Circuit Response
every model has an edge, and the figure carries it
Concept

Common-mode rejection — where it appears

The ratio of what a difference amplifier does to a difference to what it does to a voltage arriving equally at both of its inputs. In a three-amplifier instrument four resistors decide the architecture's share of it, and the input amplifiers' mismatch sets a ceiling no gain reaches past.

Named by 13 essays across 3 fields — each of them below, with the objects they name alongside it.

The same millivolts, subtracted — and what four resistors leave behind. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz. The second curve is the same measurement made differentially: a unity-gain difference amplifier across the sensor, its four resistors carrying the worst-case skew 0.1% allows. It divides the error by 500.5 — 53.99 dB against a closed-form (1 + G)/(4·tolerance) of 53.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 998.9 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.

The rejection four resistors decide

Measuring a ten-millivolt sensor against a ground that somebody else's hundred milliamps is also using puts five hundred microvolts of their current into the reading. Subtracting the two ends of that conductor with a difference amplifier removes it — by 53.99 decibels with one-tenth-per-cent resistors, against a closed form of 53.98, which is a factor of five hundred and not a removal. The amplifier has nothing to do with it: the number is one plus the gain over four times the resistor tolerance.

instruments · Common-mode rejection A first stage of 100 buys 40.0 dB of rejection, and gives it back above 10.0 kHz. computed by solving, not by drawing. The rejection of a three-amplifier instrumentation amplifier against frequency, beside the one-amplifier difference stage it is built around. At low frequency the two differ by 39.99 dB against 20 log 100 = 40.00 dB, and the reason is that the input stage passes a common-mode voltage at exactly unity: the common-mode gain of the whole instrument is 1.998 mV/V, which is the difference stage's own. So the four resistors around the last amplifier decide the rejection and the two that set the gain do not — ten per cent between them moves it by less than a hundredth of a decibel. What ends it is bandwidth: above 10.0 kHz, which is the amplifier's gain–bandwidth divided by the gain that bought the rejection, the differential gain falls and the rejection falls with it at twenty decibels a decade.

The four resistors that decide, and the two that do not

A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.

instruments · Common-mode rejection Matched parts cost nothing; a 2 dB difference between them sets a 112 dB ceiling. computed by solving, not by drawing. The common-mode rejection of a three-amplifier instrumentation amplifier against the gain of its input stage, with each amplifier's own rejection in the netlist as an input-referred error of the common-mode voltage over the rejection. The architecture's own figure rises decibel for decibel with the gain, because the difference stage sees a larger differential signal beside the same common-mode one. The parts' contribution does not rise with anything, and the part of it that matters is not their rejection but the difference between their rejections: two amplifiers of 98 dB that are identical cost 0.000 dB, while 100 dB against 98 dB leaves a ceiling of 111.7 dB with no gain in it. The two mechanisms cross: below a gain of 1903 the four resistors decide everything, and above it more gain buys no more rejection at all — 111.9 dB at a gain of 100000, where the arrangement alone would have been worth 148. The one place the instrument beats its own floor is a gain of 1000, where the two errors cancel; that is a coincidence of signs and not something a design can hold.

The rejection the parts have

Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.

instruments · Common-mode rejection 95 dB of instrument, 290 Hz corner — and the corner belongs to the source. computed by solving, not by drawing. The common-mode rejection of the same three-amplifier instrument measured earlier measured, with 1 kΩ of imbalance between the two source resistances and 10 pF at each input. The instrument's own curve is drawn beside it. Below 290 Hz the two agree; above it the measurement falls at twenty decibels a decade while the instrument does not, reaching 84.0 dB at a kilohertz against the instrument's 95.0. What converts common mode into differential is the difference of the two input time constants — 10.0 ns here — and once it is differential no rejection repairs it.

The corner the instrument has no part in

Three earlier essays measured a three-amplifier instrumentation amplifier's rejection at direct current and found 95 dB, of which the resistors' matching decides one part and the amplifiers' own mismatch another. Connect it to a source with a kilohm of imbalance and ten picofarads at each input and the rejection has a corner at 290 Hz and falls twenty decibels a decade after it — reaching 84 dB at a kilohertz on an instrument that is still doing 95. What converts common mode to differential is the difference of two time constants, and the cure is a capacitor on the quiet input.

instruments · Common-mode rejection One part, two corners: 3.98 kHz to the common mode and 7.86 MHz to the signal. computed by solving, not by drawing. Two windings on one core with a coupling of 0.999, driven twice from the same netlist — once with the two conductors in opposition, which is the signal, and once with them in parallel, which is everything the cable picked up. The mode that goes the same way round both windings meets (1+k)L and is down three decibels by 3.98 kHz; the mode that goes opposite ways meets the leakage, (1−k)L, and is untouched until 7.86 MHz. The ratio is 1975, which is 2/(1−k) and contains no inductance at all. Neither number is computed here: both modes are driven and the answer is read.

