Circuit Response
every model has an edge, and the figure carries it
Lines, where a wire has a length

The termination that runs out

A 50 Ω resistor at the end of a line terminates every edge and draws current for as long as the line is held high: 75.6 mW on an alternating pattern from a 3.3 V, 10 Ω driver. Put a capacitor in series with it and the standing current is gone, and so, eventually, is the termination: the far end reflects nothing as an edge arrives and the whole of it once the capacitor has charged, by one less the exponential of −t/2Z₀C exactly. Under two round trips of time constant the receiver overshoots — 54.8 per cent at a quarter of one — and above it the receiver waits on the capacitor to reach the swing. On a pseudo-random pattern the levels are tightest at three round trips, and on a fast pattern the capacitor holds the line's mean and the resistor still burns half of what a plain one does.
15 min read 5 figures The parasitic is the componentEvery model has an edgeTwo routes to a number

Assumes: The resistor at the wrong end · The staircase in time

Terminated at both ends found that a series resistor at the driver and a parallel one at the receiver cost exactly half the swing and no reflection ever gives it back, and it priced what the half buys. The resistor at the wrong end had already found the two single terminations looking identical at the receiver and different everywhere else. Neither looked at the cost every designer of a bus meets first: a parallel resistor at the far end carries current whenever the line is held away from the resistor’s own return, and it carries that current for as long as the level is held. A 50 Ω termination on a 3.3 V line is fifty-odd milliamps, all the time.

The standard answer is a capacitor in series with the resistor. At the instant an edge arrives the capacitor is a short and the resistor is all the line sees, so the edge is absorbed; once the edge has passed and the level is held, the capacitor charges and no standing current flows. The question that leaves is the one this essay measures: a termination made of a resistor and a capacitor terminates for a while and then stops, and how long a while has to be — against the line’s own round trip, against the pattern of bits, and against the power it was supposed to save.

A 3.3 V edge from a 10 Ω driver down a 50 Ω line of 1 ns, received through 50 Ω in series with four capacitors, beside a plain 50 Ω and an open end. computed by solving, not by drawing: the line marched by its characteristics. The open end rings up to 5.50 V; the plain resistor holds 2.750 V, the launched level, and never reaches the swing. With the capacitor the far end arrives at the same 2.75 V and then climbs as the capacitor charges and stops terminating: with a time constant of half a round trip (20.0 pF) it overshoots to 4.48 V; with one, 3.83 V; with three (120 pF) it rises without overshoot to 3.30 V by 24 ns; with ten, 3.13 V by 24 ns, still on its way.
Fig. 1 A 3.3 V edge from a 10 Ω driver down a 50 Ω line of 1 ns, received through 50 Ω in series with four capacitors, beside a plain 50 Ω and an open end. The open end rings up to 5.50 V; the plain resistor holds 2.75 V, the launched level. With a time constant of half a round trip the receiver overshoots to 4.48 V; with one, to 3.83; with three it rises without overshoot to 3.30 V by 24 ns; with ten it is at 3.13 V by 24 ns, still rising.

The line, marched by its characteristics

The line is lossless, fifty ohms, a nanosecond long, driven by a ten-ohm source — the output impedance of an ordinary logic driver, not the series-terminated fifty ohms of the earlier essays — with an edge of a tenth of a nanosecond. It is marched by its characteristics: at each instant the wave leaving the driver is fixed by the wave arriving there and the source behind its resistance, and the far end is the incident wave doubled behind fifty ohms, driving the resistor and capacitor in series. The capacitor’s charge is integrated by the trapezoid, and an open end and a plain resistor are the same march with the capacitor removed or the resistor made infinite.

The driver’s ten ohms matter. They launch Z0/(Z0+rs)=50/60Z_0/(Z_0 + r_s) = 50/60 of the swing, 2.75 volts, and a plain fifty-ohm resistor at the far end absorbs that wave and holds the receiver there for good: it never sees the full 3.3 volts, because the resistor and the driver form a divider for as long as current flows. An open end doubles the wave to 5.5 volts and rings down against the driver’s ten ohms, which reflect at −2/3-2/3. The capacitor’s termination starts like the resistor and ends like the open end, and everything this essay finds lives in between.

