The small leg of a large ratio

What a ratio does not see built a divider from four resistors that share part of their error and found that the shared part cancels: four 1% parts sharing ninety per cent of their variance made a ratio that spread 0.091 per cent, six times better than independent parts would give. Every divider in it was tapped at the middle. Two resistors above the tap, two below, every part the same value, every part carrying the same current and dissipating the same power. That symmetry did a great deal of quiet work, and the essay said at its close that a divider of ten to one was the next thing to measure, because the circuits that use resistor ratios most are almost never halving ones.

A non-inverting amplifier with a gain of eleven sets that gain with a ten-to-one divider in its feedback path. A gain of a hundred and one uses a hundred to one. A reference divided down to an ADC’s input, a current-sense amplifier’s gain network, a bridge with a ratio arm — nearly every precision ratio in a circuit is far from a half. Whether the cancellation the symmetric divider showed survives the asymmetry, and what the asymmetry costs that the symmetric divider never paid, is the question here. The answer has three parts: one about leverage, one about area, and one about heat.

What the parts do not share reaches the output nearly whole #

Take two resistors, the top one N times the bottom, so the output ratio is $k = 1/(N + 1)$. Let each part carry a fractional error, and let those errors share a fraction ρ of their variance, built exactly as the symmetric essay built them: a common draw scaled by $\sqrt{\rho}$ and a draw of the part’s own scaled by $\sqrt{1 - \rho}$, both normalised to the 0.577 per cent standard deviation of a uniform 1% part.

The first-order sensitivity of the ratio is the whole of the arithmetic. A fractional change in the bottom resistor moves $k$ by $N/(N + 1)$ of itself, and a fractional change in the top moves it by the same amount in the opposite direction. So the output ratio moves by $N/(N + 1)$ times the difference between the two parts’ errors. The common part is identical in both and its difference is zero, whatever N is. The own parts are independent, and their difference has $\sqrt{2}$ times the spread of either. So the output spreads as

$$\sigma_k = \frac{N}{N+1}\,\sqrt{2}\,\sigma\,\sqrt{1-\rho}.$$

At a half, N = 1, the leverage is one half. At ten to one it is 0.909. At a hundred to one, 0.990. The leverage is what the halving divider was quietly getting for free: each part in it moves the ratio by only half its own error, and the asymmetric divider gives that half back.

The population agrees with the law at every one of the seven ratios, to within its own sampling error, and the numbers read straight off it: 0.129 per cent at a half, 0.236 at ten to one, 0.258 at a hundred to one. Ten to one is 1.83 times worse than a half, from the same two parts with the same correlation, and nothing about the parts has changed. All of the difference is the leverage. Going from ten to one to a hundred to one costs only another nine per cent, because the leverage is already nearly one at ten — the expensive step is the first one away from the middle, and it is mostly taken by the time N reaches five, where the factor is already 0.833 against the half’s 0.5.

The slider runs through the four shared fractions the symmetric essay used. At ρ = 0, two independent parts, the output spreads 0.409 per cent at a half and 0.753 at ten to one. At ρ = 0.5 the figures are 0.292 and 0.523. The ratio between the two columns is 1.84 and 1.79 — the same 1.82 the law gives, to within the populations’ sampling error — because the correlation multiplies the whole curve by $\sqrt{1 - \rho}$ and the leverage shapes it in N, and the two factors never meet.

Correlation helps an unequal ratio exactly as much as an equal one #

That separation is the first answer to the question the symmetric essay left open, and it is worth stating plainly because the opposite is easy to believe. The shared error cancels completely at every ratio — not partly, not in proportion to how symmetric the divider is. A resistor array whose parts share 99 per cent of their error improves a ten-to-one divider by exactly the factor $\sqrt{1 - 0.99}$ = 0.1 by which it improves a halving one.

The array divider spreads 0.041 per cent at a half and 0.075 at ten to one, and the ratio of those is 1.83, the same as before. So the correct statement of what correlation does is narrow: it scales the error that reaches the output, and it says nothing about how much of that error reaches it. The leverage and the correlation are independent levers. A designer who has chosen an array has used the second one fully, and the first is still sitting there.

