The resistance that bends the signal

The floor below any load found a limit on a switch that no choice of load can move. Closed, a switch of half an ohm into a load delivers all but $R_\mathrm{on}/(R_\mathrm{on} + R_L)$ of the drive; open, a hundred megohms lets $R_L/(R_\mathrm{off} + R_L)$ of it through. One error falls as the load grows and the other rises, and they meet at the geometric mean of the two resistances, 7.07 kilohms, at 70.71 parts per million. That is 13.79 bits, with no frequency in it.

The closed error in that argument is a strange thing to call an error, though. A switch of exactly half an ohm into exactly 7.07 kilohms delivers 0.999929 of every signal, whatever the signal is. A converter behind it reads every level 71 parts per million low, and one measurement of a known reference fixes all of them. The error is a gain, and a gain is the easiest thing in a measurement chain to calibrate.

What calibration cannot fix is an on-resistance that is different at different levels, and real ones are. A switch made of a transistor conducts through a channel whose resistance depends on the voltage it carries, and one made of two complementary transistors in parallel is flattest at the ends of its range and highest in the middle. The capacitance a third switch moves and every switch before it took the on-resistance as a number. This essay gives it a shape, and asks how many of the thirteen bits were ever the on-resistance’s to take.

The floor, as it was counted #

The lone switch’s curve is a V: the closed error on the left, falling as the load grows, the open leak on the right, rising with it. The floor is the bottom of the V. Both arms are counted as errors in the same currency, and for an uncalibrated path that is correct: a reading 71 parts per million low is 71 parts per million wrong.

The two arms are not the same kind of wrongness, and the distinction matters for everything below. The open leak is another channel’s signal arriving at this one’s output. It is crosstalk, it depends on what the other channel is doing, and no calibration of this channel can remove it. The closed error, for a constant on-resistance, is the same fraction at every level, so it is a pure gain.

An on-resistance with a shape #

The switch here has an on-resistance that moves across its signal range by a fraction δ of its smallest value, in one of two shapes. A bump is highest at mid-range and falls to its smallest at both ends, $R_\mathrm{on}\bigl(1 + \delta(1 - u^2)\bigr)$ with $u$ the level as a fraction of full scale: the shape of a complementary pair. A slope rises steadily from one end to the other, $R_\mathrm{on}\bigl(1 + \delta(u+1)/2\bigr)$: the shape of a single transistor. Both move by the same δ from their smallest value to their largest.

At direct current the closed switch into a load is a divider whose top resistor depends on the voltage across it, and its output at level $u$ is $u\,R_L/\bigl(R_L + R_\mathrm{on}(u)\bigr)$. The figures solve that network at every level they draw — 41, 81 or 201 of them — and hold each solve to the closed form to a part in a million million.

Two readings of the closed error then follow. Uncalibrated, the error is the largest shortfall at any level, which is the old error at the largest on-resistance. Calibrated, a straight line is drawn through the outputs at the two ends of the range — a gain and an offset, which a two-point calibration measures and removes — and what is left is the largest departure from that line, which is the integral nonlinearity. The open leak is the same in both readings, counted for the calibrated path against the gain the calibration has restored.

What calibration cannot remove #

Into a kilohm, with the on-resistance moving by ten per cent, a bump leaves an S-shaped residual peaking at 19.23 parts per million and a slope a bow peaking at 24.99. Both are tiny beside the 500 parts per million of plain gain error half an ohm costs into a kilohm, and neither can be calibrated.

The shapes are worth the algebra because the coefficients are exact. Expand the output for a load large beside the switch and the bump adds a term $r\delta u^3$ to a straight line, with $r = R_\mathrm{on}/R_L$. The straight line through the two ends absorbs the linear part, and the residual is $r\delta(u^3 - u)$, which is odd in the signal and peaks at $u = \pm1/\sqrt3$ with a height of $2/(3\sqrt3) = 0.385$ times $r\delta$. The slope adds $-\tfrac12 r\delta u^2$ instead; the line through the ends takes an offset out of it, and the residual $\tfrac12 r\delta(1 - u^2)$ is even and peaks at the middle, at one half of $r\delta$. The figure checks both coefficients against the solve at every level.

The slope’s residual is larger by the ratio of the coefficients, 1.30, and it is also the easier one to spot: a bow is a second-harmonic distortion and an S a third, and an instrument that measures harmonics sees the slope’s first.

The floor, recounted #

Now put the two readings against the load.

