The node that does not care how many

The node that is at ground for a while measured a summing junction with one input on it and found three regions: a tenth of an ohm at direct current, a rise of a decade per decade, and a ceiling of 909 ohms which is the input and feedback resistors in parallel, with the amplifier no longer participating. It put a number on what “virtual earth” is worth — the node stops being one at a thousandth of the amplifier’s gain–bandwidth — and it ended with a figure for the crosstalk between two signals summed into the same node: twenty decibels per decade of separation, starting from eighty. The loop doing the holding is the one the ideal amplifier, and where it stops being one prices, and the division by one plus it is what how much of the amplifier gets through measures in the gain.

A summing junction with two inputs on it is an unusual thing to build. A mixer has eight or sixteen or thirty-two, and the arithmetic in that essay contains the count nowhere. Putting it in changes one of the three regions and not the other two, and which one is the finding.

What the count does to the ceiling #

The ceiling is the easy part and it is the part everybody would predict. With the amplifier gone, the node has $N$ input resistors going to ground — the sources are zeroed, which is what an impedance at a node means — and the feedback resistor going to an output that is also at ground through its own fifty ohms. So the ceiling is $R_\mathrm{in}/N$ in parallel with $R_f$:

Thirty-two times the inputs and 29.19 times the stiffness, the shortfall being the feedback resistor, which is in the parallel combination at every count and stops being negligible at neither end. At one input it is ten times the input resistor and contributes nine per cent; at thirty-two it is three hundred and twenty times $R_\mathrm{in}/N$ and contributes a third of one per cent.

That is a comfortable-looking result. A node that is thirty times stiffer sounds like a mixer that is thirty times quieter about its channels, and it is not, because the ceiling is the one region of the curve where the amplifier has stopped working.

What the count does to the rest of it, which is nothing #

Read the curve at any frequency below the ceiling and the channel count is absent.

Four per cent across a factor of thirty-two in the count at a kilohertz, and four parts in ten thousand at direct current. The node’s impedance in the band where the loop is working is a property of one channel and of the amplifier, and the number of other channels on it does not enter.

The cancellation is exact and worth writing out, because it is the whole essay. In that band the node is the open-loop impedance divided by the loop gain,

$$Z \approx \frac{R_\mathrm{in}/N \parallel R_f}{A\beta}, \qquad \beta = \frac{R_\mathrm{in}/N}{R_\mathrm{in}/N + R_f}$$

and both the numerator and $\beta$ are proportional to $R_\mathrm{in}/N$ when that is small against $R_f$. They cancel. The stiffer node came with a weaker loop in exactly the same proportion, and the two effects leave the node where it was.

Which is a specific instance of something what is left at crossover states generally: everything a loop does well it does in proportion to its loop gain, so a change that improves the open-loop quantity and weakens the loop by the same factor changes nothing at all. A designer adding inputs to a summing junction has done exactly that, and would be right to expect it if they had asked which of the two the change touches. The usual expectation is that it touches only the first.

Where the count actually arrives #

So the effect of the count is to bring the ceiling down to meet the rise earlier. At one input the node rises for five decades before flattening at 909 ohms; at thirty-two it flattens at 31.2 ohms after three. And that changes the crosstalk arithmetic in a way neither of the two obvious guesses gets right.

The leakage from one input into one other is the node’s impedance over the input resistance — a signal on one channel drives $v/R_\mathrm{in}$ into the node, that current develops $v \cdot Z/R_\mathrm{in}$ on it, and that voltage pushes a current back out down every other channel’s resistor. Since $Z$ does not move with the count and $R_\mathrm{in}$ per channel does not either, the leakage from one channel into one other is unchanged by the channel count below the ceiling, and saturates above it:

The second column improves with the count, which is the reassuring reading, and the third is the one that matters. The total leakage into every other channel is $(N-1)$ times the per-pair figure, and since the ceiling is very nearly $R_\mathrm{in}/N$, that product is $(N-1)/N$ — it tends to one, from below, whatever the channel count is.

A summing junction above its loop passes essentially the whole of one input’s signal out to the other inputs collectively. Thirty-two channels divide it thirty-two ways and two channels divide it two ways, and the total is the same. The count does not change how much leaks; it changes how finely the leak is shared out.

The two resistors are not the only two numbers #

There is a second way to read the cancellation, and it makes the result less surprising without making it less useful.

The node’s impedance below the ceiling is $R_\mathrm{in}/(N \cdot A\beta)$ near enough, and what $N\beta$ is, for small $R_\mathrm{in}/N$ against $R_f$, is $R_\mathrm{in}/R_f$ — the ratio of the two resistor values, with the count divided out. So the impedance in the working band is $R_f/A$: the feedback resistor divided by the amplifier’s open-loop gain, and nothing else.