There is a subtlety in “share 99 per cent of their error” that the unit construction below makes concrete. The parts in this model are two values, the top N times the bottom, drawn from a common error. That is a fair description of two resistors from one thin-film process of one geometry. It is a poor description of a 10 kΩ and a 1 kΩ element in one package, which are usually made with different numbers of squares of the same film, or with the same squares and different film thicknesses, and so carry different fractions of whatever error the process shares. The common part cancels in a ratio only if it moves both legs by the same fraction. A process shift that moves a long narrow element by more than a short wide one is a shared cause with an unshared effect, and it reaches the output with the full leverage. The reliable way to make two legs share their error by the same fraction is to make them of the same thing.

Built from equal units, a large ratio costs area #

Monolithic dividers and precision resistor networks do exactly that. Every element is one unit of one geometry, and a ratio of N to one is built as $m$ units in series below the tap and $N\cdot m$ above it. The shared part of every unit’s error is then genuinely common, cancels exactly, and drops out of the problem. What remains is each unit’s own error, independent, with some standard deviation $\sigma _u$. Here it is 500 parts per million, a matched array’s own-element spread after the shared error has gone.

Now the bottom leg’s own error is the average of $m$ units, with spread $\sigma _u/\sqrt{m}$, and the top’s is the average of $N\cdot m$, with spread $\sigma _u/\sqrt{Nm}$. The output ratio carries their difference at the leverage $N/(N + 1)$. Put the total number of units $M = (N + 1)m$ in and the algebra collapses to something short:

$$\sigma_k = \sigma_u\,\sqrt{\frac{N}{M}}.$$

At a fixed number of units, a ratio of N to one spreads $\sqrt{N}$ times a halving divider. The leverage is still in there; so is the fact that the top leg, being N times longer, averages its own error down. The bottom leg does not.

The lines are parallel on a log-log plot, because every ratio’s spread falls as one over the root of the units spent on it, and they are offset by $\sqrt{N}$. A halving divider of two units spreads 353 ppm. A ten-to-one divider cannot be made with fewer than eleven units, and at eleven it spreads 475. It needs 44 units to come down to 239 and 176 to reach 120. Matching a halving divider of M units needs a ten-to-one divider of ten times as many, which is the $\sqrt{10}$ in the law read the other way. At thirty to one the minimum chain is 31 units, it spreads 488 ppm, and it takes 496 units to get to 122.

Read the law’s derivation once more and the answer to “where does the cost come from” is visible. Of the two legs, the top has N times the units of the bottom. Its own error is averaged over all of them and contributes $1/(Nm)$ to the variance; the bottom’s contributes $1/m$. At ten to one the bottom leg is ten-elevenths of the variance. The long leg is cheap precision; the short leg is the whole problem.

The small leg is what is being bought #

That says where the units should go. If the bottom leg sets the spread, the way to improve a large ratio is to give the bottom leg more units without changing its value — which is what a series-parallel block does. Four units as two strings of two in parallel are worth one unit, and their average error has half a single unit’s spread. Nine as three by three, a third. The top leg is then built of N such blocks in series so that the ratio is unchanged and every element is still the same unit.

The blocks are combined exactly here, strings summed and parallel paths taken as reciprocals, so nothing in the population leans on the first-order law it is being compared with. It agrees with it all the same: 475 ppm with a single unit below the tap, 239 with a 2 × 2 block, 159 with 3 × 3, 120 with 4 × 4. The halving dividers beside them make the price explicit. A halving divider spreads 353 ppm on two units and 175 on eight. A ten-to-one divider needs 44 units to beat the first of those and 176 to come near the second.

So the design rule for an unequal ratio from an array is the opposite of the instinct a schematic gives. A schematic of a gain-of-eleven stage draws a large feedback resistor and a small gain resistor, and the large one looks like the important part, since it is where most of the voltage drops and most of the power goes. For matching, the small one is the important part. It has the fewest units, and the ratio’s error is mostly its error.

That is also why precision networks sold for fixed gains are laid out as they are. A network offering ratios of one, ten and a hundred to one does not use one unit for the smallest element; it uses a block of many, and lets the top elements be long chains of the same blocks. The data sheet quotes one ratio tolerance across all the ratios, and it can only do so because the small legs were bought down to match.

