Measurement, which is a circuit on a circuit

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.

Assumes: The millivolts in the wire · Two terminals measure the leads as well

The instruments field has already established the problem this essay is about. Ten millimetres of one-ounce copper shared between a hundred-milliamp load’s return and a ten-millivolt sensor’s reference puts five hundred microvolts in series with the sensor — five per cent of the reading — and the error rises a decade per decade above 79.6 kHz until, at 1.59 MHz, it is the whole signal.

The repair everybody reaches for is to measure differentially: read the sensor between its own two terminals rather than against a ground that other people are using, and the common part subtracts out.

It does subtract out. It subtracts out by a factor that has nothing to do with the amplifier and everything to do with four resistors, and the factor is smaller than most people expect.

The same millivolts, subtracted — and what four resistors leave behindcomputed 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.100n10µ100µ1m10m100m1101001k10k100k1M10M100Mfrequency of the interfering current (hertz)error added to the reading (volts)after a difference amplifier with 0.1% resistors10 mV — the signal being measured79.6 kHz — Rg/2πLg1.59 MHz — the error is the signalshared run10 mmits resistance5.00 mΩits inductance10.0 nHinterfering current100 mAthe signal10 mVerror at 100 Hz500.0 µVas large as the signal1.59 MHzresistor tolerance0.1%rejection, solved53.99 dB(1+G)/(4·tol)53.98 dBerror left at 100 Hz998.9 nVsolved, then checked — the return in the netlist54 dB of rejection from 0.1% resistors
Fig. 1 The same interference and the same sensor, with a unity-gain difference amplifier across the sensor’s own terminals. The upper curve is the single-ended error and the lower one is what is left after the subtraction — a factor of five hundred, flat with frequency, set by the tolerance of the four resistors and by nothing else.

What a difference amplifier is, and where the tolerance enters

A difference amplifier is one amplifier and four resistors: two forming a divider on the non-inverting input, two forming the feedback path around the inverting one. It subtracts its two inputs and multiplies by the ratio, and it does so exactly when the two ratios are identical.

They are not identical, because resistors have tolerances. Skewing all four by the worst case a stated tolerance permits — two high and two low — and solving the network gives a common-mode gain that is not zero, and a rejection ratio of

CMRR=1+G4δ\mathrm{CMRR} = \frac{1 + G}{4\,\delta}

where GG is the differential gain and δ\delta is the fractional tolerance. For a unity-gain difference amplifier that is 1/2δ1/2\delta.

resistor tolerance rejection, solved (1+G)/4δ error left of 500 µV
1% 34.07 dB 33.98 dB 9.90 µV
0.1% 53.99 dB 53.98 dB 999 nV
0.01% 73.98 dB 73.98 dB 100.0 nV

The two routes agree to a hundredth of a decibel at a tenth of a per cent, and the agreement loosens at one per cent — 34.07 against 33.98 — for the reason such agreements usually loosen: the closed form is first order in the tolerance and one per cent is where the second-order term becomes visible. The figure’s tolerance on that comparison is set to the second-order term rather than chosen.

The millivolts in the wire, which are nobody's signal. 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.
Fig. 2 The problem without the repair, which is the essay this one continues. The error is 500 µV at low frequency and rises a decade per decade above 79.6 kHz, because the shared conductor is a resistance and an inductance rather than a wire.

Fifty-four decibels is less than it sounds

One-tenth-per-cent resistors are what a careful designer specifies, and they buy a factor of five hundred. Applied to this measurement:

  • the single-ended error is 500 µV on a 10 mV signal, which is 5.0%;
  • the differential error is 999 nV on the same signal, which is 0.010%.

That is a real improvement and it is not a removal. A hundred parts per million of error from somebody else’s current is larger than the resolution of a sixteen-bit converter on the same ten-millivolt span, so the interference is still the limiting term for a measurement that is trying to be accurate.

And the number is a tolerance, not a matching. Two resistors from the same reel are far better matched than their absolute tolerance suggests, which is why four-resistor networks in a single package exist and are specified by their ratio match rather than by their individual accuracy — 0.01% ratio match in a package whose absolute tolerance is one per cent. The figure’s worst-case skew is the pessimistic reading, and it is the right one for four loose resistors on a board.

