Circuit Response
every model has an edge, and the figure carries it
Measurement, which is a circuit on a circuit

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.
18 min read 6 figures Two routes to a numberEvery model has an edgeRefused, not extrapolated

Assumes: Two terminals measure the leads as well · The divider, and the thing it does not know about · The source that is not a source

Two terminals measure the leads as well ends with a claim, and it is the one claim in this field that is stated as an elimination rather than as a bound: of the errors the instruments field measures, the four-terminal arrangement is the only one that removes its error instead of computing it and living with it. The essay is careful about the residual — it names the voltmeter’s input resistance, says the four-wire error is the ratio of the sense-lead resistance to it, and reports the number as six orders below the effect removed.

Then it draws that residual as a magnitude, on an axis that stops at a kilohm, and everything interesting happens above there.

Four wires against a 10 MΩ voltmetercomputed 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.-8-6-4-2021m10m100m1101001k10k100kresistance being measured (ohms)log₁₀ of the reading's errortwo wiresfour wiresone per centequally wrong at 707 Ωtwo wires exact at 1000 Ωsolved, then checked — the errors with their signs kepttwo wires beat four above 707 Ω
Fig. 1 The same two arrangements with the signs kept and the axis carried to a hundred kilohms. The four-wire error is not small and getting smaller — it is the voltmeter’s own divider, −(R+2Rlead)/(R+2Rlead+Rm)-(R + 2R_{lead})/(R + 2R_{lead} + R_m), and it grows with what is being measured. The two-wire curve plunges through the floor at 1000 Ω because it passes through zero there. The slider is the voltmeter’s input resistance, which is the number that decides all of this and which nothing in this collection had ever varied.

What the four-wire arrangement actually leaves

The two netlists are the ones the rung below builds. A current source, two force leads, the resistance under test, and a voltmeter of ten megohms placed either across the force pair or across a second pair tapping the device’s own terminals.

Solved, the four-wire reading is low by

R+2RleadR+2Rlead+Rm\frac{R + 2R_\mathrm{lead}}{R + 2R_\mathrm{lead} + R_\mathrm{m}}

which is the divider the voltmeter forms with everything in front of it, and agrees with the solved netlist to eight figures. Two things follow immediately and neither is in the usual summary.

The leads are still in it. At a device of one milliohm with fifty-milliohm leads the four-wire error is 1.0100 × 10⁻⁸, and 99.01 per cent of that is the two sense leads rather than the device. So the sense-lead resistance has not left the answer; it has been divided by the voltmeter’s ten megohms. That is a reduction by eight orders and it is worth having, and it is a different statement from “the leads do not appear”.

And the error grows. The two-wire error is 2Rlead/R2R_\mathrm{lead}/R and falls as the resistance rises, which is why the rung below is drawn at low resistance. The four-wire error is R/RmR/R_\mathrm{m} once the device is larger than its leads, and rises. The two arrangements therefore fail at opposite ends, and there is a resistance between them where they meet.

The reproduction, which comes before the departure

Before either of those is quoted, the difference between the two readings is checked against something that knows nothing about a netlist.

Subtracting the two solved errors leaves exactly

2RleadRmR (R+2Rlead+Rm)\frac{2R_\mathrm{lead}R_\mathrm{m}}{R\,(R + 2R_\mathrm{lead} + R_\mathrm{m})}

which is the lead term and nothing else — the four-wire arrangement removes that and leaves the rest untouched. One side of that comparison is two solved networks per resistance; the other is a ratio of four resistances. They agree to better than two parts in a thousand at all eighty-one resistances on the axis, and the figure asserts it at every one rather than at a convenient point.

That agreement is the instrument. It establishes that the only difference between the two arrangements is the term the technique was invented to remove, which is what entitles the next paragraph to be read as a statement about measurement rather than as a numerical accident.

The axis has an upper end and it is set by the arithmetic rather than by the argument, which is worth saying because it is the kind of limit that usually goes unsaid. The netlist contains a one-microhm return resistor, so its nodal matrix carries a conductance of 10610^6 siemens beside the voltmeter’s 10−710^{-7}, and thirteen decades of range is what double precision has. Above 1.111 MΩ the factorisation reports the matrix as singular and refuses — with the smallest pivot quoted, at 2.00 × 10⁻¹³ of the matrix norm — rather than returning a plausible number from a decomposition that has lost its meaning. Every sweep here stops an order inside that, and the boundaries it draws are all well below it. A solver that says which of its numbers it no longer believes is the reason the hundred-kilohm end of these curves can be quoted at all.

