Devices, and the amplitude they stop being linear at

The resistance that is below zero

An emitter follower's output resistance is 5.5 Ω at direct current and −21.9 Ω at 257 MHz, and the sign is not the transistor's: with an ideal source at the base there is no negative band at all, and a hundred nanohenries of wire between the source and the base produces one from 110 to 301 MHz. A capacitance resonating inside that band is a resonator with loss of the wrong sign, so 4.7 to 100 pF on the emitter oscillates while 1 pF and 470 pF do not — a band of load capacitance with quiet ground on both sides of it.

14 min read 8 figures The schematic is a label The parasitic is the component Every model has an edge

Assumes: The buffer that is not a buffer · The frequency a device sets for itself

The buffer that is not a buffer measured what an emitter follower costs at direct current: its output resistance is the source resistance divided by β plus one, plus one over the transconductance, so a low source impedance is repeated with a small series resistance and a high one is not. The input that pushes back measured what the same stage does to whatever is driving it.

Both are single numbers. This rung asks the first question at every frequency rather than at one, and the answer is not a resistance in series with anything: over a band it is below zero.

A follower fed through 100 nH has an output resistance of -21.9 Ω. computed by solving, not by drawing. The real part of the impedance looking into the emitter, driven by a current source and read, at every frequency. At direct current it is 5.50 ohms, which is the first rung's r_s/(β+1) + 1/g_m. Between 110 MHz and 301 MHz it is negative: r_π and C_π delay the current the transistor sources into the emitter, and past a quarter of a cycle of delay pushing the emitter up makes the device push it up as well. The dashed curve is the same follower with no inductance between the source and the base, and it never goes below zero — the sign belongs to the wire and the transistor together, and to neither alone. — Fig. 1 The real part of the impedance looking into the emitter, driven by a current source and read. At direct current it is 5.5 Ω; between 110 and 301 MHz it is negative, worst at −21.9 Ω.

The sign belongs to the wire and the transistor together

The first thing the sweep establishes is what the effect is not. With an ideal voltage source at the base — no source resistance in series with anything reactive — there is no negative band anywhere, at any frequency, for any current. The dashed curve in that figure stays above zero throughout.

The algebra says why, and it is short enough to be worth writing. With Z_π the parallel combination of r_π and 1/sC_π, and R the total resistance in the base circuit,

Z_{out} = \frac{R + Z_\pi}{1 + g_m Z_\pi} = \frac{R + r_\pi + j x R}{(1+\beta) + jx}, \qquad x = \omega r_\pi C_\pi

whose real part, after multiplying by the conjugate, is (R + r_π)(1 + β) + x²R — a sum of positive terms. A follower fed through a resistance cannot present a negative resistance, however fast it is.

Put an inductance in that path and the numerator gains a term −βxωL, which is negative and grows as the square of frequency. That is the whole mechanism: an inductance in the base circuit, which is a wire.

A follower fed through 300 nH has an output resistance of -86.3 Ω. computed by solving, not by drawing. The real part of the impedance looking into the emitter, driven by a current source and read, at every frequency. At direct current it is 5.50 ohms, which is the first rung's r_s/(β+1) + 1/g_m. Between 57.1 MHz and 195 MHz it is negative: r_π and C_π delay the current the transistor sources into the emitter, and past a quarter of a cycle of delay pushing the emitter up makes the device push it up as well. The dashed curve is the same follower with no inductance between the source and the base, and it never goes below zero — the sign belongs to the wire and the transistor together, and to neither alone. — Fig. 2 Three hundred nanohenries — a longer wire, or a ferrite bead fitted for other reasons. The band opens downward and the worst value goes to −86 Ω.

Twenty nanohenries produces no negative band; fifty produces one 0.18 decades wide; a hundred produces one 0.44 decades wide reaching −21.9 Ω; three hundred reaches −86 Ω. Ten centimetres of wire is about a hundred nanohenries.

