The R–2R ladder, built for the rest of itself

The stage that is wrong is the far one cascaded three ten-kilohm voltage dividers with nothing between them and got 0.076923 of the input where the product of their ratios promised 0.125. Taken apart stage by stage with the rest of the chain in place, the three ratios were 0.3846, 0.4000 and 0.5000, and their product was the answer exactly. The last stage was exactly its own ratio, because nothing was connected to it; the error was at the front, which is the far end from where a probe goes.

The branch the other resistance decides then set the voltage divider beside its dual, the current divider, and ended by asking the chain’s question of the dual: cascade three current dividers, read the last one, and see which end the error is at. The duality ought to answer that without a solve. It does, and the answer leads somewhere the question did not point: to the ladder everyone actually builds, which is a chain designed from the start for the error that the equal chain has.

The dual, built and solved #

Duality exchanges every quantity in a planar network for a partner: voltage for current, resistance for conductance, a series connection for a shunt one, a mesh for a node, an open circuit for a short. Applied to the chain, it turns a voltage source into a current source; each stage’s series resistor into a shunt resistor to ground; each stage’s shunt resistor into a series resistor on to the next node; and the chain’s open far end, where the output was read with nothing connected, into a short circuit, where the output is the current flowing into it.

So the dual is a ladder of current dividers. At each node the current arriving divides between a resistor to ground and a resistor that carries it on down the ladder, and the fraction carried on is that stage’s ratio. With every resistor equal, a stage on its own — a shunt resistor against a series resistor into a short — passes on exactly half.

The numbers are identical, not merely similar: 0.384615, 0.400000, 0.500000 for both, and 0.076923 out of each. The reason is that the nodal equations of the ladder are the mesh equations of the chain with the symbols renamed. A resistance of ten kilohms in the chain becomes a conductance of “ten kilo-siemens” in the dual, and with every element equal, scaling every conductance by the same factor changes no ratio. So the equal ladder is the dual of the equal chain, whatever the resistor value.

The error is at the same end, and for the same reason reflected. The chain’s last stage was exact because nothing loaded it. The ladder’s last stage is exact because it ends in a short, and a short is to a current divider what an open is to a voltage divider: the condition under which the divider’s own two elements are the only two elements. Every stage before the last has the rest of the ladder hanging on it, and the stage nearest the source has the most.

The duality also says what reading the dual costs. The chain’s output was a voltage read with nothing connected, and a real voltmeter connected there is a resistance across the last shunt — a fourth divider stage that is not quite nothing. The ladder’s output is a current read into a short, and a real ammeter there is a resistance in series with the last series element — a fourth stage that is not quite a short. The two costs are duals of each other in exactly the sense the earlier essay measured on a single divider: a voltmeter costs one per cent at ninety-nine times the resistance it looks back into, and an ammeter at that resistance over ninety-nine. The only difference the dual makes to the instrument is which way round the inequality runs.

The repair, turned round #

The chain’s repair was a staircase. Raise each stage’s resistance by a factor over the one before and each stage loads the one behind it less, and the departure from the product of the ratios falls in proportion to the factor. Duality says what the dual repair must be, and it is the opposite.

A decade up a stage takes the chain from 38.46 per cent low to 4.976, and takes the ladder to 94.41: the same change that nearly cures one ruins the other. A decade down a stage does the reverse, exactly. The two curves are mirror images about a factor of one, because a resistance rising in the chain is a conductance rising in the dual, and a conductance rising is a resistance falling.

The physical reading is short. In the voltage chain, a stage loads the one behind it by drawing current from its tap, and a larger resistance draws less. In the current ladder, a stage loads the one behind it by presenting a resistance in series with that stage’s through path, and a smaller resistance diverts less of the current away from where the divider wanted it to go. The general rule behind both is the one the divider, and the thing it does not know about started from: the ratio discards the magnitude, and the magnitude decides the loading. The dual simply decides it the other way up.

A chain that is wrong alone and right in place #

Both repairs share an assumption: that each stage should be right on its own, and that the job is to stop the rest of the chain disturbing it. The chain’s stages are designed for an open far end and the ladder’s for a short, and a staircase approaches those conditions by making every following stage look more like one.

