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

Reservoir capacitor — where it appears

The capacitor a rectifier charges in pulses and a load discharges continuously, whose size sets the ripple and whose first charge sets the inrush. Its first charge carries twenty-six times the repetitive peak, and the instant of switching that decides the factor belongs to whoever flicked the switch.

Named by 9 essays across 2 fields — each of them below, with the objects they name alongside it.

1000 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 15.69 V with 1.331 V of ripple, against the 1.569 V the expression I/2fC gives — 15.1% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 28.8° of each half cycle and carries 2.10 A at the peak, which is 13.4 times the 157 mA the load draws.

The direct voltage that is a sawtooth

A rectifier and a reservoir capacitor make what everybody calls a direct voltage. Marched with the diodes in the netlist, a thousand microfarads across a hundred ohms gives 15.69 volts with 1.33 volts of ripple on it, against the 1.57 the textbook expression predicts. The expression is high by the fraction of the cycle the diode conducts for — measured at 0.94 to 0.96 of it across two sweeps — and it has no opinion at all about the quantity that actually sizes the transformer, which is a peak diode current 13.4 times the current the load draws.

applied · Unregulated supply The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance.

The first cycle, which no steady state contains

Every number this field computes about a rectifier — the ripple, the crest factor, the conduction angle, the power factor — is read from the settled state, and the march that produces them starts from an empty capacitor and throws the first cycle away. That first conduction carries 32.4 amperes against a repetitive peak of 1.23, it is 26 times larger than anything the circuit ever does again, and how large it is depends on when somebody's hand closed the switch.

power · Inrush The inductance divides the current by 7 and leaves the capacitor 29% above the mains peak. computed by solving, not by drawing. The first conduction of a rectifier whose transformer has a leakage inductance as well as a winding resistance, marched from an empty capacitor at the worst instant of the mains. The peak falls from 36.4 A at 20 µH to 5.3 A at 5000, and — unlike the winding resistance the rung below measured, which limits the current and leaves ∫i²dt exactly where it was — the inductance takes the energy down with it, from 0.370 to 0.073 A²s. What it costs is the second curve: the inductor's current cannot stop at the instant the two voltages are equal, so the capacitor overshoots to 21.86 V at 1000 µH — 28.6% above the 17 V peak of its own supply — and the diodes will not let the charge back out. The overshoot has an interior maximum, because past it the mains reverses before the ring has finished.

The inductance that limits, and lifts

Adding winding resistance to a rectifier limits the first peak and does not reduce the energy at all — the essay below measured ∫i²dt as two per cent apart over a factor of four in the resistance. Adding leakage inductance does both: it divides the peak by seven and the energy by five, and dissipates nothing to do it. What it buys instead is a rectifier whose output sits 29 per cent above the peak of its own supply, permanently, which every steady-state expression in this field says cannot happen.

power · Inrush 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 The crest factor is 13.4 and the winding is sized by 3.01. computed by solving, not by drawing. Three ratios of the same settled march, against the reservoir. The crest factor — the peak diode current over the load's direct current — runs 6.43 to 24.25. The form factor, which is the root-mean-square current over the same direct current and is what a winding heats by, runs 2.114 to 4.077. Its square is the copper loss against a winding carrying the direct current alone, and that runs 4.47 to 16.62. The first ratio is 3.04 times the second at 220 µF and 5.95 times at 4700, so quoting one of them tells a reader nothing about the other.

The current that sizes the transformer

A reservoir's crest factor is 13.374 at a thousand microfarads and the winding is not sized by it. The root-mean-square of the same marched current is 3.0069 times the load's direct current, so the copper dissipates 9.0417 times what it would carrying the direct current alone — and the two ratios diverge, from 3.04 apart at 220 microfarads to 5.95 apart at 4700. The expression for the mean output is wrong in three places whose signs differ, and at 313.9 microfarads they cancel to six microvolts while the ripple expression inside it is still 32.3 per cent high.

applied · Unregulated supply The ripple 1000 µF leaves, through the regulator: 60.62 mV rather than 21.11 mV. One settled cycle of the reservoir's output — 1.331 V peak to peak across 1000 µF — split into 1000 Fourier lines, each passed through the solved regulator's rail-to-output response, and summed. The output is 60.62 mV peak to peak. The ripple times the rejection at 100 Hz, -36.00 dB, gives 21.11 mV, which is 2.872 times too little: the rejection worsens at twenty decibels a decade, so each harmonic of the sawtooth arrives at nearly the size of the first. The 100 Hz line carries 23.4% of the output's mean square and the lines above 1 kHz 10.4%; the first three arrive at 7.731 mV at 100 Hz, 7.200 mV at 200 Hz, 6.388 mV at 300 Hz. The dashed curve is the 100 Hz line alone.

