Field

Measurement, which is a circuit on a circuit

An instrument is not an observer; it is an element, it goes in the netlist, and every reading is a reading of the circuit that includes it. A probe is a capacitance with a bandwidth. A divider with capacitance in it has two ratios and one equation that makes them agree. Two terminals measure the leads as well, and below ten ohms that is most of the answer.
Two probes on a 2.0 kΩ source. computed by solving, not by drawing twice per frequency: the node alone, and the node with the probe's elements across it. The one-to-one probe's 115.0 pF makes the reading one per cent wrong at 6.79 kHz. The ten-to-one probe puts 12.8 pF in series with the cable, so its tip sees 11.5 pF and the same error arrives at 69.2 kHz — 10 times further up, bought with a factor of ten in signal — the two edges stand in the ratio of the tip capacitances, 10.00. At direct current neither probe is capacitive at all and the ten-to-one still reads 0.02% low, because 10 MΩ across 2.0 kΩ is a divider.

The probe is part of the circuit

A one-to-one oscilloscope probe on a two-kilohm source gives a reading that is one per cent wrong at 6.8 kHz. Not because the instrument is inaccurate — it is reading correctly — but because the hundred and fifteen picofarads at the end of the cable are across the node, and above that frequency the trace on the screen is a picture of a circuit that only exists while the probe is attached.

A 10:1 divider with 12.8 pF across its top resistor. computed by solving, not by drawing. The divider's resistors set a ratio of 0.10000 and its capacitors 0.10000; the step starts at the second and relaxes to the first over 115 µs. The balance R₁C₁/R₂C₂ is 1.0000, and the edge lands 0.00% away from where it settles. At 12.8 pF the two ratios are the same number and the response is flat.

A divider with two ratios

Put capacitance in a resistive divider and it divides by resistance at direct current and by capacitance at high frequency, and those are two different numbers unless one equation holds. The adjustable trimmer on every oscilloscope probe exists for that single equation, and the square wave on the instrument's front panel is a display of which of the two ratios is currently winning.

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.

Two terminals measure the leads as well

Fifty milliohms in each lead makes a two-wire measurement one per cent high at ten ohms, ten per cent high at one ohm, and a hundred per cent high at a tenth. Not approximately — the reading is the resistance plus the leads, and below about ten ohms most of what is being reported is the wire between the instrument and the thing.

How much faster an instrument must be for 10% of inflation. computed by solving, not by drawing. The quadrature rule answers 2.182× and gives the same answer for every instrument, because it contains no instrument. Measured on the solved network, a one-pole front end needs 2.79×, two poles need 3.97×, three need 4.87× and four need 5.62× — between 28% and 215% more than the rule asks for. The rule errs optimistic at every pole count, which is the wrong direction.

The instrument's own rise time

Rise times add in quadrature, so ten per cent of inflation needs an instrument 2.18 times faster than the edge. That constant contains no instrument. Measured on the solved network it is 2.79 for a one-pole front end, 3.97 for two, 4.87 for three and 5.62 for four — the rule is optimistic at every pole count, which is the wrong direction for a rule of thumb to err in.

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.

The millivolts in the wire

Ten millimetres of one-ounce copper is five milliohms and ten nanohenries, and if a hundred-milliamp load and a ten-millivolt sensor both return through it, half a millivolt of somebody else's current is added to the reading — five per cent of it, before anything has been amplified. Above 79.6 kilohertz the error rises a decade per decade with no ceiling, and shortening the shared run moves the whole curve down and the corner not at all.

A 100 ms pulse through a 0.159 Hz corner, 9.52% shorter by the end of it. computed by solving, not by drawing. Marched. The dashed line is the pulse that was sent. The solid line is what a 1 MΩ input with 1 µF in front of it receives: the top decays as exp(−t/RC) for the whole 100 ms, ending 9.515% down, and the trailing edge undershoots by exactly the same amount. The closed form for the same two components gives 9.516%. The input's specification is a corner at 0.159 Hz; a top flat to one per cent needs a pulse shorter than 10.1 ms, which is a rate 625.2 times the corner — a constant with no component in it, and the reason a low-frequency specification says nothing useful about an edge.

The corner that says nothing about an edge

An AC-coupled input is sold on a low-frequency corner, and a corner is a statement about steady sinusoids. What an instrument is usually shown is a pulse, and for a pulse the number is a sag: a hundred-millisecond pulse through a 0.159 hertz corner comes out 9.5 per cent shorter than it went in. A one per cent flat top needs a pulse rate 625 times the corner, which is a constant with no component in it — and above fifty per cent duty an AC-coupled pulse never reaches half its own height at all.

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.

