The stage that is wrong is the far one

The divider, and the thing it does not know about established what a divider’s ratio is a statement about: a divider with nothing connected to it. Connect anything at all and what decides the answer is the quantity the ratio was built to discard — the magnitude of the two resistances — so two dividers of identical ratio give six volts and one volt into the same load.

That essay connects a load. This one connects a second divider, which is the commonest load a divider ever has, and the interesting thing is not that the answer is wrong. It is where the answer is wrong.

The number, and the decomposition that is exact #

Three identical dividers, each a pair of ten-kilohm resistors, each halving. The product of their ratios is $0.5^3 = 0.125$.

The solved answer is 0.076923, which is 38.5 per cent low.

Thirty-eight per cent is a large error for an arrangement that looks like three copies of something right, and the size of it is worth a sentence before the location. Each stage is loaded by the input resistance of the next, which is ten kilohms in series with the ten kilohms of the one after it in parallel with its own lower resistor — a resistance comparable with the divider’s own, so the loading is not a perturbation. It is half the circuit.

The useful part is the decomposition. Taking each stage’s own ratio in place — the voltage at its tap divided by the voltage at its input, both read off one solve of the whole chain — gives

and the product of those three is $0.3846 \times 0.4000 \times 0.5000 = 0.076923$, which is the answer to the last digit. That is what makes it a decomposition rather than an estimate: the per-stage ratios multiply exactly, because each one is a genuine ratio of two node voltages on the same solve.

The last stage is exactly right and the first is not #

Read the table again. The third divider is at 0.5000 exactly — not to within a per cent, exactly, because nothing is connected to its output and a divider with nothing connected to it is its own ratio. That is that essay’s statement, unchanged.

The second is at 0.4000, loaded by the third’s input resistance of twenty kilohms. The first is at 0.3846, loaded by the second’s input resistance of 16.667 kΩ — which is itself smaller than twenty because the second is loaded in turn.

So the error accumulates from the far end and the stage furthest from the output is the worst one. That is the finding, and its consequence is practical: somebody debugging the chain with a probe starts at the output, where everything is exactly as designed, and works backwards. Each successive node is worse than the last, and the fault is at the point where the probe was least likely to go first.

It is also why the fault is usually misdiagnosed as an offset or a gain error in whatever is driving the chain. The first stage is 23 per cent low on its own, before any of the others have contributed, and a measurement at its tap against a calculation at its tap shows exactly the discrepancy a bad source would show.

The steepest part of the curve is at the bottom, where a single doubling is worth more than the two decades after it.

What the chain looks like from each end #

The input resistance and the output resistance of the chain are the two quantities the loading is built out of, and computing them once makes the whole table above readable.

Working backwards from the output, each stage’s input resistance is its upper resistor plus its lower resistor in parallel with whatever the next stage presents:

Both columns converge and both limits are the same number in different clothes. Reading the input resistances from the output backwards — twenty kilohms, then 16.67, then 16.25 — each stage adds less than the last, because the parallel combination is dominated by the ten-kilohm lower resistor rather than by what follows it. A chain of ten identical stages presents essentially 16.18 kΩ, and the limit is the positive root of $R = 10 + (10 \parallel R)$ in kilohms, which comes out at $5(1+\sqrt5)$ — ten times the golden ratio. The tap resistances converge downstream instead, on $10(\sqrt5-1)/2 = 6.180$ kΩ, which is ten times the golden ratio less one: the same irrational arriving from the other end of the same recursion, because a continued fraction of equal terms has nowhere else to go.

That convergence is why the error stops accumulating. Each stage’s ratio approaches $(10\parallel 16.18)/(10 + 10\parallel 16.18) = 0.3820$ from above, so a chain of many identical dividers has a ratio of $0.3820^n$ rather than $0.5^n$, and the discrepancy is a constant factor per stage rather than a growing one. Three stages give $0.5^3/0.382^3 = 2.24$ — and the measured 0.125 over 0.076923 is 1.625, which is smaller because the last two stages have not yet reached the limit.

So the chain has a fixed point and it is reached quickly. That is a more useful thing to know than any single number: a long chain of identical dividers behaves like a chain of identical 0.382 dividers, not like a chain of halves that gets progressively worse, and a design that measured the first three stages and extrapolated the trend would predict a collapse that does not happen.

