Field

Circuits that do a job, and the range they do it over

Thirteen fields measure an element and the edge it has. This one composes several of them into a circuit with a purpose and asks the same question of the whole, where the answer is almost never the worst of the parts: an oscillator whose design condition is an exact equality no resistor can hold, a regulator that is a voltage source below three kilohertz and a capacitor above it, a threshold whose hysteresis is set by the noise underneath it, and a bridge that is linear near one point. It needed machinery the other twelve did not — a netlist with a nonlinearity in it, marched forward in time — because an oscillator's frequency comes from the linear part and its amplitude from the nonlinear part, and no analysis that drops either one returns both.

The closed-loop poles at a gain of 3.05. The locus of the two poles as the amplifier's gain runs from 2.7 to 3.3. It crosses the imaginary axis at a gain of 3.000000 — bisected on the netlist, not quoted — and at 3.05 the real part is 2.500e+2 radians a second, which is an envelope multiplying by 1.17015 every cycle. The crosses are the closed form ω₀(k−3)/2 and they sit on the measured circles.

The gain that is exactly one

An oscillator is designed by making the loop gain one at the frequency where the phase is zero. The gain at which this circuit's poles reach the imaginary axis is 3.000000000000, bisected on the netlist — an equality, not a range. A gain three per cent high multiplies the envelope by 1.0987 every cycle and reaches the rails in 61 milliseconds; three per cent low divides it by the same factor. A one per cent resistor cannot hold the condition, and neither can any other component.

Started from a millivolt at a gain of 3.20. The output grows by 1.8804 a cycle — the factor the poles give — and then stops, at 858 mV of amplitude and 1.58 kHz. The lower panel is the envelope on a logarithmic axis, where the linear model's prediction is the straight line that keeps going. What ends it is the diode pair across the feedback resistor, and no direct-current analysis of this circuit returns that number.

The amplitude nothing linear predicts

A linear model's poles say the envelope grows by 1.88 a cycle and never say when it stops. Two diodes across the feedback resistor stop it, at 858 millivolts — a number no pole, no transfer function and no bias point contains. Getting it needs a netlist with a nonlinearity in it, marched forward in time, and that pairing is what the whole field is built on.

The limit cycle's spectrum at a gain of 3.20. Ten lines of the settled waveform. The third is 5.692% of the fundamental and the fifth 0.982%; the second and fourth are 6.8e-7, which is the arithmetic's floor and not a small residue. Two diodes facing opposite ways make a symmetric characteristic and a symmetric characteristic produces no even harmonic at all.

What the limiter charges for

The diodes that set the amplitude are the only nonlinear thing in the loop, so every harmonic in the output is theirs. Across the gain slider the amplitude rises by a factor of 1.51 and the distortion by 15.4 — an amplitude that goes as the 0.11 power of the excess gain and a distortion that goes as the 0.74 power. Two diodes facing opposite ways produce no even harmonic at all, at seven parts in ten million, which is the arithmetic's floor rather than a small residue.

The frequency that is not the formula

The Wien network's zero-phase frequency is one over two pi RC to every digit the arithmetic has. The circuit does not run there. With a perfect amplifier it runs 0.771 per cent low, because the limiter's harmonics are part of the waveform whose period is being measured; with a real one it runs lower still, by an amount inversely proportional to the gain-bandwidth product. The two are equal at a ratio of 578, and above that a faster amplifier buys nothing.

1000 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 15.69 V with 1.331 V of ripple, against the 1.569 V the expression I/2fC gives — 15.1% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 28.8° of each half cycle and carries 2.10 A at the peak, which is 13.4 times the 157 mA the load draws.

The direct voltage that is a sawtooth

A rectifier and a reservoir capacitor make what everybody calls a direct voltage. Marched with the diodes in the netlist, a thousand microfarads across a hundred ohms gives 15.69 volts with 1.33 volts of ripple on it, against the 1.57 the textbook expression predicts. The expression is high by the fraction of the cycle the diode conducts for — measured at 0.94 to 0.96 of it across two sweeps — and it has no opinion at all about the quantity that actually sizes the transformer, which is a peak diode current 13.4 times the current the load draws.

A five-volt regulator's output impedance, with 1.00 Ω of series resistance. 0.430 mΩ at direct current, 1.95 Ω at 10.0 kHz — a factor of 4.52e+3 — and it has already doubled by 4.81 Hz. The upper curve is the same circuit with its loop opened, and the ratio between them is the loop gain. A regulator is a voltage source below a frequency and the datasheet's milliohms are the value at the bottom of it.

A source below a frequency

A five-volt regulator's output impedance is 0.43 milliohms, which is the number a datasheet quotes. It has doubled by 4.8 hertz, is ten times worse by 27, and reaches 1.95 ohms at ten kilohertz — four and a half thousand times its own specification, and higher there than the same circuit with its feedback loop cut. Two routes to that curve, sharing only the netlist, agree to a part in ten to the thirteenth.

The window a 10 µF capacitor leaves. Above 939 mΩ the loop holds 45° of margin; below it the regulator rings and then oscillates. The droop after a 100 mA step is smallest at 817 mΩ — 129 mV — and by 19.9 Ω it is 1.468 V, because at the first instant of a step the capacitor cannot move and the whole step falls across its series resistance. Two requirements, opposite directions, and the useful values are between them.

Two requirements pulling one capacitor

The output capacitor's series resistance is a stability requirement and a transient requirement at once, and they pull it in opposite directions. Below 939 milliohms this loop has less than 45 degrees of margin; above about 850 the droop after a load step starts to grow, because at the first instant of a step the capacitor cannot move and the whole step falls across that resistance. The droop is smallest 10 per cent inside the unstable region, which means the best transient this design can have is one it must not be built with.