The best shunt drops 7.75 mV, whatever the current is. computed by solving, not by drawing. Two errors on one axis, both from solved networks: the shunt's own drop, which lowers the current that was to be measured, and the amplifier's 5.0 µV of offset divided by the voltage the shunt develops. The first rises with the burden voltage and the second falls, so the worst case has an interior minimum at 7.7460 mV — the geometric mean of the offset and the 12 V supply — where the error is 0.1291%, being twice the root of the offset over the supply. Neither the shunt's resistance nor the current appears in either number: at 1 A the answer is 7.75 mΩ, and at a hundred times the current it is the same burden voltage across a hundredth of the resistance. What does depend on the current is the 7.7 mW the shunt then dissipates, and 40 K of self-heating at 50 ppm/K is 0.2000% on its own. The third curve is the shunt's own Johnson noise in a kilohertz of measurement bandwidth, as a fraction of the current: it is 4.5e-6% at the best burden and is the only line here that moves with the current at all, falling as one over its square root — so above about an ampere it leaves the bottom of these axes entirely and is drawn nowhere rather than flattened onto the floor.

The ammeter that is a resistor

Every direct measurement of a current is a measurement of a voltage across something the current was made to flow through, so the instrument has two errors pointing opposite ways: a larger shunt changes the current, a smaller one leaves less for the amplifier's offset to be compared with. Written in the burden voltage they are the burden over the supply and the amplifier's offset over the burden, and the best of them is at the geometric mean — 7.75 millivolts on a twelve-volt rail, with a worst-case error of 0.129 per cent. Neither number contains a resistance, and neither contains the current: ten milliamps and a hundred amps want the same 7.75 millivolts.

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.

A part in ten thousand of ratio, and half a degree that costs 18% of a power reading. computed by solving, not by drawing. The two errors of a 1000:1 current transformer into a 10 Ω burden, read off one solve of a netlist driven by a current source. The magnetising inductance puts a high-pass under the measurement with a corner at 0.456 Hz, so the ratio error falls as the frequency rises — and stops falling at 100 parts per million, which is 1 − k and is a coupling rather than a frequency. The phase error falls too and is worth far more: 0.522° at 50 Hz is 0.0142% of the current and 18.2% of the power at a power factor of 0.05. What bounds it from above is not on this plot: the burden voltage is integrated by the core, so the largest primary current is 916 A at 50 Hz and proportionally more at 400.

The ammeter that is not in the circuit

A shunt measures a current by putting a resistance in the circuit, and every objection to it follows from that. A current transformer puts nothing in the circuit at all — a thousand-turn secondary reflects twelve microhms into the primary — and charges for it in a different currency: no response at direct current, a ratio error that stops falling at one minus the coupling, and a phase error of half a degree at fifty hertz that costs eighteen per cent of a power reading at a power factor of 0.05.

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.

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.

Where an amplifier's reading comes from, against the source it is reading. computed by solving, not by drawing. Three errors with three different dependences on the source, each measured by a solve with the other two set to zero. The offset voltage is flat — 50 microvolts wherever the source is. The bias current times the imbalance is linear in the source and is what balancing removes. The offset current times the source is linear too and is what balancing leaves. Unbalanced, the current overtakes the voltage at 1.77 kΩ; balanced, at 10.0 kΩ, which is the offset voltage divided by the OFFSET current and is the ratio of the two currents further along. Below about a kilohm, balancing makes the reading worse — the feedback network is already the larger resistance, and equalising means adding to the source.

The current the instrument draws

Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get.

A clean board is 14.0 mV of error, and a guard makes it 1.0 nV. computed by solving, not by drawing. Two boards differing by one wire, solved with the leakage present and again with it removed, so the number plotted is the leakage's own contribution and nothing else. Unguarded, a teraohm across the laminate from a 15 V rail through a 1 GΩ source is 13.99 millivolts — and a humid morning takes that resistance down two decades, which is the left-hand end of this axis. Guarded, the ring is held at the input's own potential by the amplifier, so what is across the leakage is the amplifier's own error and the result is 1.00 nanovolts. The guarded line is flat in the rail and proportional to the signal: the offset has become a gain error of 1.00 parts per billion, and the reading is low by it rather than high: the ring sits a little below the input, so the last of the leakage pulls the input down.

The current that does not reach the input

A teraohm across a board from a fifteen-volt rail is fourteen millivolts of error through a gigohm source, and a humid morning takes that resistance down two decades. A ring held at the input's own potential leaves a nanovolt — proportional to the signal rather than to the rail, so an offset has become a gain error of one part in a billion — and the same wire multiplies the input resistance by the loop gain, which makes it 10¹⁸ Ω at direct current and 10¹² Ω at a megahertz.

What each factor of attenuation buys on a 2.0 kΩ source. computed by solving, not by drawing at 12 probe ratios: the one-per-cent frequency bisected on the node with and without the probe, against the frequency a tip capacitance alone would predict. A one-to-one probe reaches 6.79 kHz and a hundred-to-one 692 kHz. The first step, from 1× to 2×, multiplies the bandwidth by 2.03 for a factor of two in signal; the two routes differ by at most 1.8% across the sweep, and they differ at all only because the probe's 1.0 MΩ is already 0.20% of the reading before any frequency is applied.