That is worth having as a formula because it names the two quantities a designer can change and excludes the two they cannot usefully. The node’s impedance at a frequency is set by the feedback resistor and by the amplifier’s gain there. It is not set by the input resistors, by how many of them there are, or by the closed-loop gain. Checked against the measurement: $R_f = 10$ kΩ and $A = 10^6/1000 = 1000$ at a kilohertz gives 10 Ω, and the table above reads 10.049 at one input and 9.5372 at thirty-two.

The departure at the top of the count is the approximation showing, and in the honest direction: at thirty-two inputs $R_\mathrm{in}/N$ is 31.25 Ω against the feedback resistor’s ten kilohms, so the “for small” is good to three parts in a thousand, and the five per cent of departure at a kilohertz comes from the ceiling already being within a factor of three of the curve there. Below a hundred hertz, where the ceiling is far away, the six counts agree to four parts in ten thousand.

The practical form is short. Halving the feedback resistor halves the node’s impedance at every frequency in the working band, at every channel count — and it also halves the gain, which is why nobody does it, and why the summing junction’s impedance is the quantity in an inverting stage that is hardest to improve without changing what the stage is for. It is the same exchange the step that is too big finds at the other end of the range, where what a stage can deliver is bounded by something no transfer function contains.

Why this is not the answer to a mixer’s channel separation #

The figures above are worst-case numbers taken above the loop, and a real mixer works below it, so it is worth saying what a design actually gets.

At a kilohertz on a megahertz part the node is ten ohms whatever the count, so the per-pair leakage is −40 dB and the total into thirty-one other channels is −30 dB. At ten kilohertz the node is a hundred ohms and the two figures are −20 and −10. Those are audible numbers on a console, they are unchanged by building a bigger console, and the only thing in the circuit that moves them is the amplifier’s gain–bandwidth — which is the conclusion the essay before it reached for two inputs and which survives the count exactly.

What does not survive is the rule of thumb about the thousandth. That essay found the node passing one per cent of the input resistor at a thousandth of the gain–bandwidth, and the frequency is unchanged here because the rise is unchanged. But a thirty-two-input node’s ceiling is 3.1 per cent of one input resistor, so the node never passes ten per cent of $R_\mathrm{in}$ at any frequency whatever. Past a certain channel count the node’s worst case is set by the resistors and not by the amplifier, and the crossover is at the count where $R_\mathrm{in}/N \parallel R_f$ falls below whatever fraction of $R_\mathrm{in}$ the design cares about.

That is a boundary worth stating because it inverts which component to change. Below the crossover a faster amplifier moves everything; above it a faster amplifier moves the frequency at which the node reaches its ceiling and does not move the ceiling, so the remedy is a lower input resistance — which costs the sources that drive it, and is the reason a passive summing network followed by one gain stage sometimes measures better than an active summing junction.

What a bench would see, and what it would blame #

The measurement is a curve of impedance against frequency, and nobody puts a network analyser on a summing junction. What a mixer’s designer sees is a channel-separation figure, so it is worth translating.

Sum thirty-two channels into one node on a megahertz part and measure the separation between two of them. At a kilohertz it is forty decibels, at ten kilohertz twenty, and the slope is twenty decibels a decade with no ceiling until the node’s own — which at this count arrives at about three kilohertz, so the slope flattens and the separation stops degrading at −30 dB for the pair and near nothing for the total.

Now take the same measurement on a sixteen-channel console built from the same part. The separation between two channels is the same number at every frequency below three kilohertz and one channel better above it. A designer comparing the two consoles, having doubled the channel count and found the crosstalk unchanged, would reasonably conclude that the crosstalk is not in the summing junction at all — and would look for it in the wiring, where it is not.

That is the diagnostic value of the cancellation, and it is worth stating as a test. A summing-junction crosstalk scales with the amplifier and not with the channel count. Halve the part’s gain–bandwidth and every figure moves a decade; double the channel count and nothing moves. Two experiments, and between them they say whether a console’s channel separation is the junction’s or the layout’s — which is the same shape of two-experiment test the millivolts in the wire uses on a shared return, where the reading scales with the interfering current and not with the signal.

The one input that is not like the others #

Every channel in the netlist above is a kilohm, and that symmetry is doing quiet work. Break it and the cancellation survives in a form that is more useful than the symmetric one.

Suppose one channel is a kilohm and thirty-one are ten kilohms — a microphone input among line inputs, which is the ordinary arrangement. The node’s total input conductance is set almost entirely by the one low-impedance channel, so the ceiling is close to that channel’s own value in parallel with $R_f$, and the node behaves as though it had about four channels rather than thirty-two.

The leakage arithmetic then stops being symmetric, and it is worth stating which way. A signal on the kilohm channel drives ten times the current into the node that a signal of the same size on a ten-kilohm channel does, and what leaks out of the node into a channel is divided by that channel’s resistor. So the loud channel leaks into the quiet ones a hundred times more strongly than they leak into it — ten in the current it drives and ten again in the resistor it leaks back through. Crosstalk between two channels of a summing junction is not reciprocal, and the ratio is the square of the ratio of their resistors.

