The load that has two voltages or none

The far end that rises measured a line’s regulation for loads that are impedances — a resistance at an angle — and found the percentage quoted for it carrying neither of the two things that decide it. Every load in that essay was the same kind of object: double its impedance and it takes half the current.

A great many loads are not that kind of object. The regulator of a source below a frequency holds its output voltage against its load, and seen from its input instead it holds its output power, so the rail that feeds it sees a load that takes a fixed power. A switching supply delivers the power its output needs, and if its input voltage falls it draws more current to deliver the same power. So does a motor drive holding a speed against a load, an inverter, an electronic load set to a wattage, and every regulated rail in a piece of equipment seen from the plug. Seen from the line, each is a load that takes a fixed power, and for a load like that the question regulation asks — how far the voltage falls — has a prior question underneath it: whether there is a voltage at which the line can deliver that power at all.

The same line, read as power against voltage #

Take a load of fixed angle and sweep its impedance from open circuit to short circuit, solving the network at every step. Open circuit takes no power at the full source voltage. Short circuit takes no power at no voltage. In between the power rises to a maximum and falls again, while the voltage falls the whole way. Plotted as power against voltage, that sweep is a curve with a nose.

The nose of the resistive curve is a familiar point. It is where the load’s resistance equals the line’s, the load takes the most power the line can give it, and the voltage across it is half the source’s. That is the load that takes the most, read from the other side. The maximum-power theorem says which load draws the most; the nose says that for a load which insists on its power, the matched load is the heaviest it can ever be, and it is drawn at fifty per cent efficiency with half the voltage gone.

On a resistive line the angle only costs power. Lagging and leading by thirty degrees reach the same nose, 0.9282 of the matched power at 0.5176 of the voltage, because with no reactance in the line nothing can tell which way the load’s current is turned. Leading by sixty degrees costs a third of the power. Nothing on this line is ever above the source.

Two voltages for every power below the nose, and none above it #

Everything to the left of the nose is reached twice. A load that takes nine tenths of the most power can take it from a light impedance at a high voltage or from a heavy one at a low voltage, and the two operating points are the two roots of one equation. Writing the line as R + jX and the load’s power as P + jQ, the far-end voltage satisfies

$$|V|^4 - \big(V_s^2 - 2(PR + QX)\big)\,|V|^2 + (P^2 + Q^2)\,|Z|^2 = 0,$$

a quadratic in $|V|^2$ with two positive roots below the nose, a double root at it, and a negative discriminant past it. Past the nose there is no voltage at which the line delivers the power, and a load that insists on it does not settle at a lower voltage — it has no steady state.

That is also why a regulation percentage has nothing to say about such a load. For an impedance, doubling the load roughly doubles the drop. For a load of fixed power on the resistive line above, the upper root is $(1 + \sqrt{1 - f})/2$ of the source at a fraction $f$ of the nose’s power: half the limit leaves 0.8536 of the voltage, nine tenths leaves 0.6581, and the limit itself 0.5. The first half of the available power costs 14.6 per cent of the voltage and the last tenth costs another 15.8, because the curve arrives at its nose with a vertical tangent. A drop measured at half load and scaled up is not a pessimistic estimate of the drop at full load; it is wrong in kind, since it assumes a slope the curve does not have.

The nose itself has a closed form. Its power is $V^2\cos\varphi/\big(2|Z|(1 + \cos(\theta - \varphi))\big)$ and the voltage left there is $V/\sqrt{2(1 + \cos(\theta - \varphi))}$, with $\theta$ the line’s angle and $\varphi$ the load’s, positive when it lags. And it falls where the load impedance’s magnitude equals the line’s — the same condition the load that takes the most found for the best resistive load on a reactive source, now true at every load angle.

None of that belongs to the load. A source behind a line is a Thévenin pair — an ideal voltage and an impedance — and the nose’s power and voltage contain only those two and the load’s angle. The source that is not a source found the ideal source’s edge as a current, set by an internal resistance nobody draws; the nose is the same pair’s edge for a load that holds its power, and it is a power.

