Where an open switch leaks to

The width no load can change found that a switch’s band — the range of loads over which it is within one per cent of ideal both open and closed — has the same width in decades whether it drives a resistor or a hold capacitor, and closes at the same 6.43 megahertz for both. The reason was a product. Closed, the switch is wrong by about its on-resistance over the load; open, it lets through about the load over its off-impedance; multiply the two and the load cancels.

That essay ended on the assumption the product rests on. The open switch’s current was taken to flow through the load, and that is true only of a switch standing alone between one source and one load. Put the same switch in a bank that shares an output and the current an open channel lets through has somewhere else to go.

A multiplexer is that bank: eight switches, one output, one of them closed and seven open. The question A band rather than an edge first asked of it was how much the seven open ones cost, and the answer it first gave was the natural one and the wrong one.

The lone switch’s curve rises with the load, because a larger load gives the leaked current a larger resistance to develop a voltage across. The multiplexer’s does not move across six decades of load. Whatever limits a multiplexer, at a hundred kilohertz it is not the resistance it drives.

Forty picofarads is the wrong sum #

The arithmetic that comes first to hand counts capacitances. Seven open switches share the output node, each has its off-capacitance connected to that node, and eight channels together make forty picofarads where one switch had five. A switch with forty picofarads across it has a band, and the band closes somewhere.

Eight hundred and four kilohertz is a plausible number with the right shape — more channels, lower frequency — and it is the number this collection first printed for the case. The count behind it is not wrong about the capacitance. It is wrong about what the capacitance is connected to.

A lone switch’s off-capacitance sits between the source and the load. Driving it from the source pushes a current through it and into the load, and the load resistance turns that current into the error. The seven open channels of a multiplexer are each connected between their own source and the shared output, and the shared output is not floating on the load. It is tied, through half an ohm of closed switch, to the source of the selected channel. The leaked current has two ways to ground — through the load, or back through the closed channel into its source — and it takes the one with less impedance, overwhelmingly.

The node the closed channel holds #

Treat the output node as the one voltage the circuit is solving for. The selected channel connects it to its source through the source impedance and the on-resistance in series, Rₛ + Rₒₙ. The load connects it to ground. Each open channel connects it to that channel’s own source through an off-impedance, which above the switch’s corner is a capacitance.

Drive the open channels, hold the selected source still, and the voltage that appears at the node is the open channels’ current divided by everything the node sees. With a load large against the closed path, that is very nearly a ratio of two admittances: the open channels’ (n − 1)·jωCₒff against the closed channel’s 1/(Rₛ + Rₒₙ). So the leak is (n − 1)·ωCₒff·(Rₛ + Rₒₙ), and it reaches a fraction ε at

$$f = \frac{\epsilon}{2\pi\,(n-1)\,C_{off}\,(R_s + R_{on})}$$

There is no load in it. There is a source impedance in it, and the channel count enters as n − 1 rather than n: the selected channel’s own off-capacitance is shorted by its own on-resistance and takes no part.

The figure solves the network rather than the expression, and a separate nodal solution of the same circuit, written without reference to the figure, puts the eight-channel limit from a buffered source at 90.95 megahertz against the expression’s 90.95. That is fourteen times above the lone switch’s closure rather than eight times below it, and the count of capacitances was low by a factor of a hundred and thirteen.

From a fifty-ohm source the same eight channels reach one per cent at 901 kilohertz, and there the count of capacitances is within eleven per cent. That is not the count being roughly right. The two land near each other only because fifty and a half ohms of source and switch happens to be close to a hundred on-resistances, which is the factor the lone switch’s closure hides in it. From a kilohm of source the multiplexer leaks one per cent at 45.5 kilohertz and the count still says 804.

The two-channel case is worth a sentence of its own, because a changeover is the smallest multiplexer. From a buffered source its limit, 637 megahertz, is ε/(2πCₒffRₒₙ) — the frequency at which the one open switch’s reactance has fallen to a hundred on-resistances — and it is ninety-nine times the lone switch’s closure. From fifty ohms it falls to 6.30 megahertz, just below the lone switch’s. A changeover is better than a lone switch or no better than one according to what drives it, and nothing about the changeover decides which.

No load resistance in the edge #

A hundredfold change of load and nothing moves. That is the absence the first figure showed at one frequency, now shown at every frequency and every source: a lone switch’s upper edge is a load resistance, and a multiplexer’s is not a load resistance at all.