What the capacitor reflects, exactly

With a matched driver there is only one reflection to see, the far end’s, and it has a closed form. The termination’s impedance is R+1/sCR + 1/sC, its reflection coefficient with R=Z0R = Z_0 is 1/(1+2sZ0C)1/(1 + 2sZ_0C), and a step reflected through that is the incident step times 1−e−t/2Z0C1 - e^{-t/2Z_0C}.

What a 50 Ω resistor in series with a capacitor reflects, as the capacitor charges: the march against the closed form. computed by solving, not by drawing: a matched driver, so the only reflection is the far end's, marched at a two-thousandth of the delay and held to the closed form, the incident step times 1 − e^(−t/2Z₀C), to two parts in a thousand at every point. With 80.0 pF the time constant 2Z₀C is 8.00 ns: the reflection is nothing as the edge arrives, 0.632 of the incident one time constant later, and the whole of it eventually. The resistor terminates the edge; the capacitor decides for how long.
Fig. 2 What a 50 Ω resistor in series with 80 pF reflects as the capacitor charges, from a matched driver: the march against the closed form, the incident step times 1−e−t/2Z0C1 - e^{-t/2Z_0C}. The time constant 2Z₀C is 8.00 ns; the reflection is nothing as the edge arrives, 0.632 of the incident a time constant later, and the whole of it eventually.

The march agrees with the closed form to two parts in a thousand at every point, which is the check on the march before anything else is read from it. And the closed form says the whole story in one line. The resistor terminates the edge; the capacitor decides for how long; and the time constant that matters is 2Z0C2Z_0C, twice the resistor’s own, because the capacitor charges through the line’s impedance and the resistor in series. A termination made with eighty picofarads terminates for about eight nanoseconds and then becomes an open end.

Too small, and the reflections come back

The capacitor's size against what it does to one edge: the overshoot it lets through and the time the receiver takes to reach the swing. computed by solving, not by drawing, each marched to thirty delays. With RC under a round trip the capacitor charges before the reflections are spent and the receiver overshoots: 54.8% at 0.25, 28.7% at 0.63, 6.1% at 1.58, none at 3.98, none at 10.00, none at 25.12 round trips. Overshoot falls below 5% from about 2.0 round trips (79.8 pF). Past that a larger capacitor holds the receiver on the launched level for longer: the time to come within 10% of the full swing grows with it, 1.94 ns at 2.5, 2.77 ns at 5.0, 8.36 ns at 10.0, 20.9 ns at 20.0, past 30 ns at 39.8.
Fig. 3 The capacitor’s size against what it does to one edge, with the 10 Ω driver: the overshoot at the receiver and the time it takes to come within 10 per cent of the swing. Overshoot is 54.8 per cent at a quarter of a round trip, 28.7 at 0.63 and 6.1 at 1.58, and below 5 per cent from about two round trips, 79.8 pF. Past that the time to within 10 per cent grows: 1.94 ns at 2.5 round trips, 2.77 at 5, 8.36 at 10, 20.9 at 20, and over 30 ns at 40.

The time constant has to be measured against the line’s round trip, because the reflections that a termination is meant to absorb take one round trip to come back. With RCRC a quarter of a round trip the capacitor is charged before the first reflection from the driver arrives, the termination has already become an open end, and the receiver overshoots by 54.8 per cent — most of what an open end would do. At 0.63 round trips it overshoots by 28.7; at 1.58 by 6.1; and below five per cent from about two round trips, a capacitor of eighty picofarads on this line. The rule of thumb that the time constant should exceed the line’s round trip is the right shape and a factor of two short of the number.

