Which way the noise goes

Everything these essays has measured so far has been a voltage. The floor a resistor sets computes $\sqrt{4kTR}$ and calls it a density; the resistor the noise comes from apportions a filter’s output voltage among the resistors that produced it; only the real part is warm generalises the resistance to an impedance and leaves the voltage where it was.

A voltage is a natural thing to compute because it is what an amplifier’s input responds to. It is not the natural currency of the underlying statement, and one quantity makes that obvious: the available power, which is what a warm resistance can deliver to the best load for it.

$$P_{\text{av}} = \frac{\overline{v^2}}{4R} = \frac{4kTR}{4R} = kT \quad \text{per hertz}$$

The resistance cancels. A kilohm and a gigohm at the same temperature have available noise powers that are identical, at every frequency, to the last digit — and the answer is Boltzmann’s constant times a temperature, which is why the whole field ended up quoting noise as a temperature rather than as a voltage.

This essay takes that one step further, into what happens when two such resistors are connected to each other.

Two resistors, four quantities, one net #

Joined together, each resistor is a source and a load at once. The noise of $R_1$ at $T_1$ delivers to $R_2$

$$P_{1\to2} = \frac{4kT_1R_1R_2}{(R_1+R_2)^2} \quad \text{per hertz}$$

and $R_2$ at $T_2$ delivers the same expression with the temperatures exchanged. The net flow is their difference, which has one temperature difference in it and one geometric factor:

$$P_{\text{net}} = 4k(T_1-T_2)\frac{R_1R_2}{(R_1+R_2)^2}$$

Two things fall out of that expression and only one of them is usually said.

At $R_1 = R_2$ the geometric factor is exactly $\tfrac14$ and the net is $k\Delta T$. No resistance appears. A kilohm pair at 400 K and 290 K exchanges 1.5187 zeptowatts per hertz; a gigohm pair at the same two temperatures exchanges 1.5187 zeptowatts per hertz, and the figure computes both rather than claiming one from the other. That is the available-power statement applied twice, and it is why a noise temperature is a complete specification of a source where a noise voltage is not.

Away from the match the exchange falls, and it falls exactly as a signal’s would. The factor $R_1R_2/(R_1+R_2)^2$ is the same one that appears in the load that takes the most, which is the maximum-power-transfer statement for a signal source. Noise is not exempt: a source can deliver its available power only to a matched load, and a mismatched load takes less. At a ratio of a hundred the exchange is 0.0596 zW/Hz against 1.5187 — a factor of 25.5, which is $(1+100)^2/(4\times100)$ exactly.

The zero that is the second law #

At $T_1 = T_2$ the net is zero. That is arithmetic — the expression has $(T_1 - T_2)$ as a factor — and it is worth doing as a measurement anyway, because of what it would mean if it were not.

Two resistors of different values inside one box at one temperature are a thermodynamic system in equilibrium. If the net flow between them were anything but zero, one of them would be warming and the other cooling, spontaneously, with no work done, which is the thing that cannot happen. Every correct expression for thermal noise has to produce that zero at every resistance ratio, and a wrong one — one where the delivered power depended on the two resistances asymmetrically — would not.

The figure measures it at twenty-four ratios spanning six decades and gets a worst case of $7.5\times10^{-37}$ watts per hertz, which is the floating-point residue of subtracting two numbers of order $10^{-21}$. It is set up as a refusal rather than reported as a fact, because a check that can reject is the thing that would catch a model that had drifted.

This is the sharpest available argument for why $4kTR$ has the form it does. Nyquist’s derivation is exactly this thought experiment run in reverse: a warm resistance connected to a matched load must deliver the same power the load delivers back, whatever the resistances, and the only expression for the noise voltage that satisfies that for all pairs is one proportional to $R$. The resistance in $4kTR$ is there so that it can cancel.

The mismatch factor, written as the thing a designer already knows #

The geometric factor $R_1R_2/(R_1+R_2)^2$ is not usually met in that form. It is met as a return loss, and the translation is worth making because it turns the curve above into a number off a test set.

A load $R_2$ on a source $R_1$ reflects $\Gamma = (R_2-R_1)/(R_2+R_1)$, and the fraction of the available power it absorbs is $1 - |\Gamma|^2$. Multiplying that out gives exactly $4R_1R_2/(R_1+R_2)^2$, which is four times the geometric factor — so the net exchange is

$$P_{\text{net}} = k\Delta T\,(1 - |\Gamma|^2)$$

and the whole of the mismatch is one number that a network analyser reads directly.

