The boundary that is a rise time

Kirchhoff’s own frequency measures the lumped-element boundary as two frequencies — 3.97 megahertz where ten centimetres of track is one degree long, 143 megahertz where it is a tenth of a wavelength — and then lists three things the boundary is not. The first is the one that matters most and it is the only one the essay does not measure:

It is not about the frequency of the signal, but about its bandwidth. A one-kilohertz square wave with a one-nanosecond edge contains significant energy up to several hundred megahertz, and it is that content which decides whether the lumped model applies. Reading the boundary against a nominal clock rate is the single commonest way to conclude wrongly that a circuit is electrically small.

It also states the time-domain version and stops: if the round-trip time is short compared with the rise time, the staircase happens entirely within the edge and is invisible; if it is comparable, the edge has steps in it.

This measures that, and the boundary comes out as a rise time with no frequency of any kind in it.

The staircase, and what it converges on #

Send a step into the line. A wave of $Z_0/(R_s + Z_0)$ of the source’s voltage sets off, arrives at the far end one transit time later, and finds a load that cannot absorb it — so a fraction $\rho_L$ turns round and comes back. That one reflects again at the source, and the far end receives a smaller wave every round trip.

What the far end sees is a staircase: steps at odd multiples of the transit time, each one $\rho_S\rho_L$ times its predecessor, each contributing $(1+\rho_L)$ of itself. The series converges, and it converges to the resistive divider the circuit was going to be all along — required here to a part in $10^9$, which is the check that the two models describe the same circuit when they describe anything.

That limit is what the lumped model returns, immediately and with no staircase. So the difference between the two models is not a difference in the answer; it is a difference in when, and the staircase is entirely a transient.

Which is why the boundary cannot be a frequency. A transient’s visibility is decided by how long it lasts against how long the thing that provoked it took, and both are times.

The overshoot, against the edge #

Drive the same line with a ramp of duration $t_r$ rather than a step — which is what any real signal is — and the staircase is smeared over the ramp. What is left on the edge is an overshoot:

Two ends of that table are worth reading separately.

At a fast edge the overshoot is 40.0 per cent and stops changing, because the edge is over before the far end’s answer comes back. Forty per cent is the first wave’s own overshoot — $(1+\rho_L)Z_0/(R_s+Z_0)$ against the final divider — and it contains no rise time at all. A faster edge cannot make it worse because there is nothing left to smear.

At a slow edge it falls in proportion to the edge, because the whole staircase happens inside the ramp and what survives is the ratio of the round trip to the rise time. That is that essay’s own sentence, with a number on it.

And the boundary is between: one per cent of overshoot at a rise time of 41.6 nanoseconds, ten per cent at 3.4. The conductor’s round trip is 1.4 nanoseconds, so the one-per-cent boundary is thirty times it — a useful figure, because a round trip is a length divided by a velocity and a designer has both.

The two frequencies, converted #

That essay’s two numbers are correct and they are in the wrong units for the question, so it is worth doing the conversion rather than leaving them as a contrast.

Its one-degree boundary is $v/(360\,\ell)$ and its tenth-of-a-wavelength boundary is thirty-six times it. Both are the transit time inverted and scaled, which is to say both are the same quantity this essay measures, in different clothes:

Read as periods, that essay’s two frequencies bracket this essay’s boundary — 252 nanoseconds and 7.0 — and the useful comparison is neither directly. A sinusoid’s period is not a rise time: a sinusoid at the one-degree frequency rises from nothing to its peak in a quarter of its period, 63 nanoseconds, which is within a factor of two of the 41.6 this essay measures. So that essay’s gentler boundary is, to a factor of two, the right answer expressed in the wrong variable — and its tenth-of-a-wavelength boundary, which is the one everybody uses, corresponds to a rise time of 1.8 nanoseconds and around thirty per cent of overshoot.

Which is that essay’s own complaint, sharpened. It says the tenth-of-a-wavelength rule marks the point at which the model is already thirty per cent wrong, and this measurement says thirty per cent of what: not an impedance at a frequency, but the overshoot left on an edge — a quantity a designer can see on an oscilloscope and a specification can be written against.

So the conversion that makes the two essays one is: a boundary stated as a frequency should be read as a quarter-period and compared with a rise time. That is the most useful sentence available from the pair, and neither essay could produce it alone, because one has frequencies without edges and the other has edges without a sinusoid to compare them with.

What the clock is doing in the answer, which is nothing #

That essay’s warning is that a clock rate is the wrong variable, and the measurement makes that stronger than a warning: the clock rate is absent.

