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

Saturation — where it appears

The condition of a magnetic core whose flux density can rise no further, so that its inductance collapses and its winding current is limited only by resistance. The limit is an integral of voltage against time rather than a frequency, which is why a winding rated at one frequency fails at a lower one.

Named by 10 essays across 4 fields — each of them below, with the objects they name alongside it.

The voltage a winding may carry, which is a volt-second limit read at a frequency. computed by solving, not by drawing. The dots are bisections on a marched flux — the voltage integrated sample by sample until the peak excursion reaches 0.35 T — and the line is N·Ae·Bsat·2πf. They agree to 0.001% over three decades, and the fitted slope is 1.000000: exactly proportional, because flux is the integral of voltage and nothing else. The quantity that belongs to the core is the 3.500 mWb-turn, which has no frequency in it. A transformer "rated for 50 Hz" is a transformer whose volt-second product was divided by 2π × 50 once.

A boundary in volt-seconds

A core saturates on the integral of the voltage applied to it, not on the current through it and not at a frequency. The quantity that belongs to the core is N·Ae·Bsat — 3.500 mWb-turn here — and it has no frequency in it at all. Everything a data sheet says about a transformer's frequency rating is that one number divided by 2πf once: measured on a marched flux, the voltage a winding may carry is proportional to frequency to a fitted exponent of 1.000000.

magnetics · Saturation A 1% imbalance, and the 64 cycles it survives. computed by solving, not by drawing. The upper panel is the peak flux density, marched cycle by cycle, under a square drive whose positive half is 1% larger in area than its negative half. It does not settle. It walks, by the same area every cycle, and reaches 0.35 T after 64 cycles — 1280 ms at 50 Hz — against a closed form of 63.7. The lower panel is the count against the imbalance, and it rises without bound and never becomes infinite. Halving the drive gives 128 cycles, which is exactly twice: reducing the amplitude buys time and not safety, and there is no amplitude at which this design is inside a limit.

The flux that walks

The previous essay's saturation limit is an amplitude, and an amplitude can be respected. This one cannot. A drive whose two half-cycles differ in volt-seconds by one per cent adds the same small area to the flux every cycle, so it reaches saturation after 64 cycles — and halving the drive gives 128, and a tenth of it gives 637. Reducing the amplitude buys time in exact proportion and removes nothing. There is no amplitude at which the design is inside a limit.

magnetics · Saturation Two of the three are one mechanism at 19.0°; the third arrives at ninety. computed by solving, not by drawing. The cosine between each pair of error waveforms at 30.0 kHz, against how much output the ideal model is asked for. Bandwidth and slewing sit at 0.9454 — 19.0 degrees — and close only slowly, reaching 0.8813 at 16 V. Clipping does not exist below 12.00 V, where its error is 4.0e-9 per cent of the signal and its direction is the direction of rounding; above it the mechanism is real — 18.1 per cent at the top of the sweep — and its cosine against both of the others stays under 0.0025. The bandwidth error is 0.615 per cent at every amplitude here, unchanged to 5.3e-15, because a linear stage's fractional error has no amplitude in it.

Where the mechanisms are one mechanism

An amplifier is said to run out of three separate things — bandwidth, slew rate and rails — and errors from separate mechanisms add in quadrature while errors from one mechanism add as magnitudes. Measured as waveforms rather than as numbers, two of the three sit 18.4349 degrees apart, which is exactly the angle between a sinusoid and its own cube, and the third sits at ninety: its cosine against both of the others stays under 0.0025 wherever it exists. So the arithmetic is neither of the two anybody reaches for, and a budget built the right way is within 2.9 per cent where quadrature is 17 per cent low and a straight sum 37 per cent high.

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

The delay that is two delays

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

applied · Hysteresis One part, three saturation currents: 1.90 A, 2.40 A, 2.69 A. computed by solving, not by drawing. B(H) is μ₀H plus a saturating magnetisation, written as a flux linkage, and the inductance drawn here is dλ/di — the slope of that flux, which is what a small signal on a direct current actually meets. It is 18.92 µH at no current, 16.73 µH at two amps and 10.43 µH at three, and it never reaches zero: the vacuum is still there, so the part falls to its air-core 0.05 µH and stays. The three marks are the ten, twenty and thirty per cent drops different manufacturers print as the saturation current — 1.896, 2.395, 2.694 amps, a spread of 42 per cent on one part.

