The ammeter that is not in the circuit

The rung below this one measured a shunt: a resistance placed in series with the thing being measured, turning its current into a voltage. Every objection to it comes from the same fact. It is in the circuit, so its voltage is subtracted from the supply; it dissipates $I^2R$, which at a hundred amperes is real power in a small part; and to keep both of those tolerable it has to be a milliohm, at which point the voltage it produces is fifty millivolts and the amplifier reading it has to be very good indeed. The essay ended with the resistance at 500 microhms and the two errors it leaves — the lead resistance in a two-terminal connection, and the amplifier’s offset against a signal that small.

A current transformer removes all three objections by not being in the circuit. The primary is one turn — which in practice is the conductor itself, passing through a hole — and the whole measurement is the flux around it. With a thousand-turn secondary into a ten-ohm burden, what the primary sees is that burden divided by the turns ratio squared: twelve microhms, which is a fortieth of the shunt the rung below settled on, and which dissipates nothing worth measuring.

What it charges instead is the subject of this essay, and it is charged in a currency the shunt has no entry for.

Why the model has to be driven by a current #

The netlist here differs from every other transformer in this collection in one element, and the difference is not a convenience.

Everywhere else a transformer is driven by a voltage source through a source resistance, because that is what a signal transformer sees. A current transformer’s primary is fed by whatever the load at the end of the wire happens to be drawing, and by nothing the transformer does. Modelling it with a voltage source and a resistance would let the secondary’s reflected impedance change the primary current — the precise error a current transformer is built not to make — and would produce a plausible ratio error that was really an artefact of the source.

So the source is a current source, the primary is one turn of the same core the secondary is wound on, and the coupling is a mutual term in the solved matrix rather than a decomposition into magnetising and leakage inductances. The T-model’s predictions are then something to check the answer against rather than the route to it, which is the arrangement the magnetics field uses throughout.

The magnetising inductance is where the answer goes #

The mechanism of the low-frequency error is one sentence. The burden and the secondary’s own resistance sit across the winding’s inductance, so the current available to the burden is the ideal one times a high-pass:

$$\frac{I_b}{I_p/N} = \frac{j\omega L_2}{R_b + R_2 + j\omega L_2}$$

with a corner at $(R_b + R_2)/2\pi L_2$.

For this core and a thousand turns the secondary inductance is 4.19 henries and the corner is 0.456 hertz. At fifty hertz the magnitude error is a part in seven thousand and the phase error is 0.52 degrees.

At direct current the response is exactly zero, which is not an error to be bounded but a property to be stated: a current transformer cannot measure direct current at all, and a direct component in the primary does something worse than not being measured — it walks the core’s flux, which the magnetics field has already measured as its own failure.

The slider is the burden resistance and it moves the corner in proportion: one ohm gives 0.114 hertz and 0.13 degrees at fifty; a hundred ohms gives 3.88 hertz and 4.43 degrees. Which is the first half of the trade — a large burden gives a large output voltage and a large error.

The floor that is a coupling rather than a frequency #

The interesting part of the measured ratio error is not the corner. It is what happens above it.

The closed form says the error falls as $1/f^2$ and goes to nothing. The solved netlist says it falls and then stops, at 100 parts per million, and stays there for four decades. That floor is $1 - k$ — the coupling coefficient — and it is not a frequency effect at all: it is the fraction of the primary’s flux that does not link the secondary, and it is the same at a hertz and at a hundred kilohertz.

Nothing in the high-pass expression contains it. It appears in the answer because the answer is a solve of the whole network rather than an evaluation of the expression, which is the reason this collection solves rather than evaluates.

The practical consequence is a ceiling on accuracy that no amount of magnetising inductance improves. The ratio error at fifty hertz here is 142 parts per million, of which 100 is the coupling and 42 is the magnetising high-pass — so doubling the turns, which would quarter the second term, would take the total from 142 to 110 and no further. Getting past that means a better-coupled winding, and a toroidal core with a fully distributed secondary is exactly what a metering-class current transformer is.

The phase error is the expensive one #

Half a degree sounds negligible and it is the largest error in the instrument, because of what the instrument is usually for.

