The width no load can change

A band rather than an edge found that a switch is a switch only between two loads. Half an ohm closed, a hundred megohms open and five picofarads across it is within one per cent of ideal for load resistances from 49.5 Ω to 1.01 MΩ, and the top of that range falls with frequency until, at 6.43 MHz, it meets the bottom and no load resistance will do.

That essay drew every number against a resistor, and the obvious question is whether the band belongs to the switch or to the switch and its resistor together. The load most switches in a converter actually drive is not a resistor. It is a capacitor — the hold capacitor of a sample-and-hold, the sampling capacitor of a switched-capacitor stage, the input of the next stage sitting on a trace — and a capacitor’s impedance has a frequency in it, which a resistor’s does not. So the band might reasonably be expected to take a different width, a different closure, or a different shape entirely.

It takes a different shape. It does not take a different width, and it does not close anywhere else. What it does do is expose a choice the resistive band never had to make, about what “one per cent wrong” means when the error has a direction as well as a size.

The same part with a capacitor for a load #

The part does not change and neither does the tolerance. What changes is what is on the far side of the switch, and with it both of the ways the switch can fail.

Closed, the switch and the capacitor are a single-pole low-pass. At low frequency the capacitor follows the drive exactly, and as the frequency rises it falls behind. The larger the capacitor, the sooner it falls behind, so the closed switch sets an upper edge on the capacitance: the largest capacitor it can still track to one per cent. That edge is a straight line on logarithmic axes, because the error depends only on the product $\omega R_{on} C$, and it runs from 3.18 millifarads at one hertz to 3.18 nanofarads at a megahertz.

Open, the switch still has its five picofarads across it, and the hold capacitor is the other half of a capacitive divider. Well above the corner of the off-resistance and off-capacitance — 318 Hz, the same corner as before — the fraction that arrives is $C_{off}/(C_{off}+C_H)$, which has no frequency in it at all. One per cent of it arrives when the hold capacitor is 99 times the off-capacitance, 495 pF, and that is a floor: a smaller capacitor is disturbed by more than one per cent of whatever happens on the other side of the open switch, at any frequency.

Below the corner the off-resistance takes over, and the floor stops being flat. The hundred megohms charge the capacitor towards the drive, and the slower the drive, the longer the leakage has to do it. So the smallest usable capacitor rises as the reciprocal of frequency — 159 nF at one hertz — and at direct current there is no capacitor large enough. A leak through a hundred megohms has all the time in the world.

The shape is not the resistive shape. The resistive band had a flat floor at 49.5 Ω and a ceiling that was flat at direct current and fell above the corner; this one has a ceiling that falls everywhere and a floor that falls below the corner and then stops. At low frequency the two edges run parallel, so the band there has a width but no fixed position: it slides upward towards infinite capacitance as the frequency falls. That is what a hold time looks like on this plot. A capacitor holds a value for as long as its own leakage takes to move it by the tolerance, and the axis is saying the same thing in the frequency domain.

And yet the two bands close at the same place.

Where the resistive band was drawn in phase #

The closure frequencies agree to four figures, and establishing that exposed an error in the band as it was first drawn, which is worth recording because the error and the agreement are the same fact.

The upper edge of the resistive band is the load resistance through which the open switch lets one per cent of the drive arrive. It was computed as the off-impedance’s size divided by 99, which is right if the off-impedance is a resistance: a resistor in series with a resistor divides in phase, and 1/(1 + 99) is one per cent. Above 318 Hz the off-impedance is a capacitance. The current it passes is at right angles to the voltage across a resistive load, so the load’s voltage and the switch’s voltage add as the root of the sum of their squares, and one per cent arrives when the off-impedance is $\sqrt{100^2 - 1} = 99.995$ times the load rather than 99 times.

The difference is half a per cent in the edge and one per cent in the closure. Solved on the network, the band closes at 6.43 MHz, not the 6.50 MHz the divider arithmetic gave. The band in A band rather than an edge now carries the solved edge, and its numbers have moved accordingly: 304 kΩ at a kilohertz rather than 306, 318 Ω at a megahertz rather than 322.

Nothing had caught it for a simple reason. The only place the upper edge had been checked against a solved network was direct current, where the off-impedance is a resistance and the two rules give the same answer to every figure. An edge checked where two rules agree says nothing about which rule is being used. The check that tells them apart has to be made where they differ — at 45.2 kHz, say, where the solved edge of 7.04 kΩ passes one per cent and the in-phase edge of 7.11 kΩ passes 1.00998 per cent.

