Where the models stop

The floor below any load

A switch of half an ohm closed and a hundred megohms open is within one per cent of ideal for loads between two edges, and the edges close on each other as the tolerance tightens. At direct current they meet at 70.71 parts per million, into 7.07 kilohms: no load makes that switch better, which is 13.79 bits and a boundary with no frequency in it. A T of three such switches has no best load at all. Its error falls with the load towards five parts per billion — the square of the lone switch's resistance ratio rather than its root — and reaches ten only into two hundred megohms. Into the 7.07 kilohms that suited one switch, the T is worse than one switch.

16 min read 5 figures Every model has an edge The parasitic is the component The constants decide

Assumes: A band rather than an edge · Every model has an edge

A band rather than an edge found that a switch is a switch only for loads between two edges, both set by the same part: below a lower edge the closed switch’s resistance costs the load too much of the drive, above an upper edge the open switch passes too much of it. The band’s width in decades was the room between them, and the essays that followed measured how a hold capacitor, a multiplexer and a third switch move the edges around.

Every one of them treated the band as something that closes with frequency, because the off-capacitance does most of the leaking. At direct current the capacitance is gone and the band is widest. But it does not get arbitrarily wide there, and the capacitance a third switch moves ended on the reason: the two edges at direct current are ninety-nine on-resistances and a ninety-ninth of the off-resistance, and they meet when the tolerance falls far enough. Below that tolerance no load at any frequency makes the switch ideal enough. That is a boundary in resolution rather than in frequency, and the question left open was whether a T, whose open node is held by a switch, removes it.

A band that closes without any frequency

Where a switch is a switch: a band, and the 6.43 MHz at which it closes. computed by solving, not by drawing. A switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1.0% of being ideal only for loads between 49.5 Ω and 1.01 MΩ — 4.31 decades, and both edges are the same part. The upper edge is a frequency as well as a resistance, because the off-capacitance shunts the open switch: it falls a decade per decade above 318 Hz and meets the lower edge at 6.43 MHz, where the band closes and no load at all will do. Checked by scanning every load at 1.3 times that frequency and finding the best possible error to be 1.17%. — Fig. 1 The band the question starts from: a switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1% of ideal for loads between 49.5 Ω and 1.01 MΩ at direct current, 4.31 decades, and the upper edge falls a decade per decade above 318 Hz until it meets the lower at 6.43 MHz.

At one per cent the band is more than four decades wide at direct current, and it closes only because the off-capacitance brings its upper edge down. Tighten the tolerance instead of raising the frequency and the band closes anyway.

Closed, a switch into a load $R_L$ delivers $R_L/(R_\mathrm{on} + R_L)$ of the drive, so its error is $R_\mathrm{on}/(R_\mathrm{on} + R_L)$ , falling as the load grows. Open, it lets through $R_L/(R_\mathrm{off} + R_L)$ , rising as the load grows. One error wants a large load and the other a small one, so the two are equal at one load, and at that load both are as small as they can be made at once:

\varepsilon_{\min} = \frac{1}{1 + \sqrt{R_\mathrm{off}/R_\mathrm{on}}}, \qquad R_L^{\ast} = \sqrt{R_\mathrm{on}R_\mathrm{off}}.

For this part that is 70.71 parts per million into 7.07 kilohms. Any tolerance tighter than that and the two edges have crossed: there is no load, at any frequency, into which this switch is that good.

One switch is never better than 70.71 ppm; a T of three reaches 10 ppb only into 200 MΩ. computed by solving, not by drawing, at direct current. The worse of a switch's two errors — closed, the fraction the load fails to receive; open, the fraction it receives anyway — against the load, for one 0.5 Ω, 100 MΩ switch and for a T of three, from a buffered source. The lone switch is best at 7.07 kΩ, the geometric mean of its two resistances, where both errors are 70.71 ppm, 13.79 bits: no load does better. The T has no best load. Its worse error falls with the load towards Rₒₙ/(Rₒₙ + Rₒff) = 5 ppb, the square of the lone switch's resistance ratio rather than its root; it is within twice that from 200 MΩ, it passes the lone switch's floor only above 14.1 kΩ, and into 7.07 kΩ it is 141.4 ppm, worse than one switch. Solved on the network up to 1000 MΩ and continued, dashed, from the closed form it matches. — Fig. 2 At direct current, the worse of a switch’s two errors against the load, for one 0.5 Ω, 100 MΩ switch and for a T of three. The lone switch is best at 7.07 kΩ, where both errors are 70.71 ppm, 13.79 bits. The T has no best load: its worse error falls towards $R_\mathrm{on}/(R_\mathrm{on} + R_\mathrm{off}) = 5$ ppb, is within twice that from 200 MΩ, passes the lone switch’s floor only above 14.1 kΩ, and into 7.07 kΩ is 141.4 ppm. Solved to 1 GΩ and continued, dashed, from the closed form.

