The resistance colder than the room

The readings that come back the same measured which readings of a passive network are reciprocal and found three: transimpedance, transadmittance and a crossed pair of a voltage gain and a current gain. The first of them says something that sounds like a convenience and turns out to be a law of physics. The transfer from a current injected anywhere in a network to the voltage at a port equals the transfer from a current injected at the port to the voltage at that place.

Every resistor’s thermal noise is exactly such an injected current, so that sentence is a statement about noise. Only the real part is warm measured the result it leads to for a single lossy capacitor — the noise across any passive two-terminal is $4kT\,\mathrm{Re}\{Z\}$, and a reactance contributes nothing however large — and the resistor the noise comes from found how unevenly the resistors of one filter share an output noise. Neither said why the shares of many resistors, each filtered differently, must always add up to a quantity that knows only the port’s own impedance.

Reciprocity is why. This essay shows it, then breaks it with a controlled source, and then uses the breakage on purpose to build a resistance that is quieter than any resistor at the same temperature.

Three routes to one noise #

The network is the ladder network of the earlier measurements — 470 ohms, a millihenry and 22 nanofarads, a kilohm and 4.7 nanofarads — now terminated as it would be used: a 50-ohm source resistance at its input and a 10-kilohm load at its output. The output port is the load. Four resistors, each at 290 kelvin, each contributing a noise current of density $4kT/R$ across itself.

The first route treats each resistor as a source in turn: a one-ampere current across that resistor, one solve, the voltage it produces at the output, squared and weighted by $4kT/R$. Four resistors, four solves, and the sum is the output noise density. Nothing in this route uses reciprocity; it is superposition of uncorrelated sources, and it is the definition.

The second route makes one solve. An ampere is injected at the output, and the voltage that appears across each of the four resistors is read. By reciprocity — the transimpedance identity — each of those voltages is the same transfer the first route needed, read in the other direction. So one solve gives all four contributions, whatever the number of resistors. At every frequency from 100 hertz to a megahertz the two routes agree to 4.4 parts in 10¹⁶.

The third route is Nyquist’s: the real part of the output impedance, which the same single solve also provides, times $4kT$. It agrees with the sum to the same 4.4 parts in 10¹⁶.

Why the third route has to be right #

The agreement between the second and third routes is not an accident of this ladder network, and the argument is short enough to give. Inject an ampere at the output. The voltage across resistor k is $v_k$, and the power that ampere dissipates in resistor k is $|v_k|^2/R_k$. The total power the ampere delivers into the network is the real part of the output impedance, $\mathrm{Re}\{Z_{\text{out}}\} \cdot 1^2$, and in a network of resistors and reactances all of that power ends up in the resistors. So

$$\mathrm{Re}\{Z_{\text{out}}\} = \sum_k \frac{|v_k|^2}{R_k}.$$

Multiply by $4kT$ and the right-hand side is the second route’s sum of noise contributions — but only because reciprocity let the second route read $v_k$, a transfer from the output to resistor k, as the transfer from resistor k to the output. Without reciprocity the power argument still gives the real part of the output impedance, but it is no longer a statement about noise. Nyquist’s theorem for a network is energy conservation plus reciprocity.

The single solve also says where the noise comes from, which the total never does.

The shares move with frequency because each resistor’s noise is filtered by everything between it and the output. At a kilohertz the ladder network is transparent and the larger resistors dominate, as their noise currents develop voltage across the larger impedances around them. By 300 kilohertz the first section’s capacitor short-circuits everything before it: the source resistance and the 470 ohms have no path to the output and contribute nothing, and the kilohm next to the output supplies 90.8 per cent. That is the pattern the resistor the noise comes from found in a filter — noise is filtered by what comes after it — computed here with one solve per frequency instead of one per resistor.

On a network of a hundred resistors the saving is a hundredfold, and it is the basis of how circuit simulators perform noise analysis: they solve the transposed network once, with the output as the source, and read every element’s transfer from that.

A controlled source breaks both shortcuts #

Put a transconductance into the ladder network and the matrix is no longer symmetric. The transfer from a resistor to the output is no longer the transfer from the output to the resistor, so the one-solve route is reading the wrong transfers. And the power argument, which still gives the real part of the output impedance, now includes power delivered by the controlled source, so it no longer equals the sum of the resistors’ noise.

