The ammeter that is a resistor

A voltmeter measures a voltage by drawing almost no current. An ammeter cannot do the corresponding thing, because there is no way to measure a current without the current going through something. So every ammeter is a resistor with an amplifier across it, and the resistor is in the circuit being measured.

That gives the instrument two errors that point in opposite directions along one axis, which is the shape this collection met once before in a switch: a quantity that must be small compared with one thing and large compared with another at the same time. The answer is a band with an interior optimum, and the optimum turns out to contain neither the resistance nor the current.

The two errors, written in the same currency #

Let the circuit be a supply $V$ and a load, drawing a current $I$ that somebody wants to know. Put a shunt $R_s$ in series with it and measure the voltage across it with an amplifier whose input offset is $V_{os}$.

The insertion error. The shunt is now part of the circuit, so the current is $V/(R_{\text{load}} + R_s)$ rather than $V/R_{\text{load}}$. As a fraction, the current is low by $R_s/(R_{\text{load}} + R_s)$, which for a small shunt is $R_s I/V$. This is not an error in the reading; the reading is correct. It is an error in the quantity, because the current the circuit would have had is not the current it has.

The offset error. The amplifier reports $I R_s + V_{os}$ instead of $I R_s$, so the current it reports is high by $V_{os}/R_s$. As a fraction of $I$, that is $V_{os}/(I R_s)$.

Both are naturally written in one variable, the burden voltage $u = I R_s$ — the voltage the instrument costs the circuit:

$$\varepsilon_{\text{ins}} = \frac{u}{V}, \qquad \varepsilon_{\text{off}} = \frac{V_{os}}{u}.$$

One rises with $u$ and one falls, so there is a best $u$, and it is where their sum of magnitudes is least:

$$u^\star = \sqrt{V_{os}\,V}, \qquad \varepsilon^\star = 2\sqrt{\frac{V_{os}}{V}}.$$

The figure does not evaluate that. It golden-sections the worst case on the solved network and then asserts the two agree: 7.7485 mV measured against $\sqrt{V_{os}V} = 7.7460$ mV, and 0.1291% measured against $2\sqrt{V_{os}/V} = 0.1291\%$.

Neither answer contains the current #

The striking part of that pair of expressions is what is missing from them.

$u^\star$ is the geometric mean of the amplifier’s offset and the supply voltage. It contains no resistance, no load, and no current. So a design measuring ten milliamps and a design measuring a hundred amps on the same rail with the same amplifier both want 7.75 mV across the shunt — and only the resistance differs, by a factor of ten thousand:

The figure asserts this across four orders of current, because a claim that a quantity does not depend on something cannot be made by evaluating it once.

$\varepsilon^\star$ is stranger still: twice the root of the ratio of two voltages, one of which is a property of the amplifier and the other of the power supply. The accuracy a series shunt can reach is set by the rail it is working on. Doubling the supply improves the achievable error by $\sqrt2$ with no change to any component; halving the amplifier’s offset does the same.

The cancellation that is real and not usable #

At exactly $u^\star$ the figure’s total error passes through zero, and the reason is worth stating because it is a trap.

The two errors have opposite signs. The insertion error makes the current low; a positive offset makes the reading high. At the burden voltage where their magnitudes are equal, their sum is exactly zero, and the solved network says so to the last digits.

It is a real cancellation and it is not a design. An input offset’s sign is not known — it is a manufacturing accident, it drifts with temperature, and on a batch of parts it is as likely one way as the other. So the quantity a design has to quote is the sum of the magnitudes, which is what the figure plots as the third curve and what $2\sqrt{V_{os}/V}$ is.

The figure draws both anyway, because a plot that showed only the cancellation would be showing an accident and a plot that showed only the worst case would be hiding a fact about the circuit. The cancellation is also the reason the optimum is unusually flat: near $u^\star$ the worst case has a minimum with zero derivative, so a shunt a factor of two away from ideal costs only 25% more error than the best one. That flatness is what makes the answer usable at all, since 7.75 mΩ is not a value anybody stocks.

Reading the picture the other way round #

There is a second reading of the two curves that is worth taking, because it turns the figure into a specification rather than an optimisation.

Fix the error a design is allowed — say a quarter of a per cent — and the two curves cut the axis at two burden voltages rather than one. Below the lower one the amplifier’s offset dominates and the reading is out; above the upper one the shunt’s own drop has changed the current by too much. Between them is a band of acceptable burden voltages, and the design question is not “what is the best shunt” but “does a stocked resistance land inside the band”.

