A star is half a millimetre long

The millivolts in the wire names its repair and states it in one sentence: the repair is called a star ground, and what it does is make the shared length zero — not smaller, zero — by giving each return its own path to the reference point. Nothing else on this slider reaches zero.

That sentence is the one thing in the essay with no number attached, and it is the one thing in it that is not true. A star ground is a pad, and the pad reaches the plane through a via, and a via is a resistance and an inductance like every other piece of copper. Every return still shares something; what the star changed is how much, and — more usefully — what the error is a function of.

The via, measured as a length #

The honest way to say how good a star ground is, on a site that has already put a length of track in the netlist, is to say what the via is worth in millimetres of it.

One-ounce copper a millimetre wide is half a milliohm and about a nanohenry per millimetre — the numbers the essay before it uses. A plated via through an ordinary board, with its two pads, is about a quarter of a milliohm and about 0.35 nanohenries. So:

A star ground is a half-millimetre daisy chain, and that is the whole of the improvement it offers over a ten-millimetre run: a factor of twenty, not a factor of infinity.

The two lengths in that table are not the same length, and the difference is the one thing about a via that is not simply “a shorter run”. A track’s resistance and inductance both scale with its length, so their ratio does not and its corner is 79.6 kilohertz whatever it is cut to — which is that essay’s second finding and the reason shortening a run moves the whole curve down and the corner not at all. A via is a different geometry: a short fat cylinder rather than a long thin strip, so its resistance is worth more millimetres than its inductance is, and its corner sits at 114 kilohertz instead. A star ground is therefore slightly better than a half-millimetre track above 114 kilohertz and slightly worse below it, and the two numbers being different is the only way to say so.

Two laws, not two numbers #

The interesting part is not the factor of twenty. It is that a star and a chain have different dependences on the number of circuits, so the comparison is a pair of laws rather than a ratio.

A star’s error is linear in the count. Every circuit’s return crosses the same via, so a sensor referenced at the star sees the other $N-1$ currents through one impedance: $(N-1)\,I\,Z_\mathrm{via}$.

A chain’s error is quadratic. Each circuit adds its own hop, and the current in hop $j$ is that of every circuit from $j$ outwards — so the error at the far end is $I\,Z_\mathrm{hop}\sum_{j=1}^{N}(N-j+1) = I\,Z_\mathrm{hop}\,N(N+1)/2$. Adding a board to a chain does not add one board’s worth of error; it adds a board’s worth to every hop behind it.

So the advantage grows with the count rather than staying put, and the reason is a power rather than a constant. Fitted over those five counts the star’s error goes as $N^{1.22}$ and the chain’s as $N^{1.87}$, approaching the exact 1 and 2 as the count grows — the fits differ from the limits because $(N-1)$ and $N(N+1)/2$ are not pure powers, which is worth saying because it is the reason the closed forms are required and the exponents are only reported.

That difference of one power is the practical content. A two-board system with a shared return is an annoyance; a thirty-two-board system with a shared return is unusable, and not because each board is worse but because the arithmetic is quadratic. And the star does not merely reduce the constant: it changes what the designer is buying when they add a board.

The position that only one of them has #

A chain has a property a star has none of, and it is the most useful diagnostic in this essay because no real fault should have it.

In a chain, a circuit’s error depends on where it sits. The error at position $k$ is the sum of the currents in the hops between it and the star, and the hops nearer the star carry more — so the far end is worse than the near one by a factor that grows with the count: 2.5 times at four circuits, 8.5 at sixteen. Moving a board along the chain changes its reading, and nothing about a genuine fault in a board should depend on the order of the wiring.

In a star every circuit crosses the same via, so there is no position at all. Every circuit reads the same error and swapping two of them changes nothing.

That gives a two-minute experiment that distinguishes the two topologies without tracing a single conductor, which matters because on a real system nobody can see where the returns go. Swap two boards’ positions and re-read. If either reading changes, the returns are a chain and the schematic’s single ground symbol is describing something else. If neither does, the returns meet at a point — or at least at one shared impedance, which is what a star is.

The test also has the right failure mode. A chain whose hops happen to be nearly equal still shows the position dependence, because the dependence comes from the currents rather than from the lengths. A star whose via is unusually bad shows none of it, and reads uniformly wrong — which is a different symptom and points at a different repair.

The schematic’s one symbol, and the four topologies it stands for #

A schematic draws a ground symbol several times and requires that every point carrying it is the same node. That essay’s whole argument is that this is false on a board, and it is worth being precise about how many different things the same drawing can mean, because the four differ by orders of magnitude and the drawing distinguishes none of them.

A daisy chain. Returns joined to each other in a line, reaching the reference at one end. Quadratic in the count, position-dependent, and what gets built when nobody decided.

A star. Returns joined at one point, which reaches the reference through that point’s own via. Linear in the count, position-independent, and a half-millimetre chain.