The lone switch’s curve is a V on logarithmic axes: the closed error’s slope on the left, the open error’s on the right, and a point at the bottom where they meet. Every value on it is solved on the network and matches the two expressions above to a part in a million, and the bottom is the floor. Expressed as a converter’s resolution it is 13.79 bits — a fourteen-bit multiplexer made of this switch is short of its last bit before the capacitance or the source impedance has done anything.

The same floor is visible in the band’s width. The width no load can change found that the room between a switch’s two edges is set by the product of its two errors, $R_\mathrm{on}/|Z_\mathrm{off}|$ , with the load cancelled out of it: both errors must be under the tolerance, so their product must be under its square. At direct current $|Z_\mathrm{off}|$ is the off-resistance, the room between the edges is a factor of roughly $R_\mathrm{off}\varepsilon^2/R_\mathrm{on}$ , and that factor reaches one — no room at all — when $\varepsilon$ is the square root of $R_\mathrm{on}/R_\mathrm{off}$ . The width argument and the balance argument are one argument, and the floor is where the band’s width in decades reaches zero without any frequency having been involved.

What a T does instead

A T puts two series switches in the path and a third from the node between them to ground, closed when the series pair is open. Its errors come from somewhere else.

Closed, the load sees two on-resistances in series and the shunt switch’s off-resistance hanging from the middle. The first costs $2R_\mathrm{on}/R_L$ , twice the lone switch’s error, and falls with the load; the second is the shunt’s leakage, a fraction $R_\mathrm{on}/R_\mathrm{off}$ that stays however large the load becomes. Open, the first series switch’s leak lands on the shunt’s on-resistance and puts a fraction $R_\mathrm{on}/R_\mathrm{off}$ on the middle node; the second series switch passes that on to the load, and into a load large beside its off-resistance it passes all of it.

So both of the T’s errors head for the same number from opposite sides as the load grows, and neither crosses the other on the way:

\varepsilon_T \to \frac{R_{on}}{R_{on} + R_{off}}.

The T has no best load. Its worse error falls at every larger load, towards five parts per billion. That is the square of the lone switch’s resistance ratio where the lone switch’s floor was its square root — about 27.6 bits against 13.8 — and it is reached only as the load becomes comparable with the off-resistance itself. The T is within twice its floor, ten parts per billion, only into two hundred megohms.

Into the load that suited one switch, the T is worse. At 7.07 kilohms its two series on-resistances cost 141.4 parts per million, exactly twice the lone switch’s floor, and it passes that floor only above 14.1 kilohms. A T bought to improve a switch driving a few kilohms makes it twice as bad at direct current, and the improvement it was bought for lives several decades of load away.

The capacitance a third switch moves found the T’s price at high frequency: the shunt switch’s off-capacitance hangs on the closed path and fails it at about the frequency at which the same capacitance leaks. At direct current the price is paid in a different coin, the second on-resistance, and it is paid on the load axis instead of the frequency axis. A T is a better switch in two regions — very high frequency, very high impedance — and a worse one in the region where most signal switches spend their lives.

Where the floor meets the frequency

The floor has no frequency in it, and the off-capacitance does, so at some frequency the capacitance takes over and the floor rises.

The floor against frequency: flat at 70.71 ppm for one switch and 6 ppb for a T, until the off-capacitance arrives. computed by solving, not by drawing. The smallest worse-of-two error any load up to 1000 MΩ gives, against frequency, for one 0.5 Ω, 100 MΩ, 5 pF switch and for a T of three, from a buffered source. Both floors have no frequency in them at first: 70.71 ppm for the lone switch and 6 ppb for the T into the largest load drawn. The lone switch's doubles by 1.23 kHz, a few times above the 318 Hz at which its off-capacitance's reactance equals its off-resistance, and reaches one per cent at 6.43 MHz, the frequency at which its band closes. The T's doubles by 662 Hz, sooner, because its tiny floor is overtaken by a small reactive leak first. — Fig. 3 The smallest worse-of-two error any load up to 1 GΩ gives, against frequency, for the same switch alone and as a T. The lone switch’s floor is 70.71 ppm at low frequency, doubles by 1.23 kHz, a few times above the 318 Hz at which its off-capacitance’s reactance equals its off-resistance, and reaches 1% at 6.43 MHz, where its band closes. The T’s floor is 6 ppb into the largest load drawn and doubles by 662 Hz.

The lone switch’s floor holds flat to a few hundred hertz and then climbs, and it reaches one per cent at exactly the frequency its band was found to close at, 6.43 megahertz: a band closes where its floor reaches the tolerance, which is the same statement made from the floor’s side. The T’s floor, six parts per billion into a gigohm, begins to rise sooner, at 662 hertz, because a floor that low is overtaken by the first femtoamperes of capacitive leak long before a floor of seventy parts per million notices anything.

