Circuit Response
every model has an edge, and the figure carries it
Where the models stop

The load spread along the track

A hundred picofarads at the end of a track ring with its inductance as a series circuit, and a 1 ns edge from a 10 Ω driver may cross only 10.9 mm inside 1% of overshoot. The same hundred picofarads spread over 32 inputs along the track allow 29.5 mm, and 300 pF spread allow 74.2. Spread capacitance is not a load at the end; it lowers the line's own impedance to Z₀ divided by the square root of one plus the ratio of the added to the line's own capacitance, and slows it, and once the loaded line's impedance falls below the driver's — 8.2 Ω for 150 pF over 30 mm — every reflection at the source returns with the same sign and the far end climbs a staircase that cannot overshoot. The valley the lumped load has at a picofarad or two survives, shallower: 1.14 mm at its lowest, against 0.88. Two taps are no better than one; from eight the track is a loaded line.
15 min read 5 figures Every model has an edgeThe parasitic is the componentMeasured, not tabulated

Assumes: Kirchhoff's own frequency · The staircase in time

The heavier input that allows a longer track put a capacitance at the far end of a track and asked how long the track may be before a one-nanosecond edge from a ten-ohm driver overshoots by more than one per cent. The answer had a valley and then a rise: 1.80 millimetres with nothing on the end, 0.86 at a picofarad and a half, then 2.71 at twenty picofarads and 10.9 at a hundred. A heavy load turns the track into the inductor of a series resonant circuit, whose damping the driver’s resistance supplies, and a heavier capacitance makes that circuit better damped.

That essay ended on three questions, and the third is this one. A real bus does not have its load at the end. A memory bus, a clock net to a row of chips, an address line across a backplane — each has many inputs tapped off along its length, and at the limit of many small inputs the load is a capacitance per unit length. That is not a load on the end of a line; it is a change to the line itself, and the question is whether the valley, the damping and the rule the earlier essay built from them survive.

Lumped at the end, and spread along

The track is fifty-ohm copper on FR-4, driven by a one-nanosecond edge through ten ohms, and the criterion is the same: the largest overshoot at any input stays under one per cent. The total capacitance is now divided equally among taps along the track, the last at the far end, and the track between taps is a line segment. The whole network is marched by its characteristics — a forward and a backward wave on every segment, the node equations settled at every tap — which with a single tap at the end reproduces the earlier essay’s march.

The longest track a 1 ns edge from a 10 Ω driver may cross inside 1% overshoot, with its load lumped at the end and spread over 32 tapscomputed by solving, not by drawing: a network of line segments marched by its characteristics, the load spread over equal taps with the last at the far end, the length bisected at each total capacitance. Lumped at the end (faint, the earlier essay's curve) the limit falls from 1.83 mm to 0.88 mm and rises to 10.9 mm at 100 pF. Spread over 32 taps (solid) the valley bottoms at 1.14 mm, and the limit is 29.5 mm at 100 pF — 2.70 times the lumped length — and 74.2 mm at 300 pF.1m10m100m1p10p100ptotal capacitance on the track (farads; the left edge is none)longest track for 1% overshoot (metres)32 taps: 29.5 mmat the end: 10.9 mmsolved, then checked — a load along the line29 mm at 100 pF, spread
Fig. 1 The longest track a 1 ns edge from a 10 Ω driver may cross inside 1% overshoot, against the total capacitance on it, with the load lumped at the end (faint, the earlier essay’s curve) and spread over 2, 8 and 32 taps (32 solid). Lumped: 1.83 mm bare, 0.88 mm at 1 pF, 10.9 mm at 100 pF. Spread over 32: 1.14 mm at the valley, 29.5 mm at 100 pF, 74.2 mm at 300 pF. The slider is the number of taps.

With one tap the march reproduces the earlier curve to two per cent: 1.83 millimetres bare, the valley at 0.88 near a picofarad, and 10.9 millimetres at a hundred picofarads. Spread over thirty-two taps the curve keeps its shape and changes its scale. The valley is shallower — 1.14 millimetres at its lowest — and the rise is much steeper: 29.5 millimetres at a hundred picofarads and 74.2 at three hundred, nearly three times and seven times the lumped figures.