What the delay is doing

The mechanism has a physical reading that is easier to hold than the algebra. r_π and C_π together are a lag: the base-emitter voltage responds to what the base circuit does with a time constant of its own, and the current the transistor sources into the emitter follows that voltage.

Push the emitter up and the transistor should respond by pulling current out of it, which is what a positive output resistance means. Past a quarter of a cycle of lag, the response arrives inverted: the device sources current into the emitter at the moment the emitter is already high. The inductance is what supplies the extra phase, because it makes the base circuit’s impedance rise with frequency rather than staying put, and a rising source impedance means the base node is progressively less held.

A common-emitter stage with 2.0 pF from collector to base. computed by solving, not by drawing. The stage's midband gain is 144.7 and its −3 dB point is at 504 kHz. The Miller approximation lumps 311 pF at the input and predicts 643 kHz — 21.6% high. The network's second pole is at 336 MHz and its right-half-plane zero at 3.08 GHz, both of which the approximation has no room for. — Fig. 3 The same two internal capacitances doing their other job, from the device-bandwidth rung: C_π and C_μ setting where a stage stops responding. At the two picofarads of C_μ this follower carries, the solved bandwidth is 504 kHz where the Miller approximation says 643 — optimistic by 21.6 per cent. Here the same C_π sets where the follower responds with the wrong sign.

That is why the effect is a band rather than a threshold. Below it there is not enough phase; above it C_π has shorted the base-emitter junction, the transistor has stopped amplifying, and what is left is the ohmic path.

A band of capacitance, with quiet ground on both sides

A negative resistance is a statement about an impedance and not yet about a circuit. Hang a capacitance on the emitter and the two together are a resonator whose loss has the wrong sign, and the question becomes where the network’s own poles are.

A band of load capacitance oscillates, and both ends of it are quiet. computed by solving, not by drawing. The real part of the network's worst natural frequency — recovered by rooting the determinant, so it is a property of the matrix and not of any output — against the capacitance hung on the emitter. Small capacitances are stable, large ones are stable, and 4.7, 10, 22, 47, 100 picofarads are not: they resonate with the wire's inductance inside the band where the output resistance is negative, and a resonator with negative loss oscillates. The frequencies are 291 MHz, 247 MHz, 198 MHz, 156 MHz, 122 MHz. This is why adding decoupling to a ringing follower sometimes cures it and sometimes starts it. — Fig. 4 The real part of the worst natural frequency against the capacitance on the emitter, recovered by rooting the determinant. Small is stable, large is stable, and the middle is not.

One picofarad is stable. Four hundred and seventy picofarads is stable. 4.7 to 100 picofarads oscillates, at 292 down to 122 MHz. The unstable region is a band with quiet ground on both sides of it, which is exactly what the impedance sweep predicts: the load has to resonate with the wire’s inductance inside the frequency band where the resistance is negative, and a capacitance too small or too large resonates outside it.

This explains a piece of bench folklore that is otherwise contradictory. Adding decoupling to a ringing follower sometimes cures it and sometimes starts it — and both are true, because the cure is to move the resonance out of the band and there are two directions to move it in.

Two poles at ζ = 0.1, recovered from the matrix. The poles are at -159.2 ± j1584 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows. — Fig. 5 Where a network’s behaviour is written down. The poles here are the same object, and the question is only which side of the axis they are on.

How a bench meets this

The symptom is a sine wave at a few hundred megahertz on the output of a stage that has no business producing one, usually of a few hundred millivolts, usually stable in amplitude, and usually disappearing when a probe is put on the emitter — because a probe is ten picofarads and a metre of cable, and it moves the resonance.

That last property is the reason the fault survives to production. An oscilloscope probe changes the circuit it is measuring, which the probe is part of the circuit measures in general, and here the change is not a loading error of a few per cent: it is the difference between a circuit that oscillates and one that does not. A fault that vanishes when observed is a fault that gets shipped.