The noise summed one resistor at a time is still correct — superposition does not care about symmetry — so it is the reference. Against it, the one-solve shortcut is wrong by $5 \times 10^{-4}$ at a millisiemens and Nyquist’s expression by six per cent. Both errors start in proportion to the transconductance, and both are symptoms of the same asymmetric entry that the readings that come back the same found breaking all three reciprocal pairs at once.

The second failure is the important one, and it is physical rather than computational. A network with gain in it is not a set of resistors in equilibrium at one temperature: the amplifier is fed power from a supply that the noise calculation does not see, and it can move noise from one place to another or suppress it. So the noise at its port is not determined by its port impedance. A circuit simulator handles this by solving the transposed network — the adjoint, with every controlled source’s control and output swapped — which restores the shortcut’s arithmetic. Nothing restores Nyquist’s expression, because for an active network it is simply not true.

A resistance made by feedback #

That last fact can be used. Take an inverting amplifier of gain A and connect a resistor $R_f$ from its output back to its input. A current into the input node has to flow through $R_f$, and the amplifier drives the far end of $R_f$ to $-A$ times the input voltage, so the voltage across $R_f$ is $(1+A)$ times the input voltage and the input sees a resistance $R_f/(1+A)$. Choose $R_f = 50(1+A)$ and the input is a 50-ohm termination at every gain.

Now ask for its noise, with the amplifier itself noiseless and only $R_f$ warm. $R_f$’s noise current is $4kT/R_f$, which at large gain is far smaller than a 50-ohm resistor’s — and that current, arriving at the input node, sees the 50-ohm input impedance.

The input presents 50.000 ohms at every gain, and its noise is exactly that of a 50-ohm resistor at a temperature of $290/(1+A)$ kelvin: 145 kelvin at a gain of one, 29 at nine, 2.90 at ninety-nine. The arithmetic is two lines. The input noise voltage density is the noise current times the input resistance, $\frac{4kT}{R_f} \cdot R_{\text{in}}^2 = 4kT R_{\text{in}} \cdot \frac{R_{\text{in}}}{R_f} = 4k\frac{T}{1+A}R_{\text{in}}$, which is Nyquist’s expression for $R_{\text{in}}$ at a temperature reduced by $1 + A$.

This is a real technique, called active or electronically cooled termination, and it is used where a 50-ohm input is needed for matching but a 50-ohm resistor’s noise is unaffordable — the input of a sensitive receiver, the termination of a cryogenic line. A noise temperature of 2.9 kelvin from a room-temperature circuit sounds like a violation of the second law. It is not, because the network is active: the amplifier is not in thermal equilibrium with its surroundings, and which way the noise goes was a statement about two passive resistors exchanging power, not about a resistance synthesised from a power supply. Couple the cold termination to a warm resistor and net power does flow into it, as it would into a cold object — the amplifier’s supply is what carries that heat away.

Only where the gain is #

The amplifier in that argument had the same gain at every frequency. A real one has a gain–bandwidth product, and the synthesis depends on the gain at the frequency of interest.

Below a megahertz the termination is 50 ohms at 29.0 kelvin, as the flat-gain case predicted for a gain of nine. Above the amplifier’s closed-loop corner the gain falls, the synthesised resistance rises toward the bare 500-ohm feedback resistor, and the noise temperature rises with it toward 290 kelvin. At a gigahertz the termination is 495 ohms at room temperature: the amplifier has stopped doing anything, and what is left is a warm resistor.

So the cold is bought with gain, frequency by frequency, and the match is bought with it too — the two fail together. The same gain–bandwidth limit the tee that charges for its own compensation spends on a transimpedance amplifier’s stability decides here how far up in frequency a cold termination stays cold. And a real amplifier is not noiseless: its own voltage and current noise add to the input, and in practice they set the floor that the feedback resistor’s reduced noise uncovers, which is why the technique is quoted with an amplifier noise figure rather than with a temperature alone.

The same trick as a derivative #

The one-solve route has a twin in a different part of circuit analysis, and seeing the two together explains why both work. Every derivative, and the one that is zero measured the sensitivities of a network’s output to every one of its elements from two solves: the circuit itself, and an adjoint circuit driven from the output. The adjoint of a network of resistors, capacitors and inductors is the network itself, because its matrix is symmetric; the adjoint of a network with controlled sources is its transpose, with every source’s control and output exchanged.