The probe that takes a tenth

A ten-to-one probe buys an order of bandwidth for a tenth of the signal, and on a two-kilohm source the bandwidth is exact: 6.79 kHz becomes 69.2 kHz. The tenth of the signal is not a tenth of the signal-to-noise ratio. Solved resistor by resistor, the noise referred to the tip goes from 1.782 µV to 55.78 µV — a factor of 31.3 — because the divider that does the attenuating is nine megohms and a megohm, and √(n(n−1)kT/C) on the cable's own capacitance has no source resistance in it at all.

Four wires against a 10 MΩ voltmeter. computed by solving, not by drawing at 81 resistances, twice each, with a voltmeter of 10 MΩ and 50 mΩ in every lead. The four-wire error is not zero: it is the voltmeter's own divider, −(R + 2R_lead)/(R + 2R_lead + R_m), which grows with the resistance being measured rather than shrinking. The two-wire error is that same quantity plus the leads, so it passes through zero at 1000 Ω — where the reading is right to 1.8e-12 while the four-wire reading is 0.0100% low — and above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

The voltmeter four wires do not remove

A four-terminal measurement is described everywhere as removing the leads from the answer. It moves them. What is left is the voltmeter's own input resistance, and it grows with the resistance being measured rather than shrinking: with a ten-megohm voltmeter and fifty milliohms of lead, the four-wire reading is 0.0100 per cent low at a kilohm, where the two-wire reading is exactly right — 1.8 × 10⁻¹² — because its lead error and its loading error cancel. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

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.

The cure changes shape at 909 Ω, which is a property of the feedback network and of nothing else. computed by solving, not by drawing. What balancing actually does to the circuit, against the source resistance it is done for, at a gain of 11 with a 1.0 kΩ bottom resistor. The inverting input looks back into 909 Ω — the bottom resistor times (G−1)/G — and that number is the whole of the knee. Below it the cure is a resistor in series with the source and the feedback network is untouched. Above it there is no resistor to add, and the network is scaled up to meet the source instead: 1100× at 1.0 MΩ, which puts 11 MΩ in the feedback path. The scaled feedback resistor is the source resistance times the gain exactly, so the network's own size has left the answer — it decided where the knee was and nothing after it.

The cure that becomes a different circuit

The classical cure for an amplifier's input current is to make the two resistances its inputs look back into equal, and it reads as one instruction. Solved, it is two circuits meeting at 909 ohms — the feedback network's bottom resistor times (G−1)/G — and above that knee there is no resistor to add: the network is scaled to the source, which at a megohm means 11 megohms of feedback and at a gain of 1001 means 1001. Above the knee three different networks become one instrument to twelve figures, the noise penalty settles at 1.41420 against a √2 of 1.41421, and the benefit at 10.49 against two currents whose ratio is ten.

The guard leaves a negative resistance, and it reaches −1.59 kΩ. computed by solving, not by drawing. The magnitude of the conductance a source sees looking into the input, guarded and not, with 100 pF of cable and a 1.00 MHz amplifier. The unguarded input's conductance is positive everywhere — a capacitance to ground and a leakage to a rail are both losses. The guarded one is negative above 0.0404 Hz, and its magnitude rises as the square of frequency: −15.9 MΩ at 10 kHz, −161 kΩ at 100 kHz, −3.18 kΩ at a megahertz. Above the amplifier's gain-bandwidth product it flattens at ωₜ·C, which is −1.59 kΩ. That is the same input the guard raises to 10¹⁸ Ω at direct current, and nothing about the leakage the guard was installed for appears in it: the negative resistance is a product of the amplifier's bandwidth and the cable it is driving.

The sign of what the guard gives back

A guard ring is sold on two numbers and they are both about magnitudes: a teraohm of leakage multiplied to 10¹⁸ ohms, and a hundred picofarads of cable bootstrapped out of the way. The guard is also driving that capacitance with a copy of the input that lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with the voltage across it. What the guarded input presents is a negative conductance rising as the square of frequency — −15.9 megohms at ten kilohertz, −3.18 kilohms at a megahertz, flattening at the gain-bandwidth product times the capacitance — and a faster amplifier makes it worse.

The corner a 2 nH shunt has, against the current it is sized for. computed by solving, not by drawing. A shunt held at the best burden voltage of 7.75 mV has R = u⁄I, so its own 2 nH of series inductance puts a corner at u⁄(2πLI) — 616 kHz at an ampere and 6.16 kHz at a hundred, for the same piece of metal. The optimum that contains no current at all therefore hands the bandwidth a current dependence: the corner falls in exact proportion. At 1 A the shunt is 7.75 mΩ with a time constant of 258.2 ns, so a 10 ns edge is read 2.58e+3% high and a 1 ns edge 259.20 times too large. A resistor and a 258.2 pF capacitor across it — the value found by search on the solved response, agreeing with L/(R·Rc) to 7.9e-5% — flatten the reading to 7.8e-5% across six decades, and a fifth too much makes it ten times worse. The dots are the corner bisected on the solved impedance rather than taken from R/2πL.

