₂/(R₁+R₂)² per hertz. At the match that is kΔT exactly — 1.5187 zeptowatts per hertz between 400 K and 290 K — and a kilohm pair and a gigohm pair at the same two temperatures exchange the same, which is why noise is quoted as a temperature. At one temperature the net is zero at every ratio to a part in 10³⁷, and that zero is the second law rather than a tolerance.
Named alongside it
The objects these essays reach for when they reach for this one.
Design tradeoffModel rangeCurrent noiseNoise gainShot noiseSpectral densityDynamic rangeEquivalent noise bandwidthNoise bandwidthNoise figurePhase marginVerification