Above two round trips the overshoot is gone and a different cost appears. The receiver arrives at the launched level, 2.75 volts, as it would with a plain resistor, and then waits for the capacitor to charge before it reaches the full swing. With two and a half round trips that wait is under two nanoseconds; with ten, 8.36; with twenty, 20.9; with forty, more than thirty. A larger capacitor terminates for longer and holds the receiver short of its final level for longer, and the receiver’s threshold decides whether that matters: a level of 2.75 volts on a 3.3-volt line is well above most logic thresholds and is a different matter for a receiver with hysteresis set near the rail.

The pattern decides the level

A single edge is the wrong test for a termination whose state depends on its history. On a stream of bits the capacitor charges and discharges with the pattern, and how far it gets depends on how long each run of ones or zeros lasts.

A 250 Mb/s pseudo-random pattern received through 50 Ω and a capacitor: the spread of the levels the receiver sees mid-bit, against the capacitor. computed by solving, not by drawing: a seeded pattern of 160 bits of 4 ns marched down the line, each bit read at its middle after the first eight. At 0.5 round trips (20.0 pF), ones from 3.07 to 4.69 V and zeros from −1.39 to 0.23 V; at 1 round trip (40.0 pF), ones from 3.29 to 3.82 V and zeros from −0.52 to 0.01 V; at 3 round trips (120 pF), ones from 3.16 to 3.30 V and zeros from 0.00 to 0.14 V; at 10 round trips (400 pF), ones from 2.92 to 3.25 V and zeros from 0.05 to 0.38 V; at 30 round trips (1.20 nF), ones from 2.91 to 3.13 V and zeros from 0.12 to 0.34 V. A small capacitor lets reflections through on every edge, and the levels ring past both rails; a large one lets its charge follow the pattern, so a long run of ones settles the receiver differently from a short one. The levels are tightest at 3 round trips, 0.14 V of spread, with an eye of 3.03 V.
Fig. 4 A 250 Mb/s pseudo-random pattern of 160 bits received through 50 Ω and a capacitor, each bit read mid-bit. At half a round trip ones range from 3.07 to 4.69 V and zeros from −1.39 to 0.23 V; at one round trip, 3.29 to 3.82 and −0.52 to 0.01; at three, 3.16 to 3.30 and 0.00 to 0.14; at ten, 2.92 to 3.25 and 0.05 to 0.38; at thirty, 2.91 to 3.13 and 0.12 to 0.34. The levels are tightest at three round trips, 0.14 V of spread.

Read at the middle of each four-nanosecond bit, the received levels are tightest with a capacitor of three round trips: ones between 3.16 and 3.30 volts, zeros between 0.00 and 0.14. Smaller capacitors let every edge’s reflections through and the levels ring past both rails — at half a round trip ones reach 4.69 volts and zeros −1.39, which is an input clamp’s problem before it is a threshold’s. Larger ones fail differently. A capacitor of ten round trips holds its charge over several bits, so a long run of ones leaves it charged and the next zero arrives at a different level from a zero after a short run; the spread grows to a third of a volt. That is intersymbol interference made by the termination itself rather than by the line, and it is the same sag the corner that says nothing about an edge measured on a pulse through an AC-coupled input.

So the capacitor has a band, and the band is set by two different times. Below it the line’s round trip wins and the reflections return; above it the bit pattern’s run lengths win and the level wanders. For this line and this rate the band is centred near three round trips; a slower bit rate moves its upper edge up, a longer line moves its lower edge.

The power it was supposed to save

The power a termination dissipates on an alternating pattern, against the bit rate: a plain 50 Ω resistor and the resistor with a capacitor. computed by solving, not by drawing, from the marched current in the resistor. The plain resistor dissipates 75.6 mW at every rate — half of (2.75 V)² over 50 Ω, since it carries current whenever the line is high. With the capacitor the resistor carries current only while the capacitor charges: 5.86 and 17.79 mW at 10 Mb/s; 17.40 and 34.06 mW at 30 Mb/s; 38.28 and 39.62 mW at 100 Mb/s; 40.77 and 40.80 mW at 300 Mb/s for capacitors of three and ten round trips. At low rates it is proportional to the rate — a fixed energy per edge. As the bits shorten towards the capacitor's time constant the capacitor stops following each bit and holds the line's mean, and the resistor dissipates only the swing about it: 40.8 mW, near the 37.8 mW of half the launched level over 50 Ω, about half the plain resistor's.
Fig. 5 The power a termination dissipates on an alternating pattern, against the bit rate. The plain 50 Ω resistor dissipates 75.6 mW at every rate. With 120 pF in series it dissipates 5.86 mW at 10 Mb/s, 17.4 at 30 and 40.8 at 300; with 400 pF, 17.8, 34.1 and 40.8. At high rates the capacitor holds the line’s mean and the resistor dissipates only the swing about it.