It changes what a designer can do about it. For a lone switch the load decides where in the band the two errors balance and nothing about how much room there is. For a multiplexer the load does not even decide that. Provided it is large against the closed path — which the closed state requires anyway, from 49.5 ohms up — its value is simply absent from the leak. The levers are the three quantities in the expression, and one of them appears on nobody’s data sheet.

This is the divider, and the thing it does not know about in a different position. There a ratio designed to be blind to magnitude turned out to depend on nothing else once a load was connected. Here an isolation that a data sheet describes as a property of the part turns out to depend on a magnitude outside it: the output impedance of whatever happens to be driving the channel that is selected.

The first power of the tolerance #

A tenfold tighter tolerance moves the multiplexer’s limit tenfold and the lone switch’s closure a hundredfold. 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, and here two exponents name two arrangements of one part.

The lone switch’s closure is a meeting of two edges — the closed switch’s error falling as the load rises and the open switch’s rising with it — and both have to be inside the tolerance at once, so the frequency at which no room is left goes as the square of the tolerance. The multiplexer has no such meeting. Its leak does not depend on the load, so there is no trade to exhaust: there is one error, proportional to frequency, and it reaches ε at a frequency proportional to ε.

So the advantage is not a fixed factor either. At one per cent eight buffered channels have fourteen times the lone switch’s frequency; at a tenth of a per cent they have 143 times it, and each further decade of precision multiplies the ratio by ten. A front end asked to settle to a few parts per million is far into the regime where the arrangement of the switches, rather than the switches themselves, decides the frequency.

The closed channel fails at the same frequency #

The figures so far have asked about the open channels’ leak. The closed channel has an error of its own, and the leak arithmetic points straight at it.

The seven open channels hang their off-capacitances on the output node whether or not their sources are moving. With those sources still, the seven capacitances are a shunt to ground on the node the selected channel drives through Rₛ + Rₒₙ, which makes a low-pass, and the selected signal arrives slightly late and slightly small. Its vector error is the same ratio as the leak — (n − 1)·ωCₒff against 1/(Rₛ + Rₒₙ) — because it is the same two admittances meeting at the same node, one of them now loading instead of driving.

The separate nodal solution confirms it to every figure it prints. For eight channels the closed path is one per cent wrong as a waveform at 90.95 megahertz from a buffered source, where the leak reaches one per cent at 90.95, and at 900.6 kilohertz from fifty ohms against a leak at 900.5. Across two, eight and sixteen channels the ratio of the two frequencies is 1.0000 from buffered and fifty-ohm sources, and 1.0010 from a kilohm, where the megohm load has begun to be visible beside the closed path.

So a multiplexer’s band is a different kind of object from a lone switch’s. A lone switch has a range of loads that narrows with frequency until it shuts. A multiplexer driving any load above its lower edge has, to first order, a single frequency, and at that frequency it stops being a multiplexer in both states together: the selected channel is one per cent wrong and the others are one per cent present. Below it the load may be anything the closed path tolerates; above it no load helps and neither state is good.

Alike, unrelated, and one #

The leak figures drive all seven open channels with the same signal, which is the worst case and not the usual one. A multiplexer scanning eight sensors usually has eight unrelated voltages on its inputs.

The three cases are three readings of one fact. The network is linear, so the leak from seven channels is the sum of seven leaks, and each is the same because each open channel is the same network — which is two solves that add applied to the quantity that is safe to add, a voltage. When the seven signals are identical the seven leaks are identical phasors and add as numbers, to seven times one channel’s. When they are unrelated the voltages still add, but what a measurement averages is their power, and independent powers add, so the total goes as the root of the sum of squares: √7 times one channel’s.

That is the distinction where the mechanisms are one mechanism drew for an amplifier’s errors, arriving here with the arithmetic already settled by what the signals are. Because the leak is proportional to frequency, a factor in amplitude is the same factor in frequency. Unrelated channels reach one per cent √7 = 2.647 times later than correlated ones, and a single driven channel seven times later, at 638 megahertz — within a fifth of a per cent of the changeover’s 637, as it should be, since one driven channel beside the selected one is a changeover whose six idle neighbours contribute only their share of the node.

