Concept

Maximum power transfer — where it appears

The condition under which a source delivers the most power to a load, being a match to the magnitude of its own impedance and half efficient there. Its stationarity is what makes a doubly terminated ladder second order in its component tolerances where a buffered cascade is first order.

Named by 21 essays across 4 fields — each of them below, with the objects they name alongside it.

A 9 V source with 500 mΩ inside it. The ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

networks · Source model
The floor an amplifier adds, against the source it is given. computed by solving, not by drawing. A part with 4.00 nV/√Hz of voltage noise and 0.60 pA/√Hz of current noise is quietest into 6.67 kΩ, where its noise figure is 1.138 dB. That resistance is the ratio of the two generators and the floor there depends only on their product. Matching the same part for maximum power into its own 1 MΩ input instead — a resistance 150 times larger — costs 12.57 dB.

The floor a circuit has

A resistor's noise is 4kTR and there is nothing to choose about it. An amplifier adds two generators that belong to the device — 4 nV/√Hz in series with its input and 0.6 pA/√Hz across it — and because one matters most into a small source and the other into a large one, there is a source resistance at which their sum is least. It is 6.67 kΩ, it is the ratio of the two, and it is not the resistance that transfers maximum power.

noise · Device noise
One per cent on one component, in two realisations of the same order-5 filter. computed by solving, not by drawing. The two realisations agree to 3e-14 dB before anything is moved. Moving each element in turn by 1%, the worst deviation at the 2 ripple peaks is 0.1621 dB for the cascade and 0.0030 dB for the LC ladder — 54 times smaller. Across the whole passband the two are within 3% of each other, because near the band edge both are dominated by the response's own steepness rather than by the realisation.

A ladder is not a cascade

The same fifth-order Chebyshev, built two ways. A one per cent component moves the buffered cascade's passband by 0.162 dB at the ripple peaks and the doubly terminated LC ladder's by 0.003 dB — fifty-four times less. The number is not the claim: the claim is the exponent. Fitted over two decades of tolerance the cascade's error grows as the 0.99 power and the LC ladder's as the 2.00 power, because at maximum power transfer the response is stationary in every element it contains.

filters · Filter tradeoff
The load that takes the most power, and the load that wastes the least. computed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, which is the magnitude of the source impedance to six figures, and the efficiency there is 0.500000 — the source dissipates as much as the load receives. Ninety per cent efficiency needs a ratio of 9.7 and delivers 34% of the available power.

The load that takes the most

A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.

power · Power transfer
50 Ω + j100 Ω of line, and the load angle past which the far end rises. computed by solving, not by drawing at 71 load resistances and four load angles. A source of 50 Ω + j100 Ω feeding loads of the same resistance and different power factor: at unity power factor the voltage across the load climbs towards the source's and stops there, reaching 0.9675 of it at the largest load drawn. A lagging load leaves less. A leading one leaves more, and past a computable angle it leaves more than the source has: the condition is 2Rₗ(Rₛ + Xₛ·tanφ) + |Zₛ|² < 0, which for the largest load here is 28.35° of lead — bisected on the solve at 28.35° — tending to atan(Rₛ/Xₛ) = 26.57° as the load grows. So the edge is a property of the line and the load angle together, and "voltage regulation" quoted as a percentage carries neither.

The far end that rises

Voltage regulation is quoted as a percentage: how far the voltage at the end of a line falls when the load is applied. The percentage carries neither of the two things that decide it. Past a computable angle of leading load the voltage at the far end goes above the source's — 28.35 degrees for a line of fifty ohms and a hundred of reactance — and with no reactance in the line there is no such angle at all, because the rise is a partial resonance and needs both halves.

power · Voltage regulation
An order-5 ladder driven from 2× the 1 kΩ it was designed between. computed by solving, not by drawing. A doubly-terminated Butterworth ladder is a two-port designed between two stated resistances, and the resistances are part of the design rather than the environment it happens to be used in. At match it loses 6.021 dB — exactly half the voltage — and its passband has no peak anywhere in it. Driving the same five reactances from 2× that resistance moves the shape by 2.345 dB and the insertion loss to 9.542 dB. The band inside which the shape is right to 0.5 dB runs 0.8857× to 1.1371× — a window of 25 per cent on a quantity usually written down as a round number. And the two sides are not alike: at twenty times the design resistance the departure has settled at 5.33 dB, while at a twentieth of it the passband is 15.51 dB out with 7.21 dB of peaking on it, so driving a ladder from too low an impedance is worse than driving it from too high a one.