The inductance the current decides

The rung below bounded the flux and the boundary is exact: the volt-seconds decide the flux swing whatever the material does, and a cycle whose current ripple runs from 519 mA to 75 A has the same flux excursion to better than two per cent. What saturation breaks is the relationship between that flux and the current — so a ripple the design expression puts at 514 mA is 1,022 mA at 2.56 A of load, the peak reaches 3.57 A where the part is at a fifth of its nameplate inductance, and the same part has three saturation currents depending on which per cent it was quoted at.

magnetics · Saturation The winding current a hysteretic core asks for, and the current the same circuit's single-valued core asks for. computed by solving, not by drawing. One netlist — a 22.6 volt peak sinusoid at 5.00 kHz through 4 ohms into a sixty-turn winding — marched twice, once with the core as a superposition of play operators and once with the single-valued saturating curve every earlier figure in this field used. The two currents differ in shape and not only in size: the hysteretic one leads the flux by an angle that is not ninety degrees, which is the whole of the core loss, and its peak is 59.45 milliamperes against 50.12. The energy the source delivers over a complete cycle is 17.19 microjoules for the loop and 2.5e-5 for the curve, which is zero to the resolution of the march. The slider takes the flux the drive demands from a twentieth of the material's saturation to most of it.

The core the solver has to remember

A saturating inductor's state is one number, because the current is a function of the flux linkage. A hysteretic core's is not: half an amp is one flux on the way up and a different flux on the way down, and there is no function of the current that returns the flux. So the march's state vector gains twenty-four more numbers, one per play operator, and the Newton loop is forbidden to touch them. The circuit and the bench then agree on the loss to three parts in ten million — one integral taken at two wires, the other inside the material.

magnetics · Magnetic loss 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 One output in saturation takes 6.5% off another that is at five volts. computed by solving, not by drawing. A three-transistor mirror: a reference, an output taken down into saturation, and a third output held at five volts throughout. The upper curve is the saturating output's own loss and the lower one is the sibling's. At 50 mV the saturating output is 27.3 per cent down and the sibling, which is nowhere near saturation, is 6.54 per cent down. The reference current moves by -0.0188 per cent, which is nothing: a base current is a hundred and fiftieth of a collector current and the reference is set by a resistor from the supply. What does move is the base-emitter voltage every output shares — by -1.75 millivolts — and every output is an exponential of it.

The refusal, and what it was protecting

A current mirror's model has declined to answer below two hundred millivolts of collector-emitter voltage since it was written, because it has no base-collector junction and would return a forward-active current for a saturated transistor. Put the junction in and the refusal turns out to have been placed where the model it protects is still right to seven parts in ten thousand — and the mirror's real failure is somewhere else entirely: a saturated output takes six and a half per cent off an output that is sitting at five volts.

semiconductors · Current mirror The temperature a part cannot come back from, and how long it takes to leave. computed by solving, not by drawing. The same fixed-point equation as the rung below, marched in time with a thermal capacitance rather than solved for its steady states: C dT/dt = P(T) − (T − T_a)/R_th, stepped adaptively on the temperature change because dT/dt goes through zero at each fixed point. Every trajectory starting below 191.1 °C returns to 88.8, however far above the operating point it began; every one starting above it leaves the material's range entirely, the closest in 0.2 minutes. The two nearest starts are 3.0 kelvin apart. The ignition temperature is a boundary in the STARTING CONDITION, and no steady-state analysis contains one.

The boundary that is a starting point

A wound part with a stable operating point at 88.8 degrees and an ignition temperature at 191.1 will never reach the second, because nothing takes it there. Marched in time rather than solved for its steady states, the same equation says what does: a trajectory starting at 189.6 degrees settles back and one starting at 192.6 leaves the material's range in twelve seconds — two starts three kelvin apart. And an overload of four times the normal loss is survivable for ever, while seven times is survivable for seventeen minutes.

magnetics · Thermal feedback The same imbalance at five winding resistances, and the fixed point three of them reach. computed by solving, not by drawing. The peak flux density in each cycle, against the cycle, under a square drive whose positive half carries 1 per cent more volt-seconds than its negative one, for winding resistances of 0, 0.05, 0.15, 0.4, 1.5 ohms. With none the flux walks to 0.35 T in 15 cycles, which is the boundary this ladder's second rung measured. With 1.5 Ω it settles at an offset of 5.67 millitesla and stays there, because the offset draws a direct magnetising current and that current's drop across the winding opposes the imbalance. The resistance at which the two outcomes change places is 0.1014 ohms, bisected on whether saturation is reached at all. Every curve carries the identical drive; the resistance is the only difference between them.

The walk that stops

A drive whose two half-cycles differ in volt-seconds walks the flux to saturation in a count of cycles, and the rung that measured it concluded that no amplitude puts the design inside a limit. That model has a stiff source and a winding of no resistance. With resistance in the loop the walk has a fixed point, held by an identity the material is not in — the mean magnetising current is the drive's direct component divided by the resistance, to seven parts in 10¹³ — and the boundary becomes a resistance rather than a time: 0.1014 ohms bisected, at a one per cent imbalance and half the volt-second limit.

magnetics · Saturation

Named alongside it

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

MarchingModel rangeVolt-secondsCoreDesign tradeoffFixed pointFlux densityMagnetic lossB h loopFlux linkageMagnetic pathNewton raphson

All concepts