The third row is the one worth internalising. A ten-to-one mismatch — 50 Ω into 500 Ω, which nobody would call a disaster — passes only a third of the available noise power, and passes only a third of the available signal power with it. The two losses are identical and they cancel in the ratio, which is the reason a mismatched connection between a source and an amplifier costs no signal-to-noise ratio at all while costing 4.8 dB of both.

That cancellation stops the moment a second noise source appears after the mismatch, which is what an amplifier is. The amplifier’s own noise is not attenuated by the mismatch, so a mismatched source delivers less signal and less source noise into an unchanged amplifier floor, and the ratio gets worse. That is the whole mechanism behind an optimum source resistance being different from the matched one, and it is why the two matches exist as separate ideas.

What a noise temperature actually specifies #

The cancellation is what makes a noise temperature the right currency, and the consequence is worth stating in the form a designer meets it.

An antenna, a cable, an attenuator and an amplifier input are all sources of noise with quite different resistances. Quoting each of them as a voltage density requires the resistance to be quoted alongside, and the numbers cannot be added without the network between them being worked out. Quoting each as a temperature makes them directly comparable and directly addable through the loss in front, counted twice’s cascade rule, and the reason that works is exactly the cancellation above: available power does not depend on impedance, so the arithmetic of a chain does not either.

A number worth holding: $kT$ at room temperature is $-174$ dBm per hertz. That is the available noise power of any resistance whatever at 290 K, and it is the floor every receiver’s sensitivity is quoted against. A 50 Ω termination and a 10 MΩ probe input have the same one.

The same statement gives noise figure its meaning. A two-port’s noise factor is the ratio by which it degrades the signal-to-noise ratio of a source at 290 K, and the definition only makes sense because “a source at 290 K” is a complete description — it would be incomplete if a resistance had to be named as well. The convention of standardising on 290 K rather than on any prettier number is the residue of it: 290 K makes $kT_0$ come out at $4.00\times10^{-21}$ joules, so $kT_0B$ is a round number in a round bandwidth.

The temperature in the expression is the part’s, not the room’s #

One more consequence of the exchange being driven by a temperature difference: the temperature that belongs in it is the resistor’s own, and a resistor carrying current is not at the room’s.

A quarter-watt film resistor has a thermal resistance to still air of two hundred kelvin per watt or so. Dissipating a hundred milliwatts it sits twenty kelvin above ambient, and its noise is $\sqrt{310/290}$ times what an ambient calculation gives — 0.29 dB. At a quarter of a watt it is fifty kelvin up and 0.69 dB.

Under a decibel is a genuinely small correction and it is worth having measured rather than assumed, because the same arithmetic in the same units gave a very different answer in a companion essay: excess noise at ten volts across a hundred kilohms is twenty-four decibels above the thermal floor at a hertz. Two mechanisms that both arrive with the current, one contributing a fraction of a decibel and the other tens, and only the small one has any physics behind it.

The reason the self-heating term stays small is that noise goes as the square root of an absolute temperature, and absolute temperatures near room temperature are large numbers. Doubling the noise power needs 580 K; doubling the density needs 1160. A resistor that hot has stopped being a resistor.

Where it stops being negligible is at the other end. A cryogenic front end at 20 K has a noise temperature fifteen times below ambient, so a milliwatt of dissipation in a load that is meant to be at 20 K is a catastrophe in decibels even though it would be invisible at 290 — the same twenty kelvin of rise that costs 0.29 dB at room temperature costs 3.0 dB at 20 K. The fractional sensitivity of $\sqrt T$ to $\Delta T$ is $\Delta T/2T$, so the colder the reference, the more every milliwatt costs.

Which is the general form and is worth carrying past this field: a quantity that goes as the square root of an absolute temperature is insensitive to heating at room temperature and violently sensitive to it in the cold, and the crossing between those two regimes is not at any particular temperature but at a particular ratio of rise to reference.

Where the resistance comes back #

Available power is impedance-free and almost nothing else in this field is, so it is worth being precise about when the resistance returns.

It returns the moment a real load is named. An amplifier does not present a matched load to its source — the floor a circuit has finds an optimum source resistance of 6.67 kΩ for a particular amplifier and observes that it is not the resistance that transfers maximum power — so the power the source actually delivers is below its available power by the mismatch factor computed above, and the signal is attenuated by the same factor. The two attenuations cancel in the signal-to-noise ratio, which is why a noise match and a power match are different points and why matching for power is the wrong thing to do to an amplifier.