The inductor one mode cannot see

Two windings on one core present a millihenry to a current that goes the same way round both and a microhenry to one that goes opposite ways, so the same component has a corner at 3.98 kHz and another at 7.86 MHz — a ratio of two thousand, which is 2/(1−k) and contains no inductance at all. It is bought to remove the conversion the previous essay measured, and five picofarads across each winding turn it over at 1.59 MHz and leave it worth 0.02 decibels by ten gigahertz.

lines · Common-mode choke Two of the four errors are divided by the gain; the best the instrument gets is 110.1 dB, at ×776. computed by solving, not by drawing. The four mechanisms that limit a three-amplifier instrumentation amplifier's common-mode rejection, each measured alone against the gain of its input stage and then all four together, at 50.0 Hz with 1 kΩ of imbalance between the source resistances and 10 pF at each input. The difference stage's four resistors and the difference amplifier's own rejection are injected after the gain, so their common-mode gain is a constant — 1.998 mV/V and 0.0100 mV/V — and the rejection they allow rises decibel for decibel with the gain. The input pair's mismatch and the source's time-constant gap are injected before it, so they are amplified by exactly the gain the signal is and the rejection they allow is flat. The four add as complex numbers: the two largest are real and of opposite sign, they cancel at a gain of 776, and what is left there is the source's 3.142 µV/V, which is purely imaginary because it is ωΔτ. The instrument's best is 110.07 dB against the source's own 110.06, and above that gain more of it buys nothing.

The errors that arrive before the gain

Four earlier measurements each found a different owner of one instrument's common-mode rejection and each measured it alone. Solved together, the four are complex numbers that add — to 1.25 parts in ten thousand — and two of them carry a factor of the first stage's gain while two do not. The two that do cancel the two that do not at a gain of 776, and what is left there is 110.066 decibels, which is exactly the number the cable sets. Better-matched amplifiers move that gain from 93 to 3392 and do not move the ceiling by a hundredth of a decibel.

instruments · Common-mode rejection What a high-side shunt's optimum is made of, at 100 dB of rejection. computed by solving, not by drawing. The same two errors as a low-side shunt, with the amplifier now standing at the rail rather than at the return. 100 dB of rejection turns 12 V of common mode into 120.0 µV of equivalent input error, which is 24 times the amplifier's own 5.0 µV of offset. The optimum keeps its form — the geometric mean of an input error and the supply, golden-sectioned on the solved worst case rather than substituted — and changes its value: 38.73 mV of burden and 0.6440% of error, against 7.75 mV and 0.129% low-side. A shunt sized by the low-side answer reads 1.678% wrong. With the common-mode term removed the optimum returns to the low-side value exactly, which is what says the term is the whole of the difference.

The rail that is an input error

The best burden voltage across a shunt is the geometric mean of the amplifier's offset and the supply, and moving the shunt to the high side does not change that form — it changes what the offset is. A hundred-decibel amplifier on a twelve-volt rail turns the rail into 120 µV of equivalent input error, twenty-four times its own five, so the optimum moves from 7.75 mV and 0.129 per cent to 38.73 mV and 0.644. The two are equal only at 128 dB, and with the common-mode term removed the optimum returns to the low-side value exactly.

instruments · Current sensing Two sensors on one return: 511 µV in each reading, 100 nV in the difference. computed by solving, not by drawing. Two sensors referenced to the same local node, each reaching it through 2 mm of its own copper and each drawing 1 mA through that copper, with a 100 mA load returning through the 10 mm they all share. The interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it: 511.0 µV in each reading against 100.0 nV in the difference at a kilohertz, which is 74.2 dB. No resistor tolerance enters it — with the two stubs equal the rejection is exact to the solver, 166 decibels. What is left is the 10 per cent MISMATCH between the stubs times the sensors' own current, which the closed form puts at 100.0 nV, and it is proportional to the mismatch: 10.0 nV at 1%, 30.0 nV at 3%, 100 nV at 10%, 300 nV at 30%.

