What the cure at the base costs

The resistance that is below zero found that an emitter follower fed through a hundred nanohenries of wire presents a negative output resistance between 110 and 301 megahertz, that a capacitance resonating inside that band oscillates, and that the cure is a resistor in series with the base. It then bisected the smallest such resistor that puts every pole back into the left half plane and reported eight to seventy-nine ohms.

That number is correct and it is not a design value, and the difference between those two statements is the whole of this rung.

What a bisection on a sign converges to #

A bisection is defined by its predicate, and the predicate here is does any pole have a positive real part. The routine therefore converges on the resistance at which the worst pole’s real part is zero — and it converges well, to the arithmetic’s floor.

At forty-seven picofarads it returns 69.7 ohms, and at that resistance the worst pole pair has a real part of about 10⁻⁷ radians a second beside an imaginary part of 9 × 10⁸. The quality factor of a pole pair is $\omega_n/2\sigma$, so that pair’s is 1.7 × 10¹⁵ — which is not a property of the circuit at all. It is the number a double-precision root-finder produces when it has been asked to put something on an axis and has succeeded.

Whether the last bisection step lands a hair above zero or a hair below is also arithmetic. Either answer describes the same circuit: a resonator, at 146 megahertz, with no loss in it.

The single most useful reading in that figure is not the ratio of sixteen. It is that thirty per cent above the bisected value the quality factor is still 11.1. A design margin of the size anybody would apply to a resistor — a preferred value, a tolerance, a round number — leaves a follower whose worst pole pair still has a Q of eleven. The bisected value is not a low estimate that a margin repairs; it is the wrong quantity, and no margin on it converges on the right one.

What eleven reads as on a bench #

A quality factor is a distance from the axis and it is also a count of cycles, which is the form a bench meets it in. The pole pair’s envelope decays as $e^{-\sigma t}$ and it rings at $\omega_d/2\pi$, so the number of cycles a disturbance takes to fall to a per cent of itself is $\ln(100)\,\omega_d/2\pi\sigma$, and the pole locations give it directly.

At the thirty-per-cent margin above the bisected value — 90.6 ohms, with 47 picofarads on the emitter — the pair sits at 139.6 megahertz with a Q of 11.10, and a disturbance takes 16.25 cycles to settle to a per cent. At the Q-of-one resistor it takes 1.27 cycles, at 44.4 megahertz, and the envelope’s time constant has gone from 25.28 to 6.21 nanoseconds. At the Q of one over root two that puts the response flat it takes 0.73 cycles.

Sixteen cycles of ringing at 140 megahertz on the output of a unity-gain buffer is not a stability problem in the sense that word usually carries — nothing grows without bound, and the circuit passes any test that asks whether it oscillates. It is a stage that turns every edge presented to it into a burst, and on a board where the follower drives a comparator or a sampler it is the thing that shows up much later as an intermittent extra transition. The rung below’s number does not merely fail to include a margin; it is compatible with a circuit nobody would ship.

Q is not measured from the response, and that matters here #

There are two routes to how close a network is to oscillating, and the reason for choosing one of them is on the page above.

The first is a swept response: drive the input, read the output, find the peak. That is the measurement a bench makes and it is the one this collection uses whenever it can. Near the stability edge it is useless, because the peak is arbitrarily tall and arbitrarily narrow, and any number a sweep returns for its height is a statement about the grid the sweep was taken on. Two different frequency grids over the same circuit gave 39.5 and 61.6 decibels here, and neither is a fact about the follower.

The second is the poles themselves, rooted from the determinant. That has no grid in it, so it stays meaningful right up to the axis — and it is the same instrument the rung below used, pointed at the magnitude of the real part rather than at its sign.

The two agree away from the edge, which is what makes the first one usable where it is usable. At a Q of one the swept peak is 1.20 to 1.25 decibels across every load drawn, against the $Q/\sqrt{1 - 1/4Q^2}$ that a second-order pair gives, which is 1.249. At a Q of two the sweep reads 5.72 to 6.22 against 6.30, and at five, 12.3 to 13.3 against 14.02. The disagreement grows with Q because the network has three poles rather than two — a real one at −1.49 × 10⁹ per second beside the pair — so the peak is not built by the pair alone — and it grows in the direction that says the sweep is the approximate route, which is why the pole locations are the ones quoted.