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 1% allows. It divides the error by 50.50 — 34.07 dB against a closed-form (1 + G)/(4·tolerance) of 33.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 9900 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.
Fig. 3 One per cent resistors, which is what a board gets if nobody specifies otherwise. Thirty-four decibels: the five hundred microvolts becomes ten, which is a tenth of a per cent of the signal and is very likely to be the dominant error in the measurement.
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 0.01% allows. It divides the error by 5001 — 73.98 dB against a closed-form (1 + G)/(4·tolerance) of 73.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 99.98 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.
Fig. 4 And one part in ten thousand, which means a matched network rather than four resistors. Seventy-four decibels, and a hundred nanovolts left — at which point the sensor’s own Johnson noise and the amplifier’s own input noise are the things to worry about instead.

The rejection is flat with frequency, and the interference is not

The most useful property of the curve is that the two lines are parallel.

Four resistors have no frequency in them, so the rejection is the same at ten hertz and at a megahertz — measured across the whole sweep, the ratio varies by less than a part in a thousand. The interference, meanwhile, rises a decade per decade above the shared conductor’s own corner, because its inductance takes over from its resistance there.

So the difference amplifier does not become less effective at high frequency; it becomes less sufficient, because the thing it is dividing by five hundred is getting larger. At 1.59 MHz the raw error equals the signal, and the differential error is 0.2% of it — still small, and heading in the wrong direction at twenty decibels per decade.

That is the honest form of the boundary. The repair has a fixed gain, the problem has a slope, and they cross somewhere. Where they cross is a frequency computed from the shared conductor’s geometry and the resistors’ tolerance together, which is a pleasing thing for two such unrelated quantities to decide between them.

Why the skew is applied to all four

The figure sets two resistors high and two low by the full tolerance, which is the worst case a tolerance allows and is not what four random resistors will do.

Four independent errors of standard deviation σ\sigma combine in quadrature, so a typical build is about half as bad as the worst case — a factor of two, or six decibels. That is worth having in mind when reading the table, and it is worth not designing against, because the worst case is what the specification permits and a production run will contain builds near it.

The more important reason for the worst case is that the four errors are not independent in the way the quadrature argument assumes. Resistors from one reel share a systematic offset, which cancels here — a common multiplicative error on all four leaves both ratios unchanged and the rejection untouched. What survives is the spread within the reel, which is much smaller than the tolerance and is not specified anywhere. So the honest bracket is: worst case if the four came from different places, and much better than the table if they came from the same one, with nothing on the data sheet to say which.

Which is the argument for a matched network in a single package, whose specification is exactly the quantity that matters and is one to two orders better than the individual tolerance of its parts.

The gain that helps

The closed form has 1+G1 + G in the numerator, and it is worth reading that as a design lever rather than as a constant.

A difference amplifier configured for a gain of ten has 14.8 dB more rejection than the same resistors configured for a gain of one — 68.8 dB rather than 54.0 at a tenth of a per cent, both solved and both agreeing with the closed form to five figures — because the differential signal has been multiplied by ten and the common-mode error has not.

That is the reason to take the gain in the difference stage rather than after it. A unity-gain difference amplifier followed by a gain of ten multiplies the residual interference by ten along with the signal and ends up exactly where it started; the same total gain taken in the difference stage does not.

It is also the reason an instrumentation amplifier is built the way it is: the two input buffers take the gain before the subtraction, so the common-mode path never sees it at all, and the rejection improves with the gain setting rather than merely keeping up.

What the amplifier contributes, which is nothing

A part’s data sheet quotes a common-mode rejection ratio, often 100 dB or better, and it is worth saying plainly that it is irrelevant to the number above.

The amplifier’s own CMRR describes how much its output moves when both of its inputs move together with the feedback network held perfect. In a difference amplifier the feedback network is four resistors and the resistors are the error. A part with 120 dB of its own rejection, surrounded by one-per-cent resistors, gives 34 dB.