Fig. 6 How a bench meets this: a base stopper, drawn at three hundred nanohenries of source inductance rather than a hundred. The resistor is placed where the negative resistance is, and what it does is add enough positive real part to keep the sum on the right side of zero over the band where the follower’s own is most negative.

The second observation that identifies it is that the frequency does not depend on anything obviously tuned. It is set by the wire’s inductance against the load capacitance — two quantities nobody chose — so it moves when the board is re-laid out, moves when a different cable is fitted, and does not move when the signal, the supply or the temperature change.

The cure, and why it is a resistor at the base

The standard repair is tens of ohms in series with the base, and the reason it works is the reason the negative resistance exists: the resistor damps the r_π–C_π lag whose delay produced the wrong sign. Bisected on the pole locations rather than taken from a rule of thumb, it is 8 to 79 Ω across the unstable range of load capacitance.

Fig. 8 The smallest base resistance that puts every pole back in the left half plane, for each unstable load. Tens of ohms, not hundreds.

It is worth noticing where the resistor is not. It is not in series with the emitter, where it would appear in the output impedance and defeat the point of a follower; and it is not across the load, where it would draw current. It is in the base circuit, where the signal current is smaller by a factor of β and the resistor’s own noise is divided by the same factor before it reaches the output.

The cost is real and small: a few tens of ohms in the base adds R/(β+1) — under half an ohm — to the output resistance, adds its own thermal noise referred to the input, and slightly lowers the stage’s bandwidth. A follower on a board with a resistor at its base that no analysis of the circuit as drawn would ask for is a circuit whose designer met this.

What this shares with the feedback field’s own load problem

The load that gets inside the loop opens a ladder about an operational amplifier driving a capacitance, and the shape of the answer there is the same: a load that is harmless in the model becomes a stability problem because a real output stage cannot hold a node instantly.

The difference is instructive rather than cosmetic. That ladder’s mechanism is a pole the load adds inside a feedback loop, which eats phase margin, and its cure is a series resistor at the output. This one is a negative resistance in an open-loop two-terminal impedance, and its cure is a series resistor at the input. Both are one component and neither is the same component.

There is a third relative worth naming. The frequency a device sets for itself found a right-half-plane zero at exactly $g_m/C_\mu$ in a common-emitter stage — a feature of the device’s own transfer function that no external component produced. The negative resistance here is not that: it needs the external inductance, and with an ideal source it does not exist at all.

A negative resistance is a component

It is worth saying plainly that this is not an anomaly but a device, because the same arrangement is used deliberately.

A two-terminal element with a negative real part supplies power at its own terminals, and putting one across a resonator is exactly how an oscillator is built: the negative resistance cancels the resonator’s loss, the amplitude grows until something limits it, and what comes out has the resonator’s frequency. The gain that is exactly one builds the same object out of a feedback network and an amplifier, and the two descriptions are the same circuit seen from different terminals.

So the follower above is a negative-resistance oscillator that nobody designed, tuned by a stray, and the reason it is a fault rather than a product is only that its frequency and amplitude were not chosen. A microwave designer builds the same thing on purpose and calls the inductance a base inductor.

Which current, and which transistor

Two of the model’s parameters move the whole picture and it is worth saying which way.

The collector current sets $g_m$ and therefore both $r_\pi$ and the transit time. More current is more transconductance, which lowers the direct-current output resistance — the number the first rung computed — and simultaneously raises C_π, which brings the negative band down in frequency. A follower biased harder is a better buffer and a worse citizen.

The base resistance of the device itself, $r_b$ , is in series with whatever the external circuit provides and is a few tens of ohms for a small transistor. It is therefore already part of the cure, and it is why the effect is worse for large devices: a power transistor’s $r_b$ is a fraction of an ohm and its C_π is nanofarads, so it has almost none of the built-in damping a small one has.

Two routes to the same instability

The poles above are recovered from the determinant, which is a property of the matrix and makes no reference to any output. There is a second route that shares none of that arithmetic, and the two are required to agree.