The staircase, and the exponent it buys #

The repair everybody reaches for first is a buffer between the stages, and it works completely: a buffered chain gives the product of the ratios exactly, at every impedance, which the figure claims as a refusal so that the departure being measured is known to belong to the loading and not to the arithmetic.

The repair that costs nothing is a staircase. Make each stage’s impedance $k$ times the one before, so that each is a light load on its predecessor:

Fitted over the upper part of the range the exponent is −0.993: the departure falls in direct proportion to the staircase. That is the answer to “how much impedance scaling is enough”, and it is a one-line rule — a decade of staircase per stage buys a factor of ten in accuracy, and a chain that has to be right to a per cent needs about fifty.

Two things about that exponent are worth noticing. It is a first power rather than a square, which means scaling is a slow repair: getting from 38.5 per cent to 0.1 requires a staircase of about four hundred, which over three stages is a span from ten kilohms to 1.6 gigohms and is not buildable. And it is a fitted exponent over the range where it holds, not a law — at a staircase of one the loading is not a perturbation at all and the linear behaviour has not started.

Two more settings of the staircase make the convergence visible, since a table of ratios does not.

What the staircase costs at the other end #

The staircase is free in parts and it is not free.

The chain’s output impedance rises with it. At ×1 the last stage is a pair of ten-kilohm resistors and its Thévenin resistance is five kilohms; at ×100 it is a pair of a hundred megohms and five megohms of source. Whatever the chain drives has to be as light a load on it as the stages are on each other, which pushes the problem one stage further along rather than solving it — the resistor that is not made of the resistors is the quantity that decides that, and it grows exactly as the staircase does.

And a megohm-class node is not a quiet node. Its Johnson noise goes as the square root of its resistance, so a hundredfold staircase costs a factor of ten in noise at the top stage. Its bandwidth falls with the stray capacitance it cannot avoid: five megohms into two picofarads is a corner at 15.9 kHz, which for a divider in a measurement path is a serious limitation and is invisible at direct current.

So the staircase converts a loading error into a noise and bandwidth problem. That is usually a good trade — an error of 38.5 per cent is not a trade at all — but it has a ceiling, and the ceiling is somewhere near two decades a stage for an ordinary board.

The honest arrangement for a chain that must be accurate is the buffer, and the arithmetic above is what says how much accuracy a buffer is worth: everything between the solved answer and the product of the ratios, which at identical stages is a third of the signal.

The three-stage chain is the worst case, not the start of one #

The convergence above changes what a long chain looks like, and it is worth stating separately because the intuition runs the other way.

A reader who has seen three stages lose 38.5 per cent expects six to lose far more, and in absolute terms they do — $0.382^6$ against $0.5^6$ is a factor of five rather than 1.625. What does not happen is the per-stage damage getting worse. The first stage of a three-chain is at 0.3846 and the first stage of a twenty-chain is at 0.3820, and the difference between those two is seven parts in a thousand. Every stage of a long chain is at the fixed point except the last two.

So the error is not cumulative in the sense that would make a long chain unusable. It is a constant multiplier per stage, and a chain of identical dividers is simply a chain of dividers with a different ratio from the one drawn on the schematic — one that can be designed with, if it is known.

Which gives the third possible repair, after the buffer and the staircase, and it is free: design for the loaded ratio. A chain that is meant to divide by eight and is built from three halves gives 0.0769; built from three stages whose unloaded ratios are 0.6, chosen so that the loaded product comes out at 0.125, it gives the right answer with the same six resistors and no extra parts. The resistors are not the ones on the schematic and the chain is exactly as accurate as their tolerances allow.

The catch is that the correction depends on the load at the far end, so a chain designed this way is right for one load and wrong for another — which is that essay’s headline arriving one level up, and is why the buffer keeps being fitted.

Why this is not the same thing as loading #

The essay before it measured a divider loaded by one resistance and found the answer depends on the magnitude. This is the same mechanism, and the reason it deserves an essay of its own is that it composes, and composition has a direction.

A single loaded divider has one error. A chain has $n-1$ of them, they multiply rather than add, and each one depends on every stage after it. That last clause is the whole of the asymmetry. The loading on stage $k$ is the input resistance of the chain from $k+1$ onwards, and that input resistance depends on everything downstream — so the error at any stage is a function of the rest of the chain and not of its neighbour alone.