That is the practical reading. A T’s direct-current floor is spectacular and fragile; it describes a precision measurement at a few hertz into a very high impedance, and it is gone by audio frequencies. The lone switch’s floor is modest and robust, and for most of its band it is the thing that sets the resolution.

Three switches as a T: the band closes at 702 MHz rather than 6.43 MHz. computed by solving, not by drawing. The band of load resistance in which a T of three 0.5 Ω, 100 MΩ, 5 pF switches is within 1% of ideal in both states, from a 0 Ω source, against the band of one such switch. Closed, the T puts two on-resistances in series, so its lower edge is 99 Ω against 49.5 Ω. Open, it has no upper edge on the axis below 637 MHz, because the shunt switch holds the node between the two open ones and that node leaks the on-resistance over the off-impedance whatever the load (the fraction over 2π·Rₒₙ times the off-capacitance is 637 MHz). The T's band closes at 702 MHz, balanced at 99.6 Ω; the single switch's at 6.43 MHz. — Fig. 4 The band the T has at one per cent, beside one switch’s: closed, two on-resistances in series put its lower edge at 99 Ω against 49.5 Ω; open, it has no upper edge below 637 MHz because the shunt switch holds the node between the two open ones; and its band closes at 702 MHz, balanced at 99.6 Ω, against one switch’s 6.43 MHz.

At one per cent the T’s advantage is all at the top of the frequency axis, a band that closes a hundred times later. At seventy parts per million its advantage is all at the top of the load axis. The two figures are one arrangement doing the same thing — moving where the leaked current lands — and paying for it in the same coin, a second on-resistance in the path.

A floor that is a fraction, and a floor that is a voltage

Converters meet several floors, and it is worth being exact about which kind this one is. The total that has no resistor in it is a floor in volts: a sampling capacitor holds $\sqrt{kT/C}$ of noise whatever signal it holds, so a large signal rises clear of it and a small one does not. The floor a converter sets is a floor in counts, a quantisation step fixed by the number of bits. The switch’s floor is neither. It is a fraction of the signal — the closed switch loses 70.71 parts per million of whatever it carries and the open switch lets through 70.71 parts per million of whatever it blocks — so a larger signal does not rise clear of it. It limits resolution at every signal level alike, which makes it a gain error and a crosstalk specification rather than a noise specification, and it is the reason the number is quoted in bits rather than in microvolts.

It is also the one of the three that improves with arrangement rather than with size. A larger capacitor lowers kT/C; more bits lower the quantisation step; neither a larger nor a smaller load lowers the switch’s floor, because the load is the variable it has already been minimised over. Only a different part or a different arrangement moves it, and the next figure measures both.

The floor against the part

Every factor of a hundred in a switch's off-to-on ratio buys one switch 3.3 bits and a T 6.6. The direct-current floor against a switch's own ratio of off- to on-resistance, from 2 × 10⁴ to 2 × 10¹²: 1/(1 + √(Rₒff/Rₒₙ)) for one switch at its best load, and Rₒₙ/(Rₒₙ + Rₒff) for a T into a load large beside Rₒff, the first held to the solved network at the 0.5 Ω, 100 MΩ part. One switch's floor falls as the −0.4996 power of the ratio and a T's as the −1.0000 power, so each factor of a hundred buys 3.32 bits alone and 6.64 as a T. At a ratio of 2 × 10⁸ they are 13.79 and 27.58 bits. — Fig. 5 The direct-current floor against a switch’s own ratio of off- to on-resistance, from 2 × 10⁴ to 2 × 10¹²: $1/(1 + \sqrt{R_\mathrm{off}/R_\mathrm{on}})$ for one switch at its best load, held to the solved network at the drawn part, and $R_\mathrm{on}/(R_\mathrm{on} + R_\mathrm{off})$ for a T into a load large beside $R_\mathrm{off}$ . One falls as the −0.4996 power of the ratio and the other as the −1.0000 power, so each factor of a hundred buys one switch 3.32 bits and a T 6.64; at 2 × 10⁸ they are 13.79 and 27.58 bits.

The two slopes turn a switch’s datasheet into a resolution in one line. A switch’s off-to-on ratio is the number that sets its floor, and the arrangement decides the exponent. Improving the ratio a hundredfold — a part with a hundredth of the leakage, or a hundredth of the on-resistance — buys a lone switch 3.32 bits and a T twice that. An on-resistance halved buys a lone switch half a bit.

It also says what the floor is not about. There is no capacitance in either expression, no source impedance and no frequency; the floor is the resistive part of the switch arguing with itself. The tolerance that is not on any part found an R–2R ladder to be a twelve-bit converter only while its resistors are inside 0.14 per cent, a limit set by matching; a converter whose channels are selected by this switch meets a second limit set by leakage, at 13.79 bits, and whichever is lower is the one it has.