The departure: the reading that is exactly right

The two-wire error is the lead term plus the voltmeter term, and the two have opposite signs. The leads make the reading high; the voltmeter makes it low. Setting them equal gives

2RleadRm=R (R+2Rlead)2R_\mathrm{lead}R_\mathrm{m} = R\,(R + 2R_\mathrm{lead})

whose positive root is Rlead2+2RleadRm−Rlead\sqrt{R_\mathrm{lead}^2 + 2R_\mathrm{lead}R_\mathrm{m}} - R_\mathrm{lead}, and with fifty milliohms of lead and a ten-megohm voltmeter that is 1000 Ω. Bisected on the solved network the two-wire error there is 1.8 × 10⁻¹² of the reading — machine zero, not a small number.

At that same resistance the four-wire reading is 0.0100 per cent low.

A two-wire measurement of a one-kilohm resistor is therefore four orders of magnitude more accurate than a four-wire one, with the same leads, the same voltmeter and the same current. Not because anything about the two-wire arrangement is good, but because it carries two errors that cancel and the four-wire arrangement carries only the one that does not.

The cancellation is not a knife edge either. The two errors are equal in magnitude — one high, one low — at Rlead2+RleadRm−Rlead\sqrt{R_\mathrm{lead}^2 + R_\mathrm{lead}R_\mathrm{m}} - R_\mathrm{lead}, which is 707 Ω here, a factor of 2\sqrt2 below the exact point. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance, and it stays so however far the axis is carried, because the two curves differ by a positive quantity that never changes sign.

Four wires against a 10 kΩ voltmeter. computed by solving, not by drawing at 81 resistances, twice each, with a voltmeter of 10 kΩ 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 31.6 Ω — where the reading is right to 2.6e-14 while the four-wire reading is 0.316% low — and above 22.3 Ω the two-wire arrangement is the more accurate of the two at every resistance.
Fig. 2 A ten-kilohm voltmeter, which is what an analogue-to-digital converter’s front end or a moving-coil movement on a low range looks like. Every boundary moves down as the square root: the two-wire reading is exact at 31.6 Ω and the two arrangements are equally wrong at 22.3 Ω, where the four-wire reading is 0.316 per cent low. The technique has not become worse; the instrument behind it has, and the technique cannot see that.

Why the geometric mean, and where the collection has met it before

Both boundaries are geometric means, and that is not decoration. One error falls as 1/R1/R and the other rises as RR, so their sum has a minimum in the logarithm and the two terms are equal at the square root of the product of the two resistances that set them.

The ammeter that is a resistor is the same arithmetic about a shunt: a burden voltage that rises with the shunt and a resolution error that falls with it, best at their geometric mean, with the answer containing neither a resistance nor a current. The difference here is what the extremum is of. There, the geometric mean is where the total error is least and it is a design choice. Here, the geometric mean is where the total error is zero, because the two terms have opposite signs rather than merely different slopes, and it is not a design choice at all — it is a property of the leads and the voltmeter that a measurement either happens to land on or does not.

That difference has a consequence worth stating. A minimum can be designed for; a zero crossing cannot be relied on, because it moves with anything that moves the lead resistance. Copper is 0.39 per cent per kelvin, so a lead that warms twenty degrees moves RleadR_\mathrm{lead} by eight per cent and the exact point by 3.92 — a resistor sitting on the crossing at one temperature is four per cent off it at another, and its error at the same resistance goes from 1.8 × 10⁻¹² to 8.0 × 10⁻⁶, which is five orders lost to twenty degrees.

The number nothing had varied

Ten megohms is the value in every netlist on this site that contains a voltmeter, and it is one value out of a range that spans seven decades in instruments a reader would actually pick up.

Ten megohms is a general-purpose digital multimeter on most of its ranges, and it is the number the input divider in front of the converter sets. On the low direct-voltage ranges the same instrument usually bypasses that divider and presents more than ten gigohms, which is three decades higher and moves the exact point from a kilohm to 31.6 kΩ. A nanovoltmeter is higher still. In the other direction, a converter’s own input is not a resistance at all but a switched capacitor drawing charge at the sampling rate, whose equivalent resistance can be tens of kilohms; a moving-coil movement on a sensitive range is comparable; and an instrumentation amplifier’s differential input with bias-return resistors fitted is whatever those resistors are.

The consequence is that the whole picture slides by the square root of that number. A ten-kilohm front end puts the crossing at 22.3 Ω, which is inside the range four-wire measurements are actually made at — a shunt of a few ohms, a contact resistance, a winding — and a ten-gigohm one puts it at 22.4 kΩ, where nobody was going to use four wires anyway. So the same technique with the same wiring is either compromised or irrelevant depending on a specification line that the technique’s description does not mention.

