Lines, where a wire has a length

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 asserted: 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.

Assumes: The staircase in time · A ladder is not a line

The resistor at the wrong end measured three ways of terminating one net and found that two of them are worst at opposite ends of the same piece of track. A series resistor at the driver leaves an interior receiver sitting at half the swing for $2(1-x)$ delays; no resistor at all leaves it ringing for $2x$ ; a parallel resistor at the receiver has no such interval anywhere and draws sixty milliamperes for as long as the level is held.

That essay closed by naming two arrangements it could not measure. One was a stub, and the receiver that is a branch put a junction in the lattice and measured it: three lines of equal impedance meeting present each other with $Z_0/2$ , so a wave is reflected by exactly $-1/3$ and two thirds goes on, and the far end holds two thirds of the swing for twice the stub’s own delay whatever the net is terminated with.

The other was termination at both ends — a series resistor at the driver and a parallel one at the receiver. It is what gets built when neither scheme alone will do, and what it costs is the one thing everybody knows about it. What it buys has not been put a number on.

Fig. 1 The same net as the three-way comparison with a fourth trace on it. Both reflection coefficients are zero, so the wave that arrives at a receiver halfway along is already the final value and there is no second arrival to wait for: the departure after the first edge is 0.0×10⁰ per cent. What it costs is the level — 1.650 volts of 3.3, exactly half. The series scheme in the same place is undefined for 1.00 delays and the unterminated one overshoots by 82 per cent.

What is being solved, and what the half is

Nothing new. The same lattice, with the source resistance set to $Z_0$ and the far-end resistance set to $Z_0$ as well.

The launched wave is decided by a divider between the source impedance and the characteristic impedance — not the load, which the source has no way of knowing about for another delay. With $R_s = Z_0$ that divider is exactly one half, which is where the series scheme’s half-amplitude wave comes from and why the staircase in time draws a first step rather than a full one.

In the series scheme the open far end has $\Gamma_L = 1$ and doubles it back to the full swing. Here the far end is $Z_0$ , so $\Gamma_L = 0$ and there is nothing to double it with. The receiver gets the launched half and keeps it. Measured: 1.650 volts of a 3.300-volt supply, which is a half to four figures, and it is a half at every position along the net rather than at one of them.

That is 6.0206 decibels of signal, given away at the driver, in exchange for something the rest of this essay measures.

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. — Fig. 2 The three schemes the rung below compared, at the same halfway point. The series-terminated net sits in the undefined band for 4.83 nanoseconds, which is one delay; the unterminated one sits in it for the same 4.83 and then overshoots by 82 per cent; the parallel-terminated one never does and draws 60.0 milliamperes while it holds the level. This is the picture the fourth trace above is added to.

The first thing it buys is the interval, and that is not the interesting one

An interior receiver’s whole difficulty in the series scheme is that it sees two arrivals at different times and the first one is deliberately half the answer. Set $\Gamma_L$ to zero and the second arrival does not exist, so there is nothing to wait for.

Measured at a tenth of the way along the net — the worst position the rung below found, where the series scheme is undefined for 1.80 delays — the double scheme’s departure from its final value after the first edge is zero to the arithmetic’s floor.

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 10% 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 1.80 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. — Fig. 3 A tenth of the way along, where the reflected wave has furthest to come back. The series scheme holds half the swing for 1.80 delays here; the double scheme has settled at 1.650 volts the instant the edge arrives, and it is the same 1.650 volts it was at halfway and will be at the far end.

This much follows from $\Gamma_L = 0$ and could have been written down. It is also what the parallel scheme already does, at a higher level and for more current, so on its own it is not a reason to fit two resistors instead of one.

The reason is that both of those schemes are correct only if the resistances are what the design said they were, and the two of them fail differently when they are not.

A driver is not fifty ohms minus the resistor

A logic output’s own impedance is a transistor’s channel, and it moves with process corner, supply and temperature by tens of per cent. A series termination is sized as $R_{\text{series}} = Z_0 - R_{\text{out}}$ , so an error in $R_{\text{out}}$ is an error in the source match — and the source match is the whole of what makes a series termination settle in one round trip.

Swept from twenty ohms to a hundred, the series-terminated net’s departure from its final value after the first arrival is exactly the source reflection coefficient it was supposed to have removed. A driver thirty per cent low, at 36 ohms, gives $|\Gamma_s| = 0.163$ and 16.3 per cent of ringing; at twenty ohms it is 42.9 per cent, and at a hundred it is 33.3.

