The leak no switch can hold

Every open switch in these essays has leaked the same way: through a resistance and a capacitance across its terminals, so that what it lets through is a fraction of the signal across it. That is the leak the floor below any load balanced against the closed switch’s resistance to find 70.71 parts per million for a lone switch, and it is the leak a T was built to defeat. A T’s shunt switch holds the node between its two open series switches at half an ohm, so a leak that has crossed the first open switch lands there and has to cross a second before it reaches the load. That is why a T’s floor at direct current is the square of the lone switch’s resistance ratio: five parts per billion, 27.6 bits.

A switch made of transistors has a second leak, and it is not across the switch. The junctions between each terminal and the substrate conduct a small reverse current to the supply, whatever the signal is and whether the switch is open or closed. It is a current rather than a fraction, it is a property of each terminal rather than of the path, and like any junction’s reverse current it roughly doubles for every ten kelvin.

The whole question is where that current lands. For a lone switch the output terminal is on the load. For a T, the shunt switch holds the middle node — but the outer series switch’s terminal is on the load too, and its current does not have to cross anything to get there.

The T with a junction current in it #

The figures give every switch terminal the same junction current, 100 picoamperes at 25 °C unless a slider says otherwise, and solve the network at direct current with that current flowing in each direction, keeping the worse. The signal is one volt full scale, from a buffered source, and each error is a fraction of it: closed, how far the load’s voltage is from the drive; open, how much voltage appears on the load anyway.

Without the current, the T’s worse error falls at every larger load and never turns: it is 15 parts per billion into a hundred megohms and still going down, towards five. With the current, the same curve turns at a hundred kilohms and climbs, and the bottom is 9.998 parts per million. The T has a best load again, which is exactly the property the resistive analysis said it lacked.

The arithmetic is short. Open, the outer switch’s junction current $I$ flows into the load and puts $I R_L$ on it, a fraction $I R_L/V$ of full scale — rising with the load, as the lone switch’s resistive leak did, and crossing no switch at all. Closed, the T’s error is its two on-resistances, $2R_\mathrm{on}/R_L$, falling with the load. They balance at

$$R_L^\ast = \sqrt{\frac{2R_\mathrm{on}V}{I}}, \qquad \varepsilon_T = \sqrt{\frac{2R_\mathrm{on}I}{V}},$$

which for 100 picoamperes is 100 kilohms and $10^{-5}$ — the figure checks the best load against the closed form to five per cent and it lands on it. The T’s floor is a square root again, and the quantity inside it is an on-resistance times a current: the shunt switch appears nowhere in it.

The lone switch hardly notices. Its junction current adds to its open leak as a conductance, $I/V$ beside $1/R_\mathrm{off}$, and at 25 °C 100 picoamperes at one volt is a hundredth of what a hundred megohms conducts. Its floor moves from 70.71 to 71.06 parts per million. So at room temperature the T still wins, by 2.8 bits rather than by 13.8: 16.61 bits against 13.78.

What ten kelvin does #

The current doubles every ten kelvin and the resistances here do not, so the two floors respond to temperature in different ways.

The T’s floor is the square root of the current from the start, so it climbs from the start: half a bit for every ten kelvin, because a doubling of the current is a factor of $\sqrt2$ in the floor. It is below sixteen bits by 37.2 °C, twelve kelvin above room temperature.

The lone switch’s floor is flat until the current matters, and the current matters when $I/V$ reaches $1/R_\mathrm{off}$ — ten nanoamperes at one volt, a hundredfold increase from 100 picoamperes, which is $10\log_2 100 = 66.4$ kelvin of doubling. Above that it too climbs at half a bit per ten kelvin. It drops below thirteen bits at 101.3 °C and below twelve at 125.9 °C.

The two curves cross, and where they cross has a closed form of its own. A T’s floor is $\sqrt{2R_\mathrm{on}I/V}$ and a lone switch’s $\sqrt{R_\mathrm{on}(1/R_\mathrm{off} + I/V)}$; they are equal when $2I/V = 1/R_\mathrm{off} + I/V$, which is exactly when the junction current equals the off-resistance’s conductance at full scale. For this switch that is at 91.4 °C, and the figure checks that the solved crossing is within three kelvin of it.

Above 91.4 °C a T of three switches is a worse switch than one. It has two on-resistances in its closed path, which a lone switch does not, and it holds nothing the junction current has to cross, so the only thing it contributes at that temperature is its second on-resistance. A multiplexer rated to 125 °C and built as T’s for isolation is, at its rating, 0.43 of a bit worse at direct current than the same switches used singly.