The heat an unequal ratio makes for itself #

The symmetric essay’s temperature argument was about the ambient: every part in the divider shared one temperature, so only the coefficient mismatch reached the ratio. A halving divider keeps that property under current too. Its two halves carry one current and have one resistance, so they dissipate one power and rise by one temperature. An unequal divider does not. Its two parts carry one current through different resistances, so the top dissipates N times the bottom’s power, and with any thermal resistance at all the top runs hotter.

That is a temperature error a matched temperature coefficient cannot remove, because the two parts are not at one temperature. It is also an error the circuit makes for itself, at its own signal, and so it has no ambient to calibrate against. The pulse that ends before the heat measured a shunt’s self-heating against a pulse; this is the same physics arriving through the ratio rather than through one value.

Take the ten-to-one divider at 11 kΩ in all, a 10 kΩ part over a 1 kΩ one, both of 25 ppm/K and both with a thermal resistance of 250 K/W to the board, about what a small surface-mount chip has. At 50 V across it the current is 4.5 mA, the top dissipates 207 mW and rises 51.7 K, the bottom dissipates 20.7 mW and rises 5.2 K. The ratio moves by the leverage times the coefficient times the difference in rise: 0.909 × 25 ppm/K × 46.5 K, or 1,055 ppm, and the exact computation agrees with that first-order estimate to two parts in a thousand. The two parts in this calculation have identical coefficients. Nothing about them is mismatched. The ratio moves by a tenth of a per cent anyway, and it moves with the square of the voltage across it.

Built instead from eleven 1 kΩ units in series, every unit carries the same current through the same resistance and dissipates the same 20.7 mW. They all rise 5.2 K together, and the ratio moves only by however much their coefficients differ. With the units’ coefficients within ±5 ppm/K of each other the worst of two hundred builds moves 29 ppm — thirty-six times less than the perfectly matched pair. The unit construction that bought the matching also bought the heat, because spreading a ratio over equal units spreads its power equally too.

A halving divider heats evenly #

The comparison at the other end of the ratio scale is the check on that reasoning. A halving divider of two equal parts has no unequal power to offer, so with identical coefficients its ratio cannot move at all, at any voltage.

At 11 kΩ in all, the same total as the ten-to-one divider above, each 5.5 kΩ half dissipates 114 mW at 50 V and rises 28.4 K. With identical coefficients the solid line sits on zero, as it must. With coefficients within ±5 ppm/K the ratio moves by half their difference times that 28.4 K rise, and the worst of two hundred builds moves 130 ppm.

The comparison between the two figures is instructive in a way neither is alone. The halving divider’s coefficient-mismatch shift, 130 ppm, is four and a half times the eleven-unit chain’s 29, although the unit chain is the asymmetric one. Each half of the halving divider is a single part carrying half the power; each unit of the chain carries an eleventh. The temperature rise that the coefficient mismatch multiplies is set by power per part, and the unit chain has more parts to spread the power over. Self-heating in a ratio is therefore two separate errors with two separate cures. The unequal-power error, which no coefficient match removes, is cured by making the legs of equal units. The mismatch error, which survives any construction, is cured by lowering the rise per part — more parts, a lower thermal resistance, or a shared substrate that puts them all at one temperature.

A gain of a hundred, worked #

A non-inverting stage with a gain of 101 uses a hundred-to-one feedback divider, and its gain error is the divider’s ratio error with the sign reversed. From two independent 1% parts it spreads 0.80 per cent. From two parts sharing 99 per cent of their error, 0.081. From a unit array of 500 ppm elements, a hundred-to-one divider needs at least 101 units and spreads 497 ppm; a 2 × 2 block below the tap takes that to 249 ppm at 404 units.

Its self-heating depends on the impedance chosen. With 100 kΩ over 1 kΩ and 10 V at the output, the current is 99 µA, the top dissipates under a milliwatt and rises a quarter of a kelvin, and the ratio moves about 6 ppm — well inside the matching error. Scale the network down to 10 kΩ over 100 Ω, perhaps for noise, and the top dissipates 9.8 mW, rises 2.5 K and moves the gain 60 ppm. That shift follows the square of the output, so on a signal that spends seconds at full scale it is a gain that depends on the signal’s own recent history — a low-frequency distortion that no static gain measurement reveals. The bias point that heats itself found a transistor whose operating point depended on its own dissipation; this is the passive cousin, much smaller and much harder to see.

