The rail that is an input error

A shunt has to go somewhere and there are two places: between the load and the return, or between the supply and the load.

Between the load and the return — low side — both of the shunt’s terminals sit within a few millivolts of ground, and an amplifier reading it sees a small difference on a small common mode. That is the arrangement the ammeter that is a resistor solved — and the optimum that hands back a bandwidth priced in the frequency domain — and the answer it produced is the geometric mean of the amplifier’s offset and the supply: 7.75 millivolts of burden, 0.129 per cent of worst-case error, with neither the current nor the resistance appearing in either number.

Low-side sensing has a defect that has nothing to do with accuracy: the load’s return is no longer at ground, so every ground-referred signal in the load moves with the current. That is unacceptable in most systems, and the shunt goes high side.

High side, the amplifier’s two inputs sit at the supply rail. The difference it has to measure is still a few millivolts and the common mode is now twelve volts, and a finite rejection turns part of that twelve volts into a difference the amplifier cannot distinguish from the signal.

What the rail contributes, as an input error #

A rejection of $R$ decibels means a common-mode voltage $V_{cm}$ produces the same output as a differential input of $V_{cm}\cdot 10^{-R/20}$. That is the definition, and it says the rail arrives at the input as a voltage in series with the signal — indistinguishable from an offset, because nothing about it is different from an offset.

On a twelve-volt rail:

A hundred decibels is a good instrumentation amplifier and is what a purpose-built current-sense part specifies. It contributes twenty-four times its own offset, from a rail it is not measuring. The two contributions are equal only at

$$R = 20\log_{10}\frac{V}{V_{os}} = 20\log_{10}\frac{12}{5\times10^{-6}} = 127.6\ \text{dB}$$

which is above what any amplifier is specified at over temperature. So for every practical high-side part, the dominant equivalent input error is the rail rather than the offset, and the offset is what the data sheet leads with.

The optimum keeps its shape and changes its value #

Here is the part worth having, and it is a statement about the form of that essay’s result rather than about its number.

The low-side derivation balanced two errors: the burden over the supply, which rises with the burden, and the offset over the burden, which falls. Their sum is least at the geometric mean, and the error there is twice the root of the ratio. Nothing in that argument cares what the numerator of the second term is. Replace “the offset” with “the offset plus whatever else appears as an input error” and every step goes through unchanged:

$$u^* = \sqrt{V_{\text{eq}}\,V}, \qquad \varepsilon^* = 2\sqrt{\frac{V_{\text{eq}}}{V}}, \qquad V_{\text{eq}} = V_{os} + V\cdot10^{-R/20}$$

At a hundred decibels $V_{\text{eq}}$ is 125.0 µV rather than 5.0, so the optimum burden is 38.73 mV rather than 7.75 and the error there is 0.644 per cent rather than 0.129. Both are found here by golden-section search on the solved worst case, so the closed form above is being tested rather than evaluated: the search lands on 38.730 mV against $\sqrt{125\ \mu\mathrm{V}\times12\ \mathrm{V}}$ = 38.730 mV.

Five times the burden and five times the error, for moving a resistor from one side of a load to the other. That is the real price of high-side sensing and it is normally quoted as “needs a high-voltage input stage”, which is a packaging statement rather than an accuracy one.

The rail cancels out of the error, exactly as the current did #

Follow the substitution one step further and the high-side result acquires its own invariance, which is the mirror image of the low-side one and is the more useful of the two.

Where the common-mode term dominates — which is everywhere below 128 dB — $V_{\text{eq}} \approx V\cdot10^{-R/20}$, so

$$u^* = \sqrt{V\cdot10^{-R/20}\cdot V} = V\cdot10^{-R/40}, \qquad \varepsilon^* = 2\sqrt{\frac{V\cdot10^{-R/20}}{V}} = 2\cdot10^{-R/40}$$

The supply voltage has cancelled out of the error. The best burden is proportional to the rail — a 400 V rail wants a burden a hundred times larger than a 4 V one — and the error at that burden is the same number on both: $2\cdot10^{-R/40}$, which for a hundred decibels is 0.632 per cent and contains nothing but the rejection.

The right-hand column is the limit and the gap to it is the offset term, which is 2.4 per cent of the answer at 12 V and 0.06 per cent at 400. The higher the rail, the more completely the amplifier’s own offset stops mattering — a result that reads backwards until the mechanism is in view, and then is obvious: a bigger rail demands a bigger burden, and a bigger burden is exactly what makes an offset negligible.

So the two sides of the load have invariances of opposite kinds. Low side, the burden is a constant and the current does not appear. High side, the burden scales with the rail and the error does not appear to care about it. Both come from the same geometric mean and both are worth memorising in the form that has nothing in it: 7.75 mV at 5 µV of offset, and 0.63 per cent at 100 dB of rejection.

