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

Propagation delay — where it appears

The interval between a cause and the response to it, which for a comparator is the time from the threshold crossing to the output moving. It is what lengthens a relaxation oscillator's period beyond the resistor and capacitor that were supposed to set it.

Named by 15 essays across 5 fields — each of them below, with the objects they name alongside it.

A 1 V step onto 1.00 m of 50 Ω line into an open circuit. computed by solving, not by drawing as a sum of 81 arrivals. The source drives 0.8333 V into the line immediately — set by 10 Ω against the line's 50 Ω, and not by the load, which it cannot yet know about. One delay of 4.83 ns later the far end reaches 1.6666 V. The staircase settles at 0.999990 V, which is what the resistive divider gives.

The staircase in time

A source driving a metre of cable does not know what is on the far end of it for 4.83 nanoseconds. What it drives into during that time is decided by the cable's characteristic impedance and nothing else — and when the far end finally answers, the answer comes back as a staircase whose limit is the resistive divider the circuit was going to be all along.

lines · Reflections A series-terminated net holds half a swing for 1.0 delays at 50% along it. computed by solving, not by drawing. What a receiver 50% of the way along one net sees under three terminations, from a lattice evaluated at that point rather than at the ends. The shaded strip is the interval in which a logic input has no defined answer — between 30% and 70% of the swing. The series-terminated net sits in it for 4.83 ns, which is 2(1 − x) delays exactly and is zero only at the far end; the unterminated net sits in it for 4.83 ns and then overshoots by 82%; the parallel-terminated net never does, and draws 60 mA down the line for as long as the level is held.

The resistor at the wrong end

A lattice diagram is read at the two ends of a line, and that is where the two respectable terminations look identical: a clean step, one delay late, at the receiver. Anyone standing halfway along a series-terminated net sees half the swing held for a full round trip, which for a logic input is not a level at all. The interval is 2(1−x) delays exactly, it is zero only at the far end, and the scheme that never has it draws sixty milliamperes for as long as the level is held.

lines · Termination A stub holds the far end at two thirds for twice its own delay. computed by solving, not by drawing. A series-terminated net with a branch on it, marched as waves on a delay grid. Three lines of equal impedance meet at the junction, so each presents the others with Z₀/2 and a wave arriving is reflected by exactly −1/3 with two thirds going on. The far end therefore receives 66.7% of the swing at one line delay instead of all of it, and is held there for 0.400 line delays — twice the stub's own delay of 0.41 ns, being the round trip to its open end and back. The same net without the branch is drawn beside it and settles in one round trip, which is what a series termination is for. Each further round trip of the stub divides what is left of the error by three and turns it over, because the returning wave doubles at the open far end — so the receiver approaches its level alternately from below and from above.

The receiver that is a branch

A lattice diagram has two ends, and an interior receiver is not a point on a net — it is a short piece of track leading off it to a pin, open at the far end. Three lines of equal impedance meeting at a junction present each other with half the impedance, so a wave arriving is reflected by exactly minus a third and two thirds goes on: the far end receives two thirds of the swing and sits there for twice the stub's own delay, whatever the net is terminated with and wherever on it the branch is.

lines · Termination 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 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 Terminated at both ends: no interval anywhere, at 1.65 V of 3.3. computed by solving, not by drawing. The same net as the three-way comparison, with a fourth trace: a series resistor at the driver AND a parallel one at the receiver. Both reflection coefficients are zero, so the wave that arrives at a receiver 50% of the way along is already the final value and there is no second arrival to wait for — the departure after the first edge is 0.0e+0 per cent, which is the arithmetic's floor. What it costs is the level: 1.650 V of 3.3, exactly half, because two equal resistances divide the supply and nothing reflects to double it back. The series scheme in the same place is undefined for 1.00 delays and the unterminated one overshoots by 82 per cent. The hold current is 33.0 mA against the parallel scheme's 60, because the path to ground now has two resistances in it.

