The shunt switch the source sizes

A T of switches is two switches in series with a third from the node between them to ground. When the path is closed the two series switches conduct and the third is open; when the path is open the series switches are open and the third is closed, so whatever leaks through the first open switch lands on half an ohm instead of reaching the load. The capacitance a third switch moves measured what that arrangement buys and what it costs, and every number in it came from three identical switches: half an ohm closed, a hundred megohms and five picofarads open.

Nothing requires the three to be alike. On a chip a switch is a transistor of a chosen width, and a wider transistor has proportionally less resistance when it conducts and proportionally more capacitance when it does not. The shunt switch is a different job from the series switches — it carries no signal, only the leak — and there is no reason in advance to think the part that suits the signal path suits the shunt.

The question has a clean shape because the shunt’s size pulls against itself. A larger shunt holds the node between the two open switches harder, which improves the open T. A larger shunt, when it is itself the open one, hangs more capacitance on the closed path, which worsens the closed T. There is a size at which those two balance, and this essay finds it.

The band three identical switches give #

The measure of a switch here is the one a band rather than an edge set out. Closed, the switch should deliver the drive to the load to within a tolerance; open, it should let through no more than that tolerance of it. Each requirement is met by a range of loads, and the band is the loads that meet both. As the frequency rises the capacitance leaks more and loads more, the band narrows, and it closes at the frequency where no load at all meets both requirements. That frequency is the single number by which one arrangement of switches is compared with another.

Three identical switches from a buffered source close their band at 702 megahertz, a hundred and nine times later than one switch. Throughout this essay the closed state is counted as the T’s bands always have been counted so far — as a shortfall in amplitude — until a later section asks what changes when it is counted as a waveform.

Two errors that pull the size opposite ways #

Give the shunt switch a size $s$: a shunt $s$ times as wide as a series switch has $R_\mathrm{on}/s$ closed, and $s$ times the capacitance and conductance open. At any one frequency the T’s two errors can then be drawn against $s$, each at the load that suits it best.

The two errors cross, and the best a single load can do sits on the worse of them. At 1.20 gigahertz from a buffered source the least error, 0.443 per cent, needs a shunt five times the width of a series switch, and any shunt from twice to 6.3 times the width keeps the T inside one per cent. Three identical switches are outside that range at 1.20 gigahertz; they closed their band at 702 megahertz.

The mechanism is visible in the two slopes. The open T leaks through the first series switch’s capacitance into the shunt’s on-resistance, so the node between the switches carries a fraction $\omega C_\mathrm{off}R_\mathrm{on}/s$ of the drive, and the second open switch passes that on: it falls as the shunt grows. The closed T has the shunt’s capacitance $sC_\mathrm{off}$ from its middle node to ground, charged through the first series switch. Because that current is at right angles to the drive, it takes from the delivered amplitude only at second order: $\tfrac12(\omega sC_\mathrm{off}R_\mathrm{on})^2$, which rises as the square of the shunt’s size.

Drag the frequency and the optimum moves: about ten times a series switch at 150 megahertz, about eight at 300, about six at 600, and about four at 2.40 gigahertz, where the best is 1.11 per cent and no size meets the tolerance. (The sizes are read off a grid of ten per decade, which is why they are round.) The band closes at the frequency at which the optimum error reaches the tolerance, and the shunt size at that frequency is the size worth building.

The size that closes the band last #

Set both errors equal to the tolerance $\varepsilon$ and eliminate the frequency. From $\omega C_\mathrm{off}R_\mathrm{on}/s = \varepsilon$ and $\tfrac12(\omega sC_\mathrm{off}R_\mathrm{on})^2 = \varepsilon$,

$$s^\ast = \left(\frac{2}{\varepsilon}\right)^{1/4}, \qquad f^\ast = \frac{\varepsilon\,s^\ast}{2\pi R_\mathrm{on}C_\mathrm{off}}.$$

At one per cent that is a shunt 3.761 times a series switch and a closure at 2.39 gigahertz, from nothing but the part’s time constant and the tolerance.

Solved over every load at every frequency, the best shunt from a buffered source is 3.775 times a series switch, and the band closes at 2.40 gigahertz — 3.42 times later than three identical switches manage. The closed form was 3.761 and 2.39. The small difference is the load: the closed form assumes one large beside everything else, and the solved optimum is found where the best load is a finite resistance.

