Circuits that do a job, and the range they do it over

The sensor that is the capacitance

Read a capacitive displacement sensor through the same kind of cable that limited a strain-gauge bridge's carrier, and the roles exchange. On the bridge, a 1% mismatch in the cable's capacitance was a quadrature leak that grew as the square of the carrier and could be escaped by a lower one. On a sensor of two 10 pF plates it is 3.0% of full-scale displacement, in phase, at every carrier from 100 Hz to 1 MHz — because a capacitance between an excitation conductor and the sense conductor is in parallel with the measurand and is a sensor capacitance. No detector phase separates it and no carrier escapes it; only geometry does. Screen each conductor separately and there is nothing left to mismatch.

15 min read 5 figures The parasitic is the component The schematic is a label Every model has an edge

Assumes: The bridge that is linear near one point · The current that does not reach the input

Four essays have read a resistive strain-gauge bridge through a cable, and each found the cable’s capacitance mattering less than its first appearance suggested. The carrier a cable allows found the capacitance to the screen cancelling in the bridge’s difference when the two conductors match. The capacitance that is already a difference found the capacitance between the conductors limiting instead, and what a calibration at the carrier removes found a calibration taking most of that away, leaving the mismatch between the conductors as the only thing that cannot be removed. In all four the cable’s capacitance works against a resistance — the bridge’s arms — and its effect is a phase, mostly in quadrature, growing with frequency.

The same essays named the inversion without drawing it: a displacement sensor is a capacitance read by this same arrangement with the roles exchanged. The sensor is the capacitance, and the resistances are the cable’s. Whether the conclusion about cable matching survives the exchange is the question, and it does not.

The sensor and its cable

The sensor is the standard differential arrangement. Two plates, driven in antiphase at ±5 V, face a common electrode that moves between them; the capacitances are $C_0(1 + x)$ and $C_0(1 - x)$ with $C_0$ = 10 pF, so the common electrode collects a charge $2C_0 x V$ proportional to the displacement $x$ , and at the centre it collects nothing. A charge amplifier holds the sense conductor at virtual earth and turns that charge into a voltage.

The cable carries three conductors: two for the antiphase excitation and one for the sense. Each conductor has capacitance to the screen — 100 pF a metre for the sense conductor — and, in a shared cable, capacitance to the others: 30 pF a metre between each excitation conductor and the sense conductor. The cable is two metres, and every conductor has a ohm of resistance.

The cross-capacitance is a sensor capacitance

A capacitance from an excitation conductor to the sense conductor carries charge from ±5 V into the virtual earth exactly as the sensor’s own plates do. The circuit has no way of telling a picofarad in the cable from a picofarad in the sensor: both connect the same two nodes. So the two cross-capacitances, 60 pF each over two metres, sit in parallel with the sensor’s two 10 pF plates, and their difference is read as a displacement.

Fig. 1 The reading of a centred capacitive sensor — two 10 pF plates at ±5 V — through two metres of shared three-conductor cable whose two cross-capacitances differ by 1 per cent, as an apparent displacement against the carrier. Solid: the in-phase part, 3.0 per cent of full scale at every frequency. Dashed: the quadrature part the leads’ resistance makes, rising with frequency. The slider is the mismatch.

A one per cent mismatch is 0.6 pF, and against the sensor’s 20 pF of span it is 3.0 per cent of full scale: $\Delta C_x/2C_0$ , which the figure checks to one per cent. It is flat across four decades of carrier, and it is in phase with the signal, because a capacitance in parallel with a capacitance adds to it with no phase at all. The quadrature part — the leads’ ohm against the capacitances — is $7.5\times 10^{-6}$ per cent at 10 kHz and a synchronous detector rejects it.

Every property the resistive bridge’s mismatch had is reversed here. There it was in quadrature and here it is in phase. There it rose as the square of the carrier and here it is flat. There a synchronous detector with a good reference rejected its first-order part and here nothing can, because it is signal. There a lower carrier escaped it and here no carrier does.