The plain resistor dissipates 75.6 milliwatts on an alternating pattern at every rate: half of (2.75 V)2(2.75\ \mathrm{V})^2 over fifty ohms, because it carries current whenever the line is high and an alternating pattern is high half the time. The capacitor changes what is being paid for. At ten megabits a second, with a hundred and twenty picofarads, the resistor carries current only while the capacitor charges after each edge and dissipates 5.86 milliwatts — a fixed energy per edge, so proportional to the rate, 17.4 at thirty megabits. That energy is close to the one any capacitor charged through a resistor loses, half its capacitance times the square of the step: for a hundred and twenty picofarads and 3.3 volts, 0.65 nanojoules, against the 0.59 a bit that the march finds at ten megabits, where each bit is long enough for the capacitor to charge fully. The energy two capacitors cannot share found the same half lost whatever the resistance in the path, and the termination’s resistor is where it goes; the line’s impedance in series only decides how quickly.

At high rates the saving shrinks and stops at a half. Once the bits are short against the capacitor’s time constant the capacitor no longer follows each bit; it sits at the line’s mean, and the resistor sees the swing about that mean in both directions. That is 40.8 milliwatts at three hundred megabits with either capacitor, close to the 37.8 that half the launched level across fifty ohms would give. The AC termination saves all of a parallel termination’s power on a line that sits still, and about half of it on a line that toggles fast — which is exactly the line that needed terminating.

Two capacitors in one termination

The capacitor in a real AC termination is a component with its own inductance, and the edge it must pass is the fastest thing on the line. The capacitor that is an inductor found a hundred-nanofarad part inductive above 14.5 megahertz; an eighty-picofarad part in a small package is a capacitor to a few hundred megahertz and an inductance above, and an edge of a tenth of a nanosecond has its energy up to a few gigahertz. At those frequencies the termination is fifty ohms in series with a nanohenry, which reflects the fastest part of the edge however large the capacitor is. The march here has an ideal capacitor and so a perfect termination at the edge; a real one puts back a small, fast reflection that the staircase in time would recognise as a mismatch at the far end of a few per cent, lasting as long as the edge.

That sets a lower limit on how fast an edge an AC termination can absorb, and it is independent of the time-constant arithmetic above: the capacitor must be large enough for the round trip and the pattern, and small enough in package to stay a capacitor at the edge’s frequencies. A capacitor of a hundred picofarads in a small package satisfies both on this line, and that combination is why the arrangement works at all.

Why two round trips and not one

The rule that the capacitor’s time constant should exceed the line’s round trip has a reason, and the figures say why its factor is about two rather than one. The reflection from an RC termination grows as 1−e−t/2Z0C1 - e^{-t/2Z_0C}, with the time constant doubled by the line’s own impedance in series with the resistor. A designer who sizes RCRC to the round trip has given the reflection a time constant of two round trips, which sounds ample; but the driver’s ten ohms send back two thirds of whatever returns, inverted, and the far end sees the sum of every pass. By the time the first re-reflection arrives, one round trip after the edge, the termination is already reflecting 1−e−1/2=391 - e^{-1/2} = 39 per cent of what it receives, and the residual ringing compounds over several passes. Only when RCRC is about two round trips is the reflection still small, 1−e−1/41 - e^{-1/4} or about 22 per cent, when the first re-reflection arrives, and the overshoot it adds has dropped under five per cent.