It is the same lesson Where the mechanisms are one mechanism drew from the other direction: whether two errors add as magnitudes or in quadrature is a property of their directions, and a formula that assumes one arithmetic is only a formula about the case where that arithmetic is right.

One width for two loads #

With the resistive edge solved rather than assumed, the widths can be compared honestly. The width is the room between the two edges in decades, which is the quantity a designer spends when a load is moved towards one side of the band to buy margin on the other.

Above thirty times the corner the resistor and the capacitor give the same switch the same width to a thousandth of a decade, and at a hundred kilohertz they agree to all four decimals: 1.8083 decades each. Below the corner they differ by 0.0087 of a decade, which is the difference between 99 and 100 — the lower edge of one band is set by a ratio of $(1-\epsilon)/\epsilon$ and the other by $1/\epsilon$ — and nothing else.

The reason is short enough to state in one line. Whatever the load impedance $Z$, the closed switch is wrong by about $R_{on}/|Z|$ and the open switch lets through about $|Z|/|Z_{off}|$. Their product is $R_{on}/|Z_{off}|$, and the load has cancelled out of it. Both errors must be under the tolerance, so their product must be under its square, and that condition has only the switch in it. The load decides where in the band the two errors balance — at 7.07 kΩ for a resistor at direct current, at some capacitance for a capacitor — and nothing about how much room there is on either side of that point.

That argument is approximate, since it treats each error as its leading term, and the figure is what makes it a result: the edges are solved on networks, not taken from the expression.

Ninety-four times the off-capacitance, and the agreement does not move; the whole picture slides a decade and a bit to the left. The width belongs to the switch.

There is a practical sentence in that. A designer who finds a switch’s band too narrow cannot fix it by choosing a different kind of load. The divider, and the thing it does not know about is the nearest relative in this collection — a network whose answer depends on what it drives — and the switch is the opposite case: what it drives decides where the answer sits and has no say in how good it can be.

The count that invents fourteen times the room #

The third line in the width figure is the one that needs explaining, because on its face it says the hold capacitor has more than a decade of extra room at every frequency and a closure fourteen times higher. It says that because of how the error in the closed state was counted.

For a resistive load the closed switch is a divider, and a divider’s error is real: the output is $R_L/(R_L+R_{on})$ of the drive and has no phase. So “one per cent short in amplitude” and “one per cent wrong as a waveform” are the same statement, and the resistive band could count its error either way. For a capacitive load they are not the same statement.

A single pole departs first in phase. At a frequency where $\omega R_{on} C$ is small, the output is smaller than the drive by a fraction of about half its square, and late by an angle of about the product itself. So the vector error grows as the first power of frequency and the amplitude shortfall as the second, and they reach one per cent at very different places.

The factor between them is $\sqrt{1/(1-\epsilon)^2 - 1}\,/\,\big(\epsilon/\sqrt{1-\epsilon^2}\big)$, which is 14.25 at one per cent and very nearly $\sqrt{2/\epsilon}$. That factor is where the 1.1538 decades came from — $\log_{10} 14.25$ — and it is the same at every frequency because it is a property of the count and not of the circuit. At the frequency where the amplitude count first objects, the waveform is already fourteen per cent wrong.

What a held sample actually sees #

It would be too quick to call the vector count right and the amplitude count wrong. A phase lag is not automatically an error, and it is worth being exact about when it is one.

Well below the pole, a single-pole lag is very nearly a pure delay, and the delay is the time constant: half an ohm times 495 picofarads, about 248 picoseconds, the same for every frequency that is much lower than the pole. A pure delay does not distort anything. A system that samples the capacitor 248 picoseconds later than it otherwise would sees the drive exactly, and for such a system the vector error of the closed switch is almost all removable and the amplitude count is closer to the truth.

Three things stop that from being the usual case. The delay is only constant far below the pole, and at the frequencies where either count reaches one per cent the curvature is already visible. The delay is set by the on-resistance, and A band rather than an edge records that a real analogue switch’s on-resistance moves with the signal — a delay that moves with the signal is not a delay but a distortion. And several channels sampled together only agree if their delays do. None of those three is modelled here, so the essay does not claim to know which count a given converter should use. It claims something narrower: the vector count is the one that assumes nothing downstream corrects the lag, it is the one whose closure agrees with the resistor’s, and a band drawn with the amplitude count is fourteen times wider than any circuit that cannot re-time its samples will find.