One voltage added to two readings

A difference amplifier across a shared return rejects somebody else's current by 54 decibels, and the number is four resistors rather than the amplifier. Two sensors referenced to the same node do better by a different mechanism entirely: the interfering voltage is one voltage on one node, so it is added to both readings identically and the difference between them has none of it — 511 microvolts in each reading against two and a half picovolts in the difference, which is 166 decibels and is exact rather than good. What is left is not a tolerance but a mismatch between the two sensors' own return paths, ten nanovolts per per cent of it.

instruments · Common-impedance A cable's carrier limit is its conductors' matching, not its capacitance. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε, excited at a carrier frequency, with 10 metres of cable at 100 pF a metre on each output terminal. Each terminal sees the two arms meeting there in parallel against that capacitance, so each lags by arctan(2πf·(r/2)·C) with a corner at 909 kHz — and with the two conductors matched the lag CANCELS in the difference: -0.117° where each terminal has -5.71°. A mismatch does not cancel, and it arrives multiplied by the ratio of the excitation to the signal, which at 1000 µε is 2000: at 1 per cent the phase is 40.3° at a hundred kilohertz. So the excitation a 10-metre cable allows for 200 parts per million falls from 205 kHz matched to 1.82 kHz at 1 per cent. And it depends on the strain being measured — 183 Hz at 100 µε, 547 Hz at 300 µε, 1.82 kHz at 1000 µε, 5.45 kHz at 3000 µε — because the amplification is the reciprocal of the signal.

The carrier a cable allows

A carrier-frequency amplifier removes an instrument's offset drift by exciting the bridge with alternating current, and the cable that was a resistance at direct current becomes a capacitance. Both output terminals see the same arms and the same capacitance, so the lag cancels in the difference: ten metres matched leaves 0.15 degrees at a hundred kilohertz where each terminal has 6.3. What does not cancel is the mismatch between the two conductors, arriving multiplied by the ratio of the excitation to the signal — two thousand at a thousand microstrain — so one per cent of it takes the allowed carrier from 205 kilohertz to 1.8.

applied · Bridge A tenth of a per cent of screen mismatch on the gauge's side cancels the pair's own phase: 18.2 kHz becomes 143 kHz. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε read through ten metres of screened pair: 100 pF a metre from each conductor to the screen, Cs, and 50 pF a metre between the two, Cp. The loss of the synchronously demodulated reading, 1 − cos φ, against the excitation frequency: for the capacitance between the conductors alone, for the screen capacitances alone with a mismatch of 0.1% on the gauge's side, for both together, and for a matched screen with nothing between. For a 200 ppm budget they allow 18.2 kHz, 18.3 kHz, 143 kHz and 205 kHz. The pair's capacitance lags the reading; extra screen capacitance on the gauge's own terminal leads it, by an amount divided by the signal, so at a mismatch of δ·Cp/Cs the two cancel to first order.

The capacitance that is already a difference

A screened pair has capacitance from each conductor to the screen, which cancels in a bridge's difference when the two match, and capacitance between the conductors, which is a difference from the start and cannot cancel. Ten metres of it between a 350 Ω quarter bridge and its carrier amplifier allows 18.2 kHz for a 200 ppm loss, against 205 kHz for the screen alone. The two phases are ωR·ΔC/δ and ωR·C, the same slope with one divided by the signal, so a screen mismatch of δ·Cp/Cs on the gauge's own terminal cancels the pair's lag exactly: 0.10% at 1000 µε takes the limit to 143 kHz. That is what a carrier amplifier's capacitance balance finds when it nulls the quadrature — and because the mismatch's phase changes sign with the strain and the pair's does not, the balance holds in tension above half the strain it was set at and never in compression.

applied · Bridge Two coupling capacitors 1% apart reject common mode by only 46.0 dB at their corner, 0.159 Hz — and perfectly at direct current and far above it. computed by solving, not by drawing at 281 frequencies: one common-mode source driving two coupling networks, each 1 µF into 1 MΩ, with the capacitors 1% apart. The difference between the two outputs, per volt of common mode, agrees with the difference of the two first-order high-passes to 6.9e-14. It is zero at direct current, where both block, and zero far above the corner, where both pass; between, it peaks at the corner, 0.159 Hz, at 5.000e-3 — half the mismatch — which is 46.0 dB of rejection. Behind a 90 dB amplifier (grey), the coupling network decides the rejection from 0.000503 Hz to 50.3 Hz.