So the earlier rule survives in form: a little capacitance shortens the track, a lot lengthens it. What changes is how quickly the lengthening sets in, and the reason is that a spread load and a lumped one are doing physically different things to the edge.

One capacitance, two places

Look at the far end of twenty millimetres carrying a hundred picofarads, once lumped and once spread.

The far end of 20 mm of track with 100 pF on it, lumped at the end and spread over 32 taps. computed by solving, not by drawing. Lumped at the end the load and the track's own inductance ring as a series circuit: the far end overshoots 8.4% (dashed). Spread over 32 taps the same capacitance turns the track into a slower line of lower impedance, 8.2 Ω, nearly matched to the 10 Ω driver, and the far end rises with 0.04% of overshoot (solid), later because the loaded line is slower.
Fig. 2 The far end of 20 mm of track with 100 pF on it, for a 1 V edge. Lumped at the end (dashed) the load and the track’s inductance ring as a series circuit and overshoot 8.4%. Spread over 32 taps (solid) the same capacitance makes a slower line of 8.2 Ω, nearly matched to the 10 Ω driver, and the far end rises with 0.04% of overshoot.

Lumped, the far end rings: the track’s inductance and the hundred picofarads at its end form a series circuit that the driver’s ten ohms under-damps, and the first peak is 8.4 per cent high. Spread over thirty-two taps the far end rises with 0.04 per cent of overshoot, and later. Nothing rings, because there is no single resonance to ring: the capacitance is distributed in the same proportion as the inductance, and a line with both distributed is a line — slower, and of a different impedance — rather than a resonator.

The loaded line’s impedance is the place the rest of the essay turns. A line of capacitance C0C_0 per unit length carrying an extra CdC_d per unit length of spread load has impedance and velocity

Z′=Z01+Cd/C0,v′=v1+Cd/C0,Z' = \frac{Z_0}{\sqrt{1 + C_d/C_0}}, \qquad v' = \frac{v}{\sqrt{1 + C_d/C_0}},

and a fifty-ohm FR-4 track has about 140 picofarads a metre of its own. A hundred picofarads over twenty millimetres is five nanofarads a metre, thirty-six times the track’s own, and the loaded line is 8.2 ohms — below the driver’s ten.

A load that terminates its own source

That is the mechanism, and it can be drawn directly: the overshoot of a fixed length of track against the spread load, beside the loaded impedance.

30 mm of track with its load spread over 32 taps: the overshoot, and the loaded line's impedance, against the total capacitance. computed by solving, not by drawing at 10 loads. Spread capacitance lowers the line's impedance to Z₀/√(1 + Cload/C₀) (dashed, in hundreds of ohms): 44.9 Ω at 1 pF, 17.5 Ω at 30 pF, 8.2 Ω at 150 pF, 4.6 Ω at 500 pF. The overshoot (solid) falls from 12% with a light load to 0.07% at 150 pF, where the loaded impedance first falls below the driver's ten ohms, and stays below half a per cent beyond it: with the line's impedance under the source's, every reflection at the source returns with the same sign and the far end climbs a monotone staircase. A spread load does for the source end what a series resistor does.
Fig. 3 Thirty millimetres of track with its load spread over 32 taps: the far-end overshoot (solid) and the loaded line’s impedance, in hundreds of ohms (dashed), against the total capacitance. The impedance falls from 44.9 Ω at 1 pF through 17.5 Ω at 30 pF to 8.2 Ω at 150 pF; the overshoot falls from 12% to 0.07% where the impedance crosses the driver’s 10 Ω, and stays below half a per cent beyond.