Rejection is not a property of the loop. The same regulator with the same loop gain and the same 46.5° of phase margin, drawn twice. With the amplifier's output referred to ground the rail reaches the output at -67.3 dB at the bottom and 5.79 dB at 10.0 kHz — where it is amplified. Referred to the rail instead, every point is 60.00 dB lower, which is 20 log(gm·ro) for the pass device and not a design choice.

What gets through from the rail

A regulator's job is to hold its output still while its input moves, and the measurement says it does that well at ten hertz, badly at a kilohertz, and not at all at ten kilohertz — where this one puts out 1.9 times what arrives. Then the same netlist with one node moved, the same loop gain and the same 46.5 degrees of margin, rejects 60.009 decibels better at every frequency in six decades. The sixty decibels is the pass device's own intrinsic gain and it is not a design choice.

A threshold crossed once, in 20k samples of noise. With no hysteresis the comparator changes its mind 22.7 times on average and as many as 29, on a signal that crosses the threshold once. The vertical bars are the range over twelve seeds. The expected number of extra transitions falls below a tenth at 3.57 standard deviations — so the hysteresis a threshold needs is set by the noise under it and not by the signal over it.

Two thresholds because there is a floor

A comparator with one threshold, watching a slow signal cross it once, changes its mind 22.7 times on average and as many as 29 — because there is noise under the signal and no threshold is ever crossed once. Hysteresis fixes it, and how much is needed is a multiple of the noise's own standard deviation rather than a voltage: 3.57 of them here. The multiple grows with how long the threshold is watched, and slowly — a hundredfold longer record needs three times the hysteresis, not a hundred times.

A comparator's delay, at a divider ratio of 0.50. Each nanosecond of comparator delay adds 2.650 nanoseconds to the period — measured as a slope between two delays, both exact multiples of the marching step — against the 2.667 that 4/(1+β) gives and the 2 that counting the delay twice gives. A 50 ns comparator therefore holds the frequency to one per cent only below 75.0 kHz.

The period a delay lengthens

Feed a comparator's output back through a resistor to the capacitor on its own input and the two thresholds stop defending a decision and start setting a period. The closed form is two RC times the log of one plus beta over one minus beta, and a marched circuit recovers it as the step shortens. A comparator that responds fifty nanoseconds late does not add fifty nanoseconds to each half cycle: it adds 4/(1+beta) times the delay to the period, 2.65 here against the 2 that counting it twice gives, because during the delay the capacitor keeps going the way it was going.

$A 350 Ω bridge, with ideal leads. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 0.498% at 1.0% — so the tangent is worth one per cent only up to 2.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 4.040%.$

The bridge that is linear near one point

The expression for a Wheatstone bridge's output is V delta over four, and the solve says it is V delta over two times two plus delta — low by half the fractional change, exactly, at every change tested. So the tangent is worth one per cent only up to a two per cent change and a tenth of a per cent only up to two tenths. Driving the bridge from a current source instead halves the departure at every point and doubles both edges, which is a change of one component and no change at all to the four resistors being measured.

Twenty nanoseconds of skew is 0.28% of duty, not 0.40. computed by solving, not by drawing. The duty cycle of a relaxation oscillator whose comparator takes longer to go one way than the other, marched, against the expression that lengthens each half cycle by its own delay. The two part company immediately and by a constant factor of about 1.42: during a delay the capacitor keeps charging past the threshold it already crossed, so the next half cycle starts further out and takes longer, and the two halves partly cancel. At 200 ns of skew on a 2585 ns period the duty is 52.805 per cent where the expression says 54.004. The open circles are the same quantity in closed form — the overshoot is V(1 − (1 − β)e^(−d/τ)) and the next half starts from it — which the march reproduces to parts in ten thousand.

The delay that is two delays

Modelling one comparator delay applied to both transitions makes the two half cycles equal by construction, and the fix is a change of one line. Made, the duty cycle moves by 0.28 per cent for twenty nanoseconds of skew rather than the 0.40 the obvious expression gives, more hysteresis improves the duty cycle and worsens the volt-seconds at the same time, and ten nanoseconds of skew saturates a hundred-turn core in 231 cycles.

The growth per cycle, and the form that is a fifth low at the top of the range. computed by solving, not by drawing. The factor the envelope is multiplied by each cycle, against the gain. The solid curve is exp(π(k−3)/√(1 − ((k−3)/2)²)), which is what the characteristic equation gives and what the netlist's own poles return to twelve digits; the dashed one is exp(π(k−3)), which drops the denominator. The circles are the marched envelope, fitted over the cycles that are still small — 80 of them at k = 3.01 and 4 at k = 3.2, and none at all above that. The two expressions differ by 3.9e-7 at k = 3.01 and by 20.462% at k = 3.8, so the approximation fails exactly where nothing is left to check it against.

Two exponentials, and where they meet

An oscillator's envelope grows by exp(π(k−3)/√(1 − ((k−3)/2)²)) a cycle, and the form usually quoted drops the denominator — exact to four parts in ten million at a hundredth above three, and 20.462 per cent low at 3.8, which is precisely where too few small cycles are left to measure it. Where the growth stops is the diode's own exponential: 108.5 millivolts of amplitude for every decade of saturation current, proportional to the ideality to four parts in a thousand. Above 60.121 nanoamperes the limiter is already conducting at zero signal and there is no oscillation at all.