That is also why a series-terminated driver would change the answer: with no reflection at the source there is no second pass, and the capacitor only has to hold out against the first edge’s own duration. The ten-ohm driver is what makes the capacitor’s time constant a matter of round trips at all.

A lumped view, and where it fails

At the far end, after the edge has passed, the line is a fifty-ohm source driving the resistor and capacitor in series, and a lumped model of that is a single RCRC with time constant 2Z0C2Z_0C — the closed form above. The lumped model is right about the termination’s own charging and wrong about everything the line adds: it has no round trip, so it cannot predict the overshoot that comes back from the driver, and it has no delay, so it cannot see the pattern’s runs arriving at the far end in sequence. The heavier input that allows a longer track found a lumped capacitance at a track’s end behaving as part of a resonant circuit with the track’s inductance; the AC termination’s capacitor is the same lumped element on the other side of its resistor, where the resistor damps the resonance and leaves only the time constant.

Where AC termination is the right answer

The figures say when it pays. A line that is idle for long periods — a clock that stops, a bus that is driven only during transfers — spends most of its time held at one level, where a parallel resistor burns its full current and the capacitor burns nothing. A line that toggles continuously at a rate near its own time constant gets back only about half of the parallel termination’s power and pays for it in pattern-dependent levels. And a line driven by a weak driver loses part of its swing to a plain parallel resistor — 2.75 volts of 3.3 here — and gets it back from the capacitor once the capacitor charges, which is a second reason, apart from power, to put the capacitor there.

The receiver that is a branch found interior receivers on a net seeing a different waveform from the end one, and an AC termination changes that picture too: an interior receiver sees the far end’s reflection grow as the capacitor charges, so its level creeps like the far end’s, a round-trip’s distance behind.

What the march leaves out

The line is lossless and the driver a linear ten ohms. A real driver’s output resistance differs between rising and falling edges and changes with its current, which makes the launched level and the reflections at the driver asymmetric; and a real line’s loss rounds the edge, which the edge the loss delivers measured, and with it shifts the frequencies the capacitor must pass. The termination’s resistor is exactly the line’s impedance; a mismatch there leaves a small reflection at the edge that no capacitor affects. And the pattern is one seeded sequence of a hundred and sixty bits, long enough to contain a run of nine equal bits but not the longer runs a real data stream can carry, which would push the large capacitors’ wander further than the pattern figure shows.

The number to keep

A 50 Ω resistor in series with a capacitor reflects 1−e−t/2Z0C1 - e^{-t/2Z_0C} of an edge: nothing at first and everything eventually. With a 10 Ω driver it lets the receiver overshoot by more than five per cent until its time constant passes about two round trips, holds a pseudo-random pattern’s levels tightest near three, and lets them wander with the pattern above that. It saves all of a parallel termination’s power on an idle line and about half on a fast one — 40.8 mW against 75.6 at 300 Mb/s.

Still open: the capacitor’s own inductance, the driver that is series-terminated, and the termination to a mid-rail

The capacitor’s inductance, measured. A termination capacitor of a stated package has an inductance that reflects the fastest part of every edge. Putting that inductance into the march would say how fast an edge an AC termination can absorb before its own package becomes the mismatch.

A series-terminated driver. With the driver’s resistance raised to fifty ohms there is no reflection at the source, and the far end’s capacitor no longer has reflections to worry about — only the pattern. Whether the capacitor’s band then shifts to smaller values, and whether a doubly terminated line with a capacitor at the far end keeps the half-swing cost the earlier essay found, is the comparison that would settle which end the capacitor belongs at.

A termination to a mid-rail. A Thévenin termination to half the supply draws current in both states and has a mean the line naturally sits at. Comparing its power and its levels with the capacitor’s, at the same bit rates, would complete the account of what a bus pays to be terminated.

Part 4 on termination

One argument about Termination, and one of 4 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

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 Part 3 — 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.

Where this goes next

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. 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. 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.

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.

AC couplingBaseline shiftCharacteristic impedanceDesign tradeoffReflection coefficientTerminationTransmission line