Over thirty millimetres the loaded impedance falls from 44.9 ohms at a picofarad to 17.5 at thirty and 8.2 at a hundred and fifty. The overshoot falls with it, from twelve per cent with a light load to 0.07 per cent where the loaded impedance first drops below the driver’s ten ohms, and stays below half a per cent for every heavier load. With the line’s impedance below the source’s, the source reflects every returning wave with a positive coefficient, the open far end reflects with a positive coefficient too, and the far end climbs a monotone staircase towards its final value with nothing to overshoot.

A spread load does for the source end what a series resistor does. It is the same physics as source termination, approached from the other side: instead of raising the driver’s resistance to the line’s impedance, the load lowers the line’s impedance to the driver’s. The earlier essay found a weaker driver to be the cleaner one for a lumped load; for a spread load the comparison is between the driver and the loaded line, and the cleanest bus is the one whose inputs have pulled its impedance down to the driver’s. How long a track may be found the source mismatch deciding whether there is a limit at all on an unloaded track; here the spread load moves that mismatch.

How many taps make a line

A continuum is a limit. The interesting question for a layout is how many inputs it takes for a row of them to behave like a loaded line rather than like a lumped load with a few extra resonances.

100 pF spread over 1 to 64 equal taps: the longest track a 1 ns edge may cross inside 1%. computed by solving, not by drawing. 1 tap: 10.9 mm, 2 taps: 11.0 mm, 3 taps: 14.8 mm, 4 taps: 15.0 mm, 8 taps: 21.4 mm, 16 taps: 26.2 mm, 32 taps: 29.5 mm, 64 taps: 30.2 mm. The limit rises in pairs rather than smoothly — one and two taps nearly alike, three and four nearly alike — because while the taps are few the one at the far end carries enough of the load to ring with the track as a lumped capacitance would. From about eight taps the track behaves as a loaded line, and 64 taps reach within a few per cent of 32's.
Fig. 4 100 pF spread over 1 to 64 equal taps: the longest track a 1 ns edge may cross inside 1%. One tap: 10.9 mm; two, 11.0; three, 14.8; four, 15.0; eight, 21.4; sixteen, 26.2; thirty-two, 29.5; sixty-four, 30.2.

The limit rises in pairs rather than smoothly: one and two taps are nearly alike, 10.9 and 11.0 millimetres; three and four nearly alike, 14.8 and 15.0; then 21.4 at eight, 26.2 at sixteen, 29.5 at thirty-two and 30.2 at sixty-four. While the taps are few, the one at the far end carries a large share of the load and rings with the track as a lumped capacitance would; it takes about eight taps before the far tap is a small enough share that the track behaves as a loaded line, and by thirty-two the limit is within a few per cent of the continuum’s.

That gives a layout rule with a number in it. A bus of eight or more equal inputs spaced along its length can be treated as a loaded line, with Z′Z' and v′v' above; a bus of two or three inputs, however they are spaced, is a lumped load with resonances, and the earlier essay’s much shorter limit applies to it. An essay on lumped sections measured the same convergence from the other direction — how many lumped sections a line needs — and found it set by how many sections fit inside the edge.

The driver a spread load wants

The earlier essay found a weaker driver to be the clean one for a lumped load. The spread load agrees, and by how much is worth seeing.

100 pF lumped at the end and spread over 32 taps: the longest track against the driver's output resistance. computed by solving, not by drawing. Lumped at the end (dashed), a weaker driver damps the track's series ring and allows more length: 2.5 Ω, 0.8 mm; 5 Ω, 2.8 mm; 10 Ω, 10.9 mm; 20 Ω, 47.4 mm; 40 Ω, no limit. Spread (solid), the same trend at about three times the length for every driver: 2.5 Ω, 1.9 mm; 5 Ω, 7.3 mm; 10 Ω, 29.5 mm; 20 Ω, 118.8 mm; 40 Ω, no limit. At 40 Ω neither has a limit inside 30 cm: the driver is then weaker than the line it drives, and the far end climbs a staircase that cannot overshoot.
Fig. 5 100 pF lumped at the end (dashed) and spread over 32 taps (solid): the longest track for 1% overshoot against the driver’s output resistance. Lumped: 0.8, 2.8, 10.9 and 47.4 mm at 2.5, 5, 10 and 20 Ω; spread: 1.9, 7.3, 29.5 and 118.8 mm. At 40 Ω neither has a limit inside 30 cm.