The two resistors a ladder was designed between

A passive ladder filter is not a transfer function with some resistors attached; it is a two-port designed between two stated resistances, and the resistances are as much part of the design as the inductors. Driving an order-five Butterworth from anything outside 0.886 to 1.137 times its design resistance puts more than half a decibel of error on the passband — a tolerance tighter than the resistor is usually specified to — and the two sides of that window are not alike.

filters · Filter termination
The load that takes the most power is 3.67 Ω, and the open-circuit voltage over the short-circuit current is 3.97 Ω. computed by solving, not by drawing. The terminal characteristic of a photocurrent with a junction across it, with the power along it drawn on the same axes and scaled to fill them. The most power, 80.99 W, is delivered at 17.244 V and 4.696 A, which is a load of 3.67 Ω. The incremental resistance of the source there — the negative of the characteristic's own slope — is 3.67 Ω, the same number to four figures. A straight line drawn between the two end points has a resistance of 3.97 Ω and would promise 24.81 W. The ratio of the true maximum to that promise is 3.264, and its reciprocal is the fill factor, 0.8160.

The load a curve recommends

The maximum power theorem says to match the load to the source's internal resistance, and for a straight-line source the internal resistance, the slope and the open-circuit voltage over the short-circuit current are one number. On a photovoltaic panel they are three: the slope at the maximum is 3.6717 ohms, the load there is 3.6717, and the open-circuit voltage over the short-circuit current is 3.9709 — and a quarter of the open-circuit voltage times the short-circuit current, which is what a straight line would deliver, is 24.81 watts against the 80.99 that is there. The theorem survives with the incremental resistance in place of the internal one, and it survives only where the characteristic has a slope: on a supply whose current limit folds back, the most power is delivered at a corner.

networks · Source model
50 Ω + j100 Ω of line: every power below the nose at two voltages, and a leading load's nose at 1.055 of the source. computed by solving, not by drawing: a load of fixed angle swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω, with the power it takes against the voltage it is left with, for four load angles. Each curve rises to a most power and turns back while the voltage keeps falling, so every smaller power is delivered at two voltages and every larger one at none. A 30° lagging load reaches 0.4222 of a matched resistive line's power with 0.5221 of the source voltage left; a unity power factor load reaches 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left; a 30° leading load reaches 0.8240 of a matched resistive line's power with 0.7293 of the source voltage left; a 60° leading load reaches 0.9960 of a matched resistive line's power with 1.0553 of the source voltage left. Each nose is found by golden-section search on the solved network, agrees with V²cos φ/(2|Z|(1 + cos(θ − φ))), and falls where the load impedance's magnitude equals the line's.

The load that has two voltages or none

A load that takes a fixed power takes more current as its voltage falls, and on a line with impedance in it every power below a limit is delivered at two voltages and every power above it at none. The limit sits at the load the maximum-power theorem describes, half the source voltage on a resistive line. On fifty ohms and a hundred of reactance a load leading by sixty degrees reaches that limit with its far end at 1.055 of the source, and at nine tenths of it reads 1.172 — so a far end that reads high is not a far end with margin. On a direct-current bus the lower of the two voltages is not a state at all: one per cent below it the bus runs down to nothing in 3.48 milliseconds.

power · Voltage regulation
The load that may be complex, and what a resistor alone gives up. computed by solving, not by drawing, with the best load searched over BOTH of its parts on the solved network rather than substituted. Against a source of 50 Ω + j100 Ω the search returns 50.00 − j100.0 Ω — the conjugate — which delivers the available 500.00 mW at exactly 50.00% efficiency, and does so at every source reactance. A load that may only be a resistance takes 2/(1 + √(1 + x²)) of that, matching the solved answer to 4.4e-16 at ten reactances — and at x = 2 that fraction is exactly the golden ratio less one, 0.618034. The resistor-only load is the MORE efficient of the two at every non-zero reactance, rising towards one while the conjugate match sits at a half for ever, so the familiar "maximum power at fifty per cent" belongs to the conjugate and not to the load.