There is another way to get a chain right, which is to design each stage for the load it will actually have. The R–2R ladder does exactly that.

At every node the current arriving meets a shunt of 2R, which is that bit’s branch, and a series R leading on to the rest of the ladder. Looking down the ladder from the far side of that series R, the rest of it is 2R — the last node’s 2R shunt in parallel with its 2R termination is R, plus the series R before it is 2R, and the argument repeats up the ladder. So every node sees 2R against 2R and splits the current arriving at it exactly in half, and the bits carry a half, a quarter, an eighth, down to 1/256 of the source at eight bits, with the termination taking the last 1/256.

Taken alone, the same stage is not a half at all. A shunt 2R against a series R into a short sends one third on. The R–2R stage is right only because of what is connected to it, which is the exact opposite of the equal chain, whose stages are right alone and wrong in place. The staircase makes the rest of the chain disappear; the R–2R ladder makes the rest of the chain part of the design.

The comparison with the equal ladder is worth a sentence of its own. Build the same nodes with every resistor equal, the chain’s habit, and the first bit takes 0.61803 of the source, the second 0.23607, and each bit after that roughly 2.618 times less than the one before. An infinite ladder of equal resistors presents the golden ratio times R at its input, so each node passes on one over the golden ratio squared of what arrives: a geometric series in a ratio nobody would choose for a converter. The 2R is what turns that ratio into two.

Why it is a current ladder #

A converter built on this ladder steers each bit’s branch either to the output or to ground, and keeps both at the same potential — the output held at a virtual earth by an amplifier. The ladder’s node voltages then do not depend on the input code being converted at all. Each bit’s current is the same whatever the other bits are doing, and the output is the sum of the currents switched to it. A voltage-mode ladder that switched its bits between a reference and ground would instead change the current each switch carries with every code, and so the voltage each switch’s own resistance drops.

That is the current divider’s defining property doing work: it is the resistance the current does not go through that decides the split, so a branch can be switched between two points at one potential without anything upstream noticing. The readings that add up and are wrong found the same property at its most inconvenient — a meter in one branch redistributing current into the others — and here it is arranged so that nothing redistributes.

The tolerance that is at the front #

The ladder halves exactly only if every resistor is exactly its value. A real resistor is not, and the question the chain leaves behind is where an error matters most.

The first resistor dominates everything. A one per cent error in the first bit’s 2R moves the step at the major carry by −1.269 LSB, which is more than the whole step: the output, which should rise by one LSB when the input code goes from 0111 1111 to 1000 0000, falls instead, and the converter is not monotonic there. The series resistor right after it moves the step by +0.636, half as much and the other way. Each position further down the ladder matters roughly half as much as the one before, until by the last bit’s shunt the same one per cent moves the step by 0.0034 LSB.

The first resistor’s effect has a closed form, and the ladder’s halving is all it needs. The first node sees 2R(1 + e) against the 2R the rest of the ladder presents, so the first bit takes 1/(2 + e) of the source instead of a half and everything else takes (1 + e)/(2 + e). The step at the carry is the first bit less all the others, which moves by about −e/2 of the source — and the source is 2ⁿ least significant bits, so the step moves by about $-e\,2^{n-1}$ LSB. At eight bits and one per cent that is −1.28, and the solved ladder agrees with the exact expression to nine decimal places.

The pattern down the ladder has a reason too. Raising the series resistor after the first bit makes the rest of the ladder look larger from the first node, so more of the source goes into the first bit’s own branch and the step at the carry grows: the opposite sign to raising the first bit’s own 2R, which pushes current out of that branch. It is half the effect because that series resistor is only half of the 2R the rest of the ladder presents; the other half is everything beyond it. The same argument applies one node further on with every quantity halved again, which is why the bars shrink by about two at each step. Deep in the ladder the halving stops, at a few thousandths of an LSB for one per cent: there an error mostly moves current between bits that are all on the same side of the carry, and what is left is the small change it makes to the termination’s share, which the carry counts against the most significant bit.

