The resistors that sum a hundred amplifiers

One turns ratio for a band measured a transformer between a 50 Ω source and an amplifier whose optimum source resistance is 6.67 kΩ, and it also found that N amplifiers in parallel are an ideal transformer of ratio $\sqrt N$, noise for noise, to a part in $10^{12}$. Paralleling costs no copper. It said what the parallel route pays instead: parts, power and the summing network, and at the end it left the third of these open. N amplifiers summed through resistors add those resistors’ noise, scaled by the gain of the sum. For a hundred parts the summing resistors’ contribution could undo much of what the parallel connection gained, and how large a summing network may be before it costs as much as a transformer’s copper is the parallel route’s equivalent of the winding-resistance figure.

The question is put naturally, and it turns out to be the wrong place to look. The summing network is not the expensive part. It has a property that protects it, and following the same arithmetic one stage further finds a different stage in the way, one that the idealised comparison has been quietly assuming away. A parallel amplifier is not a part that makes noise until it reaches an output. The whole point of an average is what it does to the noise of what came before it, and it has no effect at all on what comes after.

The arrangement and its budget #

The parts are the amplifier of the floor a circuit has: a voltage noise of 4 $\text{nV}/\sqrt{\text{Hz}}$ and a current noise of 0.6 $\text{pA}/\sqrt{\text{Hz}}$, white, with their optimum source resistance at 6.67 kΩ. There are N of them with their inputs in parallel across a 50 Ω source. Each has a gain $A$ of its own. The outputs are averaged through a resistor $R$ from each into a single node. That node is read by a buffer with a voltage noise of its own, $e_s$.

The generators of N uncorrelated amplifiers add in power. Their voltage noises average down, and their current noises add up. So referred to the source,

$$e^2 = 4kTR_s + \frac{e_n^2}{N} + N\,i_n^2R_s^2 + \frac{1}{A^2}\left(\frac{4kTR}{N} + e_s^2\right).$$

The first term is the source’s own noise and the next two are what the last essay had, with the balance at $N = e_n/(i_nR_s) = 133$ parts and a noise figure of 1.14 dB. The last term is everything this essay measures, divided by the gain in front of it. That simple expression has one thing in it to verify rather than take on trust, the $4kTR/N$, and the bowl, and the bottom of it has already fixed the rest.

An average of noisy resistors #

Each output drives its own resistor, and those resistors meet at a node. From the node’s side they are all in parallel, so its source resistance is $R/N$, and the node’s noise should be that of a single resistor of that size. A resistor’s noise is carried to the node by a divider of $1/N$, and there are N of them, each adding in power: $N \cdot 4kTR/N^2 = 4kTR/N$.

On a netlist of up to thirty-two resistors, summed resistor by resistor, it is that to a part in a thousand. This is a small result with a large consequence. The summing network’s noise falls as $1/N$, in power, and so does the amplifiers’ own voltage noise. So the ratio between them is the same at every N. A kilohm resistor is as noisy as a 4 $\text{nV}/\sqrt{\text{Hz}}$ amplifier at every N, $e_n^2/4kT$ = 999 Ω, and there is no count at which the summing network starts to dominate if it has not already.

That is why the concern was reasonable and was still not the problem. A hundred amplifiers do not mean a hundred resistors’ worth of noise. They mean a hundred amplifiers’ worth of a hundredth each, and the same for the resistors. Designing the summing network is therefore a one-part problem, solved once, and the answer is just about what one part’s output resistor could be anyway.

The resistor the design can afford #

So how large may it be? The answer is larger than a first look suggests, and it grows with N.

For 0.1 dB the largest resistor is 24 Ω with one part, 33 Ω with eight, 103 Ω with sixty-four and 201 Ω with 133. For 0.5 dB it is 128 Ω with one part and 1,055 Ω with 133. The reason for the rise is what the resistor’s noise is compared with. It is compared with the whole budget, and a term of size $4kTR/N$ has to stand against the source’s fixed $4kTR_s$, which does not fall with N. At large N, the $e_n^2/N$ term has fallen and the current noise’s $N i_n^2 R_s^2$ has risen, so the total is dominated by things that are not falling. A resistor can then be large in comparison with what it could be at one part.