$A tenth of a per cent of lead is 999 µε of strain that is not there. computed by solving, not by drawing. A 350 Ω quarter bridge at zero load, against the resistance in each lead. With both leads in the changing arm the output is 4.99500 mV at 350 mΩ — an apparent fractional change of 0.1998%, which at a gauge factor of 2 is 999 microstrain. With one lead in that arm and one in the arm beside it the output is zero to the last bit, at every lead resistance drawn. The lower panel is what the second arrangement costs: the sensitivity falls as 1/(1 + Rlead/R), which is 0.100% at the same lead and is a calibration constant rather than a drift.$

The leads that are in the bridge

A strain gauge on the end of two long wires cannot be told from a strain gauge under load: both leads in the changing arm is bit for bit the same netlist as a quarter bridge whose fractional change is larger by twice the lead over the gauge, which at 350 milliohms on a 350 ohm gauge is 999 microstrain that is not there. Moving one of those leads into the arm beside it leaves the output at exactly zero for every lead resistance drawn, and turns a twenty-kelvin drift of 76.8 microstrain into 0.077. That factor is two over the strain being read — 999 at a thousand microstrain and 9990 at two hundred — and it costs 0.100 per cent of sensitivity.

The crest factor is 13.4 and the winding is sized by 3.01. computed by solving, not by drawing. Three ratios of the same settled march, against the reservoir. The crest factor — the peak diode current over the load's direct current — runs 6.43 to 24.25. The form factor, which is the root-mean-square current over the same direct current and is what a winding heats by, runs 2.114 to 4.077. Its square is the copper loss against a winding carrying the direct current alone, and that runs 4.47 to 16.62. The first ratio is 3.04 times the second at 220 µF and 5.95 times at 4700, so quoting one of them tells a reader nothing about the other.

The current that sizes the transformer

A reservoir's crest factor is 13.374 at a thousand microfarads and the winding is not sized by it. The root-mean-square of the same marched current is 3.0069 times the load's direct current, so the copper dissipates 9.0417 times what it would carrying the direct current alone — and the two ratios diverge, from 3.04 apart at 220 microfarads to 5.95 apart at 4700. The expression for the mean output is wrong in three places whose signs differ, and at 313.9 microfarads they cancel to six microvolts while the ripple expression inside it is still 32.3 per cent high.

The ripple 1000 µF leaves, through the regulator: 60.62 mV rather than 21.11 mV. One settled cycle of the reservoir's output — 1.331 V peak to peak across 1000 µF — split into 1000 Fourier lines, each passed through the solved regulator's rail-to-output response, and summed. The output is 60.62 mV peak to peak. The ripple times the rejection at 100 Hz, -36.00 dB, gives 21.11 mV, which is 2.872 times too little: the rejection worsens at twenty decibels a decade, so each harmonic of the sawtooth arrives at nearly the size of the first. The 100 Hz line carries 23.4% of the output's mean square and the lines above 1 kHz 10.4%; the first three arrive at 7.731 mV at 100 Hz, 7.200 mV at 200 Hz, 6.388 mV at 300 Hz. The dashed curve is the 100 Hz line alone.

The ripple that arrives as a comb

The rejection essay multiplied two numbers: 1.331 volts of reservoir ripple and the regulator's 36.0 decibels of rejection at a hundred hertz, for 21 millivolts at the output. The ripple is a sawtooth, a comb of lines at every multiple of a hundred hertz, and the rejection worsens at twenty decibels a decade — so the second line arrives at nearly the size of the first, and the third too. Summed with their phases, the output is 60.62 millivolts, 2.87 times the estimate, and the hundred-hertz line carries only 23.4 per cent of it. A reservoir twenty-one times larger cuts the rail's ripple 14.8 times and the regulated ripple 5.95.

The capacitor across the upper resistor

The rejection essay said a regulator reproduces its reference times its divider's four, and that a capacitor across the lower divider resistor brings that down to one at high frequency. Measured, the loop peaks the reference's gain to 5.68 near its crossover before any capacitor is added; a nanofarad across the lower resistor raises the peak to 11.2; and the capacitor that brings it down belongs across the upper resistor, where a nanofarad keeps the gain from ever exceeding four. The same capacitor is a lead pair in the loop: 836.5 picofarads takes the phase margin from 46.5 to 90.9 degrees, and ten nanofarads, past that optimum, still holds 66 while cutting the output ripple from 60.6 millivolts to 35.4.

The period's spread is 1.20 times what counting two crossings gives. computed by solving, not by drawing. 19999 periods of a relaxation oscillator with 5 mV rms of noise on its thresholds, computed from the exact flip instants rather than marched, at β = 0.5. The measured standard deviation is 3.4 ns and the closed form — three partial derivatives of the period with respect to the three draws it depends on — gives 3.4 ns. The estimate that counts two threshold crossings and divides the noise by the slope at each gives 2.83 ns, which is 17 per cent low. The curve is the closed form's Gaussian, drawn on the measured histogram rather than fitted to it.

The decision taken where the ramp is slowest

A relaxation oscillator decides at its thresholds, and a threshold is the one place on a charging exponential where the slope is smallest. Noise there costs 3.399 units of period against the 2.828 that counting two crossings gives, because a draw moves the crossing it is armed for and the level the next ramp starts from. Consecutive periods share that draw, so they are positively correlated and the jitter accumulates at 3.771 per root period rather than at 3.399. And at a fixed frequency there is a best hysteresis: β = 0.648, where β·ln((1+β)/(1−β)) = 1.

A capacitor across the upper divider resistor removes the output capacitor's resistance floor. computed by solving, not by drawing. Two series resistances against the capacitance across the upper divider resistor: the smallest the loop tolerates at 45° of margin (lower curve), and the one that gives the smallest droop after a 100 mA load step (upper). With no capacitor they are 939 mΩ and 885 mΩ — the second BELOW the first, which is the conflict this design has: the best transient is one the loop refuses. The floor falls as the capacitor grows and between 500 and 836.5 pF it leaves the sweep altogether, so every series resistance down to a milliohm is stable. Past about 5000 pF the floor climbs back and overtakes the optimum again. The shaded band is where the design a transient wants is one the loop allows.