A plane. Returns joined everywhere, with each current choosing its own path — and choosing it differently below and above a hundred kilohertz, which is why a plane is two topologies rather than one.

And separate returns. Each circuit’s return going to the reference on its own conductor, sharing nothing but the reference itself. This is the only one of the four with no shared impedance at all, and it is the one that is usually impossible because the reference is a finite object too.

Every one of those is drawn identically. That is the sharpest form of these essays’ claim about schematics available anywhere in it, and it is sharper than Kirchhoff’s own frequency’s version — there the drawing is missing a variable, the physical size, which a drawing genuinely cannot carry; here the drawing is missing a topology, which a drawing could carry perfectly well and does not, because the convention is to draw a ground symbol rather than a conductor.

Which suggests the practice the essay before it implies and does not state: on any board where a shared return can matter, draw the return conductors. Not as a ground symbol repeated, but as elements — a resistance and an inductance between two named nodes, with a length beside them. The netlist is then complete, every number in these three essays falls out of a solve rather than out of an argument, and the four topologies above become four different drawings, which is what they are.

What the star is not a repair for #

Two things a star ground does not do, and both are in that essay’s own list of distinctions.

It does not reduce a shared flux. The essay before it separates two faults that go by the same name: a shared impedance, which is what this essay is about and which can be computed from a netlist, and a loop of conductor enclosing an area that picks up a changing magnetic field, which cannot. They are fixed by different things, and the star is a fix for the first — the millivolts in the wire makes the separation and notes that only one of the two can be computed from a netlist at all. Worse, it can make the second worse: routing every return to one point produces long returns and therefore large loops, where a return that follows its own signal encloses very little. That exchange is real and this essay does not price it.

And it does not survive a plane. Everything above assumes the returns are tracks going to a point. Above about a hundred kilohertz a return current on a plane does not go where the layout sent it — it runs directly under its own signal, because that is the path of least inductance. Where the current comes back estimates the frequency at which that gathering happens at 106 kilohertz for any track two hundred micrometres above a half-milliohm plane, with neither the length nor the width in it, and the corner that is three decades wide solves the same plane and finds a band rather than a corner — half gathered by 283 kilohertz and nine tenths by 1.42 megahertz.

Which means a star ground drawn on a board with a plane is a star ground below a hundred kilohertz and a description of nothing above it. The currents stop taking the paths the star defined, and the shared impedance becomes whatever the plane’s own geometry makes it — a quantity the layout influences and does not choose. That is not a reason to skip the star; it is the reason the star’s own numbers, and this essay’s, are low-frequency numbers.

The two ends of the slider price the same piece of copper differently, and the difference is not the copper. A star’s error is one hop long whatever the count: every circuit’s return current crosses the via and nothing else shared. A chain’s is $N(N+1)/2$ hops, because each circuit’s current crosses every segment between it and the ground point, and those segments accumulate as the count does. At two circuits that sum is three and the quadratic term contributes a third of it, so what the star buys is close to the ratio of the two impedances and nothing more. At sixteen the sum is a hundred and thirty-six, and the same ratio of impedances is multiplied by a growing number of shared segments.

Which leaves one comparison to make, and it is the one that puts a number on the whole arrangement. A star ground is not a topology a board either has or has not; it is a shared length, and the length is the via’s. Half a millimetre of plated barrel is what every circuit on the star holds in common, so the question of what a star is worth is the question of what the shared run it replaces would have cost — which is a measurement of copper rather than of layout philosophy, and the same figure that opened this field’s account of shared returns answers it.

What a star costs, and the count at which it stops being enough #

A star’s residual is $(N-1)\,I\,Z_\mathrm{via}$, and at thirty-two circuits of a hundred milliamps that is 775 microvolts — eight per cent of a ten-millivolt reading. So a star ground is not a repair that scales indefinitely either; it is a repair that buys a factor of twenty in impedance and one power of the count, and both are finite.

The count at which it stops being enough follows directly. For an error below a stated fraction $\epsilon$ of a signal $v_s$,

$$N - 1 < \frac{\epsilon\,v_s}{I\,Z_\mathrm{via}}$$

which for one per cent of ten millivolts, a hundred milliamps and a quarter-milliohm via is four circuits. Five circuits on one star is already a one-per-cent error, which is a much lower count than the arrangement’s reputation suggests, and the arithmetic is worth having because a star ground is usually adopted as a solution rather than as a quantity. It is the same shape of answer every tooth the same height reaches about an edge rate: a remedy whose worth is a number, quoted as though it were a category.

Two ways past it, and they are the two this field keeps arriving at.

Lower the impedance: several vias in parallel, which divides $Z_\mathrm{via}$ by their number until their own spacing stops them being in parallel, or a pad directly on a plane with no via at all. A factor of four from four vias is available and a factor of a hundred is not.