And the second one is a specification that can be read straight off a part’s front page without knowing anything else about the design, which is rare enough in this field to be worth saying plainly: a high-side current measurement is about $2\cdot10^{-\text{CMRR}/40}$ accurate, and nothing else about the circuit changes that.

The refusal that says the term is the whole difference #

Everything above rests on one claim: that the only thing which changed between low side and high side is the addition of $V\cdot10^{-R/20}$ to the equivalent input error. That claim is checked by removing it.

With the common-mode term set to zero and everything else left exactly as it is — the same netlist, the same solve, the same search — the optimum returns to 7.746 mV, which is the low-side answer to four figures. Not approximately, not to within the search’s resolution: the same number, because it is the same problem.

That is worth doing because the alternative explanations are plausible and wrong. A high-side shunt is in the supply rather than in the return, so one might expect the insertion error to behave differently — the shunt is in series with the same loop and it does not. One might expect the load’s own voltage to enter, since the shunt’s drop comes off the load’s supply rather than lifting its return — it does not, because the loop current is what is being measured either way. The refusal disposes of both at once.

Where the rejection actually comes from, and why it is not a constant #

Eighty to a hundred and forty decibels is a wide slider and the value is not a property of the amplifier alone.

The rejection four resistors decide measured the dominant mechanism for a difference amplifier: the rejection of a four-resistor bridge is set by how well the two ratios match, and one per cent resistors give about 46 dB whatever the amplifier does. A 0.01 per cent network gives 86. That is the reason a discrete difference amplifier is at the bottom of the table above and a monolithic one with laser-trimmed resistors is in the middle.

The rejection the parts have is the other half: the amplifier’s own rejection, which is finite, falls with frequency, and is specified at direct current. The high-side problem is not a direct-current problem — the rail has switching ripple on it, and the corner the instrument has no part in shows what a source imbalance does to rejection above a few hundred hertz. A part specified at 100 dB at direct current may be at 60 dB at a hundred kilohertz, which puts its equivalent input error at 12 millivolts and its optimum burden at 379 millivolts, and at that point the shunt is dissipating more than the measurement is worth.

So the honest way to read the table is as a function of frequency as well as of part number, and the usable entry is whatever the rejection is at the frequency the rail’s ripple sits at rather than at direct current. That is the number that never appears on a data sheet’s front page.

The three ways a part gets its rejection #

The slider spans sixty decibels and the entries on it are three different technologies rather than three grades of one.

A four-resistor difference amplifier gets its rejection from the matching of two ratios, so its rejection is the resistor tolerance read as a number of decibels: one per cent gives about 46 dB, 0.1 per cent about 66, 0.01 per cent about 86. That is the whole of it at direct current, whatever amplifier is in the middle, and it is the finding of the rejection four resistors decide. Such a part sits below the bottom of this essay’s slider and its high-side error is several per cent.

A monolithic current-sense amplifier has the same topology with the resistors trimmed on the die, reaching 100 to 120 dB, and it is the middle of the slider. Its rejection is specified over temperature and the specification is usually twenty decibels worse than the typical figure, which is where a design’s error budget should be taken from.

A chopper or an isolated part removes the mechanism rather than trimming it. A chopper modulates the input so that the common mode appears at a frequency the output filter removes; an isolated amplifier has no galvanic path at all, so there is no common-mode term to reject. Both reach past the 128 dB crossover, at which point the rail has stopped being the dominant input error and the amplifier’s own offset is back — which is the situation the low-side analysis describes, reached from the other direction.

The progression is worth reading as a single statement: the three technologies are three answers to one term in $V_{\text{eq}}$, and the money buys decibels of rejection, which buys the square root of that in error. Twenty decibels of rejection is a factor of ten in $V_{\text{eq}}$ and a factor of 3.16 in error, so each step up the list is worth about ten decibels of measurement accuracy — which is a poor exchange rate and is why high-side sensing is expensive.

What the burden costs on the high side #

The optimum is five times larger and the consequences are not linear.

At ten amps, a 38.73 mV burden is a 3.873 milliohm shunt dissipating 387 milliwatts, against 77.5 mW at the low-side optimum. That is a physically larger part with a larger thermal rise, and what the rise does to the reading is the subject of a companion essay here. It is also five times the time constant the optimum that hands back a bandwidth computed, since $\tau = L/R$ and the resistance has gone up by five — so the high-side optimum is five times faster as well as five times less accurate, which is the one thing in this essay that moves in the welcome direction.

And it is five times more voltage for the rail to lose. A 38.73 mV drop on a twelve-volt supply is 0.32 per cent of the load’s voltage, which for a converter regulating at its output is nothing and for one regulating at its input is a specification.