The two combine as errors do — the total is the sum of the two common-mode gains, and one of them is a thousand times the other — so the specification that matters is the one on the passive components.

The exception is the instrumentation amplifier, which is a different circuit for exactly this reason: two input buffers ahead of the difference stage give differential gain without touching the common-mode path, so the ratio is raised by the buffers’ gain while the resistors’ skew stays where it was. That is why an instrumentation amplifier’s rejection improves with its gain setting, and why its data-sheet figure is a real number rather than a statement about somebody’s resistors.

Measuring with 50 mΩ of lead in each wire. computed by solving, not by drawing at 61 resistances, twice each. The two-wire arrangement measures the leads too, so its error is 2×50 mΩ over whatever is being measured: one per cent at 10 Ω, and 10000% at 1 mΩ. The four-wire arrangement senses on a separate pair that carries almost no current, and its error stays under 1.0e-2% across the whole range.
Fig. 5 The other repair for the same class of problem, from this field’s own essay. Four-terminal sensing removes the lead resistance by carrying the current and the voltage in different conductors, which is the same idea as measuring differentially — separate the path the interference takes from the path the measurement takes.

What is being subtracted, and what is not

The word “common-mode” hides a distinction worth making, because the two things it covers behave differently here.

The interference on the shared return is genuinely common to the two input terminals: the sensor floats on the local ground, so both of its terminals move together when the local ground moves. This is the case the figure measures and the case a difference amplifier handles.

A voltage that appears on only one of the two leads is not. Capacitive pickup on one wire of a pair, a thermocouple junction in one terminal, an offset current flowing in one lead’s resistance — none of those is common to both, and no amount of rejection touches them. They are differential errors and they arrive at the output at full size.

Which is why the layout matters as much as the resistors do. A twisted pair makes the pickup common by making the two wires see the same field; a screen makes the pickup common by putting it on the screen instead. Both are ways of converting a differential error into a common-mode one so that the four resistors have something to work on.

The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres.
Fig. 6 Where the interfering current actually runs, from the lines field. Below a crossover frequency it spreads across a plane and above it runs directly under its own track, so the loop it shares with a measurement — and therefore the interference it couples — is a function of frequency and geometry rather than a fixed quantity.
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.
Fig. 7 The same ten millimetres with one-tenth-per-cent resistors rather than one per cent or a hundredth. Five hundred microvolts of shared-impedance error arrives and 999 nV of it is left. What is being subtracted is the common part of that error, and what is not is whatever the four resistors fail to match — so the rejection is the tolerance, and nothing else on the board changes it.

The whole chain, priced

Putting the figures of this field together gives the error budget for a small differential measurement, and every term in it is a number this collection has measured rather than quoted.

The shared impedance puts 500 µV of somebody else’s current into the reading, at low frequency, for ten millimetres of shared copper.

The difference amplifier divides that by 500 with 0.1% resistors, leaving 999 nV.

The source resistance and the amplifier’s input put a further error in through loading, which the probe essays measure as a function of the ratio of the two.

The resistors’ own Johnson noise puts a floor under the whole thing: four ten-kilohm resistors in a ten-kilohertz bandwidth contribute about 1.8 µV of noise, which is larger than the 999 nV of residual interference.

That last line is the one worth carrying. Beyond about 0.1% resistors, further matching does not improve the measurement, because the same resistors’ thermal noise is already the larger term. The tolerance and the noise are properties of the same four components, and they meet.

The same millivolts, subtracted — and what four resistors leave behind. computed by solving, not by drawing at 145 frequencies. A 100 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 50.0 mΩ and 100 nH put 4995 µV in series with the sensor at low frequency — 50.0% of the reading — rising a decade per decade above 79.6 kHz until at 138 kHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 1.5e-5. 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 9980 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.
Fig. 8 And the whole chain priced at a hundred millimetres of shared return: 50.0 mΩ and 100 nH give 4,995 µV of error, of which 9.98 µV survives one-tenth-per-cent matching. Ten times the shared length is ten times the error before rejection and ten times after — the resistors decide a ratio and the copper decides what the ratio is applied to.