The resistive analysis this overturns was built up over several measurements, and each of them survives in its own domain. A band rather than an edge drew the two load edges a switch lives between; the width no load can change showed the room between them fixed by the product of the two errors; and the floor was where that room closed. A junction current changes none of those arguments. It adds a conductance, $I/V$, to the open arm — and for a T it adds it to the one arm the shunt switch cannot reach. Temperature, which the edges that move with the room found moving model edges through a device’s own constants, here moves an edge through a leakage that doubles every ten kelvin.

At 85 and 125 °C #

At 85 °C, the top of the usual industrial range, the current is 6.4 nanoamperes. The T’s best load has fallen from 100 kilohms to 12.5, because the junction current into the load now costs more at every resistance, and its floor is 80 parts per million. The lone switch’s is 90.55 at 5.52 kilohms. The T is still ahead, by 0.18 of a bit, and the difference is no longer worth three switches and the charge they inject.

At 125 °C the current is 102 nanoamperes and the order has reversed: the T is at 320.1 parts per million into 3.13 kilohms and the lone switch at 237.1 into 2.11. The best loads have fallen by factors of four and two and a half from 85 °C, because both floors are now increasingly set by the current, and the ratio of the two floors, 1.35, is closing on $\sqrt2$ — the second on-resistance, and nothing else, which is what it reaches once the off-resistance is forgotten.

The load that suits a switch therefore moves with temperature, and in the direction a designer would not choose. A precision path sized for its best load at 25 °C into a hundred kilohms is, at 85 °C, eight times past its new best load on the side where the junction current dominates, and its error there is the current times the load the designer chose.

A best load that walks #

Between room temperature and 85 °C the T’s best load does not jump; it walks. At 50 °C, with 566 picoamperes, it is 42.1 kilohms and the floor 23.78 parts per million. At 25 °C it was 100 kilohms, at 85 °C 12.5. The closed form says it falls as the reciprocal square root of the current, so by $\sqrt2$ for every ten kelvin, and the three readings fall by 2.38 over twenty-five kelvin and by 3.37 over thirty-five, against $2^{1.25} = 2.38$ and $2^{1.75} = 3.36$.

The lone switch’s best load barely walks at all over the same range — 7.04, 6.88 and 5.52 kilohms — because until 91 °C its open arm is still mostly the off-resistance, which does not care about temperature. That difference is the practical content of the two analyses. A lone switch can be sized for its load once. A T sized for its best load at one temperature is sized for the wrong load at every other, and on the side where its error grows as the load does.

The mechanism also explains why the resistive analysis could not have seen this coming. It found the T’s error falling at every larger load, and concluded that a T wants a large load. A large load is exactly what a junction current punishes, because the current’s error is the current times the load. The two conclusions are both correct, for two different leaks, and the load that is right for a real T is decided by which of the two is larger at the temperature it runs at.

A leak measured against the signal, not the scale #

Every error here is a fraction of full scale, one volt, and that choice flatters the junction current in one way and hides it in another.

The resistive leak scales with the signal on the other channel: a channel carrying a tenth of full scale leaks a tenth as much, and the error it causes is a fixed fraction of whatever is being switched. The junction current does not scale with anything. It puts $I R_L$ on the load whether the signal is a volt or a millivolt, so as a fraction of the signal actually present it grows as the signal shrinks. A measurement taken at a tenth of full scale sees the junction current’s error ten times larger relative to what it is measuring, and the resistive leak’s no larger at all.

That is why a sensor multiplexer, which switches millivolts from a thermocouple or a bridge, cares about junction leakage out of all proportion to what the full-scale floors above suggest. In the terms of this essay, the crossing at which a T becomes worse than one switch is where the current equals the off-resistance’s conductance at the signal voltage. At one volt that is ten nanoamperes; at ten millivolts it is a hundred picoamperes, which this switch reaches at 25 °C. For a ten-millivolt signal, a T of these switches is no better than one switch at room temperature.

The half-bit-per-ten-kelvin rate is unchanged by the signal level, because it comes from the square root of a doubling. What moves is where on the temperature axis the curves start to climb, and a smaller signal moves it down by $10\log_2$ of the ratio of full scale to the signal — sixty-six kelvin for a hundredfold smaller signal.

What a T is still for #

None of this makes a T the wrong arrangement; it makes it an arrangement for a frequency. The capacitance a third switch moves found the T’s band closing at 702 megahertz against a lone switch’s 6.43 from a buffered source, because what leaks across an open switch at high frequency is capacitive and a held node stops it. A junction current is direct current: at a megahertz the capacitive leak through five picofarads is thirty-one microamperes per volt, and a hundred nanoamperes of junction current is not in the same conversation.

