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

Charging exponential — where it appears

The approach of a capacitor's voltage to its final value through a resistance, complete to a stated fraction after a count of time constants. The energy the resistance takes is half the energy stored whatever its value, so the resistance decides the time and not the heat.

Named by 8 essays across 3 fields — each of them below, with the objects they name alongside it.

What a charge through 1 kΩ costs, against how long it is given. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. A step loses 1.00000 of ½CV² — 12.5 µJ here — and that number is the same through 10 Ω and through 100 kΩ, which is what the faint curves show. What the resistance decides is where the fall starts: the loss goes as 2τ/T once the ramp is long against τ = 1 ms, with a fitted exponent of -0.994 and no floor beneath it. The circles are marched; the line through them is a closed form that never sees a netlist.

The half that never arrives

Charging a capacitor from a step loses exactly as much energy as it stores, and the resistance it is lost in does not appear in the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms, to nine figures. Drive the same network with a ramp instead and the loss falls as two time constants over the ramp, with no floor beneath it at all.

transients · Switching energy A 100 ms pulse through a 0.159 Hz corner, 9.52% shorter by the end of it. computed by solving, not by drawing. Marched. The dashed line is the pulse that was sent. The solid line is what a 1 MΩ input with 1 µF in front of it receives: the top decays as exp(−t/RC) for the whole 100 ms, ending 9.515% down, and the trailing edge undershoots by exactly the same amount. The closed form for the same two components gives 9.516%. The input's specification is a corner at 0.159 Hz; a top flat to one per cent needs a pulse shorter than 10.1 ms, which is a rate 625.2 times the corner — a constant with no component in it, and the reason a low-frequency specification says nothing useful about an edge.

The corner that says nothing about an edge

An AC-coupled input is sold on a low-frequency corner, and a corner is a statement about steady sinusoids. What an instrument is usually shown is a pulse, and for a pulse the number is a sag: a hundred-millisecond pulse through a 0.159 hertz corner comes out 9.5 per cent shorter than it went in. A one per cent flat top needs a pulse rate 625 times the corner, which is a constant with no component in it — and above fifty per cent duty an AC-coupled pulse never reaches half its own height at all.

instruments · AC coupling A comparator's delay, at a divider ratio of 0.50. Each nanosecond of comparator delay adds 2.650 nanoseconds to the period — measured as a slope between two delays, both exact multiples of the marching step — against the 2.667 that 4/(1+β) gives and the 2 that counting the delay twice gives. A 50 ns comparator therefore holds the frequency to one per cent only below 75.0 kHz.

The period a delay lengthens

Feed a comparator's output back through a resistor to the capacitor on its own input and the two thresholds stop defending a decision and start setting a period. The closed form is two RC times the log of one plus beta over one minus beta, and a marched circuit recovers it as the step shortens. A comparator that responds fifty nanoseconds late does not add fifty nanoseconds to each half cycle: it adds 4/(1+beta) times the delay to the period, 2.65 here against the 2 that counting it twice gives, because during the delay the capacitor keeps going the way it was going.

applied · Hysteresis A staircase costs one Nth, computed rather than quoted. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. The charge is broken into N equal risers, each held for 16 time constants so that it completes. The measured losses are 1.00000, 0.500001, 0.250000, 0.125000, 0.0625000, 0.0312500 of ½CV² — which is 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 times 1/N, so the law is exact to four parts in a million at the worst rather than approximately true. The fitted exponent is -1.00000 and the energy account closes to 2.17e-6 at the worst.

The half a switch keeps

Charging a capacitor from a step loses half the delivered energy whatever the resistance, and a ramp takes the loss down as 2τ/T with no floor. A staircase of N settled risers costs one Nth of the step, exact to four parts in a million, and the law ends at a dwell of 5.272 time constants. A switch is the other half of the same product and buys nothing at all: with the supply held at five volts and the channel conductance ramped over a thousand time constants, the loss is 1.00000000 of ½CV².

transients · Switching energy A second coupling stage with the same time constant: the top sags to 0.303 instead of 0.607, and the tail crosses zero at 2.27 s and overshoots to +0.025. computed by solving, not by drawing, marched and checked against the closed form to 3.0e-4: a half-second pulse through a 1 µF, 1 MΩ coupling stage (dashed), and through that stage followed by a buffered second one with a time constant of 1 s (solid). One stage sags to 0.6065 by the pulse's end and its undershoot decays monotonically to zero. Two stages sag to 0.3033, undershoot to −0.6966, cross zero 2.271 s after the pulse began and overshoot positive — the second capacitor removes the first moment of the waveform as well as its area, so the tail has to change sign.

The second capacitor, and the tail that changes sign

One coupling stage turns a pulse into a sagging top and an undershoot that returns to zero from below. Add a second stage with the same one-second time constant and a half-second pulse sags to 0.303 of itself instead of 0.607, undershoots to −0.697, crosses zero 2.27 s after it began and overshoots positive before settling. The step response becomes (1 − t/τ)·exp(−t/τ), through zero at exactly one time constant and down to −e⁻² at two. The reason is arithmetic: one capacitor removes the waveform's area, two remove its area and its first moment, and a pulse with a one-signed tail cannot have both. Built to the same stated corner, two stages still sag 1.29 times as fast, and a 1% flat top needs a pulse rate 806 times the corner rather than 625.

instruments · AC coupling Without the buffer, a second stage of 1 times the first's impedance: the top sags to 0.175 instead of 0.303, and the tail crosses at 2.00 s and overshoots to +0.010. computed by solving, not by drawing, marched and checked against the closed form to 4.5e-4: a half-second pulse through two one-second coupling stages, buffered (dashed) and with the second hung directly on the first (solid), its resistor 1 times the first's and its capacitor 1/1 of it, so that each stage alone keeps its one-second time constant. Buffered, the top sags to 0.3033 and the tail crosses zero 2.27 s after the pulse began. Unbuffered, the denominator gains the cross term R1·C2 and its poles move to −0.382 and −2.618 per second; the top sags to 0.1751, undershoots to −0.8245, crosses zero at 2.002 s and overshoots to +0.0103. The slider is the second stage's impedance relative to the first's.

Two coupling stages without a buffer

Hang a second coupling stage directly on the first, with equal parts, and the pair's denominator gains a third time constant — the first resistor carrying the second capacitor's current. The poles spread from a double pole to −0.382 and −2.618 per time constant, a half-second pulse's top sags to 0.175 instead of the buffered pair's 0.303, and the sag rate is exactly 2 + 1/k per time constant for a second stage k times the first's impedance — the sum of the two short-circuit time constants. The tail still crosses zero, because the two zeros at the origin are untouched, and a step's lowest point is still at exactly twice its zero crossing. Then a reversal: specified by the overall corner rather than by the parts, the loaded pair sags only 1.12 times as fast as one stage and undershoots 7.6% rather than 13.5% — the buffer the cascade essay assumed makes the specified input worse.

instruments · AC coupling A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV².

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

transients · Switching energy A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it.

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

transients · Switching energy

Named alongside it

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

Energy-storageReal powerAC couplingBaseline shiftCorner frequencyModel rangeSwitching lossClosed formDuty cycleOn-resistanceResiduesStep response

All concepts