The regulator that pushes past the nose

The load that has two voltages or none marched a direct-current bus into a load that takes a fixed power and found that the lower of its two operating points is not a state. A load that draws more current as its voltage falls has an incremental resistance of $-V^2/P$; below the lower point that negative resistance beats the bus’s own, and one per cent below it the bus runs down to nothing in 3.48 milliseconds.

Real loads are rarely constant power on the scale of milliseconds. A heater, a lamp or a resistive process is a resistance: pull its voltage down and it takes less power. It becomes a constant-power load only on a longer scale, through something that restores its voltage — and on a supply network that something is very often a regulator. A transformer with a tap changer, or an automatic voltage regulator, watches the voltage at the load and raises its ratio when the voltage sags. With a resistive load behind it the regulator restores the load’s power along with its voltage, and on the timescale of the regulator the combination is a constant-power load.

The same essay named the question that follows. Above the nose, raising the voltage the load sees is benign. Past it, raising it makes the line deliver less, and the regulator keeps pushing. This essay marches a regulator against a line and finds how, when and how fast the correction becomes the collapse.

The nose itself has been measured on this subject from several directions, and each is a piece of what the regulator runs into. The load that takes the most found the resistive load that draws a line’s most power, which is the ratio at the peak here; the load that may be complex freed its reactance and found the conjugate; the far end that rises found a far end reading above its source; and the headroom that is the line’s own charge moved the nose with the line’s capacitance. A load that restores its power turns each of those static limits into a dynamic one, and the incremental resistance of $-V^2/P$ that makes it do so is a relative of the negative resistance the resistance that is below zero found in a follower — a device whose current rises as its voltage falls, placed where a positive resistance was expected.

The bus that could not hold its lower point #

The earlier result is the starting point and the contrast. There the constant-power behaviour was built into the load, and the instability lived in the bus’s own dynamics, with a time constant set by a capacitor and a resistance. Here the load is an ordinary resistance and the constant-power behaviour is manufactured by a control loop, so the instability lives in the regulator, and its timescale is the regulator’s.

What the ratio buys #

The circuit is as plain as the argument allows. A ten-volt source behind one ohm of line feeds a load resistance through an ideal transformer of ratio $n$ — or anything that multiplies voltage and divides current. Seen from the line, the load is $R_L/n^2$, and the voltage at the load is

$$V = \frac{n V_0 R_L}{n^2 R + R_L}.$$

That expression rises with $n$, reaches a maximum, and falls. The maximum is at $n = \sqrt{R_L/R}$, where the load reflected through the ratio equals the line’s resistance, and the voltage there is $V_0\sqrt{R_L/R}/2$: the ratio that gives the load the most voltage is the ratio that makes the line deliver its most power, $V_0^2/4R$, which is 25 watts here. Past that ratio, further raising makes the reflected load smaller than the line, the line delivers less power, and the load’s voltage falls.

The load resistance in each curve is chosen so that at the ten-volt setpoint it takes a stated fraction of those 25 watts. For a load that takes 90 per cent, the voltage peaks at 10.54 volts at a ratio of 2.108, and the setpoint is met at two ratios: 1.519, on the rising side, and 2.925, on the falling side. The figure checks that the two multiply to $R_L/R$ — they are the two roots of one quadratic — so they sit symmetrically about the peak on a logarithmic axis. For a load that takes exactly the nose power the peak is the setpoint, at a ratio of 2. For a load that takes 110 per cent the voltage never reaches ten volts at any ratio: the most it can be given is 9.535.

This curve is the whole argument drawn once. A regulator does not know the curve; it knows only whether the voltage is below the setpoint, and if it is, it raises the ratio. On the rising side that is the right thing to do. On the falling side it is exactly the wrong thing, and nothing in the regulator’s sensing distinguishes the two sides.

The regulator, marched #

The regulator integrates its error: its ratio changes at $k$ per volt-second of difference between the setpoint and the load’s voltage, $dn/dt = k(V_{set} - V)$. The march is a fourth-order rule on the ratio, with the load’s voltage solved from the ratio at every step, and it starts from unity ratio, where the voltage is sagging.

Asked for 90 per cent of the nose, the regulator does its job: the voltage rises and settles at the setpoint with the ratio at 1.5195, the smaller of the two roots, which the figure checks to a part in ten thousand. At 99 per cent it settles too, at 1.8182, still below the nose ratio of 2.0101 but much closer to it; the settling is slower, because near the peak the voltage responds to the ratio only weakly.