For both arrangements a weaker driver allows a longer track, and the spread load allows about three times the length at every driver: 0.8 against 1.9 millimetres at 2.5 ohms, 10.9 against 29.5 at ten, 47.4 against 118.8 at twenty. At forty ohms neither has a limit inside thirty centimetres: the driver is then weaker than the line it drives, loaded or not, and the far end climbs a staircase that cannot overshoot.

The strong driver is the one that suffers, and it is the one a designer reaching for speed would choose. A 2.5-ohm driver into a hundred picofarads lumped at the end may cross under a millimetre of track before it rings; spread along thirty-two inputs, under two. A fast driver on a heavily loaded bus is buying edge rate it cannot use, and the ringing is the bill.

A bus, worked

Put the numbers on a bus. Twenty inputs of five picofarads each, a hundred in all, spaced evenly along a track driven by a ten-ohm output with one-nanosecond edges. Lumped at the far end, a hundred picofarads would allow the earlier essay’s 10.9 millimetres, which is a short route for twenty packages. Spread over twenty taps the limit sits between the sixteen- and thirty-two-tap figures, about 27 millimetres; and over that route a hundred picofarads is 3.7 nanofarads a metre, twenty-six times the track’s own, so the loaded impedance is about 9.6 ohms — at the driver’s ten, which is exactly why the limit is so generous.

Two changes would each buy more length. A twenty-ohm driver, from the driver figure, roughly quadruples it, because the loaded line is then well below the driver and the staircase has no overshoot to give. And the bus is better behaved the more evenly its inputs are spread: a route that crowds the twenty inputs into its last third has a fifty-ohm stretch in front of them and a low-impedance stretch at the end, which is closer to the lumped case the earlier essay measured. A bus designer who knows the loaded impedance can choose the driver to sit at or above it, and then the length limit for one per cent of overshoot disappears into the staircase.

Why the valley survives, and why it is shallower

The rise at heavy loads is the loaded line terminating its source. The valley at light loads is a different mechanism, and it survives the spreading for a reason worth stating.

A picofarad or two on a track whose own capacitance is a fraction of a picofarad a millimetre is a small perturbation of the line, and the line is still close to fifty ohms — five times the driver’s ten. The source reflects a returning wave with a negative coefficient, the far end reflects positively, and the far end’s staircase overshoots, as the staircase in time drew for an unloaded line. A small capacitance at the far end slows the far end’s response just enough that the first reflection’s overshoot and the edge’s own rise add rather than smear, which is why the lumped valley sits at a picofarad and a half. Spread along the line, the same small capacitance is shared among taps that each slow their own stretch a little, the far tap carries only its share, and the overshoot’s peak is smeared over the arrival times of many small reflections. The valley is still there because the line is still badly matched at the source; it is shallower because the extra capacitance is not all in the one place where it adds most efficiently to the overshoot.

That also says where the earlier essays’ rules sit relative to this one. The boundary that is a rise time found the lumped-element boundary for an unloaded track to be a rise time proportional to its length; Kirchhoff’s own frequency put the same boundary in the frequency domain. Both describe the left end of these curves, where the load is too small to change the line. The spread load moves the right end, and moves it a long way.

The same loading, met elsewhere

A capacitance spread along a conductor is not peculiar to buses. A via barrel hanging below a connection is a short stub of capacitance at one point, which the pad the stub comes out of measured as a resonance of its own; a row of vias along a track is a spread load, and lowers the track’s impedance the same way twenty inputs do. A backplane connector every few centimetres loads a long route periodically, and the backplane’s designers account for it by designing the unloaded track above the target impedance, so that the connectors bring it down to the target — the loaded-line formula used deliberately.

