The capacitance that is already a difference

The carrier a cable allows read a strain-gauge bridge through ten metres of cable with an alternating excitation, and found that the cable’s capacitance mostly does not matter. Each output terminal sees the two arms meeting there in parallel against its own capacitance to the screen, so each lags; but both lag by the same amount, and a bridge reads the difference between them. Matched, ten metres of cable at a hundred picofarads a metre leaves 0.15 degrees at a hundred kilohertz where each terminal has 6.3, and allows a carrier of 205 kHz for two hundred parts per million of loss. What does not cancel is a mismatch between the two conductors, and it arrives multiplied by the ratio of the excitation to the signal — two thousand at a thousand microstrain — so one per cent of mismatch takes the allowed carrier down to 1.8 kHz.

That essay named what it had left out. A screened pair has two kinds of capacitance, not one. Each conductor has capacitance to the screen, which is what was measured. The two conductors also have capacitance to each other, typically about half as much per metre, and that one sits directly across the bridge’s output. It is differential by construction. There is nothing for it to cancel against, and it is not amplified by the signal’s smallness. So the question is how large it is against the mismatch, and whether anything can be done about a capacitance that is already the quantity being measured.

The capacitance that cannot cancel #

The bridge is the one the earlier essay used: four 350 Ω arms, one of them the gauge, read at a thousand microstrain with a gauge factor of two, so the gauge’s fractional change $\delta$ is 0.002 and the output is five millivolts on a ten-volt excitation — the $V\delta/4$ that the bridge that is linear near one point found low by half the change, which at these strains is a part in a thousand and does not matter here. The cable is ten metres of screened pair with 100 pF a metre from each conductor to the screen, written $C_s$ below, and 50 pF a metre between the two, $C_p$. The detector is synchronous, so what it loses to a phase $\varphi$ is $1 - \cos\varphi$, and the budget is two hundred parts per million.

With the conductors matched to the screen, the pair’s own capacitance allows 18.2 kHz and the screen allows 205. Together they allow 18.2: the limit is entirely the capacitance between the conductors, and it is eleven times tighter than anything the earlier essay found for a matched cable. Every carrier frequency the earlier essay said a matched cable allows is an overestimate by that factor for a real pair.

The reason is the output’s own source resistance, and it is the same mechanism the probe is part of the circuit measures for a probe’s capacitance against the node it touches. Seen from the two output terminals, the bridge is its differential signal behind two half-arms in series — 175 Ω on each side, 350 in total — and a capacitance across those terminals is a single-pole low-pass on the signal itself, with its corner at $1/(2\pi \cdot 350 \cdot 500\text{ pF})$, 909 kHz. For a loss of 200 ppm the phase may be 0.02 radians, which is reached at a fiftieth of that corner, 18.2 kHz. Nothing about the excitation or the strain enters that calculation, and nothing about the two conductors’ matching either.

Two phases with one slope #

The earlier essay’s mismatch and this essay’s pair capacitance both lag the reading in proportion to frequency at low frequency, which is what a small phase from a resistance and a capacitance does. The interesting part is what each is proportional to.

The pair’s phase is $\omega R C_p$, with $R$ the 350 Ω the output presents: a resistance and a capacitance, and no signal anywhere in it. The mismatch’s phase is $\omega R\,\Delta C/\delta$, and the derivation is short. A capacitance $\Delta C$ more on one output terminal than the other draws a quadrature current from that terminal’s half of the excitation, which is five volts, and puts a quadrature voltage of about $\omega (R/2) \Delta C \times 5\text{ V}$ into the difference. The signal it sits beside is $10\text{ V} \times \delta/4$. Their ratio is $\omega R\,\Delta C/\delta$, the common-mode lag divided by the fractional signal. It is the conversion the rejection four resistors decide prices for a difference amplifier’s own resistors, and the corner the instrument has no part in for a source’s imbalance, arriving here from the cable’s side. The solves agree with both first-order forms to 0.59 per cent up to ten kilohertz.

So the two phases have the same slope and a fixed ratio, $\Delta C/(\delta\, C_p)$, and at a kilohertz a one per cent mismatch is 0.0110 radians where the pair is 0.00110 — ten times the pair’s, because one per cent of 1000 pF is ten picofarads and the signal divides it by 0.002. Neither phase depends on how long the cable is except through the capacitances, and the same ratio holds at every frequency where a first-order form holds at all.

Where one overtakes the other #

A fixed ratio means there is a mismatch at which the two phases are equal, and below it the pair’s capacitance is the whole of the cable’s effect. Setting $\omega R\,\Delta C/\delta = \omega R C_p$ and writing the mismatch as a fraction $m$ of the screen capacitance gives

$$m^{*} = \delta\,\frac{C_p}{C_s}.$$

The measurement puts the crossings at 0.030, 0.10 and 0.30 per cent for three hundred, a thousand and three thousand microstrain, which is $\delta C_p/C_s$ with $C_p/C_s$ a half. This is the number a cable specification would need and does not print. A pair whose conductors are matched to the screen better than the signal times the ratio of its two capacitances is pair-limited, and any further matching buys nothing. A pair matched worse than that is mismatch-limited, and the pair’s own capacitance can be ignored.