It returns in a bandwidth. $kT$ per hertz becomes $kTB$ in a band, and the band is set by a network with resistances in it, which is the whole subject of the bandwidth noise sees.

And it returns whenever the load is not resistive. A reactive load takes no average power at all, so a warm resistance connected to a pure reactance exchanges nothing with it — which is the refusal from only the real part is warm seen from the power side, and the reason a lossless matching network in front of an amplifier is free.

So the resistance-free statement is narrow and exact: it is about the power a source could deliver, to the load that would take the most, in one hertz. Everything a circuit does to that number puts a resistance back in.

What a zeptowatt does not let anybody do #

That the exchange is observable between two resistors. A zeptowatt per hertz is not a measurable heat flow in any practical arrangement; over a gigahertz of bandwidth the net here is about a nanowatt. The expression is used for what it implies about a chain of two-ports, not as a thermometry technique — although it is the principle behind one, which is how noise thermometry measures a temperature from a resistor’s noise with no calibration against another thermometer.

That the geometric factor is a coincidence. It is the maximum-power-transfer factor and it is the same function for signal and for noise, because it is a property of a source impedance and a load impedance and not of what the source contains.

That every warm thing has a temperature in this sense. A resistor in equilibrium does. An amplifier’s input noise can be expressed as an equivalent temperature, and that temperature is a bookkeeping device rather than a thermodynamic one — it is often far above the physical temperature of the device, and no second law is violated by that because the device is not in equilibrium.

That the isothermal zero is a check on the arithmetic. It is a check on the form of the expression. Floating point subtracts two numbers of order $10^{-21}$ and leaves $10^{-37}$, so the measurement establishes that the two terms are identical to the last bit — which is what the second law requires and what a wrong exponent on $R$ would break.

kΔT at two resistances six decades apart, and the zero at one temperature #

The match’s net is checked against $k\Delta T$ to a part in $10^{12}$, at a kilohm — and then checked again at a gigohm, computed independently, so that “no resistance appears in it” is a comparison of two solved cases rather than a reading of an expression.

The maximum is checked to be at the match, found by scanning the solved exchange across six decades and checking that the largest entry is within twelve per cent of a ratio of one, which is the resolution of the sweep itself.

And the isothermal case is checked as a refusal. Twenty-four ratios at one temperature, every net flow required to be below $10^{-30}$ watts per hertz, coming out at $7.5\times10^{-37}$.

The currency the field settled on, and why #

Three earlier essays here have been about voltages and this one is about a power, and the change of variable is the whole content.

A voltage density carries a resistance in it and therefore cannot be added across a chain without the network being solved. An available power does not, so it can. That single property is why noise figures cascade, why a receiver’s sensitivity is a number rather than a network, why an attenuator’s noise temperature equals its own physical temperature times its loss, and why $-174$ dBm/Hz is memorised by people who have never computed $4kTR$.

The price of the change is that available power is a statement about a load that is usually not present. It is a bound, like every other one here — the most a source could give — and the moment a real amplifier is connected, the mismatch factor reappears and the bound is not attained. What makes it worth having anyway is that it separates two things that a voltage keeps tangled: what the source is, which is a temperature, and what the connection does, which is a geometric factor with no temperature in it. The two variables separate exactly, which the expression above shows and the figure’s axes demonstrate by moving one while the other stands still.

Still open: the attenuator’s own temperature, the reactive load, and thermometry #

The attenuator, which is the case this expression was invented for. A lossy two-port at physical temperature $T$ presents a noise temperature of $(L-1)T$ at its output for a loss factor $L$, and the derivation is the isothermal zero above applied to a network rather than to a pair. Building it on a solved attenuator — a resistive pad, its own Johnson noise, and a matched termination in front of it — would give a second route to a formula that is normally derived and never measured.

Exchange with a load that is not resistive. The geometric factor above assumes two resistances. A complex load takes the real part of a complex power, and the exchange between a warm resistance and a warm complex impedance has a factor that depends on the angle as well as the magnitude. Whether the isothermal zero survives for arbitrary complex pairs — it must, and the arithmetic is not obvious — is one solve away.

And thermometry, which is this expression used backwards. A resistor’s noise in a known bandwidth gives its temperature with no reference thermometer, and the practical limits are the bandwidth’s accuracy and the amplifier’s own noise temperature. Working the error budget on a solved network would put a number on how long an integration takes to reach a millikelvin, which is ten seconds, and fifteen minutes’s question asked about a temperature rather than a voltage.