A metering current transformer is measuring power, which is $VI\cos\varphi$, and a phase error $\delta$ in the current turns that into $VI\cos(\varphi + \delta)$. The fractional error is then

$$\frac{\cos(\varphi + \delta)}{\cos\varphi} - 1$$

which contains $\tan\varphi$ and therefore blows up as the power factor falls. Measured with the 0.5225 degrees this transformer has at fifty hertz:

At unity power factor the phase error costs four parts in a hundred thousand, which is nothing. At a power factor of 0.05 — a lightly loaded motor, a transformer on no load, a reactive compensator — it costs eighteen per cent of the reading, while the ratio error at the same frequency is 0.014 per cent.

So the two errors of this instrument differ in importance by three orders depending on what is being measured, and only one of them appears in the specification most people read. It is also the error that gets worse in the direction nobody expects: raising the burden to get a bigger signal raises the phase error in proportion.

An ampere limit that is really a flux limit #

The upper bound on this instrument is not on the plot, and it is the one that is most often met in practice.

The burden voltage is impressed across the secondary winding, and a winding integrates the voltage across it into flux. So the core carries $\lambda = \int v\,dt$, the peak flux density is $\lambda/NA_e$, and when that reaches saturation the transformer stops being one — abruptly, and in the direction that makes the reading too small.

With this core, $\lambda_{\max} = NA_eB_{sat}$ is 35 milliwebers, which at fifty hertz allows a burden voltage of 11.0 volts peak and therefore a primary current of 916 amperes.

The number that matters is the shape rather than the value: it is a volt-second limit read as an ampere limit, so it is proportional to frequency. The same transformer is good to 306 amperes at 16.7 hertz and 7,330 at 400. A current transformer specified on a 50-hertz supply and used on a 16⅔-hertz railway supply saturates at a third of its rating with nothing in its markings to say so.

And it is proportional to $1/R_b$, because the burden decides the voltage: the one-ohm burden above is good to 3,665 amperes and the hundred-ohm one to 108. The burden that gives the best phase accuracy is also the one with the most headroom, which is the one place in this instrument where two requirements agree.

The burden is therefore the only knob that moves everything at once, and it is worth seeing it moved downwards as well as up. A small burden buys a small phase error, a low corner and a high saturation current; what it costs is signal, and the amplifier that has to read that signal is the subject of the rung below this one.

The open secondary, which is the one dangerous failure #

Every account of current transformers carries a warning in bold and it is worth deriving rather than repeating, because the derivation says exactly how bad it is.

A current transformer’s secondary must never be open-circuited while current flows in the primary. The reason is in the same expression as the ratio error, read at $R_b = \infty$: with nowhere for the secondary current to go, the whole primary current becomes magnetising current, the core saturates within a fraction of a cycle, and the flux then changes almost entirely during the brief interval each cycle when it is not saturated. A thousand-turn winding with $d\lambda/dt$ concentrated into a small part of a cycle produces a voltage spike of kilovolts on a transformer whose normal output is a volt.

The slider on the figure above cannot be taken there — the model’s ratio error at an infinite burden is not a small quantity to be plotted, and a linear core would give a large finite answer that is wrong for the interesting reason: it has no saturation in it, and saturation is the whole mechanism.

So this is a case where the model is deliberately not asked. What can be said from inside it is the part that matters for design: the burden is not an option, it is the element that makes the instrument an instrument, and a metering current transformer is shipped with a shorting link across its terminals for exactly this reason.

Two solves, and the difference between them #

The way both errors are obtained is worth a paragraph, because it is the same manoeuvre the shunt essay used and it is what makes a reading a measurement.

Every number here is a ratio of two solves of the same netlist: the burden voltage the network produces, against the burden voltage an ideal transformer of the same turns ratio would produce. The first is a solve. The second is one multiplication. Their ratio is a dimensionless error whose real part is the ratio error and whose argument is the phase error, so both come out of one solve and are guaranteed consistent with each other.

The alternative — computing the ratio error from the high-pass expression and the phase error from its argument — would produce two numbers that agree with each other by construction and with the circuit only if the expression is right. Here the expression is checked against the solve rather than standing in for it, which is how the coupling floor was found at all: it is in the solve and not in the expression.

The two instruments, side by side #

Putting the rung below this one beside this one gives the comparison a designer actually makes.

Neither dominates. A shunt is the instrument for direct current, for wide bandwidth and for accuracy at unity power factor; a current transformer is the instrument for large alternating currents, for isolation, and for not being in the circuit. The choice is made on which of the two failure lists the application can live with, and both lists are numbers rather than adjectives.