A driver 30 per cent off costs the series scheme 16 per cent of ringing and the double scheme none. computed by solving, not by drawing. The same net with the resistance at the driver swept from 20 to 100 ohms, which is what a logic output's own impedance does over process and temperature. The series scheme's departure from its final value after the first arrival is exactly |Γs| — 16.3 per cent at 36 ohms — because the open far end returns the whole wave for the mismatched source to reflect again. Terminated at both ends there is nothing to return, so the departure is 0e+0 per cent at every point on the sweep. The price is the other pair of lines: the series scheme's level is exactly the supply whatever the driver does, and the double scheme's is the divider between the driver and the line — 58.1 per cent at 36 ohms against 50.0 at 50. The error has moved from the time axis to the amplitude axis. — Fig. 4 The departure from the final value against the resistance at the driver, for the two schemes that have a resistor there. The series curve is a V through zero at fifty ohms and is $|\Gamma_s|$ exactly; the double curve is flat on nothing across the whole sweep. The other pair of lines is the price: the series scheme’s level is the supply whatever the driver does, and the double scheme’s is the divider between the driver and the line — 58.1 per cent at 36 ohms against 50.0 at fifty.

The doubly terminated net’s departure over the same sweep is zero at every point, and the reason is one sentence: a mismatch at the source only matters if something comes back for it to reflect. With $\Gamma_L = 0$ nothing does. The driver’s impedance decides how large the launched wave is and decides nothing else.

That claim needed care to state exactly, and the care is worth recording because it is the same trap one step, computed twice is about. Asserted as “the departure equals $|\Gamma_s|$ ” the worst row of the sweep disagreed by $4.3\times10^{-8}$ — far too large for rounding, and unchanged when the number of samples was multiplied by thirty-two, so not a discretisation error either. It is the open circuit: the far end is a resistor standing in for infinity, so $\Gamma_L$ is $1 - 2Z_0/R_{\text{open}}$ rather than one, and the departure falls short by exactly that factor. Predicted in that form the agreement is a part in a million at every driver impedance. A tolerance widened until the assertion passed would have hidden a finite resistor inside a claim about an infinite one.

The error has not gone away — it has changed axis

The same sweep says the other half, and the two halves are a matched pair.

The series scheme’s settled level is the supply, exactly, at every driver impedance on the sweep, because an open far end draws no current for the series resistor to drop. Its shape is at the driver’s mercy and its amplitude is not.

The double scheme’s settled level is $V_s Z_0/(R_s + Z_0)$ , exactly, at every driver impedance on the sweep: 71.4 per cent of the supply at twenty ohms, 58.1 at thirty-six, 50.0 at fifty, 33.3 at a hundred. Its amplitude is at the driver’s mercy and its shape is not.

So the driver’s tolerance is not removed by either scheme. It is moved. One puts it in the time axis, where it appears as ringing that has to be waited out and that a receiver may sample in the middle of; the other puts it in the amplitude axis, where it appears as a swing that is smaller than expected and that stays where it is put. Those two failures are charged to different budgets and only one of them is a timing hazard — which is a different observation from “double termination is more tolerant”, and it is the one the sweep actually supports.

The load nobody specified

The far end is the other place a resistance is assumed. A series termination assumes an open, and a real far end is a receiver pin, a second receiver, a pull-up, a connector with something on the end of it, a test point with a probe on it.

An interior receiver sees $1/(1+\Gamma_L)$ of its own answer on the first arrival, so what happens to it is decided entirely by what is at the far end. Against the site’s own criterion — the band between thirty and seventy per cent of the final value, in which a logic input has no defined answer — the series scheme’s interior receiver climbs out of the band only below one far-end resistance, and that resistance is 125 ohms.