Where 50 mΩ leads and the voltmeter put the boundaries. computed by solving, not by drawing — a bisection on the two netlists at every point inside the range they are well conditioned over, against the quadratic roots drawn here, agreeing to 8.4e-4%. Two of the three boundaries go as the square root of the voltmeter's resistance and one goes as the resistance itself, so they converge: the four-wire arrangement reaches 1.00% up to 910 mΩ with a 100 Ω voltmeter and 101 MΩ with a 10000 MΩ one, while the resistance at which a two-wire reading is exact moves only from 3.11 Ω to 31.6 kΩ. They meet at a voltmeter of 990 Ω, against 1 kΩ from 2R_lead/frac².
Fig. 3 The three boundaries against the voltmeter, which is the number that sets all of them. Two go as its square root and one goes as the number itself, so they converge: the four-wire arrangement reaches one per cent up to 910 mΩ with a hundred-ohm voltmeter and 101 MΩ with a ten-gigohm one, while the resistance at which a two-wire reading is exact moves only from 3.11 Ω to 31.6 kΩ over the same eight decades. Each drawn curve is a quadratic root and each is checked against a bisection on the solved netlists wherever that lands inside the range they are well conditioned over, agreeing to 8.4 × 10⁻⁴ per cent.

The closure nothing had called

This site’s own machinery has carried, since the day the four-terminal netlists were written, a closure that returns the resistance below which a two-wire reading is wrong by a stated fraction. It is 2Rlead/frac2R_\mathrm{lead}/\mathrm{frac}, it is one line, and nothing has ever called it.

Called and compared with a bisection on the solved network, it is right to a part in ten thousand at one per cent — 10.0000 against 9.99898 Ω. At a tenth of a per cent it is one per cent high. At three hundredths of a per cent it is ten per cent high. At a hundredth of a per cent it predicts 1000 Ω where the solved boundary is 618 Ω, so it is 62 per cent high.

The reason is the essay. 2Rlead/frac2R_\mathrm{lead}/\mathrm{frac} is the exact boundary with the voltmeter dropped, and dropping the voltmeter is precisely the approximation being tested — so the closure fails exactly where the effect it ignores becomes comparable, which is the most useful way for an approximation to fail and the least likely to be noticed, because a closed form does not become noisy or refuse. It returns 1000 Ω with the same confidence at every accuracy.

What each arrangement reaches, with 50 mΩ leads and a 10 MΩ voltmeter. computed by solving, not by drawing — quadratic roots drawn against bisections on the two netlists, agreeing to 1.9e-6% over 61 accuracies. Four wires reach 0.100% up to 10 kΩ and no further, because above that the voltmeter is the error. Two wires reach it between 99 Ω and 10.1 kΩ — a band and not a floor, because the two-wire error passes through zero at 1000 Ω. The closure this site has carried unused, 2R_lead/frac, is the lower edge with the voltmeter dropped and parts from it below 0.00707% — which is √(R_lead/R_m), and is the accuracy at which the four-wire ceiling falls to 707 Ω, the resistance at which the two arrangements were last as good as each other. Below it the ceiling is under the band and a range of resistances is reachable by neither.
Fig. 4 What each arrangement reaches, against the accuracy demanded of it. Four wires reach a tenth of a per cent up to 10 kΩ and no further. Two wires reach it between 99 Ω and 10.1 kΩ — a band rather than a floor, because the two-wire error passes through zero at 1000 Ω rather than merely getting small. The faint line is the unused closure, which is the band’s lower edge with the voltmeter left out.

A band, and then a gap

The two-wire arrangement’s usable region is a band with two edges, and that is a consequence of the zero rather than a curiosity. Below the band the leads dominate; above it the voltmeter does; inside it the two are cancelling well enough.

The four-wire arrangement’s region is a ceiling, set by the voltmeter alone: it reaches an accuracy up to frac⋅Rm\mathrm{frac}\cdot R_\mathrm{m} and no further. Both regions shrink as the accuracy tightens, and they shrink at different rates — the ceiling in proportion to the accuracy, the band’s lower edge in inverse proportion to it. So they cross.

They cross at Rlead/Rm\sqrt{R_\mathrm{lead}/R_\mathrm{m}}, which for these components is 0.00707 per cent, and the resistance at which they cross is 707 Ω — the equal-error crossing again, read as a fraction instead of as a resistance. Below that accuracy the four-wire ceiling has fallen beneath the two-wire band, and a range of resistances opens between them that neither arrangement reaches.