So the T keeps its purpose where it was bought — isolation at radio frequency, where a shunt switch holding a node is worth more than a hundred times the band — and loses it where the resistive analysis promised the most, which was precision at direct current. A switch that has to do both needs the choice made by frequency and temperature together, and the two measurements now exist to make it.

How much junction leakage a T can afford #

The hundred picoamperes in every figure so far is a round number, and the answer depends on it in a way that is worth drawing out rather than rescaling in the head.

A tenth of the current buys the T $\sqrt{10}$ in its floor, 1.66 bits: 3.16 parts per million, 18.27 bits, at room temperature. It moves the crossing 33.2 kelvin later, to 124.7 °C, which is $10\log_2 10$ — every factor of ten in leakage is a third of a hundred kelvin on the temperature axis. The lone switch is unchanged at 25 °C, 70.74 parts per million, because its floor was never the current’s.

Ten times the current costs the T 1.66 bits the other way, 31.62 parts per million, and brings the crossing down to 58.2 °C. The lone switch now feels it too, at 74.16 parts per million at room temperature, and below twelve bits at 92.7 °C.

So the T’s advantage has a clean statement. At direct current it is worth $\tfrac12\log_2\bigl((1/R_\mathrm{off} + I/V)/(2I/V)\bigr)$ bits, which depends on the ratio of the junction current to the off-resistance’s conductance and on nothing else: 4.5 bits when the current is a thousandth of that conductance, 2.8 when it is a hundredth — the two room-temperature figures above — and zero where the two are equal. The 13.8 bits that the resistive analysis promised belong to a switch whose junctions do not conduct.

What the leak is, and what it is not #

It is an offset, and an offset calibrates — at one temperature. A junction current into a fixed load is a fixed voltage whatever the signal, so a zero-scale measurement removes it. But it doubles every ten kelvin, so a calibration taken at 25 °C is wrong by the whole of the current’s growth at 35. The resistance that bends the signal split the closed error into a gain that calibrates and a curvature that does not; the junction current’s calibratable part is exactly the part that does not stay put. An instrument that re-zeroes often enough can recover most of the resistive floor. One that does not carries the figures above.

Its sign is not known. The current flows to whichever supply the junction is reverse-biased towards, and in a switch made of a complementary pair the two transistors’ junctions point opposite ways and partly cancel. The figures take the worse of the two signs at every terminal, which is the budget a designer without a characterised part has to use; a part whose leakages are known to cancel would do better, until the temperature changes which one dominates.

It is a statement about a buffered source. In a multiplexer every channel’s output terminal is on the shared output node, but the selected channel holds that node through its on-resistance, so from buffered sources the summed junction current of the off channels lands on half an ohm rather than on the load. From a source with impedance, the same current lands on that impedance instead. Where an open switch leaks to found that for the resistive leak the source takes the load’s place in a multiplexer, and the junction current should follow the same rule; nothing here measures it.

And it is direct current. At a frequency the capacitive leak dominates a switch long before the junction current matters, and the capacitance a third switch moves and the shunt switch the source sizes are the measurements that apply there. The junction current is the limit on precision at low frequency and high temperature, which is where a thermocouple multiplexer or a bridge sensor lives.

Still open: the multiplexer’s source, a T whose middle leaks, and the part that is characterised #

A multiplexer from sources with impedance. Eight channels put eight junction currents on the output node, and from sources of a kilohm the selected channel’s source impedance carries them. That turns the junction current into an error that depends on the source rather than on the load, and it should give a multiplexer a temperature at which its resolution falls a bit per channel count doubled as well as half a bit per ten kelvin. Solving a bank of switches from real sources with junction currents at every terminal would put that temperature on a part.

The T’s middle node. Three junction currents land on the node the shunt switch holds, and at half an ohm they are negligible here. A shunt switch sized small for high frequency — a twelfth of a series switch from a kilohm, in the shunt essay — holds that node at six ohms, and a shunt sized small enough for its capacitance may no longer hold its junction currents. The shunt’s best width is then a three-way balance, and whether it still has a closed form is unmeasured.

A part whose leakages are known. Every figure here assumes the worst sign and one current per terminal. A real switch’s datasheet quotes the leakage of the off switch and of the on switch separately, at two or three temperatures, and they do not double at exactly ten kelvin. Fitting those numbers and replacing the doubling rule with the part’s own would test whether the 91-degree crossing is a property of switches or of the round numbers chosen here.