The rule $-e\,2^{n-1}$ says what a converter’s resolution costs in resistor matching. Twelve bits and one part in ten thousand gives −0.205 LSB from the first resistor, and every added bit doubles it. Keeping the major carry inside half an LSB needs the first resistor matched to about $2^{-n}$: a part in 256 at eight bits, a part in 4096 at twelve. The same matching is wasted deep in the R–2R ladder, where an error of the same size moves the carry by less than a thousandth as much.

So the R–2R ladder, which fixed the equal chain’s error at the front by design, still keeps its sensitivity at the front. The first stage carries half of everything, so an error there is an error in half of everything. The chain found its error at the stage nearest the source because that stage carries the whole of the rest of the chain as its load; the R–2R ladder finds its tolerance there because that stage carries the whole of the rest of the chain as its current. It is the same position for two different reasons, and a designer trimming a ladder converter trims its front resistors for the second.

The tolerance that is not on any part measures the spread of a network’s answer from the spread of its parts, and the lesson carries over: the parts are not equally important, and the ranking is a property of the network rather than of any part. Six decibels a bit, and the half step blamed on it prices a bit in signal-to-noise ratio; this ladder prices it in resistor matching at the front, and doubles the price with every bit.

Ladders elsewhere #

The same distinction — stages designed for isolation against stages designed for their neighbours — appears wherever ladders do. A ladder is not a cascade finds a doubly terminated filter ladder whose elements are chosen for each other, and whose sensitivity to any one element is lower than a cascade of isolated sections for exactly that reason. A ladder is not a line finds the lumped ladder approximating a transmission line, where every section is designed for the characteristic impedance the rest of the line presents — the R–2R ladder’s 2R, in a different subject. In each case the rest of the structure is not a disturbance to design against but a load to design for, and a stage that is wrong alone is the price of the whole being right.

How the numbers were obtained #

Every network is solved by nodal analysis at direct current. The voltage chain is three series-and-shunt pairs from a voltage source; its dual is a current source feeding three shunt-and-series pairs into a zero-volt source, whose current is the output. Each stage’s ratio is read from the same solve — the voltage at a tap over the voltage before it, or the current into a series resistor over the current arriving at its node — and the products are checked equal to the outputs. The staircase sweep solves both networks at 51 factors from 1/300 to 300 and checks that the ladder at each factor matches the chain at its reciprocal. The R–2R ladder returns each bit’s 2R to its own zero-volt source, so each bit’s current is read directly; the step at the major carry is the most significant bit’s current less the sum of all the others, in units of the source over 2ⁿ, and each tolerance case is a separate solve with one resistor changed.

What it does not measure #

The switches. Every bit’s branch here is returned to an ideal virtual earth, and a real converter steers each branch through a switch with an on-resistance that is added to that branch’s 2R. The on-resistances are a tolerance of their own, and they sit at the front of the ladder as much as anywhere else.

The amplifier. The virtual earth is perfect here; an amplifier with finite gain leaves the output node slightly off earth, which changes the ladder’s node voltages with the input code and reintroduces exactly the loading the current ladder was chosen to avoid.

And the error across several resistors at once. One resistor is changed at a time. Real resistors all have errors, and whether a set of errors drawn from a tolerance adds up to a worse or better major carry than its single worst member is a statistical question the single-resistor sweep only bounds.

Still open: the switch in each branch, the whole tolerance at once, and a segmented front #

The switch in each branch. An on-resistance in series with each 2R is a tolerance that depends on the switch’s size, and it can be compensated by scaling the switches in the same binary ratio as the currents. Solving the ladder with switches of stated on-resistance, scaled and unscaled, would say how large a switch the first bit needs before its resistance stops mattering.

Every resistor wrong at once. Drawing every resistor from a seeded tolerance and solving many ladders would give the major carry’s spread directly, and would say whether the front resistors’ dominance survives into the statistics or whether the many small errors further down add up to something comparable.

A segmented front. Replacing the first few bits of the ladder with equal current sources — a thermometer-coded segment — removes the major carry’s dependence on one resistor. Solving the hybrid would put a number on how many bits of segmentation buy back one bit of resistor matching, which is the trade the $2^{n-1}$ rule above makes necessary.