Or remove the sharing rather than reducing it. Two readings that both contain the term subtract it exactly, and one voltage added to two readings measures that: 74 decibels of rejection, flat at every frequency, limited by a mismatch between two stubs rather than by any shared impedance at all. That arrangement does not care how many circuits are on the star, because the star’s drop is common to both of its readings — which is the sense in which it is the repair this one is not.

What the star’s own return is connected to #

One node in every netlist above is called the reference and treated as an ideal point, and it is worth asking what it actually is, because the answer bounds everything else.

The via goes to a plane, and the plane goes to whatever the system’s reference is — a connector shell, a supply’s return terminal, a chassis stud. Each of those is a piece of copper with a resistance and an inductance, shared by every circuit on the board, and the arithmetic of this essay applies to it unchanged: $(N-1)$ currents through one impedance. So a star ground has a star ground of its own, and the errors add.

That regress is not a rhetorical point; it is where the numbers stop improving. A board’s plane is a milliohm or two between its far corners and a connector pin is five to ten milliohms with a nanohenry or two — which is to say a connector pin is a ten-millimetre track, the very object the essay before it measured and the very object the star was built to avoid. A star ground implemented perfectly on a board, with several vias and a solid plane, reaches a connector and acquires the error it was designed to remove.

Three consequences, and they are the reason the topology question does not end at the board’s edge.

The reference should be where the sensitive measurement is, not where it is convenient. The arithmetic has $(N-1)$ other currents in it, so the circuit that matters should be the one closest to the reference — which in a chain means first in the chain and in a star means nothing at all, and is the one respect in which a chain has a design parameter a star does not.

A high-current path across a connector is the worst case available. A hundred milliamps through five milliohms of pin is that essay’s five hundred microvolts exactly, with the board’s careful star ground sitting in series with it and contributing nothing. That is the failure a design reviews out and a manufacturing decision puts back, because the pin count is decided on cost.

And the whole chain of shared impedances is a sum rather than a maximum. Via, plane, pin, cable: each contributes $(N-1)\,I\,Z$ and they add. So the useful measurement is never of one of them; it is of the total from the sensitive node to the reference, which is one impedance a bench can read with four terminals and which no schematic shows. Two terminals measure the leads as well is the arrangement for reading it, and this is the quantity it should be pointed at.

Still open: the vias in parallel, and the loop the star makes #

Several vias, and where they stop being parallel. Dividing a via’s impedance by putting four of them side by side is the obvious next step and it has a limit nobody quotes: two vias a millimetre apart are mutually coupled, so their inductances do not divide by two. Measuring the mutual term would say how many vias are worth using and at what spacing — which is a number a stackup decision needs and which the magnetics field’s own tools for coupled inductors could supply.

The loop the star creates, priced against the impedance it removes. The exchange above is stated and not measured, and it is the one that decides whether a star is the right arrangement at all: a star’s returns are longer than a signal-following return’s, so its enclosed area is larger and its pickup worse. The corner that is three decades wide is where the area a return actually encloses is solved rather than assumed, and it is not the area the layout drew. The two mechanisms are computable on one geometry — a shared impedance from a netlist and an induced voltage from an area and a $dB/dt$ — and the crossing between them would turn “they are different faults” into a layout rule.

And the via’s two lengths, measured rather than taken. The quarter-milliohm and 0.35 nanohenries above are ordinary figures for an ordinary board and they are inputs to this essay rather than results of it. A via’s inductance is set by the board’s thickness and its barrel diameter, and the ratio of its two lengths — which is what puts its corner at 114 kilohertz rather than 79.6 — is a function of that geometry. Computing it would make the corner a property of a stackup, which is where every other boundary in this essay’s field has ended up.

What is checked #

The via’s worth as a length is required in both quantities, and both are required to be under a millimetre of the track the rest of the figure is drawn in. That is the claim the essay’s title makes and the one that would fail first if a via were a worse conductor than it is.

The two lengths are required to differ, by more than ten per cent, because a via whose two lengths were equal would have a track’s corner and there would be nothing to say about it. The corners are then reported rather than required, since they follow from the lengths.

Both growth laws are stated as closed forms at every count — $(N-1)$ for the star and $N(N+1)/2$ for the chain — to a part in $10^{12}$, and the ratio between them is required to increase at every step. The fitted exponents come out 1.22 and 1.87 rather than 1 and 2, so requiring the exponents against their limits failed on a pair of expressions that are exactly right: a bound calibrated on a limit rather than on the range, which is the defect these notes records most often.

The chain model itself had to be repaired before any of that meant anything. The first version divided a fixed ten millimetres among $N$ circuits, which makes the hop shorter as the count grows and the chain therefore better with more circuits on it — so the star’s advantage shrank with the count and the check refused it at eight. That was the model failing rather than the star: adding a board to a chain adds wire, and a comparison whose baseline improves as the problem grows is a comparison of nothing.

And the position dependence is required for the chain and its absence for the star, because the two-minute experiment at the middle of this essay rests on the pair and either alone would not support it.