And a parallel termination at the far end of a loaded bus has to match the loaded impedance, not the unloaded one. Terminated at both ends matched a fifty-ohm line with fifty-ohm resistors; a bus whose inputs have pulled it to fifteen ohms needs fifteen-ohm terminations, and a fifty-ohm terminator on it reflects more than half of every wave that reaches it — 54 per cent, with the sign that makes the far end overshoot. The loaded line also takes longer to traverse, by the same square-root factor, so a bus’s flight time grows with its population as well as its length, and a timing budget that uses the bare track’s velocity is optimistic by that factor. The loaded impedance is the line’s real impedance, and every design decision that refers to “the line’s impedance” should use it.

Where the loaded-line picture stops

The loaded-line formulas are a continuum’s, and a row of taps is a continuum only for an edge long against the spacing. Between taps the track is unloaded fifty-ohm line, and an edge short enough to resolve the spacing sees each tap as a small capacitance with a reflection of its own — a periodic structure with a cut-off frequency at which the taps reflect coherently and the bus stops passing an edge at all. With a one-nanosecond edge and taps a few millimetres apart that cut-off is far above the edge’s content; with a hundred-picosecond edge and taps a centimetre apart it is not. The load that rings with the line is the single-tap case of that effect, and it is where the earlier essay started.

The model has no loss in the copper, no inductance in the inputs’ own leads, and inputs that are pure capacitance. Real inputs have protection diodes that clamp an overshoot of a few hundred millivolts, which is why a ringing bus often works; the one per cent criterion is a signal-integrity standard, not the edge of function.

How the network was solved

Every figure is a march of a network of lossless segments of fifty-ohm line meeting at taps. Each segment carries a forward and a backward wave, delayed by its transit in whole steps; at every tap the arriving waves, the tap’s capacitance and — at the source — the driver through its resistance settle the tap voltage by the trapezoidal rule, and the departing waves are the tap voltage less the arriving ones. The step is an eighth of the shortest segment’s transit, and the march runs for sixty transits of the whole track or six ring periods of its total inductance and capacitance, whichever is longer. A single tap at the end is checked against the earlier essay’s march to two per cent, and the length limits are bisected on the logarithm of the length to a part in a thousand.

Still open: a non-uniform bus, the spacing an edge can see, and the driver whose resistance moves

A bus whose inputs are not evenly spread. Many buses run a long unloaded stretch to a cluster of inputs at the far end. The unloaded stretch is a fifty-ohm line and the cluster a short low-impedance one, and the junction between them reflects. Whether such a bus behaves more like the lumped load or the spread one, as the cluster’s length is varied, is the same march with the taps moved.

The tap spacing an edge resolves. The periodic structure’s cut-off sets a minimum rise time for which the loaded-line picture holds, and below it the bus’s response grows a ripple at the taps’ Bragg frequency. Measuring the overshoot against rise time for a fixed spacing would put a number on the edge rate a given bus geometry can carry, which no lumped or loaded-line model can.

A driver whose resistance moves with the edge. A CMOS output’s resistance is lowest in the middle of its transition and higher at the ends, so the match it makes with a loaded line is not one number. Whether the match that matters is the one at the start of the edge, which launches the wave, or the one at its end, which absorbs the first reflection, decides which datasheet figure a designer should match the bus to — the question the heavier input that allows a longer track left open, now with a loaded line to match.

Part 7 on lumped element

One argument about Lumped element, and one of 7 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