The same 1% of cable mismatch: a loss that rises as f² on a resistive bridge, a 3.0% offset at every frequency on a capacitive sensor. computed by solving, not by drawing. The error a 1 per cent mismatch in a cable makes, against the carrier: on a 350 Ω quarter bridge at 1000 µε through ten metres of screened pair, the loss 1 − cos φ from the screen mismatch (dashed), which rises as the square of the frequency; on a capacitive displacement sensor of two 10 pF plates through two metres of three-conductor cable, the apparent displacement from the cross-capacitance mismatch, as a fraction of full scale (solid), which is 3.00 per cent at every carrier. On the bridge the cable's capacitance is across a resistance and its mismatch is in quadrature; on the sensor it is in parallel with the measurand and its mismatch is in phase. The two cross at 23.7 kHz. — Fig. 2 The same one per cent of cable mismatch on two transducers. Dashed: the resistive quarter bridge at 1000 µε through ten metres of screened pair, its loss rising as the square of the carrier. Solid: the capacitive sensor through two metres of shared cable, a flat 3.0 per cent of full scale. The two cross at 23.7 kHz; below it the bridge is the better-behaved of the two, and above it the sensor is.

Why the exchange reverses everything

The reason is one sentence of network analysis, and it is worth having because it generalises. A parasitic is benign when it is in series with the quantity’s own impedance and malign when it is in parallel with it.

On the resistive bridge the measurand is a resistance and the cable’s capacitance is across it. At low frequency a capacitance across a resistance is an open circuit, so it vanishes; it appears only through the product $\omega RC$ , which is small and makes a phase. On the capacitive sensor the measurand is a capacitance and the cable’s capacitance is in parallel with it. A capacitance in parallel with a capacitance adds, at every frequency, with no product to make it small. The cable is not a parasitic on the sensor; it is part of it.

The cable’s resistance now plays the role the capacitance played before. It is in series with the sensor’s capacitance, so it appears through the same product $\omega RC$ , as a phase, small and in quadrature — the leads’ ohm against 20 pF is a corner at eight gigahertz. The resistive parasitic that dominated the leads that are in the bridge and the two leads nobody counts is irrelevant here, and the capacitive parasitic those essays could calibrate away is the whole problem.

A fixed offset, and why it is not fixed

A mismatch that never changed would be an offset, and an offset can be zeroed: centre the sensor, read the output, subtract it for ever. That is what an instrument’s zero adjustment does and it would remove the 3 per cent at installation.

It is not fixed, and the reason is that the cable’s cross-capacitance depends on things a cable does. It depends on the spacing of the conductors, which a cable flexing under vibration changes; on the dielectric’s permittivity, which drifts with temperature by parts in ten thousand a kelvin; and on how the cable lies, since a conductor pressed nearer another has more capacitance to it. Each of those moves the two cross-capacitances by slightly different amounts, and every move is read as a displacement.

Every metre of shared cable adds 30 pF across the sensor, so 0.1% of full scale needs the cable matched to 0.067% on one metre. computed by solving, not by drawing. The apparent displacement of a centred capacitive sensor, two 10 pF plates, read through a shared three-conductor cable with 30 pF a metre between each excitation conductor and the sense conductor, against the cable's length, for cross-capacitance mismatches of 0.1 and 1 per cent. The offset grows in proportion to length: at one metre a 1 per cent mismatch reads as 1.5 per cent of full scale. Holding the offset to 0.1 per cent of full scale needs the two cross-capacitances matched to 0.067 per cent on a metre and 0.0067 per cent on ten. — Fig. 3 The apparent displacement of the centred sensor through shared cable, against the cable’s length, for cross-capacitance mismatches of 0.1 and 1 per cent. The offset grows in proportion to length — 1.5 per cent of full scale for 1 per cent on one metre — and holding it to 0.1 per cent of full scale needs the cross-capacitances matched to 0.067 per cent on one metre and 0.0067 on ten.

The offset is proportional to the cable’s length, because the cross-capacitance is. On one metre a one per cent mismatch is 1.5 per cent of full scale; on ten it is fifteen. To keep the offset — or any change in it — under a tenth of a per cent of full scale, the two cross-capacitances have to match, and stay matched while the cable moves, to 0.067 per cent on a metre and 0.0067 on ten. No cable is specified to that, because no cable is made to it; a cable’s capacitance tolerance is a few per cent.

The cure is geometry

If the cross-capacitance cannot be matched, it has to be removed, and that is a question of where the field lines go rather than of any circuit. A conductor inside its own screen has capacitance only to that screen: the screen intercepts every field line that would have reached another conductor, and there is no cross-capacitance at all. Three separately screened conductors — three coaxial cables, or a cable with an individual screen on each core — carry no capacitance from excitation to sense, and the figure’s check confirms it: with separate screens the apparent displacement is zero to the solver’s precision, at every mismatch, because there is nothing left to mismatch.

