The noise a pad adds is its temperature

Which way the noise goes joined two warm resistors and measured the net noise power flowing between them: $4k\Delta T\,R_1R_2/(R_1 + R_2)^2$ per hertz, which at a match is exactly $k\Delta T$. The zero at equal temperatures is not a coincidence of the arithmetic but a requirement of thermodynamics — two bodies at one temperature exchange no net power by any route — and the essay ended by naming the case the expression was really invented for. A lossy two-port at a physical temperature $T$ should present a noise temperature of $(L - 1)T$ at its input for a loss factor $L$, and the derivation is the same isothermal zero applied to a network rather than to a pair.

That expression is usually derived and quoted, and it is the basis of every noise budget that has a cable, a pad or a filter in front of an amplifier. It is rarely measured, because measuring it needs the pad’s own resistors’ noise carried separately to its output. This essay does that, resistor by resistor, and then uses the result to answer three questions the formula alone does not: what a cold pad buys, which resistor the noise comes from, and when a pad between two temperatures stops the flow.

The pad, solved resistor by resistor #

The pad is a matched 50 Ω T: two series arms and a shunt, chosen from the matching equations for a stated loss. At 6 dB the arms are 16.61 Ω and the shunt 66.93 Ω. In front of it is a 50 Ω source resistor at a temperature of its own. Each of the four resistors is given a noise source of $4kT_rR_r$ at its own temperature, each is carried separately to the pad’s open output, and their powers are added — they are independent, so powers do add — and divided by the output’s own resistance to give a noise temperature.

The sum agrees with $T_s/L + T(1 - 1/L)$ to nine figures at every source temperature. Its slope is $1/L$, 0.2512 at 6 dB: the source’s noise is attenuated like any other signal. Its intercept is 217.2 K: with the source at absolute zero the output is still warm, because the pad’s own resistors are at 290 K and their noise reaches the output. And at a source temperature of 290 K the output is at exactly 290 K, the ringed point where the line crosses the diagonal. With everything at one temperature the pad’s output cannot be told from a 50 Ω resistor at that temperature, which is the isothermal zero in the form a pad obeys it.

That crossing is the whole derivation, run in reverse. The pad’s contribution must be whatever makes the output $T$ when the input is $T$, and since the input’s share is $T/L$, the pad’s own must be $T(1 - 1/L)$. The resistor-by-resistor sum reaches the same number without assuming anything about equilibrium: four resistors, four gains, four temperatures. The slider on the figure at the head of the page steps the loss from 1 dB, where a cold source comes out at 59.6 K, to 20 dB, where it comes out at 287.1 K — nearly the pad’s own temperature, since a heavy pad’s output is almost all pad.

Referred to the input, (L − 1)T #

A noise budget refers everything to one point, usually the input of the chain. The pad’s own output noise, $T(1 - 1/L)$, divided by its gain $1/L$, is $(L - 1)T$ at its input.

Referred to its input, a pad at 290 K adds 75.1 K at 1 dB, 288.6 K at 3 dB, 2,610 K at 10 dB and 28,710 K at 20 dB, and the sum agrees with $(L - 1)T$ to nine figures at each. Written as a noise figure, $10\log(1 + (L - 1)T/290)$, that is the loss in decibels exactly when the pad is at 290 K. The loss in front, counted twice found the same rule from the Friis cascade’s definition of noise figure — a decibel of loss before an amplifier costs a decibel of signal and a decibel of noise figure — and here it arrives from Johnson’s formula and four resistors, with no noise figure assumed anywhere.

The input-referred form is the useful one and also the alarming one. It grows without limit: a 30 dB pad adds 289,700 K. That is not a physical temperature anywhere in the pad. It is what the pad’s modest output noise, $T(1 - 1/L)$ never more than 290 K, looks like after dividing by a gain of a thousandth, and it is why nothing with a low noise temperature is ever put after a large loss.

What cooling a pad buys #

Since the pad’s contribution is proportional to its physical temperature, cooling it is the obvious repair, and cryogenic receivers do exactly that.

The two referrals behave very differently. At its output, a pad adds $(1 - 1/L)T$, which rises with the loss towards the pad’s own temperature and never passes it: a 20 dB pad at 290 K adds 287.1 K, at 77 K adds 76.2 K, at 4 K adds 3.96 K. Whatever comes after the pad sees a source no warmer than the pad. Referred to its input, the same pads add 28,710 K, 7,623 K and 396 K: cooling scales the input-referred noise by the temperature, but the factor of $L - 1$ remains.