Doubling the bank from eight channels to sixteen divides the correlated limit by 2.15, which is fifteen over seven, and the uncorrelated one by 1.47, close to the square root of that. So the worst case gets worse with the channel count as its first power and the typical case as its square root, and which of the two a design should be held to is a statement about the signals rather than about the part. A bank of slowly scanned thermocouples is the unrelated case. A bank whose inputs all carry the same interference — one supply’s ripple, one fast edge coupled onto every cable — is the correlated case for that interference, whatever the wanted signals are.

Fifty ohms, and the impedance nobody states #

A hundred times the leak for fifty ohms of source is the expression’s Rₛ + Rₒₙ moving from half an ohm to fifty and a half, with nothing about the multiplexer changed. The quantity that sets it belongs to the circuit upstream of the switch, and it is the one term of the three that a switch’s data sheet cannot state, because the data sheet does not know what will drive the part.

That is sharper than it sounds, because “buffered” is not a number. A source that holds its voltage whatever current is taken from it is the source that is not a source, and an amplifier’s output is that kind of source only below a frequency. A source below a frequency found a regulator whose output impedance is 0.43 milliohms at direct current, has doubled by 4.8 hertz and reaches 1.95 ohms at ten kilohertz, and a feedback buffer behaves the same way: its output impedance rises as its loop gain falls. The buffered curve in these figures is therefore a limit no real driver reaches at the frequencies where it matters, and the real limit is wherever the driver’s own output impedance at that frequency puts the curve. That impedance is not modelled here, and nothing on this page says what it is for any particular part.

What the expression does say is which way to lean. The leak is proportional to Rₛ + Rₒₙ, so a channel driven from a sensor with a kilohm of output impedance leaks about two thousand times more than a well-buffered one, and buffering each input is worth that factor. Lowering the on-resistance, which is the switch parameter a designer usually shops for, is worth almost nothing once the source impedance is larger than it.

What the leak arithmetic leaves out #

The switches are three passive numbers. The offset that knows the signal measures the charge a switch’s gate dumps at the instant it turns off, which in a multiplexer lands on the shared node at every change of channel. That is a transient at the switching instant rather than a leak at a frequency, and nothing here contains it. Nor does an on-resistance that moves with the signal.

The load is a resistor. A multiplexer in front of a converter usually drives a sampling capacitor, and that capacitor is a further admittance on exactly the node the leak lands on. It would lower the leak at high frequency and add the acquisition error the hold capacitor brought to the lone switch; the combination is not solved here.

And the circuit is lumped. Every number above describes a network whose connections are short against a wavelength, and Kirchhoff’s own frequency is where that stops being true of a board. A track in ordinary laminate is a tenth of a wavelength long at 650 megahertz when it is 2.2 centimetres long, so the 637-megahertz changeover limit is a statement about a part and its own pins, and the 91-megahertz eight-channel limit about a layout of no more than about fifteen centimetres. The coupling between the tracks of such a layout is the far end that cancels, a different mechanism with a frequency of its own, and a real board has both.

Still open: where the leak can be made to land #

The whole of the multiplexer’s difference from a lone switch came from giving the leaked current somewhere other than the load to go, and the somewhere was a source impedance nobody chose. The next questions choose it.

A third switch. Between two series switches, put one to ground that is closed when the path is open and open when it is closed. The current the first open switch leaks then lands on half an ohm of closed shunt switch whatever the source is, and still has a second open switch to cross before it reaches the load. The capacitance a third switch moves measures what that arrangement buys — isolation into fifty ohms to 643 megahertz from a fifty-ohm source, where a changeover manages 6.34 — and what it costs. The shunt switch’s own capacitance now hangs on the closed path, and this essay’s last measurement says that a capacitance hanging on a closed path fails it at about the frequency at which the same capacitance leaks.

The floor below any load. Each switch model so far has let the off-resistance recede behind the off-capacitance. At direct current it does not recede, and there the lone switch’s two edges are ninety-nine on-resistances and a ninety-ninth of the off-resistance. They meet when the tolerance falls to the square root of the ratio of the two resistances, 70.71 parts per million for this part, and below that no load at any frequency makes the switch ideal enough. That is a boundary in the resolution a switch can serve rather than in frequency, and the only one among these switch edges with no frequency in it.

A bank driving a capacitor. The sampling capacitor a multiplexer usually feeds sits on the node the leak lands on, which makes the leak and the acquisition two readings of one node’s admittance. Solved together they would say which of the two a fast scanning converter runs out of first, and whether the answer depends on the source impedance the way the leak alone does.