The ripple that arrives as a comb

The rejection essay multiplied two numbers: 1.331 volts of reservoir ripple and the regulator's 36.0 decibels of rejection at a hundred hertz, for 21 millivolts at the output. The ripple is a sawtooth, a comb of lines at every multiple of a hundred hertz, and the rejection worsens at twenty decibels a decade — so the second line arrives at nearly the size of the first, and the third too. Summed with their phases, the output is 60.62 millivolts, 2.87 times the estimate, and the hundred-hertz line carries only 23.4 per cent of it. A reservoir twenty-one times larger cuts the rail's ripple 14.8 times and the regulated ripple 5.95.

applied · Regulator The series resistance that makes the ripple smaller. computed by solving, not by drawing, marched with the diodes in the netlist. A 1000 µF reservoir with 30 mΩ of its own series resistance. The resistance adds a step of ESR times the diode's peak current to the output and at the same time limits that peak current, and the two nearly cancel: the ripple has an interior minimum of 1.330 V at 17.0 mΩ, BELOW the 1.331 V a perfect capacitor gives, and rises to 1.711 V at an ohm. What the resistance buys monotonically is the peak current: the crest factor falls from 13.37 to 6.550 at an ohm, which more than halves the current that sizes the transformer, for 380 mV of mean output and a root-mean-square diode current that falls from 0.4717 A to 0.3484. The textbook ripple expression says 1.568 V here and moves by 2.4% across the whole axis, because it has no term for a series resistance at all.

The resistance that lowers the ripple

Two earlier essays here marched a reservoir with a perfect capacitor. A real one has tens of milliohms of its own, and the obvious expectation — that the resistive step it adds makes the ripple worse — is wrong in an interesting direction: the resistance also limits the charging current, and the ripple has an interior minimum of 1.3303 V at 17.0 mΩ, below the 1.3312 V a perfect capacitor gives. What the resistance buys monotonically is the peak current, which falls from 13.37 times the load's to 6.55 at an ohm, for 380 mV of mean output.

applied · Unregulated supply The same 100 mΩ in the winding instead of the capacitor: 1.317 V of ripple against 1.335 V, at the same crest factor of 11.15. computed by solving, not by drawing, marched with the diodes in the netlist: a 1000 µF reservoir behind a centre-tapped rectifier, with a series resistance from 1 mΩ to 1 Ω placed either in the capacitor or in each half-winding. The crest factor is the same in both places to two parts in a thousand at every resistance — 13.34, 13.26, 13.03, 12.49, 11.15, 9.069, 6.548 — because both limit the charging current alike. The ripple is not: in the capacitor it has a minimum and rises to 1.711 V at an ohm; in the winding it falls throughout, to 1.172 V, against 1.331 V with no resistance. At an ohm the winding costs 534 mV of mean output and the capacitor 380 mV. The diode's root-mean-square current at an ohm is 0.3449 A with the resistance in the winding and 0.3484 A with it in the capacitor.

The resistance that belongs in the winding

A reservoir capacitor's series resistance lowers the ripple to a minimum of 1.3303 V at 17 mΩ and raises it past that. The same resistance moved into the transformer's winding limits the peak current by the same amount — the crest factor agrees to two parts in a thousand at every value from a milliohm to an ohm — and the minimum is gone: the ripple falls throughout, to 1.172 V at an ohm against 1.711 V in the capacitor. The two resistances each carry a current the other does not, and that one asymmetry decides where a deliberate one should go.

applied · Unregulated supply Two 1000 µF parts of 100 mΩ each against one 2000 µF part of 100 mΩ: 709 mV against 728 mV of ripple, and a crest factor of 14.41 against 12.61. computed by solving, not by drawing, marched with each capacitor behind its own series resistance. Two identical 500 µF, 60 mΩ parts give 1.330303 V of ripple and one 1000 µF, 30 mΩ part 1.330303 V — the same network. Two 1000 µF parts of a stated resistance each, against one 2000 µF part of the same resistance: at 10 mΩ, 704 mV against 704 mV of ripple and a crest factor of 17.40 against 16.93; at 30 mΩ, 704 mV against 705 mV of ripple and a crest factor of 16.51 against 15.48; at 100 mΩ, 709 mV against 728 mV of ripple and a crest factor of 14.41 against 12.61; at 300 mΩ, 758 mV against 867 mV of ripple and a crest factor of 11.45 against 9.490; at 1000 mΩ, 1.013 V against 1.334 V of ripple and a crest factor of 8.170 against 6.620. The pair's ripple is lower only where one part's resistance is past the ripple's minimum, and its peak current is higher at every resistance.

Two capacitors that are one

Two identical reservoir capacitors in parallel are not a new circuit to be marched: 500 µF at 60 mΩ twice gives 1.330303 V of ripple and so does 1000 µF at 30 mΩ once, to the sixth decimal. So a pair against one part of the same capacitance is one curve read at two resistances, and the pair always sits at the lower one — which lowers the ripple only past the curve's minimum, 709 mV against 728 at 100 mΩ a part, and raises the peak current at every value, a crest factor of 14.41 against 12.61. Mismatch the pair and the current still divides by capacitance, so a part with twice the resistance carries 2% less current and 1.93 times the heat.

applied · Unregulated supply

Named alongside it

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

Crest factorConduction angleDesign tradeoffRectificationEquivalent series resistanceInrush currentModel rangeLoop gainMeasurement conditionRectifierTransformerCompanion model

All concepts