What the transformer inherits from the magnetics field #

A current transformer is a transformer, so three of its four limits are measured elsewhere. The band a turns ratio holds over is the band itself, with a lower edge set by the magnetising inductance and an upper one by the leakage. Where the band goes entirely is what happens when the load is removed, which for this instrument is the one dangerous failure. A boundary in volt-seconds is the ampere limit read in the unit it is actually in. And on the instrument side, The ammeter that is a resistor is the alternative with the opposite list of failures, while The four resistors that decide, and the two that do not is the amplifier that reads the burden voltage either way.

What is checked #

Four assertions, and the second is the one the closed form could not have produced.

That the solved ratio is the magnetising high-pass times the coupling, to two parts in ten thousand, with the netlist knowing neither expression — the check that the model and the two closed forms describe the same object.

That above the corner the error stops falling at $1 - k$, to a part in a thousand, which is a coupling rather than a frequency and is the floor no magnetising inductance improves.

That the phase error costs a power measurement far more than the ratio error does at a low power factor — asserted as a comparison between the two, at every burden the slider offers, rather than as a value.

And that the largest primary current is a volt-second limit read as an ampere limit, so that it rises with frequency: 916 amperes at fifty hertz and 7,330 at four hundred, from one number in webers.

The core, which is where the remaining errors are #

Three of this instrument’s four error terms belong to the magnetics rather than to the measurement, and the magnetics field has each of them with a number attached.

The volt-second limit above is a boundary in volt-seconds exactly: the quantity that belongs to the core is $N\cdot A_e\cdot B_{sat}$ — 3.500 mWb-turn for the part measured there — and it has no frequency in it at all, so everything a data sheet says about a rating with a frequency in it is that one number divided by $2\pi f$ once. That is why the ampere limit here rises in exact proportion to frequency, and it is a check on this essay’s arithmetic rather than a coincidence.

The limit that cannot be respected by staying below anything is the flux that walks, and it is the one a current transformer meets in its ordinary working life. A drive whose two half-cycles differ by one per cent in volt-seconds adds the same small area to the flux every cycle and saturates after 64 of them, with halving the drive giving 128 rather than removing the problem. A primary current with any direct component at all — a half-wave load, a rectifier drawing asymmetrically, a fault — is exactly that drive, and the transformer’s answer is not a ratio error but a core that has walked to one end of its curve and stopped transforming.

And the permeability the magnetising inductance is computed from is itself two numbers. Two inductances at one current finds a core at zero bias measuring 14.90 millihenries pushed downward and 8.30 pushed upward — a factor of 1.80 — because an operator sitting inside its own backlash contributes nothing to $dB/dH$, so a small excitation sees only the operators still moving. The ratio error this essay measures below its corner is set by that inductance, so it is a different number on the two halves of the cycle and the figure quotes an average of them.

None of the three changes the instrument’s central advantage, which is that it puts twelve microhms into the circuit it is measuring. They change what happens at the edges of its range, and all three edges are properties of the core rather than of the winding ratio a specification is written in.

What the two instruments are each blind to #

Set against the ammeter that is a resistor, the comparison comes out as a straight exchange of one blindness for another rather than as a ranking.

A shunt has no lower frequency limit and no upper one worth speaking of, responds to direct current, and its error is a burden voltage that can be optimised: 7.75 millivolts on a twelve-volt rail, with a worst-case error of 0.129 per cent, and the same answer at ten milliamps as at a hundred amps. What it cannot do is stay out of the circuit, and at a hundred amps the burden it inserts is dissipating most of a watt in a component that has to be non-inductive, stable and calibrated.

A current transformer is not in the circuit at all — twelve microhms reflected — and pays for it in a frequency response with two edges, a ratio error that stops falling at one minus the coupling, and a phase error worth eighteen per cent of a power reading at a power factor of 0.05. The phase error is the term worth carrying, because it is the one that has no analogue on the shunt side and no obvious place in an error budget: half a degree at fifty hertz reads as a small number and is a large one wherever the quantity wanted is a product of two waveforms rather than one of them.

Which is the honest summary of the pair. Choose by whether the measurement can afford a burden or can afford a phase, and the answer is decided by what the current is for rather than by how large it is.