What each screen then adds is capacitance from its own conductor to the screen, and that is harmless for the two excitation conductors, which are driven by stiff sources. For the sense conductor it matters differently, and it is the one place in this circuit where the readout decides.

The sense conductor’s screen, and the readout that ignores it

The sense conductor’s capacitance to its screen — two hundred picofarads over two metres — is in parallel with the readout’s input. What it does depends on what the readout holds the sense conductor at.

A voltage readout's sensitivity is divided by the cable; a guard or a virtual earth takes the cable out. computed by solving, not by drawing. The sensitivity of the capacitive sensor's reading against the capacitance from its sense conductor to the screen, with the cross-capacitances removed, for three readouts: a voltage amplifier with an earthed screen (dashed), which falls to 3.9 per cent of its own value between 10 pF and 1 nF because the cable forms a divider with the sensor; the same amplifier with the screen driven from its own output (dotted), which bootstraps the capacitance away; and a charge amplifier (solid), whose input is at virtual earth and whose sensitivity does not move at all. — Fig. 4 The sensitivity of the reading, with separate screens so that no cross-capacitance is present, against the sense conductor’s capacitance to its screen, for three readouts: a voltage amplifier with the screen earthed, which falls to 3.9 per cent of its own value between 10 pF and 1 nF; the same amplifier with the screen driven from its output; and a charge amplifier, whose sensitivity does not move.

A voltage amplifier with the screen earthed reads the sense conductor’s voltage, and the screen capacitance forms a divider with the sensor: the sensitivity is $2C_0/(2C_0 + C_s + C_{in})$ , which falls to 3.9 per cent of its own low-capacitance value by a nanofarad of cable. Worse, the screen capacitance drifts with the cable as the cross-capacitance did, and here its drift is a gain drift.

A voltage amplifier with the screen driven from its own output holds the screen at the sense conductor’s voltage, so no current flows into the screen capacitance and it disappears. That is a guard, and the current that does not reach the input is the same trick used against leakage: drive the surroundings to the potential of the node, and nothing flows between them. The amplifier’s own input capacitance is not guarded and remains.

A charge amplifier holds the sense conductor at virtual earth. The screen is also at earth, so there is no voltage across the screen capacitance and nothing flows in it; the sensitivity is independent of it, to a part in a million in the solve. The node that is at ground for a while is how good that virtual earth is, and its limit — the amplifier’s finite gain at the carrier — is what would bring the screen capacitance back.

So the division of labour is exact. Separate screens remove the offset; a guard or a virtual earth removes the gain dependence. The first is the cable’s job and the second is the readout’s, and neither can do the other’s.

A 0.1% mismatch in the cable reads as 0.30% of full-scale displacement, at every frequency. computed by solving, not by drawing. A capacitive displacement sensor — two 10 pF plates driven at ±5 V in antiphase facing a common electrode — read by a charge amplifier through two metres of three-conductor cable with 30 pF a metre between each excitation conductor and the sense conductor. With the two cross-capacitances mismatched by 0.1 per cent and the sensor centred, the reading's in-phase part (solid) is 0.300 per cent of full scale at every carrier from 100 Hz to 1 MHz: the cable is a sensor capacitance, and its mismatch a displacement. The quadrature part (dashed), which the cable's resistance makes, rises with frequency and a synchronous detector rejects it. With each conductor in its own screen the cross-capacitance is gone and so is the error. — Fig. 5 A tenth of a per cent of cross-capacitance mismatch — better than any shared cable is made to — still reads as 0.30 per cent of full-scale displacement through two metres. The error scales exactly with the mismatch, so there is no amount of care in a shared cable that makes the problem small; only a cable with nothing to mismatch does.

Smaller sensors make it worse, in proportion

The offset is $\Delta C_x/2C_0$ , so it scales inversely with the sensor’s own capacitance, and that is the direction sensors are going. A 10 pF sensor is a large one: plates a centimetre across a tenth of a millimetre apart. A sensor for a smaller gap or a smaller target is a picofarad or less, and the same two metres of shared cable with the same one per cent of mismatch is then thirty per cent of full scale. The cable has not changed; the thing it is in parallel with has shrunk by ten, and the error has grown by the same ten without anything in the cable’s specification having moved.

The ratio worth carrying is the cable’s cross-capacitance per metre against the sensor’s span: thirty picofarads a metre against twenty here, a ratio of one and a half per metre, so every per cent of mismatch is one and a half per cent of full scale for each metre. A capacitive sensor of a picofarad a plate, read through one metre of shared cable, would need its cross-capacitances matched to about seventy parts per million to reach a tenth of a per cent of full scale, which is to say it cannot be read through shared cable at all. That is why capacitive gauging heads come with their own individually screened cable and a stated length: the cable is part of the calibration because it is part of the sensor.