The optimum that hands back a bandwidth

The best burden voltage across a shunt is 7.75 mV and contains neither the current nor the resistance, which is what made it worth having. A shunt has two nanohenries whatever it is made of, so holding the burden fixed fixes the resistance at u*/I — and the corner R/2πL then falls in exact proportion to the current: 6.16 MHz at a tenth of an ampere, 616 kHz at one, 6.16 kHz at a hundred. A resistor and a 258.2 pF capacitor across it, found by search on the solved response, flatten the reading to 8×10⁻⁵ per cent across six decades.

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.

The shunt's resistance as a function of what it is measuring. computed by solving, not by drawing, as a fixed point: the shunt dissipates I²R, its temperature rises by 20 K per watt, and at 50 ppm/K its resistance rises with its temperature — so the resistance the reading is divided by depends on the reading. Iterated to convergence it agrees with the closed form R₀/(1 − αθI²R₀) to 2.2e-16. Along the burden-voltage optimum, where R = u⁄I, the dissipation is I·u rather than I²R, so the temperature rise is 155 mK per ampere and the error is the FIRST power of the current — fitted exponent 1.0007 over five decades. That is the only one of the shunt's errors with the current in it, and it puts a term in I² into the reading, which is a curvature no single-current calibration removes. The upper curve is a shunt of fixed resistance, where the error is quadratic. The fixed point stops existing at 129 kA and never at a current a shunt will see.

The resistance that depends on the reading

Three of a shunt's errors are free of the current being measured, which is the whole content of the burden-voltage optimum. The fourth is not: the shunt dissipates, warms, and its resistance rises — so the divisor the reading uses is a function of the reading. Solved as a fixed point it agrees with R₀/(1 − αθI²R₀) to 2×10⁻¹⁶, and along the optimum, where the dissipation is I·u* rather than I²R, the error is the FIRST power of the current: 7.75 ppm at an ampere, 775 at a hundred, fitted exponent 1.0007.

A switching load's comb in a 10 mV reading: every tooth the same height. computed by solving, not by drawing. The harmonics of a 50-per-cent trapezoid of 100 mA at 100 kHz with a 10 ns edge, each multiplied by the shared conductor's impedance at its own frequency. Below 79.6 kHz the conductor is 5.00 mΩ of resistance and the teeth fall as 1/n. Above it the conductor rises as n while the harmonics still fall as 1/n, so the two cancel and every tooth from 239 kHz to 10.6 MHz is 0.828 mV — flat to 5.4 per cent across 52 of them. The comb stops at 31.8 MHz, which is the edge's own corner, so the total of 12.6 mV is decided by how many teeth there are: fitted over five edge rates it grows as the edge time to the power -0.497, against the −½ that a quadrature sum of equal teeth gives. A faster edge raises no harmonic and adds teeth. What a slow measurement keeps is a different term entirely: the zeroth harmonic, which is the 100 mA average through 5.00 mΩ of resistance — 500.0 µV, with no inductance in it and no edge, against the comb's 12.6 mV.

Every tooth the same height

A shared return conductor's impedance rises a decade per decade above 79.6 kilohertz and a switching load's harmonics fall a decade per decade, so the two cancel exactly: every harmonic of a hundred-milliamp square wave puts 0.828 millivolts into a ten-millivolt reading, from the conductor's own corner up to the edge's. The comb is flat rather than falling, so the total is decided by how many teeth there are — and since they add in quadrature it grows as the square root of the edge rate, 39.7 millivolts at a nanosecond against 4.03 at a hundred.

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.

A star ground is a 0.50 mm daisy chain, not a point. computed by solving, not by drawing. 8 circuits of 100 mA each meeting at one pad, which reaches the plane through a via of 0.25 mΩ and 0.35 nH. The via is shared by every one of them, so the star's residual is 7 other currents through it — 175.0 µV at 1.00 kHz against the 1.800e+4 µV a daisy chain of 10 mm hops would give at its far end, a factor of 103. Measured as a length of the same track, the via's resistance is worth 0.500 mm and its inductance 0.350 mm — two different lengths, so a via has a corner of its own at 114 kHz where a track's is at 79.6 kHz wherever it is cut. The faint curves are other circuit counts. The two grow differently: fitted over counts from 2 to 32 the star's error goes as N^1.22 and the chain's as N^1.87, because a chain's hop carries every circuit beyond it — so the star's advantage grows with the count rather than shrinking. And a star has no position in it: a chain's far end reads 4.5 times its near end, which is a diagnostic no real fault should have.

A star is half a millimetre long

The repair for a shared return is a star ground, and it is described as making the shared length zero. It does not: every return meets at a pad, the pad reaches the plane through a via, and a via of a quarter-milliohm and 0.35 nanohenries is worth half a millimetre of one-ounce track by its resistance and 0.35 by its inductance — two different lengths, so a via has a corner of its own at 114 kilohertz where a track's is at 79.6. What that buys is a change of law rather than a factor: a daisy chain's error grows as the square of the circuit count and a star's as the first power, so the advantage runs from sixty times at two circuits to three hundred and forty at thirty-two.