Terminated at both ends

A series resistor at the driver and a parallel one at the receiver cost exactly half the swing — 1.650 volts of 3.300 — and no reflection ever gives it back, because there is no reflection. What the half buys is measured rather than assumed: a driver thirty per cent off its assumed impedance rings a series-terminated net by 16.3 per cent and a doubly terminated one by nothing at all, and an interior receiver on a series-terminated net sits in the undefined band for every far-end resistance above 125 ohms, which is Z₀/(1−2b) and contains no length, no driver and no frequency.

lines · Termination At 44.7 Ω the via's area is zero; its reflection is a doublet falling as the 1.99 power of the edge, not the first. computed by solving, not by drawing, as a cascade. A via of 0.25 pF, 1 nH and 0.25 pF on a line of √(L/C) = 44.72 Ω, met by an edge of 59 ps, against the same capacitance alone on the same line. The via's reflection is a doublet — a dip and a bump of equal area — whose largest excursion is 24.4 mV against 166 mV for the capacitance alone. Over edges from 5.9 ps to 295 ps the capacitance's reflection falls as the −0.97 power of the edge and the via's as the −1.99 power: 24.4 mV at 59 ps, 994 µV at 295 ps. The via delays the edge going past it by 22.4 ps, against √(LC) = 22.4 ps.

The via that is a piece of line

A via of half a picofarad of pad and a nanohenry of barrel puts no area under its reflection from a line of √(L/C) = 44.72 ohms, from any edge. It has not vanished. It delays the edge going past it by 22.4 picoseconds, which is √(LC) exactly, and it still reflects: a doublet whose largest excursion falls as the −1.99 power of the edge where a lone capacitance's falls as the −0.97 — 24.4 millivolts for a 59-picosecond edge against 166 for the pads alone, and 994 microvolts for a 295-picosecond one against 35. The balanced via is a short piece of line, and the reflection it leaves is the reflection of its length rather than of its size.

lines · Reflections 20 sections imitating a 1 m line: its delay is 1% long at 318 MHz and its group delay at 185 MHz, and it passes nothing above 1.32 GHz. computed by solving, not by drawing, as a chain of 20 series inductors and shunt capacitors carrying the inductance and capacitance of a metre of 50 Ω line of delay 4.83 ns, terminated in 50 Ω at both ends, beside the Bloch phase of an endless chain, 2·arcsin of ω over the cutoff, a section. The chain's cutoff is the cutoff 2/√(LₛCₛ), 1.32 GHz. Its phase delay is too long by arcsin(x)/x − 1 with x = f over that cutoff, 1% at 318 MHz (x = 0.2417); its group delay by 1/√(1 − x²) − 1, 1% at 185 MHz (x = 0.1404). Well below cutoff the solved chain's delay follows the closed form; nearer it the fifty-ohm terminations, which are not the LC ladder's own impedance there, add a ripple. Ten sections per wavelength is 414 MHz for this chain.

The sections a wavelength needs

A ladder of inductors and capacitors is a line only below its own cutoff, 2/√(LC) of one section, and in the frequency domain how far below can be written down exactly: its delay is too long by arcsin(x)/x − 1 and its group delay by 1/√(1 − x²) − 1, where x is π over the number of sections per wavelength. One per cent of delay needs 13.0 sections per wavelength; one per cent of group delay, 22.4. The rule of ten per wavelength is 1.72 per cent slow in phase and 5.33 in group delay. Twenty sections imitating a metre of cable are a line to a per cent of group delay up to 185 megahertz and pass nothing at all above 1.32 gigahertz.

lines · Distributed vs lumped A line cut into 8 LC sections passes 10% at 0.863 and 50% at 1.043 of its time of flight. computed by solving, not by drawing, marched at 8000 steps. A lossless line of 1 µH and 100 pF — a 100 Ω line with a 10 ns time of flight — built from 8 equal LC sections and driven through 100 Ω into 100 Ω, so the ideal line would deliver a clean half-volt step at exactly one time of flight (dashed). The ladder network passes 10% of its final value at 0.8630, 50% at 1.0428 and 90% at 1.1592 of the time of flight: a delay quoted at 10% is 18.0% of a flight time shorter than one quoted at 50%. It starts as the 16th power of time, for 16 poles and no zeros, and rings after the edge because a lumped ladder network cannot pass what lies above its sections' cut-off.

