Circuit Response
every model has an edge, and the figure carries it
Series

Current sensing — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · instruments
  2. 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.

    part 2 · instruments
  3. 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.

    part 3 · instruments
  4. 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.

    part 4 · instruments
  5. 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.

    part 5 · instruments

All series