A resistor in the guard drive raises the worst negative resistance from −1.59 kΩ to −403 kΩ, and sets it itself. computed by solving, not by drawing. The parallel resistance a guarded input presents wherever it is negative, against frequency, for a 100 pF cable whose guard has 100 pF of its own to ground, driven by a 1.00 MHz follower through no resistor, 1 kΩ, 10 kΩ, 100 kΩ. Without a resistor the worst is −1.59 kΩ, −1/(ωₜC). With one it is −6.5 kΩ, −42.6 kΩ, −403 kΩ respectively: weaker, and for the largest resistor close to −R(C + Cₛ)²/(C·Cₛ) = −400 kΩ, a figure with the amplifier's bandwidth nowhere in it. The negative band still runs to about 970 kHz.

The resistor that moves the lag

A guarded input driven by a one-megahertz follower presents a negative resistance of −1/(ωₜC), −1.59 kΩ behind 100 pF of cable. The standard remedy is a resistor between the follower and the ring. It weakens the negative resistance, to −6.5 kΩ at 1 kΩ and −403 kΩ at 100 kΩ, but it does not remove it: the resistor and the ring's own capacitance to ground make a lag of their own. With a large enough resistor the worst value is −R(C + Cₛ)²/(C·Cₛ) for a drive resistor R — minus four times the resistor when the ring is as capacitive to ground as to the input — and the same at every length of one cable. The length moves somewhere else instead. The frequency where the bootstrap is given up, 1/(2πR(C + Cₛ)) for a drive resistor R, falls with every metre.

A guarded input driven through an inductance oscillates once the source's L/R passes 1/ωₜ. computed by solving, not by drawing. The real part of the rightmost natural frequency of a guarded input — a 100 pF cable bootstrapped by a 1.00 MHz follower, 10 pF of input capacitance — driven from sources of 10 Ω, 100 Ω, 1 kΩ with an inductance in series, against the inductance. Above zero the input rings with growing amplitude. Each source crosses at 1.75 µH, 17.6 µH, 187 µH, close to R/ωₜ — 1.59, 15.9, 159 µH — and oscillates at 11.4 MHz, 3.59 MHz, 1.14 MHz. The same source without the guard is stable.

The source that rings against its guard

A bootstrapped cable is a capacitance in series with a negative resistance that falls as the square of frequency, −1/(ω²·C/ωₜ). Drive it from a source with resistance R and inductance L and the loop's reactances cancel at 1/√(LC). Its resistances cancel where R = L·ωₜ, so the input oscillates once the source's own time constant L/R passes the follower's 1/ωₜ. The poles of the whole netlist say 17.6 µH for a 100 Ω source behind a 1 MHz follower, against 15.9 from the expression. The cable's length moves that by at most a factor of two and sets only the frequency it rings at. The same source without the guard is stable. A resistor in the guard drive raises the limit, to 164 µH through 1 kΩ, and pays for it in bootstrap bandwidth.

A 1 ms pulse of 100 A warms the element 98.7 mK: 4.94 ppm at its end, 2.86 ppm averaged, against 775 ppm held. computed by solving, not by drawing, from the exact step response of the shunt's thermal network — element, terminations and board, 0.2, 4 and 15.8 K/W with 5 mJ/K, 0.1 J/K and 2 J/K — expanded into its Foster modes. A 100 A pulse of 1 ms through the 77.5 µΩ shunt the burden optimum picks, dissipating I·u* = 0.775 W: the element rises 98.7 mK by the pulse's end and cools after it. At 50 ppm/K that is 4.94 ppm of error at the end of the pulse and 2.86 ppm averaged over it, where the same current held long enough for the board to settle would read 775 ppm high.

The pulse that ends before the heat

A shunt's self-heating error is a fixed point in the steady state — 775 ppm at 100 A for the shunt the burden optimum picks, through 20 K/W. A pulse never reaches it. Heat leaves the element through a ladder network — the element's own 5 mJ/K, its terminations, the board — and a pulse short against the element's millisecond heats it adiabatically, by P·t over its heat capacity alone: 0.737 ppm for 100 µs at 100 A, a thousandth of the held figure, 10.7 ppm for 10 ms. Along the burden optimum the error is linear in the current, so the current a pulse may carry for 100 ppm is exact: 12.9 A held, 936 A for 10 ms, 13.6 kA for 100 µs. A reading averaged over the pulse carries half the error of one taken at its end while the heating is adiabatic, and the section-by-section thermal model a data sheet quotes is 10% wrong at 7 ms.

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.

A second coupling stage with the same time constant: the top sags to 0.303 instead of 0.607, and the tail crosses zero at 2.27 s and overshoots to +0.025. computed by solving, not by drawing, marched and checked against the closed form to 3.0e-4: a half-second pulse through a 1 µF, 1 MΩ coupling stage (dashed), and through that stage followed by a buffered second one with a time constant of 1 s (solid). One stage sags to 0.6065 by the pulse's end and its undershoot decays monotonically to zero. Two stages sag to 0.3033, undershoot to −0.6966, cross zero 2.271 s after the pulse began and overshoot positive — the second capacitor removes the first moment of the waveform as well as its area, so the tail has to change sign.