Nothing in the arithmetic above mentions how often the signal repeats. There is a transit time, which is a length over a velocity; a pair of reflection coefficients, which are three impedances; and a rise time. A signal that steps once and never again produces exactly the same overshoot as one stepping a million times a second, because an overshoot is a property of an edge.

So the boundary restated is short. The lumped model applies when the rise time is more than about thirty round trips, and the repetition rate is not a term. For ten centimetres of ordinary board that is 41.6 nanoseconds; for a metre it is 416; for a millimetre it is 0.4.

Compare that with the two frequencies the earlier measurement reports, and the source of the wrong conclusion is visible. A one-kilohertz clock against a 3.97-megahertz boundary is four thousand times inside it, which reads as an enormous margin — and if that clock has a one-nanosecond edge, the same conductor leaves forty per cent of overshoot on it. The two readings differ by everything, and the only thing wrong with the first is that it compared the wrong number.

Two quantities were tried before this one and both are worth recording, because both are the natural next guess and both agree with the wrong conclusion.

The error in the current the signal draws — the harmonics of the square wave through the two models’ input impedances, summed — reads 0.885 per cent for that one-nanosecond edge at a kilohertz. It is dominated by the fundamental, because a square wave’s current is mostly its fundamental and the lumped model gets it exactly right.

And the fraction of the signal’s energy above the boundary is worse: two parts in ten thousand, because a trapezoid’s energy is in its low harmonics whatever its edge.

Both are correct measurements of the wrong thing. What a line does to a signal is not an energy and not an average; it is a feature on the edge, and the only quantities that see it are the ones that look at the edge.

Where the thirty round trips comes from #

The factor between the round trip and the one-per-cent boundary is worth an explanation, because it is not a constant of nature and it moves with the mismatch.

The overshoot at a slow edge is, near enough, the first wave’s own overshoot multiplied by the fraction of the ramp the round trip occupies. So one per cent of overshoot needs the round trip to be one per cent of $0.40$, which is a fortieth of the rise time — and the measured 41.6 nanoseconds against 1.4 is thirty times rather than forty, the difference being that the staircase has more steps than one.

Which says the factor is proportional to the mismatch. A line driven from its own impedance into its own impedance has no reflection at all, so it has no boundary: the staircase is one step and the lumped model and the line agree at every edge rate. A badly mismatched line has a large first overshoot, so it needs a proportionally slower edge to hide it.

So the boundary is not a property of the conductor. It is a property of the conductor and what is at each end of it, and the same ten centimetres is electrically small for one termination and not for another. That is an unusual thing for a boundary in this collection to be — every model has an edge puts four of them on one frequency axis and every one is a property of a component or a geometry — and it is why that essay’s two frequencies, which contain no impedance at all, are a conductor’s property and not a circuit’s.

The practical form is the remedy every high-speed design uses. Terminating the line removes the boundary rather than moving it: with $\rho_L = 0$ the far end sees one wave and no staircase, and the lumped model’s answer arrives one transit time late rather than never. That is not a smaller error; it is a delay, which is a different kind of thing and one a design can budget for.

The other essay’s staircase, and the number it shares #

The staircase in time is the lines field’s own account of this and it is worth saying what the two measurements share and where they part.

That essay drives a metre of cable and finds the source not knowing what is on the far end for 4.83 nanoseconds, driving into the cable’s characteristic impedance during that interval, and receiving the far end’s answer as a staircase whose limit is the resistive divider the circuit was going to be all along. Every one of those is the object marched here.

What this essay adds is the ramp. That essay’s staircase is the response to a step, which is the worst case and is a waveform nothing produces; this one asks what survives when the step has a duration, and the answer is the boundary. So the two are the same solve read for two purposes — one to show that a line has a staircase in it, the other to say when the staircase is visible — and the second needs the first to be exact before it means anything.

Which is the sense in which the lines field is on the other side of this boundary rather than beyond it. Above the boundary the lumped model is not approximately right, as the earlier measurement says: it is describing a different object, one with finitely many poles and an asymptote where the line has infinitely many resonances and no asymptote at all. What this measurement adds is where the change of object becomes visible in a signal, and it is a rise time.

Where this boundary sits among the others #

Every model has an edge puts four of these essays’ boundaries on one frequency axis and finds their ordering to be a function of the board: Kirchhoff’s laws are the third of the four to fail on a ten-centimetre board, just below the capacitor’s self-resonance, and shrinking the board to three centimetres moves them past it.

This essay’s boundary does not go on that axis, and it is worth being explicit about why rather than leaving it out silently.