From a 10 Ω driver with a 1 ns edge, a CMOS input of 1.5 pF allows only 0.86 mm of track — less than the bare open end's 1.80 — and one of 100 pF allows 10.9 mm. computed by solving, not by drawing, bisected on the march by characteristics: the longest 50 Ω track a 1 ns edge from a 10 Ω driver can cross into an open end with a capacitance across it and overshoot by less than one per cent, against the capacitance. With none it is 1.80 mm. 0.3 pF: 1.28 mm, 0.5 pF: 1.04 mm, 1 pF: 0.88 mm, 1.5 pF: 0.86 mm, 2 pF: 0.88 mm, 3 pF: 0.96 mm, 5 pF: 1.15 mm, 10 pF: 1.68 mm, 20 pF: 2.71 mm, 50 pF: 5.70 mm, 100 pF: 10.94 mm. The limit falls to its least, 0.86 mm, near 1.5 pF, and from there rises nearly in proportion to the capacitance (slope 0.87 between 20 and 100 pF), because the driver's resistance damps the ring the track's inductance makes with a heavy load. The slider is the driver's resistance. The heavier input that allows a longer track Part 6 — Put a capacitance across a CMOS input and the longest track a 1 ns edge from a 10 Ω driver may cross inside one per cent of overshoot does not shrink steadily. It falls from the bare open end's 1.80 mm to 0.86 mm at 1.5 pF, and then it rises — 2.71 mm at 20 pF, 10.9 at 100 — nearly in proportion to the capacitance. A heavy load turns the track into the inductor of a series RLC whose damping, (Rₛ/2)·√(C/L), the driver's resistance supplies, and at every heavy load the one-per-cent length sits at the same damping, about 0.8. So the limit scales as Rₛ²·C: a slower edge buys a 20 pF input only the 0.4 power of its length, and a stronger driver — 2 Ω instead of 10 — cuts the track it may drive from 2.71 mm to 0.20. Into a heavy input, a weak driver is the clean one. A 1 ns edge from a 10 Ω driver, a trunk, and a fork into two branches each ending in 2 pF. computed by solving, not by drawing: the network of trunk and two branches marched by its characteristics, the branch length bisected for 1% overshoot at either end. Forked at the driver, each branch may be 2.33 mm, against 1.05 mm for one track carrying both inputs at its end (dashed). Every millimetre of trunk costs more than a millimetre of branch: 0.0 mm of trunk, 2.33 mm branches; 0.2 mm of trunk, 1.89 mm branches; 0.4 mm of trunk, 1.43 mm branches; 0.6 mm of trunk, 1.01 mm branches; 0.8 mm of trunk, 0.57 mm branches; 1.0 mm of trunk, 0.11 mm branches. The track that forks Part 8 — Two 2 pF inputs on one track from a 10 Ω driver with 1 ns edges may sit only 1.05 mm away inside 1% overshoot if both hang from its end. Fork the track at the driver and each branch may run 2.33 mm — more than twice as far — because two fifty-ohm branches in parallel are a 25 Ω line, half as mismatched to the driver. Every millimetre of trunk before the fork costs more than a millimetre of branch: with a millimetre of shared trunk the branches may run only 0.11 mm. Equal branches are the only good ones: a 2.2 mm branch overshoots 0.91% beside an equal partner and 1.7% beside a short one. And a star reaches further the more inputs it serves — 2.33 mm for two, 11.85 for four, where four branches in parallel make 12.5 Ω — while a daisy chain of the same inputs reaches 2.54.