That is a result worth having because it names the failure it produces. A mixer whose separation measures well in one direction and badly in the other is not faulty; it has channels of different impedance, and the measurement was taken the easy way round. The non-reciprocity is a property of the network rather than of the amplifier, and it is one of the few places in this collection where a two-port’s two transmissions differ — the networks field’s own reciprocity result is about a network with no sources in it, and a summing junction has a source on every input.

The sensible design response is the one the arithmetic implies rather than a rule: bring every channel to the same impedance before the node, with the gain taken ahead of it. Which is what a mixer’s channel strip is, and this is the reason for it that is usually given as “level matching”.

The capacitance the count also brings #

One thing scales with the channel count that this measurement does not contain, and on a real mixer it is usually the larger effect.

Every input brings a resistor, and every resistor brings a track, a connector pin and whatever is across it. So the capacitance on the summing node grows with the count as surely as the conductance does — a few picofarads a channel is ordinary, which at thirty-two channels is a hundred picofarads on a node whose ceiling is thirty-one ohms.

The gain the loop closes against is what that costs, and it is not a crosstalk figure. Nine picofarads at the summing junction takes the phase margin of that essay’s stage from ninety degrees to forty-five; a hundred puts twelve decibels of peaking on a response whose designed gain has not moved by three parts in a million. So the thing a big summing junction is actually short of is not stiffness — it has more of that than a small one — but phase margin, and the mechanism is a capacitance the count brings along with the conductance.

The two scale together, which makes the arithmetic worth one line. The zero the capacitance puts in the noise gain is at $1/(2\pi C_\mathrm{node} \cdot (R_\mathrm{in}/N \parallel R_f))$, and both factors move with the count: $C$ up as $N$, the resistance down as $1/N$. The product is constant, so the noise gain’s zero sits at the same frequency whatever the channel count — which is a genuinely counterintuitive result and the same cancellation as the one in this essay’s middle section, arriving in a different quantity.

What is not here #

The inputs are resistors to ground. A real channel is a source with an impedance of its own in series with a resistor, and a source that is not stiff adds to $R_\mathrm{in}$ for that channel. The cancellation above survives it — everything scales with the same combination — but the numbers do not, and a channel driven from a high impedance is both quieter about leaking and more susceptible to being leaked into.

There is no capacitance in the netlist. The section above says what it would do and does not measure it, deliberately, because the cancellation this essay is about is a statement about conductances and putting a reactance on the node makes it a statement about two frequencies at once.

And nothing here is about the noise. A summing junction’s noise gain is one plus the feedback impedance over the parallel combination of everything at the node, so it rises with the channel count — thirty-two inputs of a kilohm each give a noise gain of $1 + R_f N/R_\mathrm{in}$, which is 321 rather than 11. The amplifier’s own input noise is amplified by that, and every one of the thirty-two resistors contributes its own. That is the real price of a big summing junction and it is a different measurement from this one.

Still open: the capacitance the channels bring, and the count at which the resistors win #

The node’s capacitance, swept with the count. Every claim in the last two sections about the capacitance is arithmetic rather than a solve. Putting $N$ channels’ worth of stray on the node and sweeping the count would say whether the noise gain’s zero really does stand still, and what the phase margin does while it is standing still — because the margin depends on where the zero is relative to the crossover, and the crossover moves with the noise gain.

The count at which the ceiling stops mattering. The node’s worst case is $\min$ of what the amplifier allows and what the resistors allow, and the two cross at a channel count that depends on the feedback resistor and the gain–bandwidth. Above that count a faster part buys nothing, which is an unusual thing to be able to say about an analogue circuit, and locating the count would turn it into a design rule.

And the same node with a current output. Every input here drives a resistor into a voltage-mode node. A photodiode array or a current-output converter drives current straight in, which removes $R_\mathrm{in}$ from both the leakage arithmetic and the loop gain — so the cancellation this essay rests on has nothing to cancel, and the count should arrive in the answer undiluted. Where the trouble is at the input is the one-channel case of that circuit, and the many-channel case is not written.

The gate #

The ceiling is held against the parallel combination at every count, to two parts in a thousand — six counts, six closed forms — because that is the claim that the amplifier has stopped participating and it is the one that would fail first if the model were wrong.

The cancellation is stated as a spread rather than as a value. The node’s impedance at a kilohertz is required to vary by less than six per cent across every channel count from one to thirty-two, and at direct current by less than one per cent. Requiring the value at one count would say nothing; the claim is that the count is absent.

The ceiling is required to fall by less than the count, as well as to fall. Both halves are the finding: the first version required only that it fell, which is true and is what anybody would guess, and leaves the feedback resistor’s contribution unmeasured.

And the total leakage is required to approach unity, between 0.9 and 1 at thirty-two inputs, which is the essay’s headline and is the one number a reader would not predict from a curve that is falling.