Where the same subtraction is done without an amplifier

Two other arrangements in this collection subtract a common quantity, and comparing what decides each of them is the useful exercise.

A four-terminal measurement subtracts the lead resistance by not carrying current in the leads that measure the voltage. What decides its accuracy is the ratio of the voltmeter’s input impedance to the lead resistance, and the field’s own essay finds a point on that grid where the two error terms cancel exactly — a coincidence rather than a design, and it names it as one.

A bridge subtracts a reference from a measurement by construction: its output is the difference of two dividers, so a change common to both arms produces nothing. What decides its accuracy is the match of the two dividers, which is the same kind of quantity as the four resistors here and is why bridge elements are specified as ratios.

And a twisted pair subtracts interference before any circuit sees it, by making the two conductors enclose the same area with the interfering field. What decides that is geometry rather than a component value, which is why it is the cheapest of the three and the first thing to get right.

The pattern across all four is that a subtraction is only as good as the symmetry of the two things being subtracted, and in every case the symmetry is a passive property — a resistor match, a divider match, a loop area — rather than anything active. The amplifier is never the limit.

Where the rejection comes from, and its ceiling

A rejection this measurement spends is bought two essays earlier and has a ceiling one essay later. The four resistors that decide, and the two that do not is where it is bought, and where the corner turns out to be the gain-bandwidth divided by the gain. The rejection the parts have is the ceiling, set by the mismatch between two amplifiers rather than by the four resistors. The probe is part of the circuit is the other instrument error on the same measurement.

What is checked

Three assertions, and the first two are what make this a measurement rather than a restatement of a formula.

That the rejection is the same at every frequency — the ratio of the two curves varies by less than six parts in a thousand across five decades — which is what says the mechanism is the resistors rather than anything with a reactance in it.

That the solved rejection matches (1 + G)/4δ to first order in the tolerance, at every tolerance drawn. The two routes share the netlist and nothing else: one is a solve of a seven-element network with an interfering current in it, the other is a ratio of two resistor errors.

And that what is left is not zero. That assertion exists because the point of the essay is a factor rather than a removal, and a figure whose lower curve went to the bottom of the axis would be making the opposite claim.

The four rungs, and what each adds to the number

This is the first of four measurements of one architecture, and the four together say that a rejection figure has at least four independent things in it.

The four resistors that decide, and the two that do not puts a two-amplifier stage in front and 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, which is the sharpest available statement of which components a specification is actually about.

The rejection the parts have gives each amplifier a rejection of its own and finds something unobvious: 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 — so the twenty log of the gain above does not go on for ever.

And the corner the instrument has no part in is the one that removes the whole architecture from the answer. Connect a 95 dB instrument 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 a part that is still doing 95. What converts common mode into differential there is the difference of two time constants, and the cure is a capacitor on the quiet input rather than anything about the amplifier.

Read together the four say that the resistor tolerance measured here is the best case: it is what limits the rejection of an instrument fed from a balanced source at direct current, and every one of the other three rungs is a mechanism that binds before it does under ordinary conditions.

Which is worth one more sentence about what a factor of five hundred is for. Five hundred microvolts becoming one is not a removal, and the residual is above a strain gauge’s resolution and below a thermocouple’s signal — so whether the arrangement on this page is sufficient is decided by the sensor rather than by the amplifier, and the same four resistors are either ample or hopeless depending on what is plugged into them.

The remaining microvolt is also the wrong shape to be treated as an error bar. It is proportional to somebody else’s load current, so it changes when that load switches and stays put when it does not — which makes it a systematic error correlated with an event rather than a random one, and the kind that survives averaging entirely. The millivolts in the wire describes the signature: a reading that moves when a different circuit changes, unlike noise, unlike drift and unlike loading. Dividing that signature by five hundred makes it smaller and does not make it look any more like noise.

Part 1 on Common-mode rejection

One argument about Common-mode rejection, and one of 5 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down, the 8 sharing most with it of 11.

What this makes readable

Essays that name this one as a prerequisite.

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Common-impedanceCommon-mode rejectionComponent toleranceDifference amplifierFour-terminal sensingLead resistanceLoadingReturn current