The second capacitor, and the tail that changes sign

One coupling stage turns a pulse into a sagging top and an undershoot that returns to zero from below. Add a second stage with the same one-second time constant and a half-second pulse sags to 0.303 of itself instead of 0.607, undershoots to −0.697, crosses zero 2.27 s after it began and overshoots positive before settling. The step response becomes (1 − t/τ)·exp(−t/τ), through zero at exactly one time constant and down to −e⁻² at two. The reason is arithmetic: one capacitor removes the waveform's area, two remove its area and its first moment, and a pulse with a one-signed tail cannot have both. Built to the same stated corner, two stages still sag 1.29 times as fast, and a 1% flat top needs a pulse rate 806 times the corner rather than 625.

A second pole at 3.00 MHz makes a 9.49 Ω source 4.1 times safer and a 949 Ω one 20% less safe. computed by solving, not by drawing. The source inductance above which a guarded input oscillates — a 100 pF cable bootstrapped by a 1.00 MHz follower with 10 pF of input capacitance — bisected on the sign of the rightmost pole of the whole netlist, against the source resistance, for a follower with no second pole and with one at 30 MHz, 3 MHz and 1.5 MHz; the thin line is R/ωₜ. With a second pole at 3.00 MHz the critical inductance is 6.85 µH at 9.49 Ω against 1.66 with one pole, and 141 µH at 949 Ω against 177: a low-resistance source rings high above the follower's crossover, where the second pole takes gain away, and a high-resistance one rings near it, where the pole adds phase.

The second pole the ringing guard sees

A guarded input driven through an inductive source oscillates once L/R passes 1/ωₜ — for a follower with one pole. Give the follower the second pole every real amplifier has and the rule stops being a bound in either direction. A 9.49 Ω source that rang above 1.66 µH now needs 6.85 µH with a second pole at 3 MHz, four times safer; a 949 Ω source that rang above 177 µH now rings above 141, a fifth worse. The dividing line is the frequency the input rings at: above the geometric mean of the follower's crossover and its second pole, the pole takes gain from the bootstrap and helps; below it the pole adds phase and hurts. On three poles a decade apart the change of sign falls within 2.4 per cent of √(fₜ·f₂).

Without the buffer, a second stage of 1 times the first's impedance: the top sags to 0.175 instead of 0.303, and the tail crosses at 2.00 s and overshoots to +0.010. computed by solving, not by drawing, marched and checked against the closed form to 4.5e-4: a half-second pulse through two one-second coupling stages, buffered (dashed) and with the second hung directly on the first (solid), its resistor 1 times the first's and its capacitor 1/1 of it, so that each stage alone keeps its one-second time constant. Buffered, the top sags to 0.3033 and the tail crosses zero 2.27 s after the pulse began. Unbuffered, the denominator gains the cross term R1·C2 and its poles move to −0.382 and −2.618 per second; the top sags to 0.1751, undershoots to −0.8245, crosses zero at 2.002 s and overshoots to +0.0103. The slider is the second stage's impedance relative to the first's.

Two coupling stages without a buffer

Hang a second coupling stage directly on the first, with equal parts, and the pair's denominator gains a third time constant — the first resistor carrying the second capacitor's current. The poles spread from a double pole to −0.382 and −2.618 per time constant, a half-second pulse's top sags to 0.175 instead of the buffered pair's 0.303, and the sag rate is exactly 2 + 1/k per time constant for a second stage k times the first's impedance — the sum of the two short-circuit time constants. The tail still crosses zero, because the two zeros at the origin are untouched, and a step's lowest point is still at exactly twice its zero crossing. Then a reversal: specified by the overall corner rather than by the parts, the loaded pair sags only 1.12 times as fast as one stage and undershoots 7.6% rather than 13.5% — the buffer the cascade essay assumed makes the specified input worse.

With a 0.30 µF coupling capacitor and a stiff driver the transformer's damping is 0.289: the pulse's tail crosses zero 8 times in sixty units of L/Rₗ. computed by solving, not by drawing, marched and checked against the closed form to 2.3e-4: a pulse two L/Rₗ long through a series coupling capacitor into a 1 mH magnetising inductance with 100 Ω reflected across it, from a stiff driver. The capacitor is 0.30 µF, x = CRₗ²/L = 3. The poles are those of 3.000e-10·s² + 1.000e-5·s + 1, a damping of 0.2887 and a natural frequency of 9.19 kHz. The top ends at 0.128 of the high-frequency gain, and after the pulse the tail crosses zero 8 times, with lobes above a thousandth of the pulse, in the sixty units drawn. The slider is the capacitor.