The floor a second capacitor removes

Two requirements pulling one capacitor found a regulator whose best transient is one it must not be built with: below 939 milliohms of output-capacitor series resistance the loop has under 45 degrees of margin, and the droop is smallest at 885. The capacitor across the upper divider resistor, added for the reference's sake, dissolves that conflict. At 836.5 picofarads the 45-degree floor leaves the sweep entirely — every series resistance down to a milliohm is stable — and the droop falls 40 per cent at the same time. The band of capacitances that do it runs from 100 picofarads to 5 nanofarads, and above it the conflict returns.

Two leads nobody counts cost 1995 parts per million, and two more leads remove all of it. computed by solving, not by drawing. The error in the reported strain against the resistance in each of the two excitation leads, at 1000 µε on a quarter bridge of 350 Ω excited at 10 V. With four wires the instrument takes the excitation to be the supply's voltage, so every reading is scaled by R/(R + 2r): 1995 parts per million at 0.35 Ω, 54029 at 10. Two more leads brought back from the bridge's own terminals, carrying only the instrument's input current, leave 3.5e-4 parts per million. Exciting with a current instead of a voltage does the same thing with no extra leads at all, because the lead resistance is in series with a source that does not care.

The two leads nobody counts

The leads that are in the bridge moved a lead out of the changing arm and turned a 999-microstrain error into nothing. The two leads carrying the excitation are still there, and they scale every reading: 0.35 ohms each on a 350-ohm bridge is 1,995 parts per million, exactly −2r/(R + 2r), the same on a full bridge as on a quarter one, and drifting 0.153 microstrain over twenty kelvin — twice what the three-wire fix left behind. Two more wires brought back from the bridge's own terminals leave 0.00035 parts per million. So does exciting the bridge with a current, which needs no extra wires at all.

The series resistance that makes the ripple smaller. computed by solving, not by drawing, marched with the diodes in the netlist. A 1000 µF reservoir with 30 mΩ of its own series resistance. The resistance adds a step of ESR times the diode's peak current to the output and at the same time limits that peak current, and the two nearly cancel: the ripple has an interior minimum of 1.330 V at 17.0 mΩ, BELOW the 1.331 V a perfect capacitor gives, and rises to 1.711 V at an ohm. What the resistance buys monotonically is the peak current: the crest factor falls from 13.37 to 6.550 at an ohm, which more than halves the current that sizes the transformer, for 380 mV of mean output and a root-mean-square diode current that falls from 0.4717 A to 0.3484. The textbook ripple expression says 1.568 V here and moves by 2.4% across the whole axis, because it has no term for a series resistance at all.

The resistance that lowers the ripple

Two earlier essays here marched a reservoir with a perfect capacitor. A real one has tens of milliohms of its own, and the obvious expectation — that the resistive step it adds makes the ripple worse — is wrong in an interesting direction: the resistance also limits the charging current, and the ripple has an interior minimum of 1.3303 V at 17.0 mΩ, below the 1.3312 V a perfect capacitor gives. What the resistance buys monotonically is the peak current, which falls from 13.37 times the load's to 6.55 at an ohm, for 380 mV of mean output.

The same 100 mΩ in the winding instead of the capacitor: 1.317 V of ripple against 1.335 V, at the same crest factor of 11.15. computed by solving, not by drawing, marched with the diodes in the netlist: a 1000 µF reservoir behind a centre-tapped rectifier, with a series resistance from 1 mΩ to 1 Ω placed either in the capacitor or in each half-winding. The crest factor is the same in both places to two parts in a thousand at every resistance — 13.34, 13.26, 13.03, 12.49, 11.15, 9.069, 6.548 — because both limit the charging current alike. The ripple is not: in the capacitor it has a minimum and rises to 1.711 V at an ohm; in the winding it falls throughout, to 1.172 V, against 1.331 V with no resistance. At an ohm the winding costs 534 mV of mean output and the capacitor 380 mV. The diode's root-mean-square current at an ohm is 0.3449 A with the resistance in the winding and 0.3484 A with it in the capacitor.

The resistance that belongs in the winding

A reservoir capacitor's series resistance lowers the ripple to a minimum of 1.3303 V at 17 mΩ and raises it past that. The same resistance moved into the transformer's winding limits the peak current by the same amount — the crest factor agrees to two parts in a thousand at every value from a milliohm to an ohm — and the minimum is gone: the ripple falls throughout, to 1.172 V at an ohm against 1.711 V in the capacitor. The two resistances each carry a current the other does not, and that one asymmetry decides where a deliberate one should go.

Two 1000 µF parts of 100 mΩ each against one 2000 µF part of 100 mΩ: 709 mV against 728 mV of ripple, and a crest factor of 14.41 against 12.61. computed by solving, not by drawing, marched with each capacitor behind its own series resistance. Two identical 500 µF, 60 mΩ parts give 1.330303 V of ripple and one 1000 µF, 30 mΩ part 1.330303 V — the same network. Two 1000 µF parts of a stated resistance each, against one 2000 µF part of the same resistance: at 10 mΩ, 704 mV against 704 mV of ripple and a crest factor of 17.40 against 16.93; at 30 mΩ, 704 mV against 705 mV of ripple and a crest factor of 16.51 against 15.48; at 100 mΩ, 709 mV against 728 mV of ripple and a crest factor of 14.41 against 12.61; at 300 mΩ, 758 mV against 867 mV of ripple and a crest factor of 11.45 against 9.490; at 1000 mΩ, 1.013 V against 1.334 V of ripple and a crest factor of 8.170 against 6.620. The pair's ripple is lower only where one part's resistance is past the ripple's minimum, and its peak current is higher at every resistance.

Two capacitors that are one

Two identical reservoir capacitors in parallel are not a new circuit to be marched: 500 µF at 60 mΩ twice gives 1.330303 V of ripple and so does 1000 µF at 30 mΩ once, to the sixth decimal. So a pair against one part of the same capacitance is one curve read at two resistances, and the pair always sits at the lower one — which lowers the ripple only past the curve's minimum, 709 mV against 728 at 100 mΩ a part, and raises the peak current at every value, a crest factor of 14.41 against 12.61. Mismatch the pair and the current still divides by capacitance, so a part with twice the resistance carries 2% less current and 1.93 times the heat.

The walk the core sees

Threshold noise in a relaxation oscillator walks its timing at 3.77 nanoseconds per root period at β = 0.5, because the draw two half cycles share adds. A transformer driven by the same square wave sees the difference of the halves instead, where the shared draw subtracts, and that walks at 1.89 — exactly β times the timing, 18.3 ns against 35.9 after a hundred periods over six hundred seeded runs. The per-period duty error does not vanish with the hysteresis and the walk does, consecutive duty errors are anticorrelated where consecutive periods are not, a comparator skew outruns the walk after 2(σB/d)² periods, and at a fixed frequency the core wants less hysteresis than the clock does.

A cable's carrier limit is its conductors' matching, not its capacitance. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε, excited at a carrier frequency, with 10 metres of cable at 100 pF a metre on each output terminal. Each terminal sees the two arms meeting there in parallel against that capacitance, so each lags by arctan(2πf·(r/2)·C) with a corner at 909 kHz — and with the two conductors matched the lag CANCELS in the difference: -0.117° where each terminal has -5.71°. A mismatch does not cancel, and it arrives multiplied by the ratio of the excitation to the signal, which at 1000 µε is 2000: at 1 per cent the phase is 40.3° at a hundred kilohertz. So the excitation a 10-metre cable allows for 200 parts per million falls from 205 kHz matched to 1.82 kHz at 1 per cent. And it depends on the strain being measured — 183 Hz at 100 µε, 547 Hz at 300 µε, 1.82 kHz at 1000 µε, 5.45 kHz at 3000 µε — because the amplification is the reciprocal of the signal.

The carrier a cable allows

A carrier-frequency amplifier removes an instrument's offset drift by exciting the bridge with alternating current, and the cable that was a resistance at direct current becomes a capacitance. Both output terminals see the same arms and the same capacitance, so the lag cancels in the difference: ten metres matched leaves 0.15 degrees at a hundred kilohertz where each terminal has 6.3. What does not cancel is the mismatch between the two conductors, arriving multiplied by the ratio of the excitation to the signal — two thousand at a thousand microstrain — so one per cent of it takes the allowed carrier from 205 kilohertz to 1.8.

Losing capacitance to bias raises this loop's margin; the corner that binds is the ceramic at full value. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5.00 mΩ ceramic output capacitor, against the capacitance that capacitor keeps at the working voltage, from the full 10 µF down to 1.5 µF, for five capacitors across the upper divider resistor. With none, the margin climbs from 18.0° at full value to 43.6° at 15% kept, so the lost capacitance is stabilising rather than the reverse. 630 pF holds at least 49.7° across the whole range. 836.5 pF, the value that removed the resistance floor at the nominal capacitance, gives 48.8° at full value and 44.3° at 15%: the capacitor that has lost the most is the one it fails on.

The capacitance the bias takes away

A ceramic output capacitor has a few milliohms of series resistance, which a capacitor across the upper divider resistor made legal, and a capacitance that falls with the voltage across it — a ten-microfarad part at five volts is two microfarads to a small signal. Swept from the full ten down to that fifth, the regulator's margin does not collapse. It rises: with no divider capacitor from 18.0° to 38.5°, because the output pole climbs away from the amplifier's. The corner that binds is the capacitor at full value. One divider capacitor, 423 to 808 pF, holds 45° at every capacitance between. The 836.5 pF that removed the resistance floor sits just above that band, and the lost capacitance is paid for in droop: 103 mV at full value, 202 mV at a fifth.

At full capacitance the margin falls with the load, and below a tenth of an ampere no divider capacitor gets it past 36.9°. computed by solving, not by drawing. Phase margin of the five-volt regulator with a 5 mΩ ceramic keeping 10 µF, against the load current from 5 mA to 1 A, for four capacitors across the upper divider resistor. 630 pF, which held 45° at every capacitance at an ampere, gives 49.7° at 1 A and 36.0° at 5 mA. With no capacitor the margin at 5 mA is 0.1°. The dotted level is 36.87°, the most phase a zero and a pole four times apart can ever lead by, and every curve ends under it. The crossover stays between 12.9 kHz and 13.2 kHz over the whole range.

The lead a divider can give

At an ampere one capacitor across the upper divider resistor holds this ceramic regulator above 45° at every capacitance the part keeps. At 5 mA nothing does. The crossover stays at 13 kHz while the output pole falls from 3.20 kHz to 31.8 Hz, so the loop arrives at unity with every lag it has and the only phase left is the lead the divider capacitor makes — and that lead has a ceiling, arcsin((G − 1)/(G + 1)), set by the divider's ratio alone. For a five-volt output from a 1.25 V reference it is 36.87°, and the best capacitor reaches 36.83°. Over the rectangle from 5 mA to an ampere and ten microfarads to two, the best single capacitor gives 32.4°. Putting 416 mΩ back in series with the ceramic restores 45° over all of it, and lowers the droop rather than raising it.

A tenth of a per cent of screen mismatch on the gauge's side cancels the pair's own phase: 18.2 kHz becomes 143 kHz. computed by solving, not by drawing. A 350 Ω quarter bridge at 1000 µε read through ten metres of screened pair: 100 pF a metre from each conductor to the screen, Cs, and 50 pF a metre between the two, Cp. The loss of the synchronously demodulated reading, 1 − cos φ, against the excitation frequency: for the capacitance between the conductors alone, for the screen capacitances alone with a mismatch of 0.1% on the gauge's side, for both together, and for a matched screen with nothing between. For a 200 ppm budget they allow 18.2 kHz, 18.3 kHz, 143 kHz and 205 kHz. The pair's capacitance lags the reading; extra screen capacitance on the gauge's own terminal leads it, by an amount divided by the signal, so at a mismatch of δ·Cp/Cs the two cancel to first order.

The capacitance that is already a difference

A screened pair has capacitance from each conductor to the screen, which cancels in a bridge's difference when the two match, and capacitance between the conductors, which is a difference from the start and cannot cancel. Ten metres of it between a 350 Ω quarter bridge and its carrier amplifier allows 18.2 kHz for a 200 ppm loss, against 205 kHz for the screen alone. The two phases are ωR·ΔC/δ and ωR·C, the same slope with one divided by the signal, so a screen mismatch of δ·Cp/Cs on the gauge's own terminal cancels the pair's lag exactly: 0.10% at 1000 µε takes the limit to 143 kHz. That is what a carrier amplifier's capacitance balance finds when it nulls the quadrature — and because the mismatch's phase changes sign with the strain and the pair's does not, the balance holds in tension above half the strain it was set at and never in compression.

An unregulated supply's output against its load: 3.38 Ω of slope with perfect parts, 3.64 with 0.3 Ω in the winding, 3.32 with it in the capacitor. computed by solving, not by drawing, marched with the diodes in the netlist: a 17 V peak centre-tapped rectifier into 1000 µF, loaded from 200 Ω down to 25 Ω. The mean output with perfect parts falls from 15.99 V at 80 mA to 14.33 V at 573 mA, an average slope of 3.380 Ω against the textbook's 1/(4fC) = 5 Ω, drawn dashed without its diode drop. With 0.3 Ω in each half-winding the slope is 3.642 Ω; with the same 0.3 Ω in the capacitor it is 3.317 Ω. The slider is the resistance.

The resistance a supply's regulation counts

An unregulated supply's output falls as its load rises, and the slope is its output resistance: 4.17 Ω at 80 mA and 2.80 at 575 for a thousand microfarads, below the textbook's 1/(4fC) = 5 Ω and falling, because the diodes conduct for longer as the load grows. A winding resistance adds to that slope at about its own value, 0.94 to 1.18 of it; a capacitor's adds −0.14 to −0.25. The guess that each should count by the share of the cycle it carries current in — a sixth and five sixths — is wrong in both places, and the right weighting is the mean current each carries: all of the load's through the winding, none through the capacitor.

Calibrated at the carrier, a matched cable allows 121 kHz where its phase allowed 16.5 kHz. computed by solving, not by drawing. A 350 Ω quarter bridge read through ten metres of screened pair, excited over leads of 0.35 Ω and 3 µH each with 500 pF between them. Solid: the worst error of a synchronous reading over strains from −2000 to +2000 µε after a calibration at +1000 µε at the operating carrier, with the reference taken from the source at the instrument; dashed, from the bridge's own terminals on sense wires; dotted, the phase measure 1 − cos φ the earlier readings of this cable used. For a 200 ppm budget: 16.5 kHz by phase, 119 kHz calibrated against the source and 121 kHz calibrated against the bridge.

What a calibration at the carrier removes

The phase a cable puts on a carrier-excited bridge is a loss of 1 − cos φ, and a loss that is the same at every strain is a gain error — which a shunt calibration at the operating carrier exists to remove. Calibrated there, a synchronous reading of a quarter bridge through ten metres of matched screened pair allows 121 kHz for 200 ppm over ±2000 µε, where the phase measure allowed 16.5; the capacitance between the conductors, which no matching could improve, stops mattering. Where the reference is taken — at the instrument or from sense wires at the bridge — moves that by 1.2 per cent. What calibration cannot touch is a screen mismatch: its leak is a fixed voltage, not a fixed gain, and a hundredth of a per cent of it takes 121 kHz back to 11.5.

A 1% mismatch in the cable reads as 3.0% of full-scale displacement, at every frequency. computed by solving, not by drawing. A capacitive displacement sensor — two 10 pF plates driven at ±5 V in antiphase facing a common electrode — read by a charge amplifier through two metres of three-conductor cable with 30 pF a metre between each excitation conductor and the sense conductor. With the two cross-capacitances mismatched by 1 per cent and the sensor centred, the reading's in-phase part (solid) is 3.00 per cent of full scale at every carrier from 100 Hz to 1 MHz: the cable is a sensor capacitance, and its mismatch a displacement. The quadrature part (dashed), which the cable's resistance makes, rises with frequency and a synchronous detector rejects it. With each conductor in its own screen the cross-capacitance is gone and so is the error.

The sensor that is the capacitance

Read a capacitive displacement sensor through the same kind of cable that limited a strain-gauge bridge's carrier, and the roles exchange. On the bridge, a 1% mismatch in the cable's capacitance was a quadrature leak that grew as the square of the carrier and could be escaped by a lower one. On a sensor of two 10 pF plates it is 3.0% of full-scale displacement, in phase, at every carrier from 100 Hz to 1 MHz — because a capacitance between an excitation conductor and the sense conductor is in parallel with the measurand and is a sensor capacitance. No detector phase separates it and no carrier escapes it; only geometry does. Screen each conductor separately and there is nothing left to mismatch.

With a leak of a hundred periods the flux settles to a spread of 1.22% of the swing at β = 0.5 — 0.759 of what independent duty errors would give. Seeded: three runs of sixty thousand periods of the event map, 5 mV of threshold noise, β = 0.5, driving a magnetising branch whose flux leaks with τ = L/R. The stationary spread of the period-end flux in volt-nanoseconds against τ in periods, quoted here as a percentage of one half cycle's swing V·T/2 = 5493 V·ns: 1 → 0.146% (closed form 0.145%), 3 → 0.233% (closed form 0.232%), 10 → 0.401% (closed form 0.397%), 30 → 0.685% (closed form 0.673%), 100 → 1.25% (closed form 1.22%), 300 → 2.16% (closed form 2.10%). The closed form linearises one period, keeping the leak inside each half and the covariance −AB between consecutive periods' imbalances; dropping that covariance gives the upper curve, 1.61% at a hundred periods. For a slow leak the spread is V·σ·|B − A|·√(τ/T), the drawn asymptote.

The spread a winding resistance holds

A relaxation oscillator's threshold noise walks an ideal core's flux without limit. Put the winding's resistance back and the walk stops at a spread: 1.22% of a half cycle's swing when L/R is a hundred periods, growing as the square root of L/R, while a comparator skew's offset grows as L/R itself. The anticorrelation between consecutive duty errors is worth 24% of that spread at β = 0.5 and 62% at β = 0.2 — but only in a core slow enough to remember several periods. And the resistance that bounds the walk is a boundary in its own right: with a millihenry and a fifth of the swing to spare, an ohm tolerates 12.9 mV of noise, and a hundred picoseconds of skew needs at least 0.14 Ω.

With a 300 ps skew and 5 mV of noise a core with 100 ns of headroom saturates after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. Seeded: 2,000 drives of an ideal core by the event map at β = 0.5, 5 mV of threshold noise and a comparator skew of 300 ps, each run until the summed high-minus-low time leaves ±100 ns. A histogram of the counts on logarithmic bins, against the closed-form first-passage distribution, between two barriers, of a walk with drift 0.200 ns a period and spread 1.89 ns per root period. Mean 501.9 against 499.9; quantiles 1% 184, 10% 275, 50% 463; the empirical and closed-form cumulative distributions differ by at most 0.014.

The count a core survives

A relaxation oscillator with a 300 ps comparator skew and 5 mV of threshold noise saturates an ideal core with 100 ns of volt-second headroom after 502 periods on average — the 500 the skew alone gives — and one drive in a hundred by 184. The mean does not see the noise at all until the noise is large enough to saturate the core the other way first; the early tail sees nothing else. Every skew and noise level falls on one curve in one ratio, the skew's count over the count at which its ramp overtakes the walk. Read as a burst, a thousand periods with one failure in a thousand allows 150 ps of skew with no noise, 93.5 ps at a millivolt, and none at all above 2.41 mV.

The gain a Wien oscillator needs, when one of its four parts is off its value. computed by solving, not by drawing: for each part moved alone from −5% to +5%, the gain at which the poles reach the axis, bisected on the netlist, against 1 + Rs/Rp + Cp/Cs (to ten digits at every point). A series resistor or shunt capacitor 1% high raises the gain needed to 3.0100; a shunt resistor or series capacitor 1% high lowers it to 2.9901. Every part moves the frequency the same way, by half its error: 1% high on any one of them, 1.58 kHz against 1.59 kHz. The gain needed moves by the error itself, a third of it relative to three.

The gain a tolerance demands

A Wien oscillator needs a gain of exactly three only when its four network parts are exactly equal. In general it needs 1 + Rs/Rp + Cp/Cs — bisected on the netlist to ten digits — so a series resistor 1% high asks for 3.0100 and a shunt resistor 1% high for 2.9901. Four thousand seeded oscillators from 1% parts set to the textbook's comfortable 3.02 leave a quarter of them silent; 99.9% start only at 3.072, and at that setting the median unit distorts 2.6% — the limiter has to absorb a margin the tolerance forced. The frequency spreads by one part's own spread, 0.58% for 1% parts, as large as the limiter's own 0.771% shift. Parts from one array, tracking to 0.05%, need only 3.0035, because the gain depends on ratios and the frequency on products.

The supply's mean output against its load, with no leakage inductance, 0.3 mH and 1 mH in the winding. computed by solving, not by drawing, marched with the diodes in the netlist: 17 V peak into 1000 µF, 50 mΩ of winding throughout. Without inductance the output falls 1.69 V from 200 Ω to 25 Ω, an average slope of 3.43 Ω. With 0.3 mH it falls 0.62 V, 1.14 Ω, from a higher start; with 1 mH, 0.42 V, 0.77 Ω. An element that carries the load's current and drops no mean voltage makes the regulation better, not worse.

The inductance that flattens the supply

An unregulated supply's output resistance counts each resistance by the mean current it carries, and an inductance in the winding carries the whole mean current and drops no mean voltage — so by that rule it should count for nothing. Marched, it counts for less than nothing: 0.3 mH of leakage takes the output resistance at 100 Ω from 3.88 Ω to 0.67, and the mean output falls 0.62 V from 200 Ω to 25 Ω instead of 1.69. The effect has an interior optimum — 0.03 mH makes it slightly worse, 0.3 mH best, 3 mH back to 2.86 Ω — because the inductance first rings the reservoir up and then turns the rectifier toward choke input.

The gain an amplifier asks for, squared

A Wien oscillator built round a real amplifier needs more than a gain of three, and the excess is easy to overrate: the amplifier's lag moves the frequency in proportion to the lag and the required gain only in proportion to its square. At a gain–bandwidth product a hundred times the oscillation frequency the frequency is 4.22 per cent low and the gain needed is 3.00257 — fifty-one times smaller in proportion. The square is also what makes the bias spread: an amplifier 30 per cent slow asks for 1.94 times the typical excess. Against 1 per cent parts the amplifier matches the network's share at a ratio of 16; against a network matched to 0.05 per cent, not until 86.

The load that rises slower than the loop

Every droop a regulator is sized against is a load step with an infinitely fast edge, and a ceramic regulator droops 103 millivolts on its full 10 µF and 203 on a fifth of it. Give the same 100 mA a rise time and the ordering does not survive: past 23 microseconds the full capacitor droops more than the fifth, and a slow ramp's droop is the loop's own 25.2 µH times the load's rate of change — a number with no capacitance in it, raised 18 per cent on the full capacitor by the impedance peak its slower loop leaves. For a 10 mV budget the load must take 298 µs to arrive on the full capacitor and 253 on the fifth.

Forty thousand periods at β = 0.5: each period's error against the same period's duty error, correlated by 0.008. Seeded: forty thousand periods of the event map, β = 0.5, 5 mV of threshold noise; one in twenty is drawn, each scaled by its own standard deviation (3.4 ns for the period, 2.49 ns for the high-minus-low time). A period is three draws with coefficients A, A + B, B and its duty error the same three with A, B − A, −B, so their covariance is σ²[A² + (A + B)(B − A) − B²], which is zero at every β. Measured, the correlation is 0.0080, inside the ±0.0100 that forty thousand samples resolve. With normal draws, uncorrelated is independent: the cloud is round.

Every draw shared, and nothing in common

A relaxation oscillator's period error and its volt-second error are made of the same threshold draws, and they are exactly uncorrelated. A period weighs its three draws A, A + B and B; the high half less the low weighs the same three A, B − A and −B; the products sum to A² + B² − A² − B², which is zero at every hysteresis. Over forty thousand periods the measured correlation is 0.008 against a resolution of 0.010. What the two do share is one draw with the period on either side, ±0.157 at β = 0.5, and over any whole burst those two lags cancel term for term. Four thousand bursts judged against twenty-five pairs of limits fail both ways exactly as often as the product of the two rates says: the margins specified separately double-count nothing.

A Wien oscillator with a lamp for its lower gain arm, τ = 100 ms: the envelope rings 63.6 times in ten time constants at 63.6 Hz and settles at 1.000 V. computed by solving, not by drawing: three state equations marched through every cycle of the 1.59 kHz carrier. The peak at the network's node, over its settled value (solid), from a millivolt with a cold filament, against time in units of the filament's time constant τ = 100 ms. Cold, the 50 Ω filament asks for a gain of 11 and the output sits on the 12 V rail until the filament has warmed. It leaves the rail after 1.99 ms, falls through the operating point, overshoots and rings at 63.6 Hz with a damping ratio of 0.0225. The dashed bound is the linear loop's decay, exp(−(1 + m)t/2τ) with m = 0.8, and it contains neither the carrier's frequency nor the amplitude. The settled amplitude, 1.000 V at the node and 3.00 V at the output, is the one at which the filament is at 250 Ω.

The lamp that has to be slow

A small lamp in a Wien oscillator's gain network adjusts the gain to whatever each unit needs, so the margin a limiter spends is no longer spent, and the price is an amplitude loop of its own. The envelope is an integrator and the filament is a lag, which makes a second-order loop that rings at the square root of 2ω₀m/τ and decays at (1 + m)/2τ. The filament's ripple at twice the carrier is the only distortion, 0.201/ω₀τ. So the slowness that lowers the distortion lengthens the settling in exact proportion: distortion times settling is 0.163 cycles at any time constant, and 0.01 per cent takes a second at 1.59 kHz and 82 seconds at 20 Hz.

A reference 0° behind the excitation at 20 kHz, after a calibration at +1000 µε: matched, the worst error is 5.79 ppm; with a tenth of a per cent of screen mismatch, 0.607%. computed by solving, not by drawing: the error of a synchronous reading of a 350 Ω quarter bridge through ten metres of screened pair, against the strain read, with the reference 0° from the excitation at the bridge and a shunt calibration at +1000 µε through the same reference, for a matched screen (solid) and for 0.01 and 0.1 per cent of mismatch. The calibration takes out cos(φ − θ), because it multiplies the signal at every strain. It cannot take out a voltage that does not scale with the strain, and the mismatch's leak is such a voltage: a hyperbola through zero at the calibration strain. Worst over ±2000 µε: 0%: 5.79 ppm, 0.01%: 606 ppm, 0.1%: 0.607%.

The reference that should be late

A carrier amplifier's reference is taken from the excitation because that is what the output is meant to be in phase with, and nothing says that is where it belongs. A screen mismatch's leak arrives late by the common-mode node's lag and the difference's together, 525 ns on ten metres of pair and a 350 Ω bridge, so a reference on the excitation reads sin 3.7° of it at 20 kHz: 0.607 per cent of error at a tenth of a per cent of mismatch, and 72.6 parts per million with the reference 3.7° late. The null is a delay rather than a phase and does not move with the mismatch, and a delayed reference allows about five times the carrier on any mismatched screen. How exactly it must be held narrows as mismatch times carrier.

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