The same sums sit inside a difference amplifier. The errors that arrive before the gain traced an instrument’s common-mode rejection to the match between two resistor ratios, and when those ratios set a gain of ten or a hundred both are unequal ratios. Their mismatch is the difference of two ratio errors, each with the leverage of its own ratio, so a difference amplifier at high gain pays the asymmetric cost twice, once in each half.

How the numbers were obtained #

Each population is twenty thousand seeded builds, and every ratio is computed exactly from its resistances rather than from the first- order sensitivity that the laws above use; the laws are then compared with the populations, not substituted for them. The two-part dividers use the shared-and-own construction of What a ratio does not see, scaled to a uniform 1% part’s standard deviation so that the four correlations compare directly with that essay’s. The unit dividers draw each unit’s own error independently and combine series-parallel blocks exactly as series sums and parallel reciprocals.

The self-heating figures solve each part’s temperature from its own dissipation through its own thermal resistance, then each resistance at that temperature, then the ratio. They do not iterate the coupling between temperature and dissipation, which at 25 ppm/K and fifty kelvin changes a part’s power by a tenth of a per cent and its shift by less. The unit chain’s coefficients are drawn uniformly within their stated band, two hundred builds, and the figure’s worst case is the worst of those, bounded by the band times one unit’s rise.

The shape a tolerance does not print showed that a population’s shape can matter as much as its width; here every population is Gaussian in its own error, which is the right model for an array’s matching and a loose one for two discrete parts from a sorted reel. The tolerance that is not on any part set the parts-counting arithmetic this essay leans on: averaging N independent errors divides their spread by $\sqrt{N}$, and the unit law is that rule applied to each leg separately.

What it leaves out #

A thermal network. Each part rises by its own power through its own thermal resistance, with no coupling between parts. Two parts side by side on a board warm each other; units on one substrate are close to one temperature. Both make the two-part shift smaller than drawn and the unit chain’s smaller still. The figures bound the effect from the uncoupled side, which is the side a discrete design is nearest.

Time. The shifts drawn are steady states. A thermal time constant of seconds turns the self-heating shift into a frequency-dependent gain error — full at slow signals, absent at fast ones — which is the distortion the worked gain-of-101 stage mentions.

Loading. Every divider here is unloaded. The divider and its load showed that a load reintroduces the absolute value a ratio is built to discard, and an unequal divider’s output resistance is set mostly by its small leg, which is the leg this essay found to be the expensive one.

The geometry of different values. The two-part model treats a 10 kΩ and a 1 kΩ part as sharing their error by one fraction. The paragraph on equal units explained why that is optimistic for parts of different geometry; it is not measured here, because the size of the effect belongs to a process rather than to the circuit.

Still open: gradients, voltage coefficients, and two ratios at once #

A gradient across the units. Every unit here draws its own error independently of where it sits. A real array has a gradient across it, and a ten-to-one divider has an awkward shape for cancelling one: a common-centroid arrangement needs the bottom leg’s units placed around the top leg’s centre, which is possible for a linear gradient with a symmetric chain but leaves the quadratic term in proportion to the top leg’s spread of positions. How much of a large ratio’s matching a given gradient consumes, and whether the small leg’s block should sit at the centre, is the next measurement.

The voltage coefficient. Thin-film and thick-film resistors change value with the voltage across them, a few ppm per volt for some films. In a ten-to-one divider the top part carries ten times the voltage of the bottom, so this is another error an unequal ratio makes for itself; an array of equal units drops one-eleventh of the voltage across each, and the comparison with self-heating would be direct.

Two ratios that must agree. A difference amplifier’s common-mode rejection depends on two ratios matching each other rather than on either being right. Both are built from one array, and the correlation between the two dividers’ errors — not their individual spreads — sets the rejection. Whether the units should be shared between the two dividers, interleaved, or kept separate is a question of the same kind as the gradient above, and it is the measurement The mismatch that cancels itself made for a current mirror’s two transistors, waiting to be made for four resistors.