Noise analysis and sensitivity analysis are the same computation for the same reason. Both ask how a small current injected at element k shows up at the output, for every k at once. Asked naively that is one solve per element. Asked through the adjoint it is one solve in total, because the answer for every k is the voltage the output’s own injection produces at k. In a reciprocal network the adjoint costs nothing extra to build, which is why the one-solve noise route above worked without any modification; in a network with gain, the transpose has to be assembled separately, and a route that skips it inherits the error the controlled-source figure measures.

This is also the deeper reason the controlled-source error is proportional to the transconductance at small values. The difference between a matrix and its transpose is exactly the asymmetric entries, and for a single controlled source that is one entry of size $g_m$. Every transfer read from the wrong matrix is wrong by a first-order perturbation in that entry, which what a network answers would recognise as the condition-number argument made about one element instead of all of them.

What the cold termination is worth at a receiver #

A noise temperature is an abstraction until it is set against a source, and the comparison that justifies the cold termination is the one it replaces. A receiver input that must present 50 ohms to its line is often built with a plain 50-ohm resistor across an amplifier with a much higher input impedance. That resistor is matched and warm. The source behind the line is also 50 ohms at 290 kelvin, so the termination contributes exactly as much noise as the source, and by the loss in front, counted twice’s arithmetic the input’s noise figure can be no better than three decibels before the amplifier adds anything.

Replace the resistor with the feedback termination at a gain of nine, and the termination’s noise temperature is 29 kelvin against the source’s 290. The noise figure it contributes is $10\log_{10}(1 + 29/290)$, or 0.41 decibels; at a gain of ninety-nine it is 0.04. The match is unchanged — 50.000 ohms in every case — and the noise penalty for having one has fallen by nearly three decibels. The floor a resistor sets treats that three-decibel penalty as the price of a resistive match; the cold termination is the circuit that declines to pay it.

What it costs instead is the amplifier’s own noise and bandwidth, which the figures above deliberately left out. The feedback resistor’s noise is divided by $1 + A$ only because the amplifier is noiseless in this model; a real amplifier’s input noise voltage and current add at the input directly, and they set the floor that the divided resistor noise uncovers. In practice the technique buys most of the three decibels with an amplifier whose noise temperature is itself a few tens of kelvin, and it buys them only up to the frequency where the amplifier’s loop gain runs out.

What the measurement leaves out #

The resistors are at one temperature. A network whose resistors sit at different temperatures has a port noise that is not $4kT\,\mathrm{Re}\{Z\}$ for any single T, though the one-solve route still gives each resistor’s share correctly, since it needs only reciprocity and not equilibrium. The noise a pad adds is its temperature uses exactly that to compute the noise of an attenuator at a different temperature from its source.

Only thermal noise is counted. A controlled source in a real device brings its own noise — shot noise in a junction, channel noise in a field-effect device — and those are uncorrelated currents with their own transfers, which the one-solve route on the transposed network handles like any other source.

And the amplifier in the cold termination is ideal apart from its gain. A real feedback amplifier’s input impedance is complex above its corner, and a termination that is 50 ohms resistive at low frequency becomes partly inductive as the gain falls, which degrades the match before it degrades the noise temperature.

The number to keep #

In a passive network the transfer from each resistor to a port equals the transfer from the port to that resistor, so one solve from the port gives every resistor’s noise contribution, and by conservation of energy they sum to $4kT\,\mathrm{Re}\{Z\}$ — here to $4.4 \times 10^{-16}$. A controlled source breaks both the shortcut and the theorem: by six per cent for Nyquist’s expression at one millisiemens on this ladder network. Used deliberately, feedback makes a 50-ohm resistance with the noise of 50 ohms at $290/(1+A)$ kelvin — 2.90 kelvin at a gain of 99 — but only where the amplifier still has its gain.

Still open: the amplifier’s own noise, the transposed network, and the cold match at high frequency #

The amplifier’s own noise in the cold termination. Adding an input voltage noise and current noise to the amplifier gives a floor under the $290/(1+A)$ curve, and the gain at which the feedback resistor stops being the limit is a design number this essay has not computed.

The adjoint as a measurement. Solving the transposed network — controlled sources with control and output exchanged — should restore the one-solve shortcut exactly for an active network. Measuring its agreement with the resistor-by-resistor sum across the transconductance range would close the account the controlled-source figure leaves open.

The match above the corner. The cold termination’s impedance grows a reactive part as the gain falls, and the mismatch it presents to a 50-ohm line costs signal before the noise temperature rises. Where the two crossings fall, for a given amplifier, decides the band over which the technique is worth having.