Both of those push in the same direction: accept a worse error than the optimum and take a smaller burden. The error curve is flat near its minimum — that is what a geometric mean does — so a burden of 15 mV instead of 38.73 costs about a third more error and saves sixty per cent of the dissipation. The optimum is a landmark rather than a target, which is a property the low-side essay noted and which matters more here because the optimum is now expensive.

What a single rejection figure leaves out #

That rejection is a single number. It is a function of frequency and of source imbalance, and the figure uses a direct-current value as a slider precisely so that the consequence of each value is visible. A real design has to read the rejection at the frequency of whatever is on the rail.

That the offset and the common-mode term add linearly in general. They add here because both are referred to the input and both are worst-case. A statistical budget would add them in quadrature and get a smaller number; the shunt analysis here is a worst-case one throughout, and mixing the two conventions is how a budget comes out optimistic.

That the shunt’s position changes nothing else. It changes the amplifier’s input common-mode range requirement, which is a packaging and process question, and it changes what happens in a fault — a shorted load puts the full supply across a low-side shunt’s amplifier and nothing across a high-side one’s. Neither is an accuracy question and both decide designs.

That 128 dB is unreachable. A chopper-stabilised part with a trimmed network can exceed it at direct current. What it cannot do is hold it over temperature and over frequency, and the crossover is a statement about the specification a designer can rely on rather than about what is achievable on a bench.

The optimum searched, the crossover named, and the term removed #

The optimum is golden-sectioned on the solved worst case at every rejection on the slider, and checked against $\sqrt{(V_{os} + V/\text{CMRR})V}$ to five parts in a thousand — a closed form tested rather than evaluated.

The error at the optimum is checked against $2\sqrt{V_{\text{eq}}/V}$ to one per cent, which is the second half of the same form and would not follow from the first if the shape of the balance had changed.

The crossover at 128 dB is checked as a number, and the measurement at the drawn rejection is required to agree with which side of it that rejection is on.

And the whole difference is refused. With the common-mode term removed, the search is required to return the low-side optimum to five parts in a thousand, which is what makes “the term is the whole of the difference” a measurement.

A result that survived by being about the right quantity #

The essay before it produced $u^* = \sqrt{V_{os}V}$ and made a good deal of the fact that no resistance and no current appear in it. This essay has changed the numerator by a factor of twenty-five and the form has not moved.

That is worth noticing because it is the test of whether a derived result is a formula or an understanding. A formula with “the amplifier’s offset” written into it would have been wrong the moment the shunt moved, and a designer carrying it would have sized a high-side shunt five times too small and been 1.68 per cent out. The balance it came from — one error rising with the burden and one falling — is what survives, and the thing to carry forward is the balance rather than the number.

The same substitution works again for anything else that arrives as an equivalent input error: the amplifier’s drift over the operating temperature range, its input bias current times the source resistance, a thermocouple at a solder joint. Each goes into $V_{\text{eq}}$ and the optimum moves accordingly, and the reason the millivolts in the wire matters to a shunt at all is exactly that: a few microvolts of thermal EMF at a junction between copper and manganin is a term in the same sum, and at a 7.75 mV burden it is not negligible.

Which suggests the general form of the rule, and it is the useful sentence: the best burden voltage is the geometric mean of the supply and everything that looks like an input offset, and the job is to enumerate everything that does.

Still open: the rejection at the ripple’s frequency, the fault case, and the sensor with no common mode #

Rejection as a function of frequency, on the same axes. Every number here uses a direct-current rejection. A rail has ripple on it at the switching frequency, where the rejection is tens of decibels worse, and the equivalent input error at that frequency is what decides whether the measurement is usable. Sweeping rejection against frequency for a real difference amplifier and reading the optimum off it would give a burden voltage that depends on the converter’s switching frequency, which is a strange and probably correct result.

What the optimum becomes in a fault. A shorted load puts the whole supply across a low-side shunt’s amplifier inputs and a whole supply of common mode across a high-side one’s. The error analysis above assumes a common mode equal to the rail; during a fault it is the rail on one input and something else on the other, which is a differential input of twelve volts into a part expecting millivolts. Whether the optimum burden should be chosen to survive that rather than to minimise error is a design question the arithmetic here does not answer and does inform.

And the arrangement with no common mode at all. An isolated amplifier or a current transformer has no common-mode term because it has no galvanic connection to the rail, so $V_{\text{eq}}$ returns to the offset and the optimum returns to 7.75 mV — at the price of the errors the ammeter that is not in the circuit measured. Putting the two error budgets on one axis against current would say at what rail voltage isolation becomes the cheaper answer.