The load that may be complex

Freed of the constraint that it be a resistance, the best load is the source's conjugate — found here by a two-dimensional search on the solved network rather than assumed — and it takes the available power at exactly fifty per cent efficiency whatever the source's reactance. A load that may only be a resistance takes 2/(1 + √(1+x²)) of that, and at a source reactance of twice its resistance that is exactly the golden ratio less one, 0.618034. The resistor-only load is also the MORE efficient of the two, rising towards one while the conjugate sits at a half for ever.

power · Power transfer
With its own capacitance at B|Z| = 0.2, a line's nose for a unity-power-factor load moves from 0.618 to 0.658 of a matched line's power, at 0.635 of the source. computed by solving, not by drawing: a unity-power-factor load swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω carrying its own shunt capacitance as a π, half at each end, with the total susceptance stated as B|Z|. At B|Z| = 0 the nose is 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left, and the unloaded far end reads 1.0000. At B|Z| = 0.1 the nose is 0.6376 of a matched resistive line's power with 0.6108 of the source voltage left, and the unloaded far end reads 1.0465. At B|Z| = 0.2 the nose is 0.6582 of a matched resistive line's power with 0.6353 of the source voltage left, and the unloaded far end reads 1.0969. At B|Z| = 0.4 the nose is 0.7025 of a matched resistive line's power with 0.6895 of the source voltage left, and the unloaded far end reads 1.2107. Each nose is found by golden-section search on the solved network and agrees with the nose of the Thevenin equivalent V/(1 + jBZ/2) behind Z/(1 + jBZ/2).

The headroom that is the line's own charge

A line of 50 + j100 ohms delivers at most 0.6180 of a matched resistive line's power to a unity-power-factor load, with 0.5878 of the source voltage left. Give the line its own shunt capacitance — a π, half at each end, B|Z| = 0.4 in all — and the nose moves to 0.7025 at 0.6895: 13.7 per cent more power and 17.3 per cent more voltage. It is headroom, but not the headroom the far end advertises, which with no load rises 21.1 per cent. The capacitance turns the source and line into a Thevenin equivalent with more voltage behind more impedance, and it moves a voltage threshold's meaning in opposite directions depending on whether it is referred to the source or to the unloaded far end.

power · Voltage regulation
The net noise power between two resistors at two temperatures. computed by solving, not by drawing. A 1 kΩ resistor at 400 K joined to a second one at 290 K, whose resistance runs over six decades. Each delivers 4kTR₁R₂/(R₁+R₂)² per hertz to the other; the net is 1.5187 zW/Hz at the match, which is exactly kΔT and contains no resistance — a gigohm pair at the same two temperatures exchanges the same. Away from the match the exchange falls, to 0.0596 zW/Hz at a ratio of a hundred, so the maximum-power argument applies to noise as it does to a signal. At one temperature the net is zero at every ratio, to 7.5e-37 W/Hz, which is the statement a wrong noise model would break.

Which way the noise goes

Two warm resistors joined together each drive the other, and the net flow is 4kΔT·R₁R₂/(R₁+R₂)² per hertz. At the match that is kΔT exactly — 1.5187 zeptowatts per hertz between 400 K and 290 K — and a kilohm pair and a gigohm pair at the same two temperatures exchange the same, which is why noise is quoted as a temperature. At one temperature the net is zero at every ratio to a part in 10³⁷, and that zero is the second law rather than a tolerance.

noise · Johnson noise
A regulator holding 10 V on a load that asks for 101% of the nose power collapses it in 256 s; at 110%, in 79 s. Marched with a fourth-order rule. A 10 V source behind 1 Ω feeds a load resistance through an ideal ratio n, and a regulator raises n at 0.05 per volt-second of error to hold the load at 10 V. The load's resistance is chosen so that at 10 V it takes the stated fraction of the most the line can deliver, 25 W. At 90% the regulator settles at n = 1.5195, below the nose ratio √(Rₗ/R) = 2.1082. At 99% the regulator settles at n = 1.8182, below the nose ratio √(Rₗ/R) = 2.0101. At 101% the voltage climbs to 9.950 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 101.7 s, and falls below half the setpoint at 256.3 s while the regulator keeps raising the ratio. At 110% the voltage climbs to 9.535 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 23.5 s, and falls below half the setpoint at 78.6 s while the regulator keeps raising the ratio. The regulator's gain, the slope of the load's voltage against the ratio, is positive below √(Rₗ/R) and negative above it.