Only one far-end resistance keeps a series-terminated interior receiver out of the band, and it is 125 Ω. computed by solving, not by drawing. What an interior receiver's first arrival is worth, as a fraction of the level the net will settle to, against whatever resistance is actually at the far end. The series scheme relies on the far end doubling its half-swing, so the first arrival is 1/(1+Γ_L) of the answer: at a real open it is 52 per cent — dead in the band a logic input cannot read — and it climbs out of the band only below 125.0 ohms, which is Z₀/(1−2b) for a band edge of 30 per cent and has no length, no driver and no frequency in it. Terminated at both ends the first arrival is already the answer, so the curve never falls below 100 per cent at any load: 102.32 per cent at a kilohm. What is bought is immunity to a load nobody specified; what is paid is that the level itself is now 1.61 V rather than 3.3. — Fig. 5 The first arrival as a percentage of the level the net will settle to, against whatever resistance is actually at the far end. The series curve enters the undefined band above 125 ohms and sits at 52.3 per cent at a kilohm; the double curve never falls below a hundred per cent at any load, reaching 102.32 per cent at the same kilohm. The boundary is bisected on the solve rather than inverted from the expression it agrees with.

That number is worth reading carefully, because it has nothing in it. It is $Z_0/(1-2b)$ for a band edge $b$ — 83.3 ohms at a twenty per cent band, 125 at thirty, 250 at forty — and it carries no length, no driver impedance, no rise time and no frequency. It is $2.5\times Z_0$ at the band this collection uses, which is far below any far end anybody builds deliberately, so the series scheme’s interior receiver is in the band essentially always. The rung below measured that as an interval in delays; this is the same statement made as a resistance, and it is the form that says the interval cannot be designed away by choosing a better far end.

The double scheme’s curve never enters the band from any direction. Its first arrival is already the answer, so a far-end resistance that is not $Z_0$ makes it an overshoot rather than a plateau — 102 per cent at a kilohm, and above a hundred per cent everywhere — and an overshoot is a level a receiver can read.

What it costs, in the currency it is paid in

Half the swing is the headline and it is not quite the right comparison, because the series scheme does not deliver the supply into a real load either.

Half the swing at an open far end, and three quarters of it at twenty-five ohms. computed by solving, not by drawing. What each of the four schemes finally delivers, against the resistance actually at the far end. With nothing there but its own terminator the double scheme gives 1.650 V of 3.3 — exactly half the 3.298 V the series scheme gives, because the wave is halved by the divider and never doubled by a reflection. It is the lowest of the four at every load on the axis. The ratio to the series curve is (Z₀+R)/(Z₀+2R) exactly, which is where the half comes from and which says the second resistor costs less as the far end gets heavier: at 25 ohms the series scheme has fallen to 1.100 V and the double scheme to 0.825, a ratio of 0.750 rather than a half. A scheme that costs a stated fraction is a budget item; one that costs an unknown interval is not. — Fig. 6 What the four schemes finally deliver against the same far-end resistance. Against an open far end the double scheme gives 1.650 volts to the series scheme’s 3.298. The ratio between them is $(Z_0+R)/(Z_0+2R)$ exactly — a half only at an open end, and 0.750 at twenty-five ohms, where the series scheme has fallen to 1.100 volts and the double scheme to 0.825.

The ratio is worth having as an expression rather than as a factor of two, because it says which way the cost moves. The second resistor is a shunt across the far end, so its effect is largest when there is nothing else there and smallest when the far end is already heavy. Against an open it halves the swing. Against twenty-five ohms it takes a quarter. The cost of double termination falls as the thing it is protecting against gets worse, which is an unusually convenient shape and is not obtainable from the word “half”.

The current is the other half of the price and it is smaller than the parallel scheme’s. Holding a level, the double scheme draws the supply through two resistances rather than through a low driver and one, so it is 33.0 milliamperes and 108.9 milliwatts against the parallel scheme’s 60.0 milliamperes and 198.0 — fifty-five per cent of it. That is a real saving and it is the wrong way round from the usual account, in which fitting a second resistor is described as costing more power.

Where that current goes matters more than how large it is. A source below a frequency measures a regulator’s output impedance at 0.43 milliohms at direct current and 1.95 ohms at ten kilohertz, and the pair that is worse than either finds a decoupling network presenting six times the impedance either capacitor does alone at the frequency between them. Thirty-three milliamperes switching at a clock rate is a current step into a rising impedance, and what it produces is a voltage on a rail everything else shares — the millivolts in the wire is what that looks like arriving somewhere it was not budgeted.

Where the half is not affordable

The scheme’s cost is a fixed fraction of the swing, and a fixed fraction of the swing is exactly what a logic family’s thresholds are quoted as.

The rung below put it plainly for the series scheme’s interior plateau: half of 3.3 volts is 1.65, which is above every input low threshold and below every input high threshold in use. That was a description of a transient there. Here it is the settled level, so the same arithmetic says that a doubly terminated net driven from a 3.3-volt output cannot present a high level to an ordinary receiver at all — not for a delay, but permanently.