The ring a blocking capacitor buys

A series capacitor in front of a transformer stops its flux from walking: a 55%-duty drive that walks a bare core by 0.40 units of volt-time a period settles within a few dozen periods once the capacitor takes the drive's average. The price is a second-order high-pass with the magnetising inductance, and an inductance lets its poles go complex where two resistive coupling stages never could. Between equal resistances the damping cannot fall below 1/√2, and the poles are complex only for a capacitor between 0.17 and 5.8 times L/R². From a stiff driver the damping is √(L/C)/2R and has no floor: a capacitor large enough to make the droop small — the droop is 1/RC, the capacitor's and not the transformer's — makes the tail ring, from one zero crossing at a tenth of L/R² to thirty-five at a hundred.

A 10:1 divider into 100 pF and 15 pF more behind 1.0 kΩ, trimmed for the calibrator, on a 1 ns edge. computed by solving, not by drawing from the residues of its four poles, on a time axis from 0.1 ns to 1 ms. The trimmer is set to 12.8 pF, where a 1 kHz square wave reads flat: the upper arm then matches the lower arm's total of 115 pF. On a 1 ns edge the output overshoots to 1.128 times where it settles, because for the first few nanoseconds the late 15 pF is not yet part of the divider, and falls back over that branch's own time constant, 15 ns, within 1% only by 35.5 ns. With a second branch in the upper arm, 9.0 kΩ and 1.7 pF — a scaled copy of the late branch — the same edge is flat to 7.0e-11 (dashed).

The time constant one trimmer cannot reach

A ten-to-one divider whose lower arm is 100 pF of instrument input and 15 pF more reached through a kilohm reads flat on a 1 kHz calibrator with its trimmer at 12.8 pF — the same setting as a lower arm of 115 pF lumped — and overshoots 12.8% on a 1 ns edge, settling within 1% only after 35.5 ns. No setting of the one trimmer is flat: the best judged on a fast edge is 6.1%, at 11.9 pF, and a trimmer set by eye so the fast edge lands flat leaves the calibrator's square wave 9.9% low with a 114 µs slope on its top. Flat at every time needs the upper arm to be a scaled copy of the lower one — a second branch of 9 kΩ and 1.7 pF — and then a 1 ns edge is flat to 10⁻¹⁰, with a window of −18 to +24% on that resistor and ±6% on its capacitor.

An integrator of 100 kΩ and 1 µF with its input grounded: what its own input currents do to it. computed by solving, not by drawing, marched for ten seconds from rest. With 50 nA flowing out of each input and the + input grounded, the output ramps at 50.0 mV/s — the bias current over the capacitance, to six figures — and reaches 0.500 V. With 100 kΩ in the + input to balance the resistances, only the 5 nA of offset current is left and the ramp is 5.00 mV/s. Neither stops: an integrator has no gain to quote, because its gain is time.

The current a capacitor adds up

An amplifier's input current in a resistor is an offset; in a capacitor it is a rate, and time is the gain. An integrator of 100 kΩ and 1 µF with its input grounded ramps at 50.0 mV/s from 50 nA, and balancing the inputs leaves the 5.00 mV/s of the offset current, still unbounded. A leak resistor bounds it at the current times the leak and costs the lowest frequency the circuit integrates. A hold capacitor's droop times its acquisition time is fixed, so 50 nA through a 100 Ω switch needs 655 nF and 0.77 ms for half a 16-bit step. A charge amplifier's offset times its low corner is Ib/2πCf, whatever its bleed.

What balancing the source resistances is worth, against how correlated the two input currents are, for a bias-cancelled input. computed by solving, not by drawing; 400 seeded parts at each correlation, each solved unbalanced and balanced at a 1 MΩ source, input currents of zero mean and 1 nA spread. The ratio of rms errors (dots) follows 1/√(2(1 − ρ)) (line): 0.72 for independent currents — balancing makes it 38% worse — one at ρ = ½, and 6.5 at 0.99. The cure removes what the two currents share, and is a cure only if they share more than half.

The cure that needs one current

Balancing the resistances an amplifier's inputs look into removes whatever its two input currents have in common and leaves their difference. On a plain bipolar input the two are one current at two pins, correlated at 0.991, and the cure is worth ten. On an input whose bias current has been cancelled they are two residues of either sign, correlated at 0.10, and over 400 seeded parts the cure makes the error 38 per cent worse. It is worth 1/√(2(1 − ρ)) for a zero-mean current, nothing at ρ = ½, and in general exactly the input current over the offset current.

A milliohm read four-terminal with a drifting thermoelectric EMF: the error against the test current, read three ways, and the shunt's own heating. computed by solving, not by drawing: an EMF of 0.5 µV settled plus 1 µV relaxing with a minute's time constant, read thirty seconds after a disturbance with readings a second apart; the shunt 20 ppm/K and 20 K/W. At 10 mA one polarity is out by 11.1%, a reversal by 501 ppm, the three-reading delta by 4.14 ppm. Every EMF error falls as 1/I; the heating, 0.400 ppm at an ampere, rises as I². The best current is 11 A for one polarity (149 ppm in all), 1.8 A reversed (4.08 ppm) and 0.37 A with the delta (0.167 ppm).