The band’s width follows from the same two expressions. Setting $u/V + V_{os}/u = \varepsilon$ and solving gives a ratio between the two roots of

$$\frac{u_{\text{hi}}}{u_{\text{lo}}} = \left(\frac{\varepsilon + \sqrt{\varepsilon^2 - \varepsilon^{\star 2}}}{\varepsilon - \sqrt{\varepsilon^2 - \varepsilon^{\star 2}}}\right)$$

which is one — no band at all — when $\varepsilon = \varepsilon^\star$, and widens quickly above it. At twice the optimum error the band is a factor of 13.9 wide; at four times, 62. So a design that can tolerate half a per cent on a twelve-volt rail with a five-microvolt amplifier has a comfortable choice of shunt, and a design that wants 0.15% has almost none.

That is the same shape as the switch’s band and the filter’s impedance band elsewhere in this collection: a tolerance and two mechanisms give a region, the region shuts as the tolerance approaches the mechanisms’ own limit, and the useful design number is the width rather than the centre.

The third axis, which finally decides it #

Dissipation is $u I$, and $u$ is fixed at 7.75 mV by the argument above, so the power in the shunt is proportional to the current: 78 µW at ten milliamps, 7.75 mW at an amp, 775 mW at a hundred.

At a hundred amps that power is in a component small enough to have a thermal resistance of tens of kelvin per watt, so the shunt runs warm. A metal-film shunt at fifty parts per million per kelvin, risen by forty kelvin, has changed its resistance by

$$50\times10^{-6} \times 40 = 0.20\%$$

which is larger than the entire electrical optimum of 0.129%. So above some current the answer stops being the one this essay derived and becomes a thermal one, and the boundary between the two regimes is where the self-heating error crosses $2\sqrt{V_{os}/V}$.

Three responses to that are used and each moves a different term.

Lower the burden voltage below the optimum, accepting more offset error to get less heat. That is why high-current shunts are specified at 25 or 50 mV full scale rather than at the hundreds of millivolts an older instrument used.

Buy a lower temperature coefficient, which is what a manganin or Zeranin element is for — a few parts per million per kelvin rather than fifty, bought at real cost.

Sense the shunt in four terminals, so the connection resistance and its much worse temperature coefficient are outside the measurement. That is the same repair as the resistance-measurement essay’s and it is why a real shunt is a four-terminal component.

High side, low side, and why the choice is not free #

The essay so far has been indifferent to where the shunt sits. In practice there are two places and they fail differently.

Low side, between the load and ground. The amplifier’s inputs sit near zero, so an ordinary amplifier will do and the common-mode problem does not exist. What it costs is that the load’s ground is no longer ground: it sits at the burden voltage above it, 7.75 mV here, and every other signal referred to that node is offset by it. That is the common-impedance fault of this field, created deliberately.

High side, between the supply and the load. The load’s ground stays clean. What it costs is that the amplifier’s inputs sit at the supply and must reject it: the rejection required is the rail voltage divided by the error allowed, and for a twelve-volt rail and a 0.1% error on a 7.75 mV signal that is $12/(7.75\times10^{-6})$, or 124 dB. A difference amplifier built from 0.1% resistors gives 54 decibels, seventy short.

That gap is the reason a high-side current sense is a specific kind of part rather than an op-amp and four resistors, and it is the same arithmetic the instrumentation-amplifier essay does: rejection is bought with gain in front, and the four resistors around the last stage are what set it.

What is on the other side of the model #

Everything above assumes the shunt is a resistance. Above a frequency it is not, and the boundary is the one this collection draws for every passive.

A shunt has inductance — a few nanohenries for a small one, more for a bar — and its impedance rises with frequency, so a current step is read with an overshoot that decays with $L/R$. At 7.75 mΩ and two nanohenries that time constant is 258 nanoseconds, which is enormous compared with the current step it might be asked to read in a switching converter. The usual repair is a compensation network across the sense terminals with a matching time constant, which is a pole cancelled by a zero and therefore has its own tail.

And the amplifier has noise as well as offset, so the error floor derived above is a direct-current one. In a bandwidth, the offset term is replaced by the amplifier’s noise in that bandwidth, and the same optimisation gives the same shape of answer with a different constant.

Two other ways to know a current, and what each gives up #

The series shunt is one of three answers, and the other two are worth naming because their boundaries are different in kind rather than in size.