Where this goes next

From a 10 Ω driver with a 1 ns edge, a CMOS input of 1.5 pF allows only 0.86 mm of track — less than the bare open end's 1.80 — and one of 100 pF allows 10.9 mm. computed by solving, not by drawing, bisected on the march by characteristics: the longest 50 Ω track a 1 ns edge from a 10 Ω driver can cross into an open end with a capacitance across it and overshoot by less than one per cent, against the capacitance. With none it is 1.80 mm. 0.3 pF: 1.28 mm, 0.5 pF: 1.04 mm, 1 pF: 0.88 mm, 1.5 pF: 0.86 mm, 2 pF: 0.88 mm, 3 pF: 0.96 mm, 5 pF: 1.15 mm, 10 pF: 1.68 mm, 20 pF: 2.71 mm, 50 pF: 5.70 mm, 100 pF: 10.94 mm. The limit falls to its least, 0.86 mm, near 1.5 pF, and from there rises nearly in proportion to the capacitance (slope 0.87 between 20 and 100 pF), because the driver's resistance damps the ring the track's inductance makes with a heavy load. The slider is the driver's resistance. The heavier input that allows a longer track Put a capacitance across a CMOS input and the longest track a 1 ns edge from a 10 Ω driver may cross inside one per cent of overshoot does not shrink steadily. It falls from the bare open end's 1.80 mm to 0.86 mm at 1.5 pF, and then it rises — 2.71 mm at 20 pF, 10.9 at 100 — nearly in proportion to the capacitance. A heavy load turns the track into the inductor of a series RLC whose damping, (Rₛ/2)·√(C/L), the driver's resistance supplies, and at every heavy load the one-per-cent length sits at the same damping, about 0.8. So the limit scales as Rₛ²·C: a slower edge buys a 20 pF input only the 0.4 power of its length, and a stronger driver — 2 Ω instead of 10 — cuts the track it may drive from 2.71 mm to 0.20. Into a heavy input, a weak driver is the clean one. A 1 ns edge from 10 Ω into 200 Ω on 50 Ω track stays inside one per cent of overshoot only up to 2.5 mm of track. computed by solving, not by drawing, the staircase of reflections marched against a ramp: the far end's overshoot against the track's length for a 1 ns edge, 50 Ω track on FR-4 driven from 10 Ω into 200 Ω. 2 mm: 0.79%; 5 mm: 1.98%; 10 mm: 3.98%; 20 mm: 8.07%; 50 mm: 22.93%; 100 mm: 40.00%; 200 mm: 40.00%; 500 mm: 40.00%. The longest track inside one per cent, bisected, is 2.51 mm — a round trip of 35 ps, the edge being 28.5 times it — and the rise-time boundary of that length is 0.994 ns, this edge given back. How long a track may be The lumped-element boundary read the way a layout is designed: given the edge, how long may the track be? For 50 Ω track driven from 10 Ω into 200 Ω, a 1 ns edge stays inside one per cent of overshoot for 2.51 mm, and every nanosecond of edge buys 2.52 mm more — proportional to within half a per cent from 0.1 ns to 30. The mismatch decides the rest. From 45 Ω the same edge may cross 23.5 mm; from 50 Ω, or anything above it into this load, no length overshoots at all; and into an open end, a CMOS input, 1.80 mm. The common rule — a track is short while its round trip is under the edge — allows 49 per cent of overshoot from a 5 Ω driver, and a tenth of its length still allows 3.3. 10 pF on the far end of 10 cm of 50 Ω track: a 1 ns edge overshoots by 47.5% and rings near 174 MHz, against 40.0% with no capacitance. computed by solving, not by drawing, by the method of characteristics: 10 cm of 50 Ω track on FR-4 (a 0.70 ns transit), driven from 10 Ω by a 1 V edge rising in 1 ns into 200 Ω with 10 pF across it, and the far end's voltage against time. With no capacitance (dashed) the far end is the staircase of the rise-time essay, overshooting by 40.0% and settling on the divider, 0.952 V. With 10 pF the reflection is no longer a number: the capacitor first holds the far end down and then lets it swing, the overshoot is 47.5%, and the ringing is near one over 2π√(LC) = 174 MHz, with L the track's own 35.0 nH and C its own 14.0 pF plus the load's. Both settle on the divider. The load that rings with the line Put 10 pF on the far end of ten centimetres of 50 Ω track and the reflection stops being a number. A one-nanosecond edge from 10 Ω into 200 Ω overshoots by 47.5 per cent instead of 40.0, and rings near 174 MHz — the track's own 35 nH against its 14 pF and the load's 10 together. The one-per-cent boundary, a rise time of 39.9 ns on a bare track, becomes 50.4 ns with 10 pF and 83.4 with 50: a rise time and a capacitance together. As the load outweighs the track the lumped model comes back, missing the line's waveform by 55 per cent of the settled value with no load capacitance and by 2.2 with 200 pF. And a source matched to the line turns every capacitance's ring into a delay of about 2.2·Z₀·C.

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Capacitive loadCharacteristic impedanceInput capacitanceLumped elementPropagation velocityReflectionsTransmission line