So a cold pad is quiet looking backwards from what follows it and noisy looking forwards from what precedes it, and which matters depends on where the signal is. A cryogenic attenuator in the input line of a very low-noise amplifier is there to stop warm noise coming down the line from above, and for that the output view is the right one: a 20 dB pad at 4 K turns 290 K of room-temperature noise arriving from above into $290/100 + 3.96 = 6.9$ K at its output. For a weak signal arriving from above the same pad is a 20 dB loss and a 396 K noise contribution, and there is no temperature at which a large loss in front of the signal is free.

Which resistor the noise comes from #

The sum was taken resistor by resistor, so it also says which of the pad’s three resistors makes its noise. The resistor the noise comes from found that identical resistors in one filter contribute very different shares because each resistor’s noise is filtered by everything after it; a pad is the same argument with attenuation in place of filtering.

At 1 dB the shunt resistor makes half the pad’s noise, 49.8 per cent, with the two arms sharing the rest. As the loss grows the output arm takes over: 44.4 per cent at 6 dB, 57.7 at 10, and 82.6 per cent at 20 dB, where the input arm contributes less than one per cent. The reason is the pad’s own attenuation. The input arm’s noise has to cross the whole pad to reach the output and is attenuated by nearly all of it; the output arm’s noise is at the output already. A heavy pad’s noise is made in its last resistor.

That has a practical edge that the formula hides. The formula assumes the whole pad is at one temperature. If it is not — a pad whose output end is bolted to a cold stage and whose input end is not — the effective temperature in $(1 - 1/L)T$ is the resistors’ temperatures weighted by these shares, and at 20 dB that weighting is almost entirely the output arm’s. Cooling the output end of a heavy pad buys most of what cooling the whole of it would.

A pad between two temperatures #

The last question is the one that joins this essay to the exchange that preceded it. Set a pad between a source at 400 K and a load at 290 K. The load receives $k(T_s/L + T_{pad}(1 - 1/L))$ per hertz from the pad and radiates $kT_2$ back, and the net flow depends on the pad’s own temperature.

A cold pad attenuates the warm source’s noise and adds little of its own; a warm pad adds more than it removes. The flow stops where the two balance, at $T_{pad} = (T_2 - T_1/L)/(1 - 1/L)$: 179.5 K for a 3 dB pad, 253.1 K for 6 dB, 277.8 K for 10 dB — always below the load’s own temperature, because the pad must make up for the warm source it lets through. And below a loss of $10\log(T_1/T_2)$, 1.40 dB here, there is no such temperature. A 1 dB pad at absolute zero still passes 317.7 K of the source’s noise, more than the load’s 290 K, so the flow into the load continues however cold the pad.

With all three at one temperature nothing flows at any loss, which is the isothermal zero once more. A pad is a thermal link with a transmission of $1/L$ and an emissivity of $1 - 1/L$, and the two always sum to one, as a body’s transmission and emission must.

A receiver, worked #

The numbers make the design rule concrete. Take an antenna looking at a sky of 50 K, an amplifier whose own noise temperature is 50 K, and a metre of cable with 1 dB of loss at room temperature between them. Referred to the antenna, the cable adds $(L - 1)T = 75.1$ K and the amplifier’s 50 K arrives multiplied by $L = 1.259$, 62.9 K, so the system temperature is $50 + 75.1 + 62.9 = 188$ K. Without the cable it would be 100 K. One decibel of warm cable has nearly doubled the noise of a receiver whose own parts contribute 50 K each — the Friis rule in a form the floor a resistor sets would recognise, since a cable’s loss is a resistance and 75 K is what that resistance’s temperature is worth once it stands in front of everything else.

Move the amplifier to the antenna and the cable after it, and the cable’s 75.1 K is divided by the amplifier’s gain: at 20 dB of gain it becomes 0.75 K and the system is 100.8 K. That is the whole case for mounting a low-noise amplifier at the antenna, reached from four resistors’ Johnson noise.

The same arithmetic prices a cooled input line. With the 1 dB of cable at 77 K instead of 290, its input-referred contribution falls to 19.9 K and the system to 133 K; at 4 K, to 1.0 K and 114 K. Cooling the cable recovers most of what it cost, and the remainder — the 12.9 K by which $L \cdot T_{amp}$ exceeds $T_{amp}$ — is the loss’s own attenuation of the signal, which no temperature removes.