A current transformer puts no resistance in the circuit at all — its burden voltage is the drop across a magnetising impedance and can be made microvolts — and pays for it by not working at direct current at all. Its lower edge is a volt-second limit: the core walks a little each cycle if there is any asymmetry, and this collection’s magnetics field measures exactly that. So the shunt’s two-sided band in resistance is replaced by a two-sided band in frequency, bounded below by the core and above by the winding’s own capacitance.

A Hall or magnetoresistive sensor reads the field rather than the voltage, so it too costs almost no burden and does work at direct current. What it gives up is offset and drift of its own, an order worse than a good amplifier’s, and a sensitivity to any other current in the neighbourhood. Its error floor is not $2\sqrt{V_{os}/V}$; it is a fixed fraction of full scale, which makes it good at the top of its range and poor at the bottom in a way the shunt is not.

The shunt’s distinguishing property, and the reason it remains the default for accuracy, is that its error is set by two voltages that a designer can buy — a better amplifier, a higher rail — rather than by a physical effect with a floor. That is what the closed form says, and it is why the closed form is worth having.

What the shunt shares with the field’s other instruments #

A shunt is a resistance placed in the circuit, so every complaint the instruments field makes about itself applies to it. The ammeter that is not in the circuit is the alternative, with a different list of failures rather than fewer. Two terminals measure the leads as well is the repair that makes a shunt a four-terminal component, and it is the one arrangement in the field that removes an error rather than bounding it. The millivolts in the wire is where the burden voltage goes once it shares a return, and The four resistors that decide, and the two that do not is the amplifier that has to read it. A band rather than an edge is the other two-sided boundary in the collection, and this optimum is the second.

What is checked #

Three assertions, and the third is the one that makes the first two more than arithmetic.

That the best burden voltage is $\sqrt{V_{os}V}$ — found by golden section on the solved worst case, and compared with the closed form to two parts in a thousand. The search knows nothing about the expression.

That the error there is $2\sqrt{V_{os}/V}$, to five parts in a thousand, which is the statement that no shunt at all does better on a given rail with a given amplifier.

And that the same burden voltage is the answer at ten milliamps and at a hundred amps — the optimisation run again at three currents four orders apart, with the answers required to agree to five parts in a thousand. Without it the first two are expressions that happen to fit one circuit; with it they are statements about what a series shunt can do.

What is under the offset, and what is beside it #

The optimum on this page is a geometric mean of two voltages, and one of the two is an amplifier’s input offset — a single number standing for several mechanisms that this collection measures separately. Each of them moves the answer, and two move it in ways an error budget written in volts would not catch.

What matching does about temperature is the one that decides how the offset behaves rather than how large it is. A differential pair’s residual offset is the thermal voltage times the logarithm of the saturation-current mismatch, which is proportional to absolute temperature and therefore drifts in proportion to itself: 3 333 parts per million per kelvin, at every mismatch, because the ratio contains nothing about the device. So the burden voltage this essay computes is optimal at one temperature, and the optimum moves as the square root of the offset — which for a sixty-kelvin excursion is about ten per cent, comfortably inside the flatness of a geometric-mean optimum and worth knowing is there.

The current the instrument draws is the term that is not an offset at all and that the arithmetic above has no place for. Fifty nanoamps of input bias current is nothing until it flows in a resistance, and a shunt amplifier’s input sees the shunt on one side and whatever sets the gain on the other — so a bias current turns into an apparent burden voltage through the gain-setting resistors rather than through the shunt. That essay’s classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten rather than a thousand, and it costs forty per cent of the noise density to get.

And the floor a circuit has is the term that decides whether the optimum is reachable at all at the small end. An amplifier contributes a voltage generator and a current generator, with a source resistance at which their sum is least — 6.67 kΩ, the ratio of the two — and a shunt is three or four decades below that, so a current measurement is always made from a source far too low for the amplifier that reads it. The offset is what limits the accuracy of a single reading, which is what this essay optimises; the noise is what limits how small a change is visible, and it is not optimised by the same burden voltage.

The alternative that has none of those three terms is the ammeter that is not in the circuit, and its price is instructive precisely because it is paid in a different currency. A thousand-turn secondary reflects twelve microhms into the primary — no burden worth the name, no offset, no source impedance for an amplifier to be noisy into — and charges instead with 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. Neither instrument is better; the geometric mean on this page is what a series element can do, and the other essay is what a magnetic one can.