Written the other way, a fixed cost in decibels means $R/N$ is nearly constant at large N. So a design can scale its summing resistors with its count and hold the cost constant. Holding $R/N$ fixed also fixes the total current the resistors draw from the outputs for a given signal at the node, which is $V/(R/N)$, independent of N. So the network’s noise, for a given total drive current, is the same whatever the count. A hundred parts each drive a hundredth of the current through a resistor a hundred times larger. The whole network can be thought of as one resistor of $R/N$ driven by one amplifier of the same total capability.

The 100 Ω in the hero figure costs 0.05 dB at the optimum count, and the best count moves slightly upward, from 134 to 140, because a little extra noise behind the amplifiers makes more of them worth having. Nothing that was feared here has happened. The summing network, however it is built, is an easy thing to get right.

The stage after the sum #

And then there is the thing that actually reads the node. Averaging has reduced the parts’ noise to 0.35 $\text{nV}/\sqrt{\text{Hz}}$ at N = 133, and the node is now a very quiet source that has to go somewhere. Whatever comes next is an amplifier, and it is not averaged by anything.

With a gain of one in each part, a buffer no better than one of the parts brings the noise figure at 133 parts to 13.28 dB. That is 12.14 dB worse than the perfect sum. It is a large number, and it is entirely due to a stage the comparison with a transformer did not mention. A follower after a hundred followers has thrown away almost the whole of the parallel connection’s benefit, because a chain has to have gain at its front to protect itself from its back. That is Friis’s old result in a different suit: the second stage is divided by the first stage’s gain, and an averaging network has no gain.

With $A = 3$ the figure is 5.47 dB, with 10 it is 1.76, and with 30 it is 1.21. Bringing the buffer’s cost under 0.1 dB needs $A = 26$. So each part cannot be a follower. It has to be a gain stage, each with its own feedback network, each of whose lower arm’s noise also counts as part of the amplifier’s voltage noise. A paralleled front end is really a parallel set of complete gain stages, and it has to be a hundred of those, or the idea does not work. That is rather more than a hundred followers and a resistor network. What it costs in supply current is the honest price of this route.

The other escape is to make the following stage quieter, and for that the same arithmetic has one more turn in it. A following stage made quieter by paralleling it is a second, smaller parallel array. Its own following stage is the same question, at less cost, and the sequence terminates quickly because each stage has gain. The first stage’s noise is what sets the floor, and the gain a stage has to have in front of everything after it is the general form of this.

Against the copper #

The original comparison can now be drawn correctly.

A transformer’s winding resistance sits in front of everything at the source’s own impedance level, so it is compared with 50 Ω directly: 1 Ω of winding gives 1.22 dB, 3 Ω gives 1.39, 10 Ω gives 1.93 and 30 Ω gives 3.18. The loss in front, counted twice gave the reason this costs as much as it does. The parallel route with a gain of 30 in each part, with 100 Ω summing resistors and a buffer as good as a part, is 1.21 dB. So parallel amplifiers match a transformer with about 1 Ω of winding, and a transformer of a modest size, on a low band, has a winding resistance of that order or above. They are a real alternative, and they come with no microphony, no magnetic pick-up and no low-frequency corner.

The comparison turns on the source’s impedance. On a 5 Ω source a transformer’s ohm of winding is 20 per cent of the source’s resistance, and on a 500 Ω one it is negligible; that changes what each route is worth by a corresponding amount. The parallel route’s balance also moves, since at a smaller source N’s optimum grows in proportion to the source’s own fall. A very low source, of the kind the source resistance the flicker moves warns about, makes the current noise less of a concern and makes the transformer’s copper a much bigger one, which is why very low-impedance sensors have such a strong record with parallel input stages rather than transformers.

Parallel parts cannot lower the bowl #

The result that N amplifiers are a transformer has a corollary that is worth stating, because the talk of a hundred quiet parts obscures it. A transformer does not change the best noise figure a part can reach. It moves the part’s optimum source resistance onto the source that is there. The minimum over source resistance is $F_{min} = 1 + 2e_ni_n/4kT$, and the product $e_ni_n$ is the same for one part as for N of them, since the voltage noise falls by $\sqrt N$ and the current noise rises by it. So the 1.14 dB at 133 parts is the part’s own bottom of the bowl, reached from a 50 Ω source. No larger count improves it, and that is why the curve in the hero figure rises again past 133.