The threshold a delay is quoted at

A step through a network of r poles starts as the r-th power of time, so a delay measured at 10% moves with the network's order more than one measured at 50%. Cut a line into more sections and the order grows without limit, and what happens to the two delays depends on the line. A lossless LC line converges on its time of flight at every threshold: the spread between its 10% and 50% delays falls from 0.656 of a flight time with one section to 0.070 with thirty-two, as the −0.68 power of the section count, the −2/3 of an Airy front. An RC line converges too, but its 10% and 50% delays converge to 0.134 and 0.391 of R·C, a third of each other for good. On a propagating line a delay belongs to the path; on a diffusing one, two-thirds of it belongs to the threshold.

transients · Value theorems A parasitic of 0.01 ms changes the start only inside itself: t³ until 30.0 µs with a pole, t until 20.0 µs with a zero. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a straight line of slope two (dashed); with one more pole of 0.01 ms (solid), slope three until the asymptotes cross at (r + 1)τp = 30.0 µs; and with a zero of the same time constant (dotted), slope one until r·τz = 20.0 µs. Well inside the parasitic's time constant the pole multiplies the bare step by t/((r + 1)τp) and the zero by r·τz/t, both checked on the expansions; outside it both curves rejoin the bare one, displaced. A parasitic a hundred times above the network's bandwidth decides the start's power law for its own few time constants and no longer.

The parasitic that moves the start

A pole far above a network's bandwidth raises its relative degree by one, so the initial-value theorem says its step now starts as tʳ⁺¹ instead of tʳ; a zero lowers it. Both are true and both expire. Added to a two-section step, a 10 µs parasitic pole holds the start at the cube of time until the asymptotes cross at (r + 1)τp = 30 µs and a 10 µs zero holds it at the first power until r·τz = 20 µs; thirty time constants later the local power is back within 0.07 of the bare step's. After that each does one thing only: it moves every threshold by its own time constant, the pole later and the zero earlier. Since the time is the same at every threshold, the fraction it costs is not — 9.1% of a 1% delay, 0.82% of a 50% one — and a pole and a zero together keep the start's power, scale it by τz/τp, and move every later threshold by τp − τz.

transients · Value theorems A 10 µs delay in the loop: the peak rises to 13.53 dB and the sensitivity above crossover becomes a comb, and the two areas are still equal — 7.14 kHz·Np each side. computed by solving, not by drawing, from the loop's closed form, held against the netlist's cut loop to a part in a billion. The three-pole amplifier of the area essay (second pole 4 kHz) with a pure delay of 10 µs in its loop. Its margin falls from 34.9° to 14.3° at an unchanged crossover of 5.73 kHz, and the peak of |1/(1 + T)| rises from +5.70 to +13.53 dB. Above crossover the delay keeps turning the loop's phase, so the sensitivity rises and falls about zero with a period of 100 kHz. The area below zero is 7.144 kHz·Np, the area above 7.144, net 9.00 × 10⁻⁵ Hz·Np — integrated to 50 MHz on a grid that takes sixty steps a tooth.

The phase a delay never stops adding

Put a pure delay in a feedback loop and Bode's sensitivity integral does not notice: with 10 µs in the three-pole amplifier's loop the logarithm of the sensitivity still has equal areas above and below zero, 7.14 kHz·Np each side. What changes is where they are. The delay keeps turning the loop's phase above crossover, so the sensitivity becomes a comb with a tooth every 100 kHz; the desensitisation shrinks from 6.74 to 6.48 kHz·Np, the main peak rises from 5.70 dB to 13.53 and overpays it by 7%, and the comb hands the excess back below zero. The margin reads the peak from below by about 1.4 dB at every delay. Past 16.9 µs the loop is unstable and the integral stops balancing — by exactly the growth rate of the pole pair that has crossed.

feedback · Desensitivity Near-end crosstalk from a 200 ps edge, 100 mm of coupled track against 10 mm. computed by solving, not by drawing, the pair solved mode by mode and returned to time from its harmonics. Over 100 mm the near end rises with the edge, holds a plateau of 12.51 mV per volt for twice the line's delay, 1.33 ns, and falls. Over 10 mm the round trip is shorter than the edge and the pulse never reaches the plateau: 8.34 mV. The plateau is Kb, a quarter of the sum of the two couplings, on the 0.5 V the driven line carries: 12.50 mV.