Working it out from the output backwards is therefore the only way that terminates. Start at the far end, where the input resistance is $R_1 + R_2$ with nothing to load it; work back one stage at a time, each stage’s input resistance being its upper resistor plus its lower in parallel with what it sees. That recursion is exactly the one a chain of loaded sections obeys, which is why a chain of sections is not a cascade is the same statement about filters: a cascade of stages that load each other is not the product of the stages, and the difference is not a small correction.

And the far end of the axis, where the accuracy is bought and something else has become the limit.

What three equal dividers do not represent #

That three equal dividers are a sensible design. They are the clearest possible demonstration and nobody builds them. What is built is a chain in which the loading was not thought about at all, and the numbers above say what that costs.

That the error is always low. It is low for a chain of voltage dividers, because every stage’s load pulls its tap down. A chain in which the loads are on the upper resistor, or one with active stages of gain greater than one between, behaves differently, and the sign follows from the topology rather than from the principle.

That the per-stage decomposition is unique. It is the decomposition into ratios of node voltages on one solve, and those multiply to the answer exactly by construction. A different decomposition — into ideal ratios times per-stage correction factors, say — is equally valid and attributes the error differently. The one used here is chosen because each factor is a measurable voltage ratio.

That a buffer is free. It has an offset, a bias current, a bandwidth and a noise, all of which enter where the loading used to be. What the arithmetic here says is how much error a buffer is being asked to be better than, which is the only way to decide whether one is worth fitting.

A decomposition that multiplies exactly, and the buffer that refuses it #

The per-stage ratios are checked to multiply to the answer to a part in $10^{12}$, so the decomposition is exact rather than approximate.

The last stage is checked to be exactly its own ratio — 0.5000 to twelve digits — at every staircase, because that is the statement the whole argument about direction rests on.

The monotone ordering is checked at the identical-stage setting: the first ratio below the second below the third, which is the claim that the error is at the front.

The staircase exponent is fitted over a decade and a half and checked against −1 within five per cent, rather than read off two points.

And the buffered chain is refused: with a buffer between the stages the answer is the product of the ratios exactly, at every impedance, which is what says the departure measured here belongs to the loading.

An error whose location is the finding #

The number in this essay — 38.5 per cent — is not the useful part. Anyone who thought about the circuit for a minute would expect a large error and could compute it.

The useful part is that the error is at the far end from the output, that the output stage is exactly right, and that the two facts together make the defect nearly invisible to the ordinary debugging procedure. A measurement at the output shows a number that is wrong; a measurement at the last tap shows a stage behaving exactly as designed; and the natural conclusion is that the fault is upstream of the whole chain rather than distributed through it with a gradient.

That shape — a correct answer at the point of observation, produced by a circuit that is wrong everywhere else — recurs here. The node that is at ground for a while is a summing junction that reads zero whether the loop is working or not. The answer that is perfect and absurd is a solve that returns a clean number for a network that has no answer. In each case the instrument is pointed at the one place the defect does not show.

The general lesson is about where to measure rather than what to compute: in a chain whose stages load each other, the informative node is the one furthest from the load, and the ordinary instinct sends a probe to the other end.

Still open: the chain measured backwards, the tolerance that compounds, and the compensated divider #

The recursion as the design tool. Working the input resistance back from the far end gives every stage’s ratio in one pass, and it is the same recursion a chain of loaded sections obeys. Building it explicitly and comparing against the solve would turn the decomposition above into a design procedure rather than a diagnosis — and would say how many stages a chain can have before the recursion’s own conditioning becomes the limit, which is the question the matrix that is ill, and the answer that is not asks of a different network.

What a tolerance does in a chain. Each stage’s ratio depends on every stage after it, so a tolerance on the last stage’s resistors moves every ratio upstream of it. The sensitivity of the total to each resistor is therefore not the same for all six parts, and the apparatus every derivative, and the one that is zero uses would give all six in two solves — and would say whether the first stage’s parts or the last’s deserve the tight tolerance.

And the same chain at a frequency. A divider chain with stray capacitance at every node is a different circuit above some frequency, and the compensation that fixes one divider — the capacitor that was right once — has to be applied at every stage with the loading included. Whether a chain can be compensated stage by stage, or whether the compensation of each depends on the rest as the ratios do, is the natural next question and is one solve away.