At 101 per cent the setpoint does not exist anywhere on the curve, and the march shows what that means in time. The voltage rises as the regulator raises the ratio, slows as it approaches the peak, and reaches 9.950 volts at 101.7 seconds — exactly the most the line allows this load, $V_0\sqrt{R_L/R}/2$, which the figure checks. The regulator, still seeing a voltage below its setpoint, keeps raising the ratio. The voltage turns over and falls, the error grows, the regulator raises the ratio faster, and the load’s voltage is below half its setpoint at 256.3 seconds. At 110 per cent the same sequence is shorter: a peak of 9.535 volts at 23.5 seconds and half the setpoint at 78.6.

Every check in the figure is about the regulator’s gain changing sign. At each load it checks numerically that the slope of the voltage against the ratio is positive a thousandth below $\sqrt{R_L/R}$ and negative a thousandth above it. The collapse is not a failure of the regulator’s tuning or a limit of its rate; it is that the plant the regulator controls reverses its gain at the nose, and an integrator with a reversed gain is an integrator running away.

What the collapse looks like from the control room #

It is worth reading the 101 per cent curve as an operator would see it, because nothing on it looks like the failure it is. For the first hundred seconds the voltage rises steadily from its sag towards ten volts, which is what a regulator is for. It slows as it approaches, which is what an integrating regulator nearing its setpoint does. At its best it reads 9.950 volts — half a per cent short of the setpoint, well inside the band most regulators are set to tolerate — and the ratio indicator shows a tap still moving in the direction that has been raising the voltage all along. Only after that does the voltage begin to fall, and for tens of seconds the fall is slower than the rise was.

Every quantity the regulator displays in the first two minutes is the display of a regulator working. The quantity that would have said otherwise is the slope of the voltage against the ratio, and it passes through zero at the moment the voltage is closest to the setpoint — so the best reading of the whole event and the loss of control happen together.

The setpoint is part of the demand #

The load in every run is a resistance, and a resistance does not ask for a power; it asks for whatever power the voltage across it gives. The regulator is what turns it into a demand, and the size of the demand is the setpoint squared over the resistance. So the fraction of the nose a regulated load asks for is set as much by the regulator’s setpoint as by the load.

The 110 per cent load makes the point with its own numbers. At a ten-volt setpoint it asks for a tenth more than the line can deliver and collapses. The most voltage the line can give it is 9.535 volts, and at that voltage the same resistance takes exactly the nose power: a regulator set to 9.53 volts would hold it, at the peak, with no margin, and one set to nine volts would hold it comfortably at 89 per cent of the nose. The load did not change and the line did not change. Lowering the setpoint by a twentieth moved the demand from past the nose to on it, because the demand goes as the square of the setpoint.

That is the arithmetic behind the practice of reducing a network’s voltage setpoints when it is heavily loaded, and it runs against the instinct that a sagging voltage should be pushed back up. The instinct is right below the nose, where the push is rewarded with voltage. Near the nose the push is rewarded with less voltage, and a lower setpoint is the one action that moves the equilibrium back onto the side where the regulator’s gain has the right sign.

A setpoint met on the wrong side #

Below the nose there is a second failure, and it does not need the load to ask for too much.

The load asks for 90 per cent of the nose, and the setpoint is met at two ratios. The larger, 2.925, is an equilibrium — the voltage there is exactly the setpoint and the regulator does not move — but it is on the falling side of the peak, where a small increase in the ratio lowers the voltage and the regulator responds by increasing the ratio further. Started one per cent below it, where the voltage is slightly above the setpoint, the regulator lowers the ratio, crosses the peak, and settles at 1.519. Started one per cent above it, where the voltage is slightly below, the regulator raises the ratio and the voltage falls below half its setpoint at 92.4 seconds. Started half as far again, at 23.9 seconds.

So the second root is an edge, exactly as the bus’s lower operating point was, and for the same structural reason: it is the equilibrium on the side of the maximum where the incremental response has the wrong sign. The difference is what puts a system there. A bus has its operating point set by its load; a regulator has its ratio set by its own history. A tap changer that was left at a high tap during a period of low demand, or that was driven up during a fault that depressed the voltage, can find itself above the second root when the load returns — and then a load comfortably inside the line’s capacity collapses the voltage anyway.