Take the impedance sweep — driven and read, one solve per frequency — and ask where a capacitance’s reactance equals the impedance’s imaginary part with the real part still negative. That is a Nyquist-style question about a two-terminal impedance rather than a root-finding question about a polynomial, and it predicts the same band of load capacitance: 4.7 to 100 picofarads, from the frequency range 110 to 301 MHz over which the real part is below zero.

The agreement is worth having because the two routes fail differently. Rooting a determinant is sensitive to the polynomial’s conditioning — a defect this field found and fixed, and which the digits the arithmetic did not have is about — while a sweep of an impedance is not, and is instead sensitive to the resolution of the grid it was swept on. Neither error mechanism is available to the other.

What is not in the model

No package. The transistor here is a hybrid-π with two capacitances and a base resistance, and its own lead inductance — a nanohenry or two of emitter lead, in particular — is absent. Emitter lead inductance is the other classic cause of follower oscillation and it acts by a different route, so a real device has two mechanisms where this model has one.

No large signal. Everything is small-signal about a bias, so the model has no slew rate, no clipping and no current limit. A follower that oscillates does so at an amplitude set by the nonlinearity that eventually limits it, and nothing here says what that amplitude is — the amplitude nothing linear predicts is the machinery for that question, applied to a different oscillator.

And no distributed effects. At 250 MHz a hundred nanohenries of wire is 30 centimetres of it, and 30 centimetres is a fifth of a wavelength — so treating it as a lumped inductance is on the edge of what Kirchhoff’s own frequency permits. The mechanism survives; the exact frequencies would not.

What the first two rungs were measuring

Seen from here, the two rungs below were both measuring the same impedance at a point where it is well behaved, and the interesting thing is that neither of them could have found this.

The first rung’s number, $r_s/(\beta+1) + 1/g_m$ , is the zero-frequency limit of the expression above. It is correct, it is useful, and it is the value at one end of a curve that changes sign in the middle. The second rung’s input impedance is the same network looked at from the other pair of terminals, and it has its own frequency dependence — a negative real part appears there too, for a capacitive load rather than an inductive source, which is the dual and is the more commonly quoted half of the folklore.

That duality is exact and worth stating: a follower with an inductive source and a capacitive load presents a negative resistance at both ends. Which one causes trouble depends on which end has a resonator attached, and a real board usually has both.

The number that changed sign

An output resistance is a ratio of two things, and this collection has spent a ladder of essays establishing that such ratios are numbers only over a range. This one is not a number over any range: it is 5.5 Ω at direct current, 20 Ω at a hundred megahertz, −21.9 Ω at 257, and 1 Ω at ten gigahertz.

What makes it worth a rung of its own is the sign. A quantity that grows or shrinks by a factor of ten outside its range is a model losing accuracy. A quantity that changes sign is a model losing its meaning: a positive resistance dissipates and a negative one supplies, and the circuit’s behaviour on the far side of the crossing is not a worse version of what it was doing before.

The other negative resistance in the collection

There is one more, in a field with no transistors in it, and the comparison says which of the two is the dangerous kind.

The inductor that is an amplifier builds an inductance out of four resistors, a capacitor and two amplifiers, finds it good to within one per cent over three and a half decades — and finds its series resistance going negative at 63 hertz, well inside the band where the inductance is still excellent. A resonator built around it there does not have a high quality factor; it has a loss of the wrong sign, and starts on its own noise.

The two are the same phenomenon with opposite provenance. There the negative resistance belongs to the arrangement — it is a property of the gyrator, present in every build, computable from the component values, and avoidable by staying above 63 hertz. Here it belongs to the wiring: an ideal source at the base produces no negative band at all, and a hundred nanohenries of lead produces one, so two units of one design can differ in whether the band exists.

Which is why this one is worth the essay and that one is worth a paragraph. A boundary that is in the design can be designed around; a boundary that appears when somebody leaves a lead long is one that passes every test on the bench it was measured on. The edges that are lengths is where the collection keeps the rest of that class, and this is the only member of it whose consequence is an oscillation rather than an