Why equilibrium is enough #

It is worth asking why a derivation that assumes nothing but equilibrium gives an answer that holds out of equilibrium, with the source at one temperature and the pad at another. The reason is that the pad’s noise and the source’s noise are independent, so they add as powers, and each can be found separately. The pad’s own contribution does not depend on what is in front of it; it can therefore be found in the one case where the answer is known without calculation, the case where everything is at one temperature and the output must be too. The resistor-by-resistor sum confirms that the contribution found that way is the same at every source temperature, which is what independence means.

The argument would fail for a pad whose resistors made noise that depended on the signal through them. The floor that is only a floor while nothing flows found exactly such a noise in real resistors carrying a direct current, an excess that grows with the current and has nothing to do with temperature. A pad passing a DC bias — an attenuator in a line that also carries power to a masthead amplifier — has that excess on top of $(L - 1)T$, and the equilibrium argument knows nothing of it.

What a designer should take #

Treat any matched lossy element as a noise source at its own physical temperature: it adds $(1 - 1/L)T$ at its output and $(L - 1)T$ referred to its input. At room temperature that is a noise figure equal to its loss. Cooling it helps what comes after it in proportion to the temperature and never removes the factor of $L$ that it puts in front of everything after it.

Put attenuation after gain, never before it, when the signal is weak. And when a pad must go before a sensitive stage — to set a level, to isolate a reflection — make it as small as the job allows, since its input-referred noise grows as the loss itself and not as the loss in decibels.

How the numbers were obtained #

The pad is a netlist of the source resistor and the T-pad’s three resistors, with the output left open. For each resistor a unit voltage source is placed in series with it, every other source is zeroed, and the voltage gain to the open output is solved; the resistor’s contribution to the output’s noise temperature is its temperature times its resistance times that gain squared, over the output’s own resistance, which is solved separately and is the pad’s 50 Ω to rounding. The contributions are independent and their powers add. Nothing is frequency-dependent in a resistive pad, so every solve is at one hertz and holds at every frequency. The closed forms are checked against the sums and not used to draw them.

What it leaves out #

Mismatch. Every pad here is matched at both ends. A pad driven from a mismatched source, or into a mismatched load, reflects part of the noise it would otherwise pass, and its effective loss and noise temperature both change; the resistor-by-resistor sum handles that without change, but the closed forms do not.

Loss that is not resistive. A cable’s loss is in its conductor’s resistance and its dielectric’s loss tangent, both frequency-dependent, and a filter’s is in its components’ losses. Each is a lossy two-port at a physical temperature and obeys the same rule at each frequency; only the real part is warm is the reason, since the reactive parts carry no noise.

Temperature gradients. A cable running from a cold stage to room temperature has a temperature that varies along it. The pad-share argument above says its noise is dominated by its warm end if the loss is concentrated there and by its output end in any case; the integral along a real cable is a separate calculation.

Still open: the reactive load, thermometry, and a cable’s gradient #

Exchange with a load that is not resistive. The net flow between a warm resistance and a warm complex impedance has a factor that depends on the angle as well as the magnitude. That the isothermal zero survives for any complex pair is required; how the non-isothermal flow depends on the reactance is the earlier essay’s second open question, and the pad’s transmission-and-emissivity reading suggests it is a question about how much of the load’s impedance is a real part at each frequency.

Thermometry, which is this expression used backwards. A pad’s output noise temperature is a thermometer for its own temperature when its source is known: $T = (T_{out} - T_s/L)/(1 - 1/L)$. Working the error budget — the loss’s uncertainty, the amplifier’s own noise, the bandwidth the bandwidth noise sees defines, the integration time — would say how precisely a pad can read its own temperature, and ten seconds, and fifteen minutes has the averaging arithmetic for the last of the three: a noise power read to a part in a thousand needs a million independent samples of it, and the bandwidth decides how long that takes.

A cable along a gradient. A line from a cold stage to room temperature is a chain of short pads, each at its own temperature. Its output noise is each piece’s $(1 - 1/L_k)T_k$ attenuated by everything after it, and the shares above say which end dominates; summing the chain for a stated gradient would put a number on how much of a cryogenic input line’s noise comes from its warm end.