For a strain gauge at a thousand microstrain the dividing line is a tenth of a per cent. The earlier essay found a tightly twisted pair matched to a fraction of a per cent over ten metres, and a connector whose pins differ by a picofarad adds a tenth of a per cent of a thousand picofarads on its own. So a well-made screened pair on a gauge at working strain sits near the crossing, and the two effects are the same size — which is the case where the next thing matters most.

A mismatch that cancels #

Both phases were drawn as magnitudes. They have signs, and the signs are not the same.

The pair’s capacitance lags the signal, as any shunt capacitance across a resistive source does, and a quadrature component in a phasor sum has a sign as well as a size, which three voltages that close on one draws for a network of three. A mismatch lags the common-mode voltage on the terminal that carries more capacitance, and what that does to the difference depends on which terminal it is. Extra screen capacitance on the gauge’s own output terminal pulls that terminal’s quadrature in the direction that opposes the pair’s lag; on the other terminal it adds to it. So a mismatch on the gauge’s side of exactly $\delta C_p/C_s$ does not merely equal the pair’s phase. It cancels it.

The curves are sharp peaks, sharper than the 0.025 per cent grid they are drawn on. At a thousand microstrain the mismatch that nulls the phase, 0.10 per cent on the gauge’s terminal, takes the carrier allowed from 18.2 kHz to 143 kHz, and the same 143 kHz at three hundred and at three thousand microstrain: at the null the first-order terms are gone and what limits is the second-order remainder, which does not depend on the strain. The same tenth of a per cent on the other terminal halves the limit instead, to 9.08 kHz, and three tenths of a per cent on the gauge’s side at a thousand microstrain overshoots the balance and allows 9.11 kHz, worse than none. The mismatch that nulls the phase moves with the strain, 0.030 per cent at three hundred microstrain and 0.30 at three thousand, and in each case it is $\delta C_p/C_s$ to a tenth of a per cent of itself.

This is what the carrier a cable allows proposed as a trim, and what it actually does. It suggested a small adjustable capacitance brought against the detector’s quadrature output, which is proportional to the residual, so that the adjustment has a null to trim to. It does — but with a pair capacitance present the null is not the match. The quadrature output is zero where the mismatch’s lead cancels the pair’s lag, and that is a deliberate mismatch of $\delta C_p/C_s$ on the gauge’s side, not an equality of the two screen capacitances. Carrier-frequency amplifiers carry exactly this control, usually called a capacitance balance, and it is set by nulling the quadrature with the bridge loaded.

When it is set matters. With the gauge unstrained the signal is zero, the pair capacitance puts no quadrature into the output, and nulling the quadrature drives the mismatch to zero: a true match, and a pair-limited cable at 18.2 kHz. Set with the gauge loaded to $\delta_0$, the null lands at $\delta_0 C_p/C_s$ and the pair’s phase cancels at that load. The same knob and the same procedure produce two different instruments depending on the strain applied at the time.

Balanced for one direction #

A balance set at one strain is read at others, and the residual phase is easy to write. With the mismatch fixed at $\delta_0 C_p/C_s$, the mismatch’s phase at a strain $\delta$ is the pair’s times $\delta_0/\delta$, with the opposite sign, so the total is the pair’s phase times $(\delta_0/\delta - 1)$.

In tension that factor is smaller than one in magnitude for every strain above half the set point, so the balance helps everywhere above 500 microstrain when it is set at a thousand, and helps most at the set point, 143 kHz against 18.2. Below half the set point it is worse than no balance, since the mismatch’s phase grows as the strain falls and eventually exceeds the pair’s.

In compression the strain changes sign and so does the mismatch’s phase. The pair’s does not, because it is a low-pass on the signal whatever the signal’s sign. So the two add, and the balance hurts at every compressive strain: at −1000 microstrain the limit is 9.11 kHz, half what the untrimmed cable allows. The slider moves the set point to 300 or 3000 microstrain and the shape is the same — a factor of two of useful range above the set point in tension, and nothing in compression.

The design rule that falls out is short. A capacitance balance suits a gauge read in one direction over a range of strain, and should be set at the harmonic mean of that range’s two ends, which makes $|\delta_0/\delta - 1|$ equal at both; the residual is then the pair’s phase times $(\delta_{\max} - \delta_{\min})/(\delta_{\max} + \delta_{\min})$. A gauge read in both directions — a bending beam, a torque shaft — cannot be balanced this way at all, and has to live with the pair capacitance or reduce it, the same sign-dependence the carrier a cable allows found in the mismatch’s amplification by the reciprocal of the strain.