The near end that stops growing

Near-end crosstalk is a length only until the coupled run's round trip equals the edge's rise time; past that it is a constant. Solved mode by mode and returned to time, a 200 ps edge on 100 mm of coupled track gives a plateau of 12.51 mV per volt — Kb, a quarter of the sum of the two couplings, on the driven line's half-volt — held for the 1.33 ns round trip. The knee is at v·tr/2: 7 mm for a 100 ps edge, 75 mm for 1 ns. Past it a longer run adds width and not height, and the pulse's area keeps growing as Kb times the round trip.

lines · Crosstalk A chebyshev low-pass of order 5: where its impulse response balances, and where its step crosses half. computed by solving, not by drawing, the step built from the network's poles and the impulse response its derivative. The impulse response (dark, scaled to its peak) balances at 0.7090 ms; the group delay at a thousandth of the corner is 0.7090 ms — the same number by a different route. The step (pale) crosses half its final value at 0.7238 ms, 2.1% after it.

The delay that is a centre of mass

A filter's group delay at zero frequency is not an approximation to when its step arrives; it is exactly where its impulse response balances. Across three families and orders one to seven the two agree to 5.7 parts in a million by routes that share only the netlist. The step's half-way point is a different thing: ln 2 of the delay for one pole, approaching it from below for a Bessel filter, and passing it — by up to 16.9 per cent — for Butterworth and Chebyshev filters whose impulse responses swing negative. Elmore's RC-tree delay is the same centre of mass, and a uniform ladder network crosses half at 0.757 of it.

filters · Group delay Both ends of the quiet line, 100 mm of microstrip-like coupling, edge 200 ps. computed by solving, not by drawing. The near end (pale) is the saturated plateau, 16.28 mV per volt. The far end (dark) is a single pulse arriving with the edge after one delay, 0.676 ns, as wide as the edge and 25.16 mV per volt deep — negative, because the inductive coupling exceeds the capacitive. Its depth is the coupling difference, the delay and the edge's slope multiplied: 24.96 predicted.

The far end that is a derivative

Where the two couplings are unequal the far end does not cancel, and in time it is the derivative of the edge: a single pulse as wide as the rise time, arriving after one delay, its height the coupling difference times the delay times the edge's slope. On 100 mm of microstrip-like coupling a 200 ps edge gives −25.16 mV per volt against 24.96 predicted, and the height falls as one over the rise time with a fitted slope of −1.000 while the saturated near end sits still. It grows with length without a knee until the two modes separate by a whole edge — 249 mm for a 50 ps edge — and then stops at half the driven line's voltage.

lines · Crosstalk A 10 µs lead-inductance resonance holds the start at t⁴ until √12·τr = 34.6 µs, at a damping of 0.1 or any other. Expanded about infinity, and by residues after. The start of the step through two buffered RC sections (τ = 1 ms and 0.5 ms) on logarithmic axes: bare, a line of slope two (dashed), and with a buffered series resistance and lead inductance into a stray capacitance of natural time constant τr = 10 µs and damping 0.1 (solid). The relative degree rises by two, so the start is the bare step times t²/(12τr²) — the faint line, checked on the expansion to three per cent a hundredth of τr in — until the two meet at √12·τr = 34.6 µs. The damping does not enter that coefficient: it is the inductance and the capacitance, through τr² = LC, and nothing else. After it the curve rejoins the bare one, displaced by a delay that the damping does set.

The delay an inductance does not add

Lead inductance against a stray capacitance is a resonance, and far above a network's bandwidth it raises the relative degree by two: a two-section step starts as t⁴ instead of t² until √12·τr — 34.6 µs for a 10 µs resonance — at any damping, because the coefficient is LC and nothing else. After that the resonance moves a half-way threshold by R·C, its first moment, and the inductance is not in it. It moves every threshold by R·C only at ζ = 1/√2, where its second moment is a pure delay's, and below that its variance is negative: at ζ = 0.1 the 1% point moves by 1.48 R·C while the 10–90 rise time gets shorter. The ring that reaches the output is the network's own gain at the resonance, 2.0 × 10⁻⁴ at ω0τ = 100, and the quality factor decides only how long it lasts. Lightly damped, the start is a staircase that touches the bare step once a period and never overtakes it.

transients · Value theorems

Named alongside it

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

Reflection coefficientRise timeTransmission lineCharacteristic impedanceLattice diagramStep responseTerminationDesign tradeoffInitial-value theoremLogic thresholdParasiticsCrosstalk

All concepts