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

Reflections — where it appears

The copies of an edge that a mismatched end of a line sends back towards its source, each scaled by that end's reflection coefficient. They decide how long a line may be before its far end rings rather than follows its driver, and they vanish only when each end matches the line.

Named by 6 essays across one field — each of them below, with the objects they name alongside it.

10.0 cm of track leaves 40 per cent on a 1 ns edge. The far end of 10.0 cm of 50 Ω track, driven from 10 Ω into 200 Ω, marched as the exact staircase of reflections it is and driven by a ramp rather than by a step. The staircase converges on the resistive divider the circuit was going to be all along, to a part in 10⁹, which is the number the lumped model returns immediately and with no staircase. What is left on the edge is an overshoot: 40.0 per cent at 0.1 ns, which is the first wave's own reflection coefficient and contains no rise time at all, falling to 0.399 per cent at 100 ns where the staircase happens inside the edge. It reaches ten per cent at an edge of 4.24 ns and one per cent at 40.0 — which is 28.6 times the 1.40 ns round trip. Nothing in the measurement is a repetition rate: the boundary is a transit time, a mismatch and a rise time, and the clock the signal repeats at does not enter.

The boundary that is a rise time

Kirchhoff's own frequency is 3.97 megahertz for ten centimetres of track, and a one-kilohertz clock is four thousand times below it — which is the reading that concludes wrongly. Marched as the exact staircase of reflections it is, the same track leaves forty per cent of overshoot on a one-nanosecond edge and 0.40 per cent on a hundred-nanosecond one, reaching one per cent at 40.0 nanoseconds. The boundary is a rise time, it is twenty-nine times the conductor's round trip, and the repetition rate is not in it anywhere.

limits · Lumped element 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.

limits · Lumped element 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.

limits · Lumped element 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.

limits · Lumped element 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 taps. computed 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.

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.

limits · Lumped element 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

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.

limits · Lumped element

Named alongside it

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

Lumped elementTransmission lineInput capacitanceRise timeCharacteristic impedanceElectrical lengthModel rangePropagation velocitySource impedanceCapacitive loadDamping ratioDesign tradeoff

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