This is the same pairing one step computed twice insists on for a transient and two measurements of one margin makes for a loop: a frequency-domain number and a time- or root-domain number, agreeing where both apply, so that the region where only one of them applies is trustworthy.

The two criteria move in opposite directions #

If the stability answer were simply a constant fraction of the damping answer, the rung below’s number would be usable with a multiplier on it. It is not, and the way it fails is the part worth carrying out of this essay.

The sign change is not monotone in the load, and by two hundred and twenty picofarads it is falling. The reason is the mechanism the rung below established: the negative resistance exists over a band of frequency, and a large enough load resonates below that band, so less base resistance is needed to keep the poles on the correct side. Every damping criterion goes the other way, because a larger capacitance makes a slower, heavier resonator that needs more loss to damp.

So the two answers diverge, and they diverge fastest exactly where a designer’s instinct says the problem is easing. A board that oscillated with 47 picofarads and stopped when a bigger capacitor went on has not been fixed; it has moved from a Q that grows without bound to a Q of 5.7, which rings for about six cycles and looks on an oscilloscope like an artefact of the probe.

One resistance covers the whole unstable band of load capacitance at a Q of one, and it is 2.21 kΩ. That is the honest single answer to “what resistor goes in the base”, and it is a factor of thirty above the range the rung below quoted.

What 2.21 kΩ costs, exactly #

A resistor in the base circuit costs three things, and two of them turn out to be small.

The output impedance is the one that is not. The first rung’s expression for a follower’s output resistance is $R_s/(\beta+1) + 1/g_m$, and a base resistor enters it in exactly the place the source resistance does, so it adds $R/(\beta+1)$. At 47 picofarads the Q-of-one resistor is 1.15 kΩ and the solved output impedance goes from 5.664 to 13.097 ohms — an increment of 7.433 against the $(1154 - 30)/151 = 7.444$ the expression gives, agreeing to a part in seven hundred. The stage’s whole reason for existing is that number, and the cure has more than doubled it.

The gain is the one that is not. It goes from 0.99965 to 0.99918, which is five parts in ten thousand, and it is small for the reason the base resistor is the received cure at all: the base node carries a signal current smaller than the emitter’s by $\beta + 1$, so a resistance there loads the signal by a factor of a hundred and fifty-one less than the same resistance at the output.

The noise is also small, and it is worth having the number because the resistance that is below zero asserted it without computing it. Every resistor in the network’s own thermal noise, integrated at the emitter from ten kilohertz to two gigahertz, is 58.46 microvolts for the bare follower and 62.51 with the 1.15 kΩ stopper and the load fitted — seven per cent more, with the base resistor contributing 38.6 of it and overtaking the source resistance as the second largest term. Referred to the input, where a buffer’s noise belongs, those are 58.48 and 62.56 microvolts.

Where that noise comes from is worth one more sentence, because the stopper changes the ranking as well as the total. On the bare follower the three largest contributions at the emitter are $r_\pi$ at 37.7 microvolts, the source resistance at 35.2 and the transistor’s own base resistance at 27.3. With the 1.15 kΩ stopper fitted and the load on, they are $r_\pi$ at 48.5, the stopper at 38.6 and the source resistance at 8.03 — the stopper has become the second largest term and has simultaneously reduced what the source resistance contributes, because it stands between the source and the base and divides the source’s noise down on the way through. The two effects nearly cancel, which is why seven per cent is the whole of the answer.

So the price is one quantity and not three. A base stopper sized to damp a follower costs output impedance and essentially nothing else — which is a clean trade, and is only a clean trade once the size is the one that damps rather than the one that stabilises.

The other resistor, which the model has always accepted #

There is a second place to put a resistor and the machinery here has carried it since the anchor was written: a resistance from the emitter to ground, left throughout at the one gigohm that stands in for an open circuit. Nothing has ever set it, and it is a genuine second cure — a conductance across a resonator damps it, whatever else it does.

It works, and it costs half the signal.

The reason is the same sentence as the one that makes the base resistor cheap, read backwards. Damping a resonator means putting real conductance across it, and the emitter is where the signal comes out — so a conductance there is subtracted from the signal at full size. The follower’s own output impedance is 5.66 ohms, so a 5.68-ohm shunt is very nearly a two-to-one divider, and the measured 0.5006 against the $R_e/(R_e + Z_{out})$ the divider gives is agreement to within the third figure.

The base resistor reaches the same pole pair through a node whose signal current is smaller by $\beta + 1$. It therefore needs to be about two hundred times larger to do the same damping and costs about six hundred times less signal, and the received wisdom about where the resistor goes is that ratio and nothing else.