It is a time, not a frequency, and the edges that are lengths gathers the boundaries of its kind — every one the same constant divided by a distance — while the conversion above shows this one and that essay’s to be within a factor of two of each other — so it could go on the axis at a quarter-period, with a caveat attached to every reading of it.

It contains an impedance, and none of the other four does. The capacitor’s self-resonance is a property of one component; the small-signal boundary is a property of a device; Kirchhoff’s own frequency is a property of a geometry. This one is a property of a conductor and what is at each end of it, so the same board has different values of it on different nets — which makes it a boundary of a circuit rather than of a model, and the other four are boundaries of models.

And it is removable. Terminating the line takes the overshoot to nothing and leaves a delay, which is a different kind of error and one a design budgets for rather than avoids. None of the other four can be removed by anything: a capacitor’s inductance is there, a transistor’s exponential is there, and a board’s size is what it is.

That third difference is the one worth carrying, because it changes what a designer does with the number. A boundary that cannot be moved is a range to stay inside, which is how the other four are used. A boundary that can be removed for the price of a resistor is a decision, and the arithmetic above is what the decision costs on each side: forty per cent of overshoot untreated, or the line’s own transit time of delay and a resistor’s worth of signal amplitude if it is terminated.

Which also explains why the lines field exists as a separate field rather than as the far side of this one’s boundary. Above the boundary the lumped model is describing a different object, as the earlier measurement says — and the different object is one a design chooses to work with, by terminating, matching and budgeting delay, rather than one it is forced into. The boundary is the frontier of a technique as much as of a model.

What is left out #

One reflection coefficient at each end. Both are real here, because both terminations are resistive. A capacitive load — which every logic input is — makes the far-end reflection frequency-dependent, so the staircase’s steps are shaped rather than square, and the overshoot becomes a ringing with a decay of its own. That is the same object a ladder is not a line meets from the other side, where forty sections of inductors and capacitors still ring through every plateau by five per cent.

The line is lossless. The delay that is not one number measures what a real trace’s series resistance does below the frequency where its reactance overtakes it — 910 kilohertz for ordinary copper — where the line is a diffusion rather than a wave. A slow edge’s content is partly down there, so the slow end of this essay’s table is the optimistic one.

And the source is a resistance. A real driver is a nonlinear output stage whose impedance differs between its two states, so the reflection at the source is different on a rising edge and a falling one — which is why overshoot and undershoot on a real board are not mirror images and why one of the two is usually the problem. How small is small signal is where the amplitude at which such an asymmetry begins to matter is priced, at 7.3 millivolts.

Still open: the capacitive load, and the boundary as a length #

A load that is a capacitance. The reflection from a logic input is not a number, and the overshoot becomes a damped ring whose frequency is set by the line’s inductance against that capacitance. The same march would give it, and the useful output is whether the one-per-cent boundary is still a rise time or becomes a rise time and a capacitance together.

The lumped model’s own waveform. This essay marches the line and compares its overshoot against its own settled value. What the two-element lumped model predicts on the same ramp is a third waveform, with an overshoot of its own from its L and C — so the honest comparison is between two time-domain responses rather than between one and its limit, and the size of their disagreement is what Kirchhoff’s own frequency reports in the frequency domain.

The boundary as a length, at a stated edge. Everything here fixes the conductor and sweeps the edge, which is the measurement a boundary is usually stated as. A design has the opposite question: given the edges the parts produce, how long may a track be? That is the same curve read the other way and it is one number a layout rule could be written from — and it would have a mismatch in it, which no existing layout rule does.

And the same question for the lumped model’s own answer. This essay measures what the LINE does. What the lumped model predicts on the same ramp is a different waveform again — an RLC transient with its own overshoot — so the honest comparison is between two time-domain waveforms rather than between a waveform and its own limit. The two should agree below the boundary and the size of their disagreement above it is the quantity that essay’s frequency-domain figure reports, in the domain where it can be seen.

What is checked #

The staircase is required to converge on the resistive divider, to a part in $10^9$, before any overshoot is measured. That is the statement that the two models describe one circuit, and it is the only thing in this essay that the lumped model and the line agree about.

The overshoot is required to fall at every step of the slider, which is the claim that the rise time is the variable.

At a fast edge it is required to equal the first wave’s own overshoot — the reflection coefficient arithmetic, with no rise time in it — to five per cent, because that is what says the fast end of the table is a plateau rather than a trend.

And the boundary is required to be a small multiple of the round trip, between two and two hundred times it. Not a fixed factor, because the factor is proportional to the mismatch and a fixed one would be a fact about this termination rather than about the family — which is the defect these notes records more often than any other.