What each arrangement reaches, with 50 mΩ leads and a 10 MΩ voltmeter. computed by solving, not by drawing — quadratic roots drawn against bisections on the two netlists, agreeing to 1.9e-6% over 61 accuracies. Four wires reach 0.00100% up to 99.9 Ω and no further, because above that the voltmeter is the error. Two wires reach it between 951 Ω and 1.05 kΩ — a band and not a floor, because the two-wire error passes through zero at 1000 Ω. The closure this site has carried unused, 2R_lead/frac, is the lower edge with the voltmeter dropped and parts from it below 0.00707% — which is √(R_lead/R_m), and is the accuracy at which the four-wire ceiling falls to 707 Ω, the resistance at which the two arrangements were last as good as each other. Below it the ceiling is under the band and a range of resistances is reachable by neither.
Fig. 5 One part in a hundred thousand demanded of the same leads and the same voltmeter. Four wires reach it up to 99.9 Ω. Two wires reach it between 951 Ω and 1.05 kΩ, an island ten per cent wide around the exact point. Between 99.9 Ω and 951 Ω there is no arrangement of these four wires and this voltmeter that reads to one part in a hundred thousand, and nothing in either technique’s description says so.

That gap is the strongest form of the essay’s claim, and it is worth being precise about what it is not. It is not a statement that the measurement is impossible: a better voltmeter closes it, and the accuracy at which it opens moves as the square root of the voltmeter’s resistance, so a gigohm instead of ten megohms takes 0.00707 per cent to 0.000707. It is a statement that the boundary is set by the instrument and not by the technique, and that no amount of care with the wiring reaches past it — the same shape as every model has an edge, with the edge belonging to a specification sheet rather than to a frequency.

What the rung below got right

Almost all of it, and the exception is instructive. Its figure reports a four-wire error under a hundredth of a per cent across the whole range, and that is true of the range it draws, which ends at a kilohm. The number is a maximum over an axis rather than a property of the arrangement, and the axis was chosen — correctly — for the resistances a four-wire measurement is used at.

That is the ordinary way a bound becomes a belief: it is measured honestly, over the region where the technique is applied, and then quoted without the region. The site’s own habit is the answer, and it is why the sweep here runs two decades further and why the sign is kept.

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. 6 The rung below’s figure, unchanged: the same two arrangements as magnitudes, over the six decades a four-wire measurement is used across. Two wires are one per cent out at 10 Ω, four wires stay within a hundredth of a per cent — 1.00 × 10⁻² per cent at the top of its axis, which is where its maximum is — and every number on it is right. What it cannot show is that the lower curve is rising and the upper one is about to cross zero, because a magnitude on a logarithmic axis has thrown both facts away.

What this does not say

It does not say to measure a kilohm with two wires. The cancellation is real and it is unusable as a technique, for three reasons that are all in the numbers above. It holds at one resistance rather than over a range. It moves with the lead resistance, which moves with temperature, with flexing and with the state of a connector. And it requires the lead resistance and the voltmeter resistance both to be known to the accuracy being claimed, which is a harder measurement than the one being made.

What it says is that four-wire is not exact, and the region where it is the better arrangement has an upper edge. That edge is at RleadRm\sqrt{R_\mathrm{lead}R_\mathrm{m}} — 707 Ω for ordinary leads and an ordinary voltmeter, 22.3 Ω for a ten-kilohm one, 2.24 kΩ for a hundred-megohm one — and above it a four-terminal connection is buying nothing and costing two conductors.

And it says nothing about the force side. The current source in these netlists is ideal: it delivers whatever voltage the chain demands, so the force leads’ resistance affects only that voltage and never the answer. A real source has a compliance limit, and the force leads’ resistance eats into it — two ohms of lead at one ampere is four volts before the device is reached, which is most of what a bench milliohm meter has. That is a genuine limit on the force side and it is a limit on the current, not on the reading, so it enters this page’s arithmetic nowhere and enters a real measurement as the largest current that can be forced. Since the voltage across the device is what has to beat the instrument’s own noise floor, and that voltage is the current times a milliohm, the compliance limit is the practical floor of the whole technique and none of these figures contains it.

It also does not model the second error at low resistance, which the rung below names and this page inherits: every junction of dissimilar metals is a thermocouple, a few microvolts per kelvin is a large signal across a milliohm, and there are no sources in these netlists but the forcing current. Reversing the current and averaging is the answer to that and it is a different technique solving a different problem.