Which is why the arrangement is not used with plain single-ended logic and its thresholds, and is used where the receiver’s reference is not a fixed fraction of the driver’s supply: a differential pair, where the comparison is between two conductors and the common level is free, or a receiver biased to the middle of the swing it is actually going to get. The trade is a signal-to-noise ratio given away for tolerance to a load nobody specified, and it is only available to a receiver that has the margin to give.

This is the same shape as two thresholds because there is a floor: a decision boundary is only as good as the amplitude presented to it, and halving the amplitude halves the distance to whatever is underneath. The difference is that hysteresis buys margin back and a resistor divider does not.

What it does not fix, and this is the part that matters

A branch is not touched by any of it.

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. — Fig. 7 The same 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 receives 66.7 per cent of the swing at one line delay and is held there for 0.400 line delays, being twice the stub’s own 0.41 nanoseconds. The junction’s coefficient is set by three conductors meeting and by nothing at either end of the net.

The junction’s reflection coefficient is exactly $-1/3$ whether the net is terminated at neither end, at the driver, at the receiver or at both, because it is decided by three lines of equal impedance meeting at a point. So the two-thirds deficit and its recovery after twice the stub delay survive double termination unchanged, and they are now two thirds of a halved swing.

That inverts the reason for fitting the second resistor. The arrangement removes the interval that belongs to the termination scheme and leaves the interval that belongs to the layout — and the second is the one that scales with how the parts are placed rather than with what is on the schematic, which puts it among the edges that are lengths. A designer who has fitted two resistors and moved on has bought immunity to the resistances and none at all to the geometry.

What is checked

Four assertions on the waveform figure and four on each of the sweeps, and none of them is a value at one setting.

That the level is exactly half the supply, to a part in $10^{12}$ , at every position the slider offers — which is a statement about a divider between two equal resistances rather than about a number.

That the departure from the final value after the first arrival is zero, and the undefined interval is zero, at every position and at every driver impedance from twenty ohms to a hundred.

That the series scheme’s departure over that same sweep is $|\Gamma_s|$ exactly, short only by the finite resistor standing in for an open far end, stated in the form that includes it.

That the interior receiver’s first arrival never falls below its own final value at any far-end resistance from twenty-five ohms to a hundred kilohms — which is the immunity claim, made as a bound rather than as an example.

And that the ratio of the two settled levels is $(Z_0+R)/(Z_0+2R)$ , to a part in a million across the whole load axis, rather than a half asserted under a tolerance wide enough to swallow the difference.

What this does not say

It does not say the double scheme is the right answer. It costs a fraction of the swing that a single-ended receiver usually cannot pay, and the fraction is fixed rather than negotiable.

It does not say the lattice is the right model. Everything here is resistive at both ends, and a real receiver is a capacitance as well — which slows the edge at the far end rather than reflecting it, and which the lumped boundary a ladder is not a line measures decides whether any of this is needed at all. On a metre of the line drawn here the round trip is 9.67 nanoseconds, so a twenty-nanosecond edge needs none of it.

And it does not say anything about the impedance the whole argument is matched to. Where the current comes back is where $Z_0$ comes from, and it is a geometry rather than a component value: a track whose return path is interrupted does not have the impedance the terminator was chosen for, and two resistors matched to the wrong number are worse than one matched to the right one.

The number worth carrying

Half the swing, and $Z_0/(1-2b)$ .

The first is what double termination costs and it is exact: two equal resistances divide the supply and no reflection ever puts it back, so 1.650 volts of 3.300 at every point on the net and at every driver impedance the part happens to have.

The second is what it buys, written as the thing it removes. An interior receiver on a series-terminated net has no defined level for any far-end resistance above $Z_0/(1-2b)$ — 125 ohms at this collection’s band — and that number contains no length, no driver, no rise time and no frequency, so it cannot be designed around by making the net shorter or the edge slower. It can only be removed by putting $Z_0$ at the far end, and putting $Z_0$ at the far end is what halves the swing.

The habit that goes with it is about where an error goes rather than whether it exists. Neither scheme removes the driver’s tolerance; the series one turns it into ringing and the double one turns it into amplitude. Deciding between them is deciding which of those two a receiver can survive, and that is a question about the receiver rather than about the line.