Length, which is not the question #

A cable’s length multiplies both capacitances, so it is natural to ask how long a cable a given carrier allows. The answer is simpler than the question.

Every limit falls as one over the length, including the matched screen’s, whose phase is second order but a function of $\omega R C$ alone, so frequency and capacitance enter together. The ratios between the limits therefore hold at every length: the one per cent mismatch allows exactly a tenth of what the pair allows, at a metre and at a hundred. Which capacitance binds is a property of the cable per metre and of the strain, never of how much cable there is.

One thing does depend on length. A balance is a ratio, $\delta C_p/C_s$, but it is made with a capacitor of a definite size: 0.10 per cent of ten metres of screen is one picofarad. On a cable twice as long the same ratio needs two, so a balance set for one cable must be reset when the cable is changed, even if the strain range is the same.

What a designer should take #

Quote a screened pair’s two capacitances and compare their ratio with the signal. The pair capacitance sets a carrier limit of about $0.02/(2\pi R C_p)$ for a 200 ppm loss that no matching can improve; the screen mismatch sets one that is $C_p\delta/(\Delta C)$ times that. If the conductors are matched better than $\delta C_p/C_s$, stop improving the match and look at the pair capacitance instead.

For a gauge read in one direction, a capacitance balance on the gauge’s terminal cancels the pair’s phase and raises the carrier limit roughly eightfold. Set it with the gauge loaded to the harmonic mean of the working range, not unloaded, since an unloaded balance finds the match and leaves the pair uncancelled. For a gauge read in both directions, use a carrier low enough for the pair alone.

The earlier essays of this sequence each converted a quantity that must be small into two that must be equal: the leads that are in the bridge by moving a lead into the adjacent arm, and the two leads nobody counts by sensing the excitation at the bridge. The pair capacitance resists that conversion because it is already a difference, and the only thing that can be made equal to it is a second difference of the opposite sign — which exists, and holds for one sign of the signal only.

How the numbers were obtained #

The bridge and cable are one netlist solved at each frequency: the excitation source, four resistors of 350 Ω with the gauge’s arm at $350(1 + \delta)$, a capacitance from each output terminal to ground standing for the screen, with any mismatch added to one side, and a capacitance between the two output terminals. The reading’s phase is measured against the same bridge’s output at a hundredth of a hertz, which removes the quarter bridge’s sign. The carrier allowed is bisected in log frequency to the point where $1 - \cos\varphi$ reaches 200 ppm, seventy iterations. Each balance is located by bisection on the sign of the phase at a kilohertz over mismatches from zero to one per cent, fifty iterations, and each crossing in the overtaking figure is interpolated in log mismatch between the two solves that straddle it.

The closed forms are the first-order expansions stated above, and they are checked against the solves rather than used to draw anything; the lines in the closed-form figure are the expressions, and the dots beside them are the netlist.

What it leaves out #

The cable as a line. Ten metres is about a twentieth of a wavelength at a megahertz and far less at the carrier frequencies here, so the capacitances are lumped. The pair’s inductance, which forms a series resonance with its capacitance well above these frequencies, is not in the model.

The amplifier’s own input. A carrier amplifier’s input capacitance adds to the pair capacitance if it is differential and to the screen capacitance if it is common-mode, and a real front end has both. The analysis is unchanged if they are folded into $C_p$ and $C_s$, which is how the earlier essay treated the connector.

The gauge’s own reactance. A foil gauge has a few microhenries and a few picofarads, both of which the earlier essay estimated to be negligible below a few tens of kilohertz. At the 143 kHz a balanced pair allows, the gauge’s inductance becomes a phase of its own in the one arm that has it, which is a differential effect of the same kind as the pair’s and would need its own balance.

Still open: the reference taken at the wrong end, a half bridge, and the capacitive sensor #

Where the detector’s reference is taken. Every phase here is measured against the excitation as the bridge sees it. A detector that takes its reference at the instrument sees the excitation leads’ own lag in the reference and not in the signal, and that lag has the same form as the pair’s — a resistance and a capacitance — so it either adds to the pair’s phase or is available to cancel it. Six-wire sensing with the reference taken from the sense wires is the arrangement that would decide which.

A half bridge, with two active arms. With the second active arm on the other output terminal, the signal doubles and the source resistance seen by the pair capacitance is unchanged, so the pair’s phase stays and the crossover mismatch doubles. Whether a half bridge in bending — one arm in tension, one in compression — can be balanced, when a quarter bridge in compression cannot, turns on which terminal each arm’s signal appears at.

And the capacitive sensor. A displacement sensor is a capacitance read by this same arrangement with the roles exchanged: the sensor is the capacitance and the cable’s resistance and the bridge’s arms are the parasitics. The signal-divided mismatch and the undivided differential term should both reappear, and whether a balance for one direction of travel is the whole of what is possible there too is the question worth drawing.