How long the collapse takes #

The time between the demand exceeding the nose and the voltage falling is the operator’s only window, and it has a law.

At a hundredth of a per cent past the nose the collapse takes 2,521 seconds; at a tenth of a per cent, 801.5; at one per cent, 256.3; at ten, 78.6. Close to the nose the time grows as the −0.497 power of the excess, and the figure checks the exponent at a half to within five hundredths.

The square root is not a fit, it is a mechanism. Just past the nose the regulator’s equation near the peak is $dn/dt \approx k\bigl(\Delta V + \tfrac12|V''|(n - n^\ast)^2\bigr)$, where $\Delta V$ is the small gap between the setpoint and the most the line allows. When the demand is at the nose that gap is zero and the regulator stops at the peak; just past it, the equilibrium has vanished but its ghost remains, and the ratio crawls through the region where the equilibrium used to be. The time to cross it is $\pi/\bigl(k\sqrt{\tfrac12|V''|\,\Delta V}\bigr)$, and since $\Delta V$ grows in proportion to the excess, the time grows as its inverse square root. The figure draws that closed form dashed and checks that at a hundredth of a per cent it agrees with the march to two per cent: 2,513 seconds against 2,521.

The practical reading is uncomfortable. A load that is just past the nose collapses slowly — for this regulator, forty minutes at a hundredth of a per cent — and the voltage during most of that time is close to its setpoint and rising towards its peak. Nothing in the voltage looks like an emergency until the regulator is well past the peak. The slowest collapses are the ones with the smallest margin to recover, and they give the most time to notice and the least evidence to notice it by.

A slower regulator stretches every time in proportion — 640.7 seconds instead of 256.3 at one per cent past the nose, with the rate cut from 0.05 to 0.02 — because the rate sets the time scale and nothing else in the equation has one. That is the argument for a slow regulator on a heavily loaded line, and the limit of it: slowing the regulator buys time to act and does not change whether the collapse happens.

What the measurement contains, and what it leaves out #

A resistive load behind an ideal ratio on a resistive line. That is the smallest system in which a regulator’s gain changes sign, and every number here is exact for it. A reactive line has the nose of the load that has two voltages or none at a different place and the peak of the voltage against the ratio with it, and a line with its own charge has the moved nose of the headroom that is the line’s own charge; the sign change happens at the nose in both, and the ratio at which it happens moves.

A continuous ratio. A tap changer moves in steps, with a deadband and a delay, and its march is a staircase rather than a curve. The steps do not remove the mechanism — each step past the peak lowers the voltage and provokes the next — and they make the bottleneck coarser, so the square-root law should hold only when many steps fit inside the bottleneck.

A load that restores its power instantly. Real loads restore over their own time constants — a thermostat’s cycle, a motor’s slip — and the constant-power behaviour exists only on timescales longer than those. When the regulator is faster than the load’s restoration, the regulator sees a resistance and there is no nose to push past; the collapse needs the regulator to be slower than the load. That ordering of time constants is the practical condition for this failure, and it is not modelled.

No limit on the ratio. Every real regulator has a highest tap. The march here raises the ratio without bound; a real one reaches its limit and stops, with the voltage wherever the curve is at that ratio — which past the nose is lower than it would have been had the regulator never acted.

Still open: the blocking rule, the reactive line, and the load that restores slowly #

A regulator that blocks. The standard protection against this failure is to block the regulator when the voltage fails to respond to a tap change in the expected direction. That rule is a measurement of the sign of the gain on the running system. Marching a regulator with such a rule — step, observe, block if the voltage fell — would say how much of the collapse time the rule needs before it can act, and whether a noisy voltage makes it block below the nose, where the gain is small but positive.

A reactive line with a leading load. The far end of a reactive line feeding a leading load can read above its source at the nose, and a regulator watching that voltage sees no sag to correct until it is far past the nose. Whether a regulator on such a line ever acts before collapse, or only afterwards, is the dynamic version of the far end that rises, and it combines the worst of both measurements.

A load with its own time constant. A load that restores its power over a time comparable with the regulator’s gives the system two slow states instead of one, and the bottleneck above becomes a question about which of the two reaches the ghost of the equilibrium first. The same march with a first-order restoring load would find the ratio of time constants below which the regulator cannot collapse the voltage at all.