Uncalibrated, a ten-per-cent bump costs almost nothing. The floor rises from 70.71 to 74.16 parts per million, which is the old floor at the largest on-resistance in the range, 0.55 ohms: $1/\bigl(1 + \sqrt{R_\mathrm{off}/1.1R_\mathrm{on}}\bigr)$, checked to two parts in a thousand. It is 0.07 of a bit.

Calibrated, the floor is a different curve with a different bottom. The closed arm is now the curvature, $0.385\,\delta R_\mathrm{on}/R_L$, still falling as the load grows but twenty-six times lower than the gain error it replaces. The open arm is the same leak. They cross at a smaller load, 1.39 kilohms, and the floor there is 13.89 parts per million — 16.14 bits, two and a third more than the switch was supposed to have.

The balance has a closed form of the same shape as the old one. Equal $c\,\delta R_\mathrm{on}/R_L$ to $R_L/R_\mathrm{off}$, with $c$ the shape’s coefficient:

$$\varepsilon_\mathrm{cal} = \sqrt{c\,\delta\,\frac{R_\mathrm{on}}{R_\mathrm{off}}}, \qquad R_L^\ast = \sqrt{c\,\delta\,R_\mathrm{on}R_\mathrm{off}}.$$

For the bump at ten per cent that is 13.87 parts per million into 1.39 kilohms; the figure checks the solve against it to eight per cent, and it holds to under a quarter of one. The old floor was the same expression with $c\delta = 1$. So a switch that has been calibrated is not limited by its on-resistance at all. It is limited by the product of the fraction by which that resistance moves and the coefficient its shape leaves — and the reason it was never obvious is that for an uncalibrated switch that product is replaced by one.

The slope gives the same uncalibrated floor, since it reaches the same largest resistance, and a calibrated floor of 15.81 parts per million into 1.58 kilohms, 15.95 bits. The ratio of the two calibrated floors is $\sqrt{0.5/0.385} = 1.14$, which is exactly what they show. A single transistor costs a fifth of a bit against a complementary pair, at the same total movement.

The concavity that makes the curvature uncalibratable is the same arithmetic how small is small signal applies to a junction: a smooth transfer curve driven across a range shifts its average by half its curvature times the spread of the drive, and no straight line through the ends can take that shift out. What changes here is only which curve is doing the bending. And the choice between the two readings is the kind of boundary a boundary is a model and a tolerance describes: the floor moves by a factor of five without the part changing, because the tolerance changed what it counts. The crosstalk arm, which calibration cannot touch, is the same leak where an open switch leaks to followed into a multiplexer’s shared node.

The load that suits a calibrated switch #

The recount moves the best load as well as the floor, and the direction is the useful surprise. An uncalibrated switch wants the geometric mean of its two resistances, 7.07 kilohms, because the gain error falls only as the load grows. A calibrated one wants $\sqrt{c\,\delta}$ of that, because the error it is trading against the leak is smaller by $c\,\delta$ and so stops mattering at a smaller load.

With a movement of one per cent the calibrated floor is 4.392 parts per million, 17.80 bits, and the load that gives it is 438 ohms — sixteen times smaller than the load the uncalibrated switch wanted, and within a factor of nine hundred of the half-ohm switch itself. At that load half an ohm is a gain error of 0.114 per cent, eleven hundred parts per million, which a designer working to the uncalibrated rule would have called hopeless. It is hopeless only for a path that is never calibrated.

A small load has consequences outside this figure, all of them good. A converter’s hold capacitor behind the switch charges through the switch’s resistance and whatever the load resistance represents, so a smaller resistance settles faster; the thermal noise of the network scales with its resistance; and the leak into a smaller load is a smaller voltage for any leakage current, which is the subject of the next measurement. The uncalibrated rule pushes a precision multiplexer towards kilohms and the calibrated one pulls it back towards hundreds of ohms, and the calibrated one is right about every other error as well.

The uncalibrated floor at one per cent is 71.06 parts per million, into 7.11 kilohms: the constant switch’s floor to within half a per cent. Uncalibrated, a movement of one per cent is invisible. Calibrated, it is four bits.

What the switch’s designer is being asked for #

Put the two floors side by side and the specification of a precision switch changes. The number that sits on the front of a datasheet is the on-resistance, and uncalibrated that is the right number to put there: the floor goes as its square root. Calibrated, the on-resistance appears only multiplied by its own movement, and a switch of a full ohm that moves by one per cent — $c\,\delta R_\mathrm{on}$ of 0.00385 ohms for a bump — is a better precision switch than one of half an ohm that moves by ten per cent, whose product is 0.0192. The second has the better headline and five times the product, so 2.2 times the floor.