The same mistake with a probe

There is a more familiar instrument in which a cable’s capacitance ends up in parallel with the quantity being measured, and it fails in the same direction for the same reason. The probe is part of the circuit measures a one-to-one oscilloscope probe on a two-kilohm source, and finds the hundred and fifteen picofarads at the end of its cable across the node, in parallel with whatever capacitance the node already has, so that the reading is one per cent wrong at 6.8 kHz while the instrument reads correctly. The probe that takes a tenth is the ten-to-one probe whose compensation network keeps its division flat, and it still cannot take its own capacitance off the node; it only makes that capacitance smaller. The remedy here is the same kind of admission: the cable’s capacitance cannot be compensated out of a capacitive measurement, because compensation adjusts a ratio and this error is an addition.

What the probe and the sensor share is the rule stated above in one sentence: a parasitic in parallel with a capacitance adds to it. The strain-gauge bridge was spared because its measurand was a resistance, and a capacitance across a resistance at a kilohertz is nearly nothing.

What the conclusion about matching became

The strain-gauge essays ended with a rule: buy the match, not the capacitance. After the exchange the rule is: buy no capacitance to match. The match that governed the resistive bridge’s carrier was between two capacitances to a screen, converting common mode into a quadrature leak; the match that governs the sensor’s offset is between two capacitances from the excitation to the signal, and it is signal. Matching was enough in the first case because the error was a leak the detector could discount. In the second nothing can discount it, and matching to the necessary precision is not available.

It is worth noticing which of the earlier results survived. The one that did is the structural one from the capacitance that is already a difference: a parasitic that has the same form as the quantity being measured cannot be separated from it by anything downstream. There it was the pair capacitance, already a difference, which a calibration eventually removed because it scaled with the strain. Here it is the cross-capacitance, already a sensor capacitance, which no calibration removes because it adds to the displacement instead of multiplying it.

What the offset is required to equal

The offset is required to equal $\Delta C_x/2C_0$ to one per cent at 10 kHz — the statement that the cross-capacitance is a sensor capacitance and nothing else — and to be the same at every carrier to two per cent below 300 kHz.

With separate screens it is required to be zero, at every carrier.

The resistive bridge’s loss is required to grow as the square of the carrier between 100 Hz and 1 kHz — a factor of a hundred, to five per cent — which is what makes a lower carrier an escape there and not here.

And the charge amplifier’s sensitivity is required not to move with the sense conductor’s screen capacitance, to a part in a million, while the earthed-screen voltage readout is required to lose more than nine-tenths of its sensitivity across the range.

Ideal plates, balanced excitation and a still cable

The sensor is an ideal pair of plates. A real capacitive sensor has fringing fields and its own stray capacitances to its housing, which are the same problem at the sensor’s end of the cable and have the same cure: a guard ring around the sensing electrode, driven at its potential.

The excitation is perfectly antiphase. An imbalance between the two excitation amplitudes makes the sensor’s own two capacitances read as an offset in the same way the cable’s do, and it is removed by trimming the excitation rather than by screening.

And the cable does not move in the solve. Everything here is a fixed cable; how much a cable’s cross-capacitance changes when it flexes is a property of its construction, and it is the number that would turn the length figure into a vibration specification.

Still open: the guard ring, the imbalanced excitation, and the cable in motion

The guard ring at the sensor. The sensing electrode’s fringing field reaches its housing, and a ring around it driven at the electrode’s potential makes the field between the plates uniform. How much of the sensor’s nominal capacitance the fringe is, and how much of it a ring of a given width removes, is a field solve — the sensor’s own version of the cross-capacitance.

Trimming the excitation instead. An antiphase excitation whose two amplitudes can be adjusted will null any fixed offset, cable and sensor together, by making the two charges balance at the centre. What it cannot null is an offset that moves, and the question is how much of a cable’s drift is common to its two excitation conductors, and so trimmed out, and how much is not.

The cable that flexes. A shared cable’s cross-capacitance moves when it bends, and a sensor read through one has a vibration sensitivity equal to that movement over $2C_0$ . Separate screens remove it by construction; a shared cable with a screen between each pair of conductors would too, and the question is how much screening a cable needs before its flexing is below the sensor’s own noise.