The factor is 203 rather than 151, and the gap is not slack in the argument. The base resistor damps through $r_\pi$ and $C_\pi$, which is a divider that is not simply $\beta + 1$ at the frequency the pair rings at — the current gain has begun to fall there, so a resistance in the base is more effective per ohm than the direct-current ratio suggests and needs to be correspondingly larger than that ratio would predict to reach the same damping. The two numbers being within a third of each other is what says the mechanism is the current gain and not something else; their being unequal is what says the ratio is a frequency-dependent one.

One thing the emitter resistor is not worse at is noise, and this is the half that could not be guessed. Its integrated output noise at 47 picofarads is 11.70 microvolts against the base cure’s 62.51, and referred to the input — which is the comparison that counts, since it has halved the signal — 23.38 against 62.56. A shunt that low sets the output node’s impedance to under three ohms and takes the noise bandwidth down with it, and its own thermal noise at 5.68 ohms is small.

So the price of the emitter cure is the signal, and only the signal. That is a worse price than the base cure’s output impedance, but it is a stated one, and a stage whose output already drives a low impedance may find it the cheaper of the two.

What this shares with the feedback field’s version #

The same shape of question is answered in a different field with a different criterion, and the comparison says what a good criterion looks like.

The resistor that buys the margin back never computes the smallest resistor that makes the loop stable, and the reason is that nobody in that field would ask for it. The criterion there is a phase margin, quoted at forty-five or sixty degrees, and it has a number in it because the whole point is to stand some distance from the edge. A stability criterion with no distance in it is not something that field has ever offered, and the result is that its cure is sized correctly by default.

The follower has no loop, so there is no phase margin to quote, and the vocabulary that came with the problem — “the smallest resistor that works” — has no distance in it at all. A quality factor is the missing equivalent: it is a distance from the axis, it is read off the same poles, and one is a number a designer can be asked for. The load that gets inside the loop and this essay are then the same question with the same answer, and only the criterion differs.

The relationship between the two is worth stating rather than left implied. A phase margin of 60° corresponds to a closed-loop pole pair with a Q near 0.7 and no peaking; 45° corresponds to a Q near 1.2 and about two decibels. So “sixty degrees” and “a Q of one” are the same request made of two different descriptions of one circuit, and the follower’s version was missing only because nobody had written it down.

What is not in this #

The load is a capacitance and nothing else. A real cable is a transmission line, and at 146 megahertz a metre of it is not a lumped capacitor — the limit Kirchhoff’s own frequency sets. The mechanism survives and the frequencies would move.

The model is small-signal, so a follower that rings does so at an amplitude nothing here computes; the amplitude nothing linear predicts is the machinery for that question.

And the emitter resistance is a small-signal shunt, not a bias network. Five and a half ohms from a 5-milliampere emitter to ground is not a circuit anybody builds; the realistic version is a resistor in series with a capacitor, which damps at the frequencies that matter and passes no standing current. That changes the arithmetic and not the ordering, because the damping still arrives at the emitter and is still subtracted from the signal there at full size.

The number worth carrying #

Sixteen times at forty-seven picofarads, and two hundred and eighty-one times at two hundred and twenty — with the two criteria diverging in the direction that looks safe. One resistor for the whole unstable band is 2.21 kΩ, and what it costs is the output impedance and nothing much else: 5.66 ohms becomes 13.10 at forty-seven picofarads, exactly $R/(\beta+1)$ added.

The habit is about what a bisection was asked. A root-finder converges on its predicate, and a predicate that is a sign converges on a boundary rather than on a design. That is the right thing to compute when the question is whether a boundary exists — which is what the rung below was asking, and its answer stands — and it is never the answer to what to fit. The tell is that the returned value has no units of margin in it: nothing in “eight to seventy-nine ohms” says how far from oscillating, because the quantity it was bisected on has no distance in it.

The general form of that appears wherever this collection bisects, and it is worth naming the difference. The one current a constant is right at and how small is small signal bisect on an error reaching a stated fraction, so the fraction is the margin and the answer arrives with it attached. A band rather than an edge bisects on two errors and returns the interval between them. Those are boundaries with a size in them. A sign change has no size, and the moment a number found that way is read as a component value, the margin has to be supplied from somewhere else — and on this circuit thirty per cent is not nearly enough of it.