A better part is therefore a different kind of purchase from more parts. A 1 $\text{nV}/\sqrt{\text{Hz}}$ part with the same current noise has a product four times smaller: its bottom is 0.31 dB, and its optimum is 1.67 kΩ, so it reaches that from 50 Ω at 33 parts rather than at 133. That is fewer parts and much less noise together. The choice of part is what sets the floor, and the count is only what brings the source up to it. On the other side, a 10 $\text{nV}/\sqrt{\text{Hz}}$ part with far less current noise, as a field-effect input tends to have, has a large optimum, and it takes thousands in parallel to reach a 50 Ω source. A count that large is why that combination is not built for this source. It would be a good reason to choose a transformer, and for a much higher-impedance source a field-effect part with no help at all.

The flicker corner, and what parallel parts cannot do #

A low-frequency band is where this choice is usually being made, and it has a limit. The corner where averaging stops working measured a single stream’s flicker refusing to average in time. N independent parts have independent flicker, and averaging across parts does work on them: each part’s flicker noise is uncorrelated with its neighbours’, and it falls as $1/N$ in power, unlike averaging one part over time. This is a real difference between averaging in space and in time, and it is a good reason for paralleling in a low band. Its limit is a shared supply or shared temperature, which correlates the parts, and on a low band the shared temperature is a real concern: a term common to all the parts does not average at all. The measurement of two generators with one cause applies inside one part, and across parts the same argument sets how much of the $1/N$ can be believed.

What this leaves out #

The parts are identical here. A real array has a spread in their input voltage noise and in their gain. An average weights each part by its gain, so unequal gains make the best average a weighted one and a plain average a slightly worse one, and the effect is second-order for a few per cent of spread. The offset of each part differs too, which is a current through the summing resistors and is exactly what limits how much direct gain a parallel array can have. An array with a gain of 30 in each part and a millivolt of spread between them has 30 mV across each resistor at direct current, and that is a real design constraint. AC coupling is the usual way out, and on a low band it has a cost of its own.

The input capacitance adds as well. N parts have N times the input capacitance across the source, so a parallel array loses its bandwidth at the source’s own impedance as the count rises, which a transformer does not do in the same way, and which at a 50 Ω source and a hundred parts of 5 pF is a 500 pF load and a corner at 6 MHz. None of these change the arithmetic of the noise figure in the band of interest.

The numbers to keep #

N resistors averaging N outputs have the noise of $R/N$, to a part in a thousand on a netlist, and it falls with N exactly as the parts’ voltage noise does. So the resistor equal to a 4 $\text{nV}/\sqrt{\text{Hz}}$ amplifier is about a kilohm at every count. For a cost of 0.1 dB, 201 Ω is allowed at 133 parts, and the summing network is not the expensive part. The stage after the sum is. It is not averaged, so a buffer as good as one part costs 12.14 dB at 133 parts unless each part has gain, and each needs a gain of 26 to bring that under 0.1 dB. With that gain, 134 parts reach 1.21 dB, which is what a transformer reaches with about 1 Ω of winding resistance.

Still open: the magnetising inductance, a capacitive source, and parts that are not alike #

The magnetising inductance as a noise source. At the bottom of the band the transformer’s magnetising inductance appears in parallel with the source and so changes the impedance the amplifier sees, frequency by frequency, as a drifting optimum would like. Whether a transformer can be designed so that its own low-frequency roll-off moves the presented impedance towards the drifting optimum, turning a defect into a correction, is the question that would decide the comparison on a low band.

A capacitive source. A piezoelectric or capacitive sensor is a reactance first, and its impedance falls with frequency as the optimum’s real part rises. For a parallel array the count’s optimum depends on the source’s size, so a count is a value at one frequency, and how far a fixed count can follow a capacitive source across a band is the drift essay’s second open question with an array added.

Parts that are not alike. With a spread of a few per cent in the parts’ voltage noise, some are worse than the mean, and a weighted average beats the plain one. How much of the budget a production spread costs, and whether it ever pays to measure and weight each part rather than simply adding a few more of them, is a small Monte Carlo on the same budget.