That is why the two shapes are worth distinguishing even though they are idealisations. A complementary pair exists precisely to flatten the movement: each transistor conducts well where the other conducts badly, and what remains is a bump whose height is set by how well they are matched. A single transistor passing a signal that swings towards its own gate voltage has a slope — a steep one, since its resistance rises without limit as the channel pinches off — and a movement of several hundred per cent across a wide range is ordinary. The figures stop at a resistance that doubles, because beyond it the expansion that gives the coefficients is no longer the whole story, and at that movement a single transistor’s calibrated floor is 50 parts per million, 14.3 bits: better than its uncalibrated floor, and nowhere near a pair’s.

The movement is also the quantity a datasheet is least likely to state. The on-resistance flatness, where it is given, is usually the largest minus the smallest resistance over a stated range, which is $\delta R_\mathrm{on}$ with no shape attached; the shape’s coefficient then decides whether that flatness costs 0.385 or one half of itself, and a flatness figure measured over a narrower range than the signal uses says nothing about the ends, where a pair’s resistance is lowest and a single transistor’s is highest.

How much movement a bit costs #

The slider on the hero figure moves δ, and the floors are worth having as a curve.

Uncalibrated, the floor is indifferent to the movement until the movement is large: 70.71 parts per million for a constant resistance, 99.99 when it doubles across the range. That flatness is the reason the movement never shows up in a gain-error budget — it changes the thirteen-bit answer by a few per cent.

Calibrated, the floor follows the square root of the movement, and the figure checks the exponent at 0.5 to three hundredths over the middle of the range. A bump that moves by a tenth of a per cent leaves 1.389 parts per million, nineteen and a half bits; by one per cent, 4.392, 17.80 bits into 438 ohms; by ten per cent, 16.14 bits; and a resistance that doubles across its range still leaves 43.92 parts per million, 14.47 bits — better, calibrated, than a perfectly constant switch uncalibrated.

Every factor of a hundred in the movement costs 3.3 bits. That is the same exchange rate the floor below any load found for the ratio of off- to on-resistance, and for the same reason: both floors are square roots of a ratio of two errors that balance at one load.

What the recount assumes #

That the calibration holds. A two-point calibration removes the gain and offset it measured, at the temperature and supply at which it measured them. An on-resistance of half an ohm with a temperature coefficient of a few thousand parts per million per kelvin moves the gain error by about a part in a million for every kelvin into a kilohm, which spends a bit of the calibrated floor for every ten kelvin or so. The calibrated floor is a floor for a calibration repeated as often as the temperature moves, and the uncalibrated one is what is left if it is not.

That the leak is the only other error. The open switch here leaks through a resistance, which makes its leak proportional to the other channel’s signal. A switch also leaks through its junctions, as a current that flows whatever the signal, and the leak no switch can hold measures what that does to both floors — and to the T, whose whole advantage was to hold a leak that has to cross an open switch.

That the switch sees the load at direct current. Everything here is static. At a frequency the on-resistance’s movement modulates the corner it makes with the load’s capacitance, and the curvature becomes a distortion that grows with frequency; the width no load can change found that corner for a constant on-resistance, and a moving one would make it a function of level.

That the shape is one of two. Real on-resistance curves are neither a clean parabola nor a straight line, and the coefficient $c$ is a property of the curve, not of the movement. A curve with a sharp knee — a transistor approaching the end of its conduction — has a larger coefficient for the same δ than either shape here, because more of its movement is curvature and less of it is a straight line a calibration can take away.

Still open: a leakage that is a current, a movement with temperature, and the T recounted #

A leakage that is a current. The next question is about the other arm of the V. A junction leakage doubles every ten kelvin and flows into the load whatever the other channels carry, so it is an offset rather than crosstalk. It calibrates at one temperature and not at another, and it does not have to cross an open switch to arrive, which is the thing a T was built to exploit. The leak no switch can hold takes it up.

An on-resistance whose movement changes with temperature. δ is itself temperature-dependent, because the two transistors in a complementary pair drift differently and the bump’s height and position move with them. A switch calibrated for its curvature at one temperature would then carry a residual from the change in shape, and the floor would be set by the derivative of δ rather than by δ.

The T, recounted. A T puts two on-resistances in series and loads the middle with the shunt’s off-resistance. Calibrated, both on-resistances’ curvatures add while the gain error that made the T worse than one switch into 7.07 kilohms disappears. The shunt switch the source sizes found the shunt’s best width at high frequency; at direct current, with calibration, the question is whether a T’s floor becomes the square of its curvature rather than the square of its resistance ratio, and at which load.