And nothing here is frequency. Every solve on this page is at direct current, which is what a four-terminal measurement is normally made at — but the millivolts in the wire measures ten millimetres of one-ounce copper as five milliohms and ten nanohenries with a crossover at 79.6 kHz, and a sense pair carrying no current still has that inductance in series with the voltmeter. A chopped milliohm meter escaping thermal voltages at a few kilohertz is inside that, and a pulsed one is not.

Where the same argument has already been made about the same anchor

This anchor’s other essay is in a different field and makes the point from the other side. Two ports from two one-ports recovers a coupled pair’s four impedance parameters by four solves, each port driven with the other open, and finds the two off-diagonal entries equal to 2.8×10−162.8\times10^{-16} — reciprocity, measured rather than assumed. Its “open circuit” is a large resistor rather than an infinity, for exactly the reason this page is about: a definition that says open has to be built as a number, and the number decides how nearly the definition is met.

A voltmeter is that same large resistor with a display on it. The four-terminal arrangement’s whole premise is that the sense pair carries no current, and what carries no current is an open circuit, and there is no such component. Ten megohms is what “no current” is built out of, and 0.0100 per cent at a kilohm is what it costs.

The rest of the field is the same sentence about other components. The probe is part of the circuit is a capacitance that cannot be made zero, and the probe that takes a tenth prices the divider that reduces it. The current the instrument draws is an input current that cannot be made zero, and the current that does not reach the input is the same structural repair as this one — an error current arranged to flow where it does not develop a voltage, rather than removed. The source that is not a source is the mirror image in the supply, and the divider, and the thing it does not know about is where all of it started.

The number worth carrying

707 Ω, for fifty-milliohm leads and a ten-megohm voltmeter: above it, four wires are the worse of the two arrangements. And 1000 Ω, where the two-wire reading is exactly right and the four-wire reading is a hundredth of a per cent low.

The habit that goes with it is one question asked of any technique described as eliminating an error. An error is never eliminated; it is exchanged for a different one, and the exchange is favourable over a range. The range has two ends, the description almost always gives one of them, and the way to find the other is to keep the sign, carry the axis past where the technique is used, and see what the residual is doing rather than how small it is.

Part 3 on Four-terminal

One argument about Four-terminal, and one of 3 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

A coupled pair as a two-port, at k = 0.8. computed by solving, not by drawing. Each port driven in turn with the other open, four solves, and the four impedance parameters read out. The diagonal terms measure each winding's own inductance — 10.0000 mH and 40.0000 mH against 10 and 40 — and both transfer terms measure the mutual inductance, 16.0000 mH against k√(L₁L₂) = 16.0000. The two transfer terms agree to 2.83e-16, which is reciprocity — a property of the device rather than of the measurement, and the first thing a coupling stamped into the wrong row would break. Neither of this site's two standing checks can see it: a coupling adds no current and dissipates nothing. Two ports from two one-ports Part 2 — An inductor is a one-port: one impedance, one number. Two of them coupled is a two-port, and the four impedance parameters that describe it are recovered here by four solves — each port driven with the other open, the definition read literally. Two of the four come out equal to 2.8 × 10⁻¹⁶, which is reciprocity, and is the first property a coupling stamped into one row instead of two would break.

Where this goes next

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. A coupled pair as a two-port, at k = 0.8. computed by solving, not by drawing. Each port driven in turn with the other open, four solves, and the four impedance parameters read out. The diagonal terms measure each winding's own inductance — 10.0000 mH and 40.0000 mH against 10 and 40 — and both transfer terms measure the mutual inductance, 16.0000 mH against k√(L₁L₂) = 16.0000. The two transfer terms agree to 2.83e-16, which is reciprocity — a property of the device rather than of the measurement, and the first thing a coupling stamped into the wrong row would break. Neither of this site's two standing checks can see it: a coupling adds no current and dissipates nothing. Two ports from two one-ports An inductor is a one-port: one impedance, one number. Two of them coupled is a two-port, and the four impedance parameters that describe it are recovered here by four solves — each port driven with the other open, the definition read literally. Two of the four come out equal to 2.8 × 10⁻¹⁶, which is reciprocity, and is the first property a coupling stamped into one row instead of two would break. 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. 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. A 10 kΩ + 10 kΩ divider, solved with its load. The unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does. The divider, and the thing it does not know about A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

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

Design tradeoffFour-terminal sensingInput impedanceKelvin connectionLead resistanceLoadingLow resistance measurementMeasurement conditionModel rangeVerification