The regulator that pushes past the nose

A regulator that raises a ratio whenever its load's voltage is low is a stabiliser only while raising the ratio raises the voltage, and on a line that stops being true at exactly the nose: the load's voltage, n·V₀ times the load resistance over n²R plus that resistance, peaks at a turns ratio of the square root of the load over the line and falls beyond it. Ask the load for 90 per cent of the nose power and the regulator settles at n = 1.519 — unless it starts above n = 2.925, where the same setpoint is met on the wrong side of the peak, and then it collapses the voltage. Ask for 101 per cent and the voltage climbs to 9.950 volts, the most the line allows, and is below half its setpoint 256 seconds later. Near the nose the collapse takes a time that grows as the inverse square root of the excess: 2,521 seconds at a hundredth of a per cent.

power · Voltage regulation
The load alternates with parity between 1 and 1.9841, at every order. computed by solving, not by drawing. The last quotient of the continued fraction that turns a reflection polynomial into element values, which is the load resistance the design demands, drawn against order. A Butterworth returns exactly one at every order. A Chebyshev alternates: one at every odd order and 1.984056 at every even one, the same number each time, and it is the closed form (√(1+ε²)+ε)² to a part in 10¹⁵. The mechanism is in the last two rows of the panel. A lossless ladder filter is two resistances at direct current, so it must deliver the maximum available power there, which requires that zero be one of the frequencies the design reflects nothing at — and an even-order Chebyshev's reflection zeros are the roots of an even Chebyshev polynomial, none of which is zero. The synthesis stalls at order 9 for a Butterworth, which is why that series stops at eight.

The termination an even order cannot have

Every even-order Chebyshev ladder terminates in 1.984056 times its source resistance at half a decibel of ripple, the same number at orders two, four, six and eight, and it is (√(1+ε²)+ε)² to a part in 10¹⁵. Building one between equal terminations instead — which is what a table of g-values and a matched pair gives — turns 0.5000 dB of ripple into 1.8123 at order four, deletes one of the passband maxima outright, and moves the worst tolerance corner from the 2.000 power of the component tolerance to the 1.109 power: a factor of 95 at one per cent parts.

filters · Filter termination
A lossy cancelling capacitor beats a resistor alone only above Q = 3.236 if it cancels everything, and only above 1.618 at all. computed by solving, not by drawing. A source of 50 Ω + j100 Ω, its reactance cancelled by a capacitor whose quality factor is Q, so that the capacitor carries a series resistance of its reactance over Q. Cancelling fully, with the load resistor matched to the source and capacitor resistances together, the load takes 1/(1 + x/Q) of the available power, at exactly half efficiency; a resistor alone takes 0.6180, 2/(1 + √(1 + x²)). They are equal at Q = 3.2361, twice the golden ratio. Cancelling only part of the reactance does better than either: it helps at all above Q = (√(1 + x²) + 1)/x = 1.6180, and at Q = 3 takes 0.6650 against 0.5971 cancelled fully.

The reactance that is not free

A conjugate match cancels the source's reactance with an opposite one and takes all the available power, where a resistor alone takes 2/(1 + √(1 + x²)), 0.618 of it when the reactance is twice the resistance. The cancelling capacitor is not free. With quality factor Q it carries a resistance of its reactance over Q, and cancelling fully then delivers 1/(1 + x/Q). That beats the resistor alone only above Q = 2(√(1 + x²) + 1)/x, 3.236 here, twice the golden ratio. But full cancellation is not the best use of a lossy capacitor. Cancelling part of the reactance beats both, and helps at all above Q = (√(1 + x²) + 1)/x, the golden ratio itself. Below that, the right amount to cancel is none.

power · Power transfer
Matched at 1 kHz, the conjugate load wins only from 730 Hz to 1.84 kHz; the resistor-only load is a low-pass, not a broadband answer. computed by solving, not by drawing. A source of 50 Ω and an inductance of 100 Ω at 1 kHz, with the conjugate load for 1 kHz — 50 Ω and a capacitor of −j100 Ω — and the resistor-only load for 1 kHz, 111.8 Ω, both held fixed while the frequency moves. The conjugate match takes all the available power at 1 kHz and stays within a decibel of it from 777 Hz to 1.29 kHz, a band of 50.9% of the centre, 1.0176/x. The resistor-only load takes 0.6180 at 1 kHz, 0.8509 at 100 Hz, and less at every frequency above: a low-pass whose peak is at direct current. The conjugate load is the better of the two only between 730 Hz and 1.84 kHz.