The voltage four wires carry in

Four wires remove a resistance measurement's leads by keeping the sense loop free of current, and that loop is exactly where a thermoelectric voltage lives. Ten milliamps through a milliohm is ten microvolts, so every microvolt where a copper lead meets a brass terminal moves a single-polarity reading by a tenth. Reversing the current removes any EMF that holds still and leaves the change over one interval; the + − + delta leaves the second difference; four readings leave the third — 501 ppm, 4.14 and less at 10 mA while a disturbed junction settles. Against the shunt's own heating each scheme has a best current, and the least error rises only as the cube root of the shunt's temperature coefficient: the reading method moves the error by orders of magnitude, the alloy by a cube root.

A 150 nH ground lead on a ten-to-one probe, at a 50 Ω source: the screen shows 49.4% of overshoot ringing at 121 MHz on an edge that has none. computed by solving, not by drawing, twice. A 1 V step from a 50 Ω source, watched through a ten-to-one probe with 11.5 pF at its tip, its ground returned through 150 nH of lead — about fifteen centimetres — and the scope reading the tip against the lead's end (solid, from residues; dots, a trapezoidal march, peak 1.4942). The same probe with its ground at the circuit's ground (dashed). The node itself steps cleanly. With the lead the tip capacitance and the lead inductance resonate at 1/(2π√(LC)) = 121 MHz, damped only by the source: ζ = (Rs/2)√(C/L) = 0.219, and the display overshoots by 49.4%.

The ring the ground lead adds

Every earlier measurement of a probe returned it to the circuit's ground directly. A real ten-to-one probe returns through fifteen centimetres of clip lead, about 150 nH, and its 11.5 pF tip and that lead are a series resonance at 121 MHz across the node being watched. The only damping it has is the source's own resistance, ζ = (Rs/2)·√(C/L), so the screen shows 49.4% of overshoot on a clean step from 50 Ω, 87.1% from 10 Ω, and none above 228 Ω — the stiffer the node, the worse the picture. Solved by residues and by a march of the netlist, the overshoot follows the second-order law at every source. A spring tip's 5 nH rings at 664 MHz and not at all from 50 Ω; a 150 Ω resistor in series with the tip brings a clip lead under 2% at the price of a 3.65 ns rise.

A ten-to-one divider with 15 pF of its arm behind a kilohm, stepped through 1 kΩ: one trimmer overshoots 0.00936% where an ideal source shows 13.2%, and the second branch that was exact no longer settles first. computed by solving, not by drawing: the step of the divider of the earlier essay, 9 MΩ and a trimmer over 1 MΩ with 100 pF direct and 15 pF behind a kilohm, through a source resistance of 1 kΩ, by residues. One trimmer at its calibrator setting (solid) overshoots 0.00936%, and the second branch that made the arm a scaled copy (dashed) rises without overshoot on the source's own lag. An ideal source (dotted) shows a 13.2% hook with one trimmer and nothing with two. A source resistance charges the input capacitance, 11.5 pF, so its lag and the hook act on one edge, and when they are comparable the hook's excess goes into shortening the lag.

The source the hook hides behind

A compensated divider with a late branch in its lower arm overshoots 13.2 per cent on an ideal step, and every measurement of that hook has been made from a source with no resistance. Through a real source the input capacitance charges first, and the hook and the source's lag act on the same edge. The overshoot is 10.5 per cent from 50 Ω, 4.13 from 300 Ω and gone by a kilohm, where the source's lag equals the hook's. The excess has not gone away: it is in the edge, which the screen shows 15 per cent faster than the source and the probe's capacitance allow. And the second branch that made the divider exact only makes its ratio exact, so from a kilohm the exact divider settles in 65.6 ns and the one with the hook in 44.

The same 15 pF spread along 2 kΩ of line, on a 1 ns edge: with 1 of its branches copied the output is flat to 1.07%, against 11.2% with one trimmer alone. computed by solving, not by drawing: the response to a 1 ns edge of a ten-to-one divider whose lower arm is 100 pF direct and 15 pF along a 2 kΩ resistive line, with one trimmer (dashed) and with the first 1 branch of the line's exact partial-fraction expansion copied into the upper arm (solid). The expansion's time constants are 12.2 ns, 1.35 ns, 486 ps and so on, as 1/(2k − 1)². Inverted numerically by Talbot's method, and the line itself checked against a netlist of sections.

The branches a square root needs

A resistive cable's late charge is a continuum and not one branch, and the partial fractions of its tanh are an exact description of it: 81.1 per cent of the capacitance behind a time constant of 12.2 ns, 9.01 per cent behind a ninth of that, 3.24 behind a twenty-fifth, all through the same kilohm. So a divider's upper arm can copy it one piece at a time. On a 1 ns edge one trimmer alone is 11.2 per cent out, one copied branch 1.07, two 0.252 and four 0.0353. The first branch does most of the work and the uncopied capacitance falls only as one over the count; what saves the rest is that a branch faster than the edge is averaged away. Faster edges ask for more branches, and the tolerance on the first ends the series before the arithmetic does.

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