That is also the reason a capacitor’s band is harder to specify than a resistor’s. A resistor’s band can be stated with one number and a tolerance. A capacitor’s needs a tolerance and a statement about what the tolerance is measured on — the condition beside the number that A boundary is a model and a tolerance found missing from nearly every edge in this collection.

A tenth of a per cent, and a gap that widens #

The two counts do not merely disagree by a constant factor. They disagree by a factor that depends on the tolerance, which means they give the tolerance different exponents.

Tightening from one per cent to a tenth moves the vector closure from 6.43 MHz to 63.7 kHz, a hundredfold, which is the square law the resistive band had: one decade off each edge. The amplitude count moves from 91.6 MHz to 2.85 MHz, only a factor of 32, because its reach grows as the square root of the tolerance while the floor grows as the tolerance itself. So the amplitude count’s closure scales as the three-halves power of the tolerance, and the gap between the two counts widens from 14.25 at one per cent to 44.7 at a tenth.

A boundary is a model and a tolerance found that the exponent with which an edge moves against its tolerance names the mechanism behind it. Here the exponent names something else — the way the error was counted — and a reader given only the closure frequency and its movement could tell the two counts apart from that alone. It is a small warning about exponents: two different measurements of one mechanism can carry two exponents.

What the capacitor adds to a sample-and-hold budget #

Read as a design statement, the hold band says three things a sample-and-hold budget usually states separately.

The floor is feedthrough. 495 picofarads is the capacitor below which the open switch’s own capacitance carries more than one per cent of the input onto the held value. It does not depend on frequency, on the hold time or on the on-resistance, and at a tenth of a per cent it is ten times larger. That number is also the reason a larger hold capacitor is attractive for reasons that have nothing to do with noise, although The total that has no resistor in it makes the noise argument for it on its own terms.

The low-frequency slope is droop. Below 318 Hz the floor is set by the leakage, and a capacitor that must hold for longer must be larger in proportion. The capacitor that remembers finds a second mechanism that bends this line — the dielectric giving back charge it absorbed — which the model here does not contain.

The ceiling is acquisition. The closed switch has to charge the capacitor, and the largest capacitor it can track to the tolerance falls as the reciprocal of frequency. Where the ceiling meets the floor the design has no capacitor left, and that happens at the same frequency a resistive load would have closed at — 6.43 MHz for this switch.

What the band leaves out is the charge the switch’s own gate dumps on the capacitor at the instant it opens, which The offset that knows the signal measures and which moves the held value by an amount that depends on the signal. That is a third error of the same part and it would add a third edge; it needs a model of the channel rather than of three passive numbers.

Why the load could not have helped #

The product argument has one assumption buried in it, and it is the one worth carrying away. The open switch’s error was taken to be $|Z|/|Z_{off}|$ — the leaked current flowing through the load. That is true of a switch between a source and a load, and it is what makes the load cancel. It is not true of every arrangement a switch is used in.

The same current can be made to land somewhere else. It can land on another switch that is closed, or on a switch placed to catch it, or on a source impedance that holds the node. In each case the product of the two errors picks up a new impedance in place of the load, and that impedance is not constrained to cancel. So the width of the band is fixed for a switch in series with its load, and only for that.

Every model has an edge put this collection’s boundaries on one axis and treated each as belonging to one assumption. The switch’s band belongs to two — the part, and the place its leak lands — and this essay has held the second one fixed.

Still open: where the leak lands #

A multiplexer. Put the switch in a bank that shares an output and the open channels no longer leak into the load. They leak into a node the one closed channel holds, through its on-resistance and its source, so the upper edge on the load should disappear and a source impedance should take its place. Where an open switch leaks to measures that, and finds the eight-channel limit a hundred times more sensitive to the source than to anything the switch’s data sheet states.

A load that resonates. The product argument assumed the load’s impedance and the off-impedance add without cancelling, which holds for a resistor and for a capacitor. An inductive load does not obey it: with a millihenry and a hundred ohms beside the five picofarads, the open switch passes 138.7 times the drive at 2.25 MHz, where the two reactances cancel. Taking inductive loads — a transformer winding, a relay coil, a length of cable below its quarter wave — would find a band with a hole in it, and the hole is where the switch is least a switch while the product argument says nothing is wrong.

An on-resistance that moves. Everything above holds the half ohm fixed. A channel whose resistance depends on the signal turns the ceiling of the hold band from a lag into a distortion, and turns the 248-picosecond delay that a re-timed system could ignore into one that no re-timing removes. That question needs a model of the channel, and answering it would say which of the two counts a real sample-and-hold is entitled to.