The load for a source that will not hold still

A conjugate match is exact at one frequency. Held fixed while the frequency moves, it stays within a decibel of the available power over 1.0176/x of the centre: 51 per cent of it when the source's reactance is twice its resistance, 12.7 per cent at eight times. It beats the resistor-only load only between 0.73 and 1.84 of the centre frequency. The resistor-only load is not a broadband answer, as it is usually called. It is a low-pass whose best frequency is direct current. And for a source whose reactance could be anywhere within ±Δx of a centre, the fixed load that guarantees the most power cancels the centre exactly and takes a resistance of √(1 + Δx²) times the source's. That guarantees 2/(1 + √(1 + Δx²)): the resistor-only rule, applied to the uncertainty instead of to the reactance.

power · Power transfer
Two 230 V sources joined by 0.05 + j0.2 Ω: the most that can be received is 194 kW, at 76.0°; with 50% of the reactance cancelled it is 262 kW, at 63.4°. computed by solving, not by drawing, at 50 Hz: two 230 V sources, the sending one leading the receiving one by an angle δ, joined by a line of 0.05 Ω and j0.2 Ω with a series capacitor cancelling a fraction of the reactance; the real power absorbed at the receiving end against δ. Uncompensated (dashed) the maximum is 194.37 kW at 75.96°, the line's impedance angle. With 50% cancelled (solid) it is 261.55 kW at 63.43°. Both maxima match V² cos β (1 − cos β)/R, found by a search on the solve to a part in a million. Past its maximum the received power falls as the angle grows: a larger angle drives more current, and past the impedance angle the extra current's loss outgrows what it carries.

Sixty degrees, and the limit compensation stops raising

Series capacitors are put in transmission lines to raise how much power can pass between two sources before the angle between them runs out. The lossless textbook limit, V²/X(1 − k), rises without bound as the reactance is cancelled. With the line's resistance kept, the limit is V² cos β (1 − cos β)/R for an impedance angle β, and it peaks where β is exactly sixty degrees: 56.7 per cent cancellation and 264.5 kW on a 230 V tie of 0.05 + j0.2 Ω, against 194.4 uncompensated — then falls, to nothing at full cancellation. A line of R/X 0.1 can nearly triple its limit; one above 1/√3 cannot raise it at all. And at the best compensation the line delivers exactly half of what it is sent.

power · Series compensation
Over ±25% of the centre, three elements deliver 97.3% at worst; Bode and Fano allow 99.8%. computed by solving, not by drawing. The fraction of the available power a source of 2 times its resistance in inductive reactance delivers to a resistive load through the best matching network of 0, 1, 2 and 3 reactive elements for a band of ±25 per cent about the centre, each network the one whose worst transfer in the band is largest; the vertical lines mark the band. The worst in-band transfer is 54.2%, 82.8%, 89.5%, 97.3%; the Bode–Fano limit for any lossless network, reached only with infinitely many elements, is 99.81%.

What each element buys against the limit

A source whose resistance sits behind twice as much inductive reactance can deliver, over ±25 per cent of its centre frequency, 54.2 per cent of its available power to the best plain resistor and at worst 82.8 per cent through the best single capacitor. That single-element answer is exactly the fixed load for an uncertain reactance: the capacitor makes the reactance swing ±1 across the band and the load is √2 times the source. Two elements lift the worst case to 89.5, three to 97.3, and a fourth, in a chain of single elements, to 97.4. Bode and Fano bound every lossless network at 99.81 per cent over that band — a limit three elements come within a tenth of a decibel of and no finite network reaches.

power · Power transfer
With 0.05 Ω of machine reactance at each end the sixty-degree peak needs 107% of the line cancelled — out of reach — so the limit rises all the way to full cancellation, 261.6 kW against 145.3. computed by solving, not by drawing, at 50 Hz, searched over the angle between the two EMFs on the solve (dots) and from V²·cos β·(1 − cos β)/R with the totals in it (line): the 230 V tie of 0.05 + j0.2 Ω with each source a machine whose EMF is held behind 0.05 Ω of its own reactance, against the fraction of the LINE's reactance cancelled. The capacitor cannot reach the machines' reactance, so the total impedance angle reaches sixty degrees only at k* = (X + 2Xₘ − √3·R)/X = 106.7%, beyond full cancellation; the limit rises from 145.34 to 261.55 kW and never turns. Dashed: the bare tie, peaking at 56.7%. The slider is the machines' reactance.

The reactance the capacitor cannot reach

A series capacitor raises a tie's transfer limit until the loop's impedance angle is sixty degrees, and on a bare 0.05 + j0.2 Ω line that is 56.7% cancellation. Put a machine at each end, its EMF held behind a reactance of its own, and the capacitor still cancels only the line: the sixty degrees now needs the line's reactance plus twice a machine's, less √3 times the resistance, as a fraction of the line's reactance, which passes 100% as soon as each machine's reactance exceeds √3/2 of the line's resistance — 0.043 Ω here. Beyond that the limit rises at every step to full cancellation, and the lossless estimate of the gain, wildly high on a bare line, becomes nearly right: 1.43 against 1.50 at 0.2 Ω. The higher limit is reached with the receiving terminal lower — 180 V falling to 148 at 0.1 Ω — and the capacitor's own resistance is negligible for any power capacitor: a Q of 222 keeps the peak within 1%.

power · Series compensation
With the receiving end at 0.9 of the sending voltage the limit peaks at 62.6% cancellation, where the impedance angle is 56.3° — and the peak is still Vₛ²/4R = 264.5 kW. computed by solving, not by drawing, at 50 Hz, searched over the angle between the sources on the solve (dots) and from (Vᵣ·cos β/R)·(Vₛ − Vᵣ·cos β) (line): the 0.05 + j0.2 Ω tie with the sending source at 230 V and the receiving one at 207 V, against the fraction of the reactance cancelled; dashed, equal ends. The limit is a parabola in cos β with its top at cos β = Vₛ/(2Vᵣ) = 0.5556, an impedance angle of 56.25°, reached at k = 62.58%; its height there is Vₛ²/4R = 264.50 kW, the same as for equal ends. Uncompensated, this tie carries 180.53 kW at most. The slider is the receiving voltage as a fraction of the sending.

The peak a sagging receiver keeps

The sixty degrees at which series compensation stops raising a tie's transfer limit is a statement about two ends at equal voltage. With the receiving end lower, the limit is a parabola in the cosine of the impedance angle whose top is at cos β = Vₛ/2Vᵣ: 56.3° and 62.6% cancellation for a receiver at 0.9 of the sender, 33.6° and 83.4% at 0.6, full cancellation at one half. But the height of that top is Vₛ²/4R whatever the receiving voltage — 264.5 kW for the 0.05 + j0.2 Ω tie from a receiver at 1.1 of the sender down to one at 0.5 — and at it the receiving end takes exactly half of what is sent, to four decimal places, for every receiving voltage tried. The best compensated tie is the sending source driving a matched load, and the receiving source's voltage only decides how much capacitor that takes.

power · Series compensation
Three trackers finding a panel's maximum power from open circuit, against the number of readings each takes. computed by solving, not by drawing: the single-diode panel of 81.0 W at 17.244 V, every tracker starting at 0.995 of the 19.85 V open-circuit voltage. Perturb-and-observe with 0.2 V steps walks down and then oscillates about the maximum, within two parts in a thousand of it from reading 39. Newton on I + V·dI/dV, two readings a step for the current and its slope, is within a part in a million from reading 10. The fractional open-circuit rule needs one reading and lands exactly, because its fraction, 0.8686, was taken at this light.

How many readings a maximum takes

A panel's maximum power sits where I + V·dI/dV is zero, and a tracker finds it by reading the panel. From open circuit, perturb-and-observe in 0.2 V steps takes 39 readings to come within two parts in a thousand and then oscillates about the maximum forever, giving up 7|P''|δ²/24 of it; Newton's method on the condition itself takes 10 readings to come within a part in a million, the digits doubling at each step. Newton converges only from a window: above 15.82 V on a panel whose maximum is at 17.24, because below it the condition is nearly the short-circuit current and nearly flat, and one step leaves the characteristic. And the one-reading rule — a fixed fraction of the open-circuit voltage — loses nothing at the light it was set at and 17 per cent at a fiftieth of it.

networks · Source model
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.

Design tradeoffVoltage regulationImpedance matchingAvailable powerInternal resistanceModel rangeOperating pointPhasorReactive powerLine impedanceReactanceReal power

All concepts