The curve is not symmetric about its peak and it is not smooth there. To the left the band is closed by the open state, which a smaller shunt holds too weakly; to the right it is closed by the closed state, which a larger shunt loads too heavily. The peak is the corner where the two limits swap, which is why the curve comes to a point.

The source moves it as a square root #

The buffered source is the one case that leaves the source out, and the figure’s other two curves say it is not the typical one. From fifty ohms the best shunt is a third of a series switch; from a kilohm, a twelfth.

The reason is where each error’s current flows. The open T’s leak lands on the shunt’s on-resistance whatever drives the first switch, because the first switch’s capacitance is a far higher impedance than any ordinary source. The closed T’s capacitive current is drawn through the source and the first switch, so the second error carries $(R_s + R_\mathrm{on})$ where the first carries only $R_\mathrm{on}$. Put that into the balance and the optimum picks up the square root of their ratio:

$$s^\ast = \left(\frac{2}{\varepsilon}\right)^{1/4}\sqrt{\frac{R_\mathrm{on}}{R_s + R_\mathrm{on}}}.$$

A larger source makes the closed state’s capacitance more expensive, and the cure is a smaller shunt.

Solved from one ohm to three kilohms, the best shunt follows the root law closely at both ends and departs from it by up to sixteen per cent in the middle — 0.225 against 0.265 at a hundred ohms. The departure is the load again. From a middling source the best load at closure is comparable with the source, and the closed form’s assumption of a load large beside everything is exactly what fails there.

The band the optimum buys is the second curve, and it is not monotonic. From a one-ohm source the best shunt closes the band 2.00 times later than three identical switches; from ten ohms only 1.22 times, because ten ohms is close to the source for which identical switches are already the best choice. The root law puts the optimum at exactly one series switch’s width where the square root cancels the fourth root, at $R_s = R_\mathrm{on}\bigl(\sqrt{2/\varepsilon} - 1\bigr)$ — 6.57 ohms at one per cent — and the solve crosses one between its three-ohm and ten-ohm points, at 1.423 and 0.817. Past that, the gain grows steadily: 5.31 times at a hundred ohms, 13.1 at a kilohm, 24.8 at three. A T driven from a kilohm with an identical shunt closes its band at 4.53 megahertz; with a shunt a twelfth the width, at 59.4.

An optimum that has to be hit exactly is not much use to anyone laying out a chip, so the width of the window matters as much as its centre. At half its best closure frequency, from fifty ohms, the T is inside one per cent for any shunt from 0.079 to 0.631 of a series switch — a factor of eight in width — and least wrong, at 0.448 per cent, near 0.40. The same reading from a buffered source at half its closure gave a window from 2.00 to 6.31, a factor of three, and from a kilohm at 29.7 megahertz the window runs from 0.032 to 0.126, a factor of four, with the least error, 0.318 per cent, at a tenth of a series switch.

None of those windows contains a shunt the same width as the series switches. That is the practical result, and it does not depend on hitting the root law to two figures: from any real source, at any frequency near where the T’s band is under pressure, the shunt that works is a fraction of the series switch and the identical one does not. The closed form’s job is to say which fraction to aim at, and the width of the window says how little the aim has to be trusted.

That last figure — the kilohm source’s gain of thirteen — is the one worth dwelling on. The shunt switch in a T is usually thought of as the part that must be good — low resistance, to hold the node. Driven from a real source, the best shunt is a small switch, and the reason is that a large one’s capacitance costs more on the closed path, through the source, than its resistance saves on the open one.

Counted as a waveform, the root goes #

The fourth root of the tolerance came from one assumption, and it is worth undoing. The closed state was counted as a shortfall in amplitude, and a capacitive current at right angles to the drive shortens the amplitude only at second order. Count the closed state as a waveform instead — the whole vector error, magnitude and phase together — and the same current costs at first order, $\omega sC_\mathrm{off}(R_s + R_\mathrm{on})$.

Now the balance is between two first-order errors, and the tolerance drops out of the size:

$$s^\ast = \sqrt{\frac{R_\mathrm{on}}{R_s + R_\mathrm{on}}},$$

which is never more than one. From a buffered source it is one: three identical switches are exactly the best T, and the solve returns 1.000. The rule every T is drawn by turns out to be the optimum — for a buffered source, and for a waveform.