The rejection two capacitors lose at their corner

AC-couple both inputs of a differential instrument and the two coupling networks become part of its common-mode path. Two 1 µF capacitors one per cent apart turn common mode into difference at exactly half their mismatch at their shared corner — 46.0 dB of rejection at 0.159 Hz, from a pair that is perfect at direct current and far above. Behind a 90 dB amplifier they decide the instrument's rejection over five decades around that corner. A 10 V common-mode step arrives as a 36.8 mV differential pulse one time constant later. Four hundred pairs of 20 per cent electrolytics reject a median 25 dB. Coupling the difference instead of each side removes the dip and hands the common mode to the amplifier.

instruments · AC coupling Three imbalances, three slopes: a 2% amplitude imbalance is 1% common mode at every frequency, 1 mm of skew overtakes it above 455 MHz, and a 5 ps rise-time mismatch climbs at forty decibels a decade. computed by solving, not by drawing, from the transforms of the two conductors' edges and checked against the sampled edges transformed by FFT to two per cent: the common mode at the far end of a pair, as a fraction of the differential, for three separate imbalances between its halves. An amplitude imbalance of 2% gives 1.00% at every frequency. A skew of 1 mm on FR-4, 7.00 ps, gives sin(ωΔτ/2), rising at twenty decibels a decade: 2.20% at 1 GHz. Edges of 50 and 55 ps, perfectly timed and equal in size, give a common mode rising at forty decibels a decade: 0.043% at 1 GHz and 5.38% at 10 GHz. The skew passes the amplitude imbalance near 455 MHz, which is a/(2πΔτ).

The common mode that has three slopes

A differential pair turns into common mode wherever its two halves differ, and they can differ in three ways — in size, in timing and in edge rate — each with its own slope. A 2 per cent amplitude imbalance is 1 per cent common mode at every frequency. A millimetre of skew rises at twenty decibels a decade and overtakes it above 455 MHz, a/(2πΔτ). Edges of 50 and 55 picoseconds, perfectly timed, rise at forty: 0.043 per cent at a gigahertz and 5.4 at ten. In time they are a step that stays, a pulse with an area, and a doublet with none — each slope is a moment that vanishes. And an amplitude imbalance and an edge-rate mismatch of the right signs cancel at one frequency, down 74 dB, which half a picosecond of skew fills back to −25.

lines · Differential line A reference 0° behind the excitation at 20 kHz, after a calibration at +1000 µε: matched, the worst error is 5.79 ppm; with a tenth of a per cent of screen mismatch, 0.607%. computed by solving, not by drawing: the error of a synchronous reading of a 350 Ω quarter bridge through ten metres of screened pair, against the strain read, with the reference 0° from the excitation at the bridge and a shunt calibration at +1000 µε through the same reference, for a matched screen (solid) and for 0.01 and 0.1 per cent of mismatch. The calibration takes out cos(φ − θ), because it multiplies the signal at every strain. It cannot take out a voltage that does not scale with the strain, and the mismatch's leak is such a voltage: a hyperbola through zero at the calibration strain. Worst over ±2000 µε: 0%: 5.79 ppm, 0.01%: 606 ppm, 0.1%: 0.607%.

The reference that should be late

A carrier amplifier's reference is taken from the excitation because that is what the output is meant to be in phase with, and nothing says that is where it belongs. A screen mismatch's leak arrives late by the common-mode node's lag and the difference's together, 525 ns on ten metres of pair and a 350 Ω bridge, so a reference on the excitation reads sin 3.7° of it at 20 kHz: 0.607 per cent of error at a tenth of a per cent of mismatch, and 72.6 parts per million with the reference 3.7° late. The null is a delay rather than a phase and does not move with the mismatch, and a delayed reference allows about five times the carrier on any mismatched screen. How exactly it must be held narrows as mismatch times carrier.

applied · Bridge

Named alongside it

The objects these essays reach for when they reach for this one.

Component toleranceModel rangeDesign tradeoffDevice mismatchMeasurement errorParasiticsDifference amplifierStrain gaugeWheatstone bridgeCommon-impedanceDevice matchingExcitation

All concepts