The exponents are also a small instance of the rule a boundary is a model and a tolerance draws: the power with which an edge moves against its tolerance names the mechanism behind it. Here the thing that moves is the floor, the thing it moves against is the part’s own ratio, and the exponent names the arrangement. A half says one switch whose two errors are balanced against each other; a one says two leaks in series, each a fraction of the one before, so that the fractions multiply. Given only a floor and how it moved when the part was changed, the arrangement could be read back.

In a multiplexer the picture has one more term. Where an open switch leaks to found the open channels’ leak landing on the selected channel’s source rather than on the load, so the direct-current leak into that source is set by the source’s own resistance against the open channels’ off-resistance, once for every open channel. The floor for a lone switch between a source and a load is therefore the best case, and a bank of them has a floor that grows with the number of channels and with the source impedance — which is the frequency-free half of the result that essay measured at a megahertz.

What a resolution asks of the part

The two closed forms the figures are held to turn a required resolution into a requirement on the part, and the requirement is steep. Fourteen bits is 61 parts per million, and a lone switch reaches it only if its off-to-on ratio is at least $(1/\varepsilon - 1)^2$ , about 2.7 × 10⁸ — this switch’s 2 × 10⁸ falls just short, which is the 13.79 bits. Sixteen bits is 15 parts per million and needs a ratio of 4.3 × 10⁹: half an ohm on and more than two gigohms off. Twenty bits needs 1.1 × 10¹², and no analogue switch sold is that. A lone switch runs out of ratio a few bits past fourteen, and six decibels a bit is the reminder of how quickly each of those bits doubles the demand.

A T asks something different. Its floor into a large load is the ratio itself, so this switch as a T could in principle serve twenty-seven bits — but only into a load that keeps $2R_\mathrm{on}/R_L$ under the tolerance. Sixteen bits from the T needs a load above about 65 kilohms, and twenty bits one above a megohm. So the T moves the requirement from the part to the load: it turns a switch that cannot reach sixteen bits into one that can, provided whatever follows it presents tens of kilohms or more, which a buffer’s input does and a converter’s sampling capacitor, at direct current, also does. A T into a few kilohms is the one arrangement that is worse than the part it was built from.

How the numbers were obtained

Each error is a solve of the switch network — one switch, or three as a T — against the same source and load with a wire where the switching is, so that every error is a fraction of what an ideal switch would deliver. The closed switch’s error is counted as the vector difference, which at direct current is the same as the shortfall in amplitude. The lone switch’s best load is found by bisecting on where its two errors cross, and it is held to the closed form for the floor and to the geometric mean for the load; at every load on the first figure both arrangements are held to their closed forms to a part in a million. Loads are solved up to a gigohm, where the spread of conductances in the matrix is still one the solver’s own current-law check certifies, and the T’s curve is continued past that from its closed form. The floor against frequency is the smallest worse-of-two error over a grid of loads at each frequency, and for the lone switch the balance is bisected.

What it does not say

It holds the on-resistance fixed. A real switch’s on-resistance moves with the signal it carries, so its closed error is a distortion rather than a constant fraction, and a floor set by a constant fraction is the best the switch could do if its on-resistance kept still.

It models the off state as a resistance and a capacitance. The leakage of a real switch is a current that depends on temperature and on the voltage across it, doubling every several degrees, and a floor set by a leakage current is a function of temperature and signal level that the resistance here stands in for at one condition.

And the T’s shunt switch is the same part as its series switches. A larger shunt switch, with less on-resistance and more capacitance, would lower the T’s direct-current floor and raise its capacitive leak, and the two pull in opposite directions.

Still open: a shunt switch that is a different part, an on-resistance that moves, and a leakage that is a current

The shunt switch sized on its own. The T’s floor is set by the shunt’s on-resistance against the series switches’ off-resistance, and its high frequency is set by the shunt’s off-capacitance hanging on the closed path. A larger shunt switch improves the first and worsens the second. Sweeping the shunt’s size against the series switches’ would find the size that makes the T’s band widest at a given tolerance, and would say whether that size moves with the source impedance.

An on-resistance that depends on the signal. Give the closed switch a resistance that varies across the input range and the lone switch’s closed error becomes a curve rather than a constant, its floor becomes a distortion floor, and the load that balances it is no longer the geometric mean. Measured, it would say how many of the 13.79 bits survive a switch whose resistance varies by ten per cent across its range.

A leakage that is a current. Replacing the off-resistance with a current source that doubles every ten kelvin turns the floor into a function of temperature. It would put a temperature on the thirteen-bit multiplexer, above which it is a twelve-bit one, and it would say whether the T’s square-law advantage survives when both of its leakages rise together.

Part 5 on ideal switch

One argument about Ideal switch, and one of 8 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it: