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

Stability — where it appears

Whether a circuit's own natural response decays, decided by the real parts of its poles and by nothing about what is applied to it. It is decided by the loop and not by the input, which is why a conditionally stable loop can be unstable at a lower gain than one it is stable at.

Named by 24 essays across 11 fields — each of them below, with the objects they name alongside it.

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.

applied · Oscillator Loop gain of a three-pole amplifier closed for a gain of 100. Unity loop gain at 5.73 kHz, where 34.9° of phase remains before −180°. The phase reaches −180° at 89.6 kHz, where the loop gain is 46.1 dB below unity.

What is left at crossover

A feedback loop is stable or not according to one number read at one frequency — how much phase remains before −180° at the point where the loop gain passes unity. The loop gain here is obtained the way it is obtained on a bench: cut the loop, drive one side of the cut, and measure what comes back to the other.

feedback · Loop gain A loop whose phase comes back, at a gain of 1.0e+7. computed by solving, not by drawing. Three identical poles take the phase to −270°, two lead sections bring it back to −158°, and their own poles take it down again — so ∠L = −180° at 189 Hz, 3.46 kHz and 23.5 kHz. The gain moves the magnitude curve and not those three frequencies, so it decides only which side of each the unity-gain point falls. At 1.00e+7 the crossover is 10.6 kHz, in the recovered band, and the loop is stable.

Stable, and unstable with less gain

A three-pole loop with two lead sections is stable for gains between 1.81 × 10⁶ and 3.36 × 10⁷ and unstable on both sides of that window. Turning its gain *down* is what breaks it. Neither margin can see this, because a phase margin describes the loop at one frequency and a gain margin at one other, and this loop's phase crosses −180° at three: 189 Hz, 3.46 kHz and 23.5 kHz.

feedback · Loop gain An inverting unity gain, and the 100 pF that only the loop can see. computed by solving, not by drawing. Two ten-kilohm resistors around a 10 MHz amplifier make a gain of 1.00, and a loop that closes against 2.01 — one plus the ratio, not the ratio. Adding 100 pF at the summing junction leaves the closed-loop gain at a kilohertz unchanged — 0.99998051 against 0.99997988, three parts in a million at the far end of the slider — and takes the phase margin from 90.0° to 14.4°, because the noise gain now rises a decade per decade and the loop closes at forty decibels per decade instead of twenty. Forty-five degrees is reached at 9.00 pF, bisected on the netlist. The capacitance is not part of the signal path and does not appear in any expression for the gain.

The gain the loop closes against

An inverting amplifier with two equal resistors has a gain of one and a loop that closes against two, so it has half the bandwidth of a follower built from the same part — 4.99 megahertz against ten. Nine picofarads at the summing junction, less than a scope probe, takes the phase margin from ninety degrees to forty-five, and a hundred picofarads puts twelve decibels of peaking on a response whose designed gain is nought decibels and whose measured gain at a kilohertz has not moved by three parts in a million.

feedback · Noise gain A follower with 1000 pF on it looks like -1182 Ω of negative resistance. computed by solving, not by drawing. The impedance looking into the base of an emitter follower carrying 5.0 mA, with 1000 pF on its emitter. The real part is negative from 1.25 MHz upward and reaches -1182 Ω at 3.40 MHz: the load's reactance multiplied by a complex current gain, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it — the total loop reactance passes through zero at a frequency the inductance chooses, and the loop resistance there goes negative above 74.9 nH with 10 Ω of source, which is a few centimetres of wire. A hundred ohms of source raises that to 913 nH: the repair is a resistor in the base, and it works by making the source worse.

The input that pushes back

An emitter follower with a capacitor on its emitter has a negative resistance looking into its base — 1182 ohms of it at 3.4 megahertz for a nanofarad, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it: with ten ohms of source the loop goes unstable above 74.9 nanohenries, which is seven centimetres of wire. The repair is the opposite of the instinct — a hundred ohms of source raises the threshold to 913 nanohenries, so the fix for a follower that oscillates is to make the thing driving it worse.

semiconductors · Emitter follower A 8th-order Butterworth at 12 bits, as a cascade and as one polynomial. computed by solving, not by drawing. The open circles are the poles as designed, on the z-plane with the unit circle drawn. Filled marks are where they go once the coefficients are stored in 12 bits. A cascade of biquads keeps two coefficients per pole pair, so a rounding error moves that pair and nothing else: 8.43e-3, largest radius 0.97478. A direct form keeps one denominator whose coefficients are symmetric functions of every pole, so one rounding error moves all of them: 6.21e-1, largest radius 1.54393 — outside the unit circle, which is not an inaccurate filter but an unstable one. The two are the same filter until they are written down.

The same filter, rounded twice

The filters field measured two realisations of one analogue response and found the passband error growing as the 0.99 power of a component tolerance in a cascade and the 2.00 power in a ladder. The digital version of that argument comes out harder. At eighth order and sixteen bits, a cascade of biquads moves its poles by 6.6 × 10⁻⁴ and stays at a radius of 0.9748; the same filter written as one polynomial moves its poles to a radius of 1.4536, which is not an inaccurate filter but an unstable one.

digital · Digital realisation A second integrator is 5.8 more decibels an octave, and a third 5.4 more. computed by solving, not by drawing. Modulators of order 1, 2, 3 marched one sample at a time with a one-bit quantiser, their in-band noise read from the transform of the error with the test tone a quarter of the way up the band. The measured slopes are 8.17, 13.93, 19.33 decibels per doubling of the oversampling ratio, against the 9, 15, 21 the white-noise argument predicts and the 3.01 plain oversampling gives. Each order is short of its own prediction by about a decibel, in the same direction, which is the error in the band not being white. The straight lines are each order's prediction drawn through its own first point, so what is compared is a slope against a slope.

The loop that is worse at full scale

A second integrator in a one-bit loop takes the shaping law from nine decibels an octave to fourteen, and a third to nineteen. What it charges is not in decibels at all: past six tenths of full scale the ratio starts falling, and a fourth integrator's states run away at seven tenths. The best input to a second-order modulator is 0.6 of the reference it is measured against — an amplitude boundary of exactly the kind this collection is built for, on the one object in the field that has no continuous output to draw.

digital · Noise-shaping An inrush limiter's steady state, and how little of it is still a limiter. computed by solving, not by drawing. A negative-temperature-coefficient thermistor in series with a supply, at 1 ampere of load current. The falling curve is what it dissipates at a temperature — I²R with R following the two-point β fit a catalogue prints — and the rising line is what its mounting removes. They cross once, at 83.1 degrees, and the loop gain there is -1.374: negative, so the part is stable at every current rather than below a boundary. What is left of its cold 10 ohms at that temperature is 1.937 — 19.4 per cent. The slider moves the load current, and more current leaves less resistance.

The protection that is gone by the second time

An inrush thermistor is ten ohms cold and holds the first cycle down; then the load current warms it and it settles at 83 degrees and 1.94 ohms, which is 19 per cent of what was bought. That is the design working. It is also a part that takes 198 seconds to recover half its cold resistance, against a reservoir capacitor that empties in tens of milliseconds — so a mains dip in that window hands the rectifier an unlimited inrush into an empty capacitor, which is the exact event the part is on the bill of materials for.

power · Thermal feedback What a cascaded modulator is worth, against how well its two paths match. computed by solving, not by drawing. Two first-order loops marched sample by sample at an oversampling ratio of 64, with the first stage's quantisation error taken as the difference between what its comparator said and what was presented to it — nothing here reads a state a real converter could not. Perfectly matched, the cascade gives 72.5 decibels against the first stage's own 47.5: second-order shaping out of two first-order loops, neither of which can be unstable. The gain with which the first error reaches the second stage is then given an error, and the flat left-hand half of the curve is the arrangement working. It costs three decibels at 5.29 per cent, which is a capacitor ratio — achievable, and not free, and not something the digital side can measure or correct.

Two loops, and the mismatch between them

Single loops of second, third and fourth order, marched alone, each stop working above an amplitude that falls with the order — which is why nobody builds a fourth-order single loop. The standard answer is two first-order loops with the first one's error fed to the second and differentiated back out, giving second-order shaping out of parts that cannot be unstable. The cancellation is between an analogue path and a digital one, and it is worth 25 decibels until the two differ by five per cent.

digital · Noise-shaping The direct form works at 28 bits and fails again at 29. computed by solving, not by drawing. The largest pole radius of an order-8 Butterworth at a 1 kHz corner and a 48 kHz clock, at every coefficient word length from 6 to 40 bits, for both realisations. Above the unit circle the filter is not inaccurate but unstable. The direct form is outside at 23 of the 35 word lengths drawn; the cascade at none of them, its worst radius being 1.000000. The failures are not an interval: 28 bits works and 29 bits does not, so the smallest word length that works is 1 below the largest that does not and a search for the crossing has no crossing to find. The direct form's curve is clipped at 1.24: a radius of 4.5 and a radius of 1.01 are the same verdict.

The word length that is not a threshold

The essay below this one measured an eighth-order direct form at five word lengths, found its poles outside the unit circle at every one, and left behind a phrase — the coefficient resolution required. Sixty word lengths later there is no such resolution. That filter is stable at 28 bits, unstable at 29 and stable again at 30, and the smallest word length a search would return is not one anybody can use. What survives is a law with a slope: 4.53 bits of coefficient for every pole added, rising to 47 bits at twelfth order, and none at all for a cascade.

digital · Digital realisation Loop gain of a two-pole amplifier closed for a gain of 100. Unity loop gain at 5.73 kHz, where 35.0° of phase remains before −180°. The phase never reaches −180° at any frequency, so there is no gain margin to quote: 2 poles contribute at most 180° and the last of it arrives only at infinity.

The loop that never crosses

Every loop this field draws carries two margins, and one of them is not always a number. Take the third pole out of the standard loop and its crossover moves by 2.45 parts per million and its phase margin by 0.164 degrees — the third pole's own arctangent there, to five decimal places — while its gain margin goes from 46.06 decibels to no number at all. The phase reaches −180° only where the magnitude has already reached −62 decibels, and the two instruments that are supposed to notice report 3.19 × 10⁻⁶ either way.

feedback · Loop gain A follower fed through 100 nH has an output resistance of -21.9 Ω. computed by solving, not by drawing. The real part of the impedance looking into the emitter, driven by a current source and read, at every frequency. At direct current it is 5.50 ohms, which is the same rₛ/(β+1) + 1/gₘ the follower's first essay computed. Between 110 MHz and 301 MHz it is negative: rπ and Cπ delay the current the transistor sources into the emitter, and past a quarter of a cycle of delay pushing the emitter up makes the device push it up as well. The dashed curve is the same follower with no inductance between the source and the base, and it never goes below zero — the sign belongs to the wire and the transistor together, and to neither alone.

The resistance that is below zero

An emitter follower's output resistance is 5.5 Ω at direct current and −21.9 Ω at 257 MHz, and the sign is not the transistor's: with an ideal source at the base there is no negative band at all, and a hundred nanohenries of wire between the source and the base produces one from 110 to 301 MHz. A capacitance resonating inside that band is a resonator with loss of the wrong sign, so 4.7 to 100 pF on the emitter oscillates while 1 pF and 470 pF do not — a band of load capacitance with quiet ground on both sides of it.

semiconductors · Emitter follower The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance.

The loss that depends on what it causes

Every thermal figure in this collection has had the power handed in. A ferrite's has no business being: its saturation flux falls with temperature, its permeability rises, and both move the loss. Closing that loop makes the temperature a fixed point rather than a product — and the fixed point has a stable root at 89 degrees whose loop gain is negative, an ignition root at 191 whose loop gain is 120, and a thermal resistance of 183 kelvin per watt at which the two touch and neither exists.

magnetics · Thermal feedback A diode thermometer measuring its own sense current, at 600 K/W. computed by solving, not by drawing. The same fixed point as the core and the thermistor, on a junction: the dissipation is I·V and V falls with temperature, so the loop gain is negative and the equation has one root at every current. What it costs is two errors. The junction sits above ambient by 0.29 millikelvin at a microamp and 4.26 kelvin at ten milliamperes, which a calibration removes; and a kelvin of ambient produces less than a kelvin of junction, by 1/(1 − R_th·dP/dT), which it does not. After a calibration at 25 degrees the reading at 85 is out by -583 millikelvin at ten milliamperes and -0.09 at a microamp. The coefficient itself moves too — -2.403 against -1.613 millivolts a kelvin — so a quoted tempco carries a sense current as well as a junction.

The sensor inside its own answer

A junction driven from a current source cannot run away, because its forward voltage falls with temperature and its loop gain is therefore negative. What that costs is a thermometer that is warmer than what it is measuring by 4.26 kelvin at ten milliamperes, and — the part a calibration cannot remove — under-reports every change in ambient by 9,584 parts per million, because the sense current's own dissipation falls as the reading rises. Calibrated at 25 degrees, it is out by 583 millikelvin at 85.

semiconductors · Thermal feedback At Q = 2 and a hundred times the corner, one arrangement is 1.97% out and the other 2.83%. computed by solving, not by drawing. The same second-order section designed twice — a follower with a capacitance ratio, and equal passives with the Q supplied by a closed-loop gain of 2.5000 — built with the same one-pole amplifier and swept over its gain-bandwidth, with both pole pairs recovered by rooting the determinant. The unity-gain arrangement's error is Q over the ratio: 2.00 per cent at a hundred times the corner. The equal-component arrangement's is K²/2 over the ratio, 3.13 per cent, because its amplifier is a gain-of-K stage and therefore has K times less bandwidth to spend. Below about thirty times the corner both laws fail, and the second one changes sign.

Where the Q comes from

Two second-order sections with no component value in common have the same transfer function to 1.8×10⁻¹⁰ of a decibel, and are not the same circuit. Against a slow amplifier the follower's Q error is Q over the gain-bandwidth ratio and the gain stage's is K²/2 over it — so the second is worse below Q = 3.08 and better above it. Against component tolerance the follower's worst element carries ½ and the gain stage's carries 2Q − ½, nineteen times as much at Q = 5. Neither oscillates at any gain-bandwidth at all.

filters · Q enhancement The resistance that just stabilises it is 17 times smaller than the one that damps it. computed by solving, not by drawing. Two consequences of one base resistor, against how much of it there is, for a follower fed through 100 nH of wire with 47 pF on its emitter. The falling curve is the Q of the worst pole pair, rooted from the determinant so that no frequency grid is involved; the rising one is the output impedance the stage presents at low frequency. Bisecting on the SIGN of the pole's real part returns 69.7 Ω — at which the pair is stable with a Q of 1.7e+15, which is to say no damping and a peak whose height belongs to the arithmetic. A Q of one needs 1.15 kΩ, 16.6 times more, and that resistor takes the output impedance from 5.66 to 13.10 Ω — exactly R/(β+1) added, which is the bare follower's own expression with the base resistance in the place of the source resistance.

What the cure at the base costs

Bisecting for the smallest base resistor that stops an emitter follower oscillating gives 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.

semiconductors · Emitter follower BDF2 loses amplitude as N⁻³ and phase as N⁻², four times the trapezoidal rule's: a hundred cycles to 1% needs 363 steps a cycle, against 182 and 196,404. Each rule's error a cycle on an undamped inductor–capacitor pair, against the steps a cycle, measured from BDF2 marches and drawn with the trapezoidal rule's and backward Euler's closed forms. BDF2 loses amplitude a cycle as the −2.91 power of N and phase as the −1.95 power, its phase loss 3.988 times the trapezoidal rule's at 128 steps a cycle. To keep 99% of the amplitude over a hundred cycles it needs 157 steps a cycle, and to stay within 1% of a cycle in phase 363; the trapezoidal rule needs 182 for the phase and loses no amplitude, and backward Euler needs 196,404 for the amplitude.

The rule between the two

The trapezoidal rule keeps an undamped ringing's amplitude and loses its phase, and it preserves a stiff pole's error with its sign alternating. Backward Euler damps the stiff error at once and damps the ringing too, keeping 0.0082 per cent after ten cycles at twenty steps a cycle. The second-order backward difference sits between them in a precise sense. It damps the stiff error by 1/√(3 + 2h/τ), 0.152 a step at twenty time constants, and settles in three steps where the trapezoidal rule needs twenty-three. It loses a ringing's amplitude only as the cube of the steps a cycle, and its phase as the square with a constant four times the trapezoidal rule's. A hundred cycles to one per cent costs 363 steps a cycle, against 182 and 196,404.

transients · Step response A guarded input driven through an inductance oscillates once the source's L/R passes 1/ωₜ. computed by solving, not by drawing. The real part of the rightmost natural frequency of a guarded input — a 100 pF cable bootstrapped by a 1.00 MHz follower, 10 pF of input capacitance — driven from sources of 10 Ω, 100 Ω, 1 kΩ with an inductance in series, against the inductance. Above zero the input rings with growing amplitude. Each source crosses at 1.75 µH, 17.6 µH, 187 µH, close to R/ωₜ — 1.59, 15.9, 159 µH — and oscillates at 11.4 MHz, 3.59 MHz, 1.14 MHz. The same source without the guard is stable.

The source that rings against its guard

A bootstrapped cable is a capacitance in series with a negative resistance that falls as the square of frequency, −1/(ω²·C/ωₜ). Drive it from a source with resistance R and inductance L and the loop's reactances cancel at 1/√(LC). Its resistances cancel where R = L·ωₜ, so the input oscillates once the source's own time constant L/R passes the follower's 1/ωₜ. The poles of the whole netlist say 17.6 µH for a 100 Ω source behind a 1 MHz follower, against 15.9 from the expression. The cable's length moves that by at most a factor of two and sets only the frequency it rings at. The same source without the guard is stable. A resistor in the guard drive raises the limit, to 164 µH through 1 kΩ, and pays for it in bootstrap bandwidth.

instruments · Guarding A 10 µs delay in the loop: the peak rises to 13.53 dB and the sensitivity above crossover becomes a comb, and the two areas are still equal — 7.14 kHz·Np each side. computed by solving, not by drawing, from the loop's closed form, held against the netlist's cut loop to a part in a billion. The three-pole amplifier of the area essay (second pole 4 kHz) with a pure delay of 10 µs in its loop. Its margin falls from 34.9° to 14.3° at an unchanged crossover of 5.73 kHz, and the peak of |1/(1 + T)| rises from +5.70 to +13.53 dB. Above crossover the delay keeps turning the loop's phase, so the sensitivity rises and falls about zero with a period of 100 kHz. The area below zero is 7.144 kHz·Np, the area above 7.144, net 9.00 × 10⁻⁵ Hz·Np — integrated to 50 MHz on a grid that takes sixty steps a tooth.

The phase a delay never stops adding

Put a pure delay in a feedback loop and Bode's sensitivity integral does not notice: with 10 µs in the three-pole amplifier's loop the logarithm of the sensitivity still has equal areas above and below zero, 7.14 kHz·Np each side. What changes is where they are. The delay keeps turning the loop's phase above crossover, so the sensitivity becomes a comb with a tooth every 100 kHz; the desensitisation shrinks from 6.74 to 6.48 kHz·Np, the main peak rises from 5.70 dB to 13.53 and overpays it by 7%, and the comb hands the excess back below zero. The margin reads the peak from below by about 1.4 dB at every delay. Past 16.9 µs the loop is unstable and the integral stops balancing — by exactly the growth rate of the pole pair that has crossed.

feedback · Desensitivity The bootstrap with 1 nF at its node: one natural frequency, against the gain. computed by solving, not by drawing: the natural frequency of the network found as the root of its determinant at each of 31 gains. It is −G/C to a part in a million at every gain, where G is the Thévenin conductance the reduction gives: −200 thousand per second at A = 0, zero at A = 2 where the resistance passes through infinity, and +100 thousand at A = 3 — a growth rather than a decay. The gain at which the pole crosses into the right half-plane is the gain at which the resistance changes sign, and the capacitor's size moves the pole's speed and nothing else.

The pole that crosses with the resistance

A bootstrap's Thévenin resistance passes through infinity and turns negative at a gain of two. Put one nanofarad at its node and the network has one natural frequency, −G/C to a part in a million: −200 thousand per second with no gain, zero at A = 2, +100 thousand at A = 3. The gain at which the pole crosses is the gain at which the resistance changes sign — for this network. Loaded by a resistor it crosses at one plus R₂ times the sum of the conductances to ground, so a negative resistance is stable across a smaller load and not a larger one; a lag in the source slows the growth and does not move the crossing; a tank on the node turns the runaway into a 30.5 kHz oscillation once its coil is below about 1.5 Ω; and with an inverting source the resistance is positive at every gain and the node still oscillates beyond A = −16.

networks · Equivalent circuit 3 equal lags in a loop closed by an inverting gain k, from k = 0 to 12.8. computed by solving, not by drawing: the closed-loop poles, rooted from the netlist's determinant at 61 gains. At k = 0 the 3 poles coincide at −1/τ; as k rises they leave along 3 straight lines at angles π(2j + 1)/3, each ∛k/τ from the start, to the imaginary axis at k = sec(π/3) to the power 3 = 8.000. A gain of a hundredth moves them 21.5% of their own frequency; a single lag would move 1%.

The root a ring of equal stages takes

Two coincident poles separate as the square root of whatever splits them; three as the cube root; m as the m-th. In a loop of m equal lags closed by a gain k, the m-fold pole leaves along an m-pointed star, each point the m-th root of k from where it started — fitted exponents 1.0000, 0.5000, 0.3333 and 0.2500 over six decades — so a loop gain of a hundredth moves a triple pole 21.5% of its frequency where a single lag would move 1%. The star's nearest points reach the imaginary axis at k = secᵐ(π/m): 8 for three lags, 4 for four, 1.884 for eight, falling towards one. Staggering three lags by a factor of four raises the gain they tolerate from 8 to 26.6, and at the same gain cuts a step's overshoot from 53.9% to 20.6% — or, at the same gain margin, buys nothing.

transients · Poles A second-order loop at a ratio of 64, with a quantiser of one to five bits: the ratio in the band against the input amplitude. computed by solving, not by drawing: the loop whose noise transfer is exactly (1 − z⁻¹)², marched with a sine inside the band, the error measured through a squared Hann window. At half scale, 1 bit 69.4 dB, 2 bits 82.5 dB, 3 bits 90.3 dB, 4 bits 96.8 dB, 5 bits 102.0 dB. The first extra bit adds 13.1 dB because a one-bit quantiser's error is larger than a step's uniform error; after that each adds six or seven. The amplitude each does best at moves up with the bits — 0.6 for 1, 0.6 for 2, 0.8 for 3, 0.85 for 4, 0.95 for 5 — because a fine quantiser is never driven past its last level by the shaped error.

The levels the loop cannot shape

Give a second-order noise-shaping loop a quantiser of several bits instead of one and every bit beyond the second buys six or seven decibels, the best input amplitude climbs from 0.6 of full scale to 0.95, and loops of third and fourth order that will not run on one bit run on three or four. What it charges is outside the loop's reach. The levels the quantiser feeds back enter where the input does, so their errors are passed rather than shaped: a three-bit loop loses its lead over a one-bit one when each level is out by 0.027 per cent, its error stops falling with oversampling, and the error arrives as harmonics. One bit is immune because two levels always lie on a line.

digital · Noise-shaping A zero on the right at 1.00 kHz: the straight lines report 46.47° of margin, and the loop has 18.19°. computed by solving, not by drawing, the closed form held to the netlist — an inverting differentiator added back to its input makes the zero — at three frequencies and to the closed loop's poles. An integrator with a zero in the right half-plane at 1.00 kHz, the integrator set so that the straight lines cross unity at 950 Hz. The magnitude, and so the lines, are those of a zero on the left; the phase falls by the zero's angle where the left zero's would rise (dashed). The loop crosses at 3.04 kHz. Read at the lines' crossing with the loop's own phase the margin is 46.47°, and 133.53° with the phase the magnitude implies; at the loop's own crossing it is 18.19°.

The zero the lines put on the wrong side

A zero in the right half-plane has exactly the magnitude of its mirror on the left, so it has exactly the same straight lines, and nothing drawn from a magnitude can say which side it is on. The phase can: it falls where the mirror's rises. An integrator and such a zero, read off its lines, reports 90° − atan x of margin on a loop that has acos x, where x is where the lines cross as a fraction of the zero — so the lines say 45° of a loop with none. Behind one real pole the over-report reaches 37.6° on a loop of 2.1° when the zero sits below the pole, and vanishes when it sits above it: the reading is unsafe exactly when the zero comes first.

frequency · Asymptotic approximation Asked for 105% of the nose power, a tap changer that blocks on a wrong-signed step stops at 9.759 V and holds it; one that does not keeps raising, and the voltage keeps falling. Marched. The 10 V source behind 1 Ω of the regulator essays, its load drawing 105% of the line's 25 W nose power at the 10 V setpoint, regulated by a tap changer that raises the ratio by 1.25% every 5 s while the load is more than 20 mV low. With no blocking rule (dashed) it climbs past the nose ratio √(Rₗ/R) = 1.9518 and keeps going, to n = 3.250 and 8.61 V by 900 s, still falling; the integrating regulator of the earlier essay (dotted) collapses the same way, faster. With the rule (solid) the tap at n = 1.9625 is the first to lower the voltage, the regulator stops at 385 s, and the load holds 9.759 V — within a tap of the most this line can give it, 9.759 V. The setpoint is out of reach either way; the rule converts a collapse into a stall.

The rule that reads the sign of the gain

A regulator that raises a ratio whenever its load is low collapses the voltage once the load asks for more than the line's nose power, because past the nose raising the ratio lowers the voltage. The standard protection is a rule: after each tap, compare the voltage with the one before, and if it fell, stop. On a 10 V line asked for 105% of its 25 W, a 1.25% tap changer with the rule stops at n = 1.9625, the first tap past the nose, and holds 9.759 V — the most the line can give — where without it the ratio climbs to 3.25 and the voltage is 8.61 V and still falling. The rule is exact without noise and fragile with it, because it reads a difference that vanishes at the nose: with 3 mV of measurement noise it wrongly stops a raise that could have reached the setpoint 0.3% of the time at 98% of the nose and 15% at 99.5%. Trusting it to 1% costs margin — 97.9% of the nose at 3 mV, 79.9% at 10 mV — or coarser taps.

power · Voltage regulation

Named alongside it

The objects these essays reach for when they reach for this one.

Loop gainPhase marginPolesDesign tradeoffModel rangeNegative resistancePole pairEmitter followerFixed pointGain marginNoise-shapingOscillation

All concepts