From anything else the shunt should be smaller, and the gain is much larger than it was for amplitude. From fifty ohms the best shunt is 0.063 of a series switch and closes the band at 103 megahertz, sixteen times later than identical switches’ 6.30. From a kilohm, 0.025 of one, at 25.5 megahertz, eighty times later than 318 kilohertz. Identical switches counted this way barely do better than a changeover from a real source, because the shunt’s capacitance hangs on the path exactly as a changeover’s open channel does; a shunt a fortieth the size removes most of that capacitance and keeps enough resistance to hold the node.

The solved optimum follows the closed form within 37 per cent from one ohm to three kilohms — closely at the ends, worst between ten and thirty ohms, where the best load and the source are again comparable. The gain it buys runs from 1.73 at one ohm to 132 at three kilohms.

So the two criteria agree on the direction — from any real source, a smaller shunt — and disagree about the size and about whether identical switches are ever right. Where the mechanisms are one mechanism is the general statement of why they must: a capacitive error is at right angles to the drive, and a tolerance that does not say whether it is counted in amplitude or as a vector has not yet said which of these two optima it is asking for.

A tighter tolerance wants a larger shunt #

The amplitude optimum carries a fourth root of the tolerance, and that is a prediction worth checking, because it says a design asked for more precision should use a larger shunt.

At a tenth of a per cent the best shunt from a buffered source is 6.690 times a series switch, where the closed form gives $(2000)^{1/4} = 6.687$, and the band closes at 426 megahertz against the closed form’s 425.7. From fifty ohms it is 0.665 times, where the closed form gives $6.687 \times 0.0995 = 0.665$, closing at 42.4 megahertz. At this tolerance the closed form is exact to three figures from both sources, because the best load at closure is larger and the load term has receded.

The ratio of the two tolerances’ optima, 6.690 over 3.775, is 1.77, and the fourth root of ten is 1.78. The band closes six times later than identical switches from a buffered source at a tenth of a per cent, against 3.4 times at one per cent: the tighter the tolerance, the more a shunt sized for it is worth. The floor below any load found the T’s floor at direct current set by the shunt’s on-resistance against the series switches’ off-resistance, and there the shunt’s size trades the same way it does here: a larger shunt holds the open node harder and its own off-resistance, hanging on the closed path, leaks more. At direct current the two meet at a shunt of exactly one series switch, because resistance scales with width on both sides of the balance; it is the capacitance’s second-order cost to an amplitude that moves the optimum above one at high frequency.

What this does not settle #

The part is a scaled transistor, and a real switch is not quite one. Every conductance and capacitance here scales exactly with width. A real switch’s capacitance has a fixed part — pads, wiring, the package — that does not shrink with the transistor, and a shunt a fortieth the width of its neighbours would be dominated by it. The optimum from a kilohm is a statement about the scalable part of the capacitance, and a layout that cannot make a switch that small has a floor under $s$ that this essay does not draw.

The charge the switches move is not here. A T changes the state of three switches at every transition, and the offset that knows the signal measures the charge a switch injects as it turns off, which scales with its size. A smaller shunt injects less into the node it holds, which is a further argument in the same direction, unmeasured.

Every number above a few hundred megahertz is a lumped statement. Kirchhoff’s own frequency puts a length on that: the 2.40 gigahertz a buffered, amplitude-optimised T reaches holds only for switches whose connections are a small fraction of a centimetre. The optimum’s size survives that caveat better than its frequency does, because it is a ratio of two errors that both scale with the same frequency.

Still open: an on-resistance that moves, a leakage that is a current, and a shunt sized for the source it will meet #

An on-resistance that depends on the signal. Everything here takes a switch’s on-resistance as a number. A real one varies across the signal range, so the closed state’s error at direct current is a curve rather than a constant, and part of it — the part that is the same at every level — could be calibrated away while the rest cannot. The resistance that bends the signal separates the two and asks what the floor becomes when only the uncorrectable part counts.

A leakage that is a current. The open switch here leaks through a resistance and a capacitance, both of which a shunt switch can hold. A switch’s junction leakage flows to its supply whatever the signal is, and the leak no switch can hold asks whether the T’s advantage survives a leak that never has to cross an open switch.

A source impedance that is not a resistance. The optimum moved as the root of a source resistance. A multiplexer’s source is often a buffer whose output impedance rises with frequency, and a shunt sized for its low-frequency resistance is sized for the wrong source at the frequency where the band closes. Sizing the shunt against an inductive source is the same balance with a frequency inside the square root, and it may not have a single answer.