Every nose drawn on this page is found on the network rather than from that formula. A golden-section search on the logarithm of the load’s magnitude, each trial a solve, returns the most power; it agrees with the closed form to a part in a billion, and the load it lands on has the line’s magnitude to a part in a hundred thousand. At nine tenths of the nose’s power the load that delivers it is then bisected on each branch of the solved network, and the two voltages it is left with are the two roots of the quartic to a millionth of the source.

A reactive line, and a nose above the source #

Put reactance in the line and the angle does more than cost power.

A lagging load is worse than on a resistive line and a leading load is better, in both coordinates at once. Leading by thirty degrees the line can deliver 1.33 times what it delivers at unity power factor and does so at a higher voltage. Leading by sixty degrees it delivers 1.61 times as much, and the nose sits at 1.055 of the source voltage. The load is at the very edge of the power the line can give it, the next increment has no voltage at all, and a voltmeter at the far end reads five per cent above the supply.

That is the same partial resonance the far end that rises found raising the voltage past a load angle, and on this line its crossing for a very large load was 26.57° of lead. The nose is a different point on the same surface and needs more lead to rise above the source, which is where the next figure comes in. Before it, one more line:

On a line this reactive, a load leading by sixty degrees is at its limit with a third more voltage at the far end than at the source, and a lagging one at 0.5431. The more reactive the line, the further apart those two readings of “at the limit” are, and the more a voltage misreports the distance to it.

Where the most power goes, and where the nose passes the source #

The nose’s two formulas can be read against the load’s angle for any line, which says what an installation’s power factor does to its limit.

Two angles come out of it, and both are properties of the line alone.

The most power is drawn by a load that leads by exactly the line’s own angle: unity power factor on the resistive line, 45° on the line whose reactance equals its resistance, 63° and 76° on the other two. A load at that angle is the conjugate of the line in angle, and it is the case the maximum-power essay describes as extracting more than any resistive load can.

The nose rises above the source’s voltage when the load leads by more than 120° less the line’s angle, because $V/\sqrt{2(1 + \cos(\theta - \varphi))}$ equals $V$ when $\cos(\theta - \varphi)$ is $-\tfrac12$. On 50 Ω + j100 Ω that is 56.57°, thirty degrees more lead than the 26.57° past which the far end that rises found light loads raising the voltage. Between those two angles a leading load’s far end is above the source at light load and below it at the limit; past the second, it is above the source all the way to collapse. The bisection on the nose voltage lands on θ − 120° to a millionth of a degree on every line.

The voltage does not say how close the nose is #

The practical question about any operating point is how much further it can go, and the reading that is available is the far-end voltage. The next figure asks how well one tells the other.

At the same distance from collapse — ninety per cent of the most the load can take — the far end reads anywhere from 0.678 to 1.172 of the source, depending on nothing but the load’s angle. A rule that treats a low voltage as the warning sign is looking at the wrong end of that range. The lagging installation at 0.678 looks alarming and has ten per cent in hand; the leading one at 1.172 looks over-voltage rather than under, and has the same ten per cent.

The usual repair makes this worse in a specific way. The capacitor that was right once sizes correction for a load and finds it over-correcting away from that load, which pushes an installation towards leading. Leading raises the nose, which is genuinely good, and raises the voltage at every margin, which removes the only symptom anyone was reading. The reactance cancelled, and the resonance it buys is the series form of the same move, and it raises the limit by shrinking θ and the line’s |Z| together — which is what series compensation is installed for, and also the reason a compensated line can be run nearer its nose than its voltage suggests.

On a direct-current bus the lower point is an edge #

The two roots of the quartic are both steady states of the algebra. Whether both are states a real circuit can sit in is a question the algebra cannot answer, and a direct-current bus is the cleanest place to ask it, because there a constant-power load is common and the answer has no phasors in it.

A 10 V source behind 1 Ω feeds a load that takes a fixed power, with a millifarad at the load as every such load has. The two operating points are the roots of $V^2 - V_0V + PR = 0$. The capacitor’s own equation is $C\,dV/dt = (V_0 - V)/R - P/V$, and its slope at each root is $-1/R + P/V^2$.

The upper point holds and the lower one does not, and the reason is the load’s own incremental resistance. A load that takes more current as its voltage falls has a $dV/dI$ of $-V^2/P$: a negative resistance. At the upper point its magnitude is larger than the bus’s one ohm and the bus’s resistance wins; at the lower point it is smaller, the negative resistance wins, and any disturbance grows. It is the same kind of object the resistance that is below zero finds looking into an emitter follower, where it makes an oscillator out of a stray capacitance. Here it makes a bus collapse out of a one per cent dip. The input that pushes back adds the condition for that kind of negative resistance to oscillate rather than run away — a reactance to cancel — and the bus here has none: a capacitor and a resistor give a real root, so the departure is a growing exponential rather than a growing ring. Put the inductance of a supply lead in series with the bus’s one ohm and the same negative resistance has something to resonate with, which is the first question this page leaves open.

So the lower branch of every nose on this page is not an alternative place to operate. It is a boundary: a load pushed below it by any transient does not come back.

Near the nose the two points close on each other. At ninety-nine per cent they are a volt apart, so a dip of eighteen per cent from the normal operating point — the kind a motor starting on the same supply produces — lands below the lower point, and the bus does not recover when the dip ends. Collapse is slower here, 11.59 ms against 3.48 at ninety per cent, because the slope through the lower point is shallower, and slower is not the same as less certain.

The same geometry explains a start-up hazard. A bus rising from zero passes through the low region before it reaches the upper point, and a constant-power load that switches on during the rise is sitting on the lower branch or below it, drawing a current the bus cannot supply. The first cycle, which no steady state contains is about a rectifier’s inrush for the same kind of reason — the interesting current is the one before the steady state exists — and it is one reason a converter does not start drawing its rated power until its input has passed a threshold.

How the numbers were obtained #

Each power–voltage curve is 181 solves of the line and a load of fixed angle whose magnitude runs over five and a half decades. Each nose is found by golden-section search on the solved network and held against the closed forms for its power, its voltage, and the magnitude of the load that draws it. The two operating points at nine tenths of the nose’s power are found by bisecting the solved network on each side of the nose and held against the roots of the quartic. The angles at which the most power is drawn and at which the nose passes the source are read from the closed forms and checked by the same search on the network at three angles per line and, for the second, by bisection on the nose voltage against θ − 120°. The direct-current bus is marched with a fourth-order rule at a microsecond from one per cent either side of each operating point.

What it does not say #

It treats a load that holds its power exactly and instantly. A real constant-power load holds it only as fast as its own control loop acts; faster than that its input looks like the capacitor across it, and whether the negative resistance of the slow part and the reactance of the line can oscillate together is a separate question with a frequency in it.

It uses a line with no capacitance of its own. A long line’s charging current is a leading load that cannot be switched out, and it moves every nose on the page upward in voltage.

And the stability argument is made on a direct-current bus. On an alternating line the same negative incremental resistance exists, but the lower branch’s behaviour then depends on how the load’s controller reads the voltage it is regulating against, which is a model of the load rather than of the line.

Still open: the loop inside the load, the line’s own charge, and the regulator that pushes #

A constant-power load with a bandwidth. Below its loop’s crossover a regulated converter’s input is $-V^2/P$; above it the input is its own capacitor. With a line inductance in front, that is a resonance whose damping is negative at low frequency, and the condition under which it grows puts a limit on the filter in front of every regulated supply. Solving the input impedance of a converter with a real loop against the line would say how much of this page’s static margin survives a disturbance with a frequency in it.

The line’s own capacitance. Adding shunt capacitance at each end of the line turns this page’s two-element source into a π network, which raises the nose voltage and moves the angle past which it exceeds the source. Measured, it would say how much of a long line’s apparent headroom is its own charging current.

A regulator that pushes the load down the curve. A tap changer or an automatic voltage regulator that raises the voltage at a load of fixed power makes that load draw less current, which is benign above the nose. Below it, raising the voltage the load sees makes the line deliver less, and the regulator keeps pushing. Marching a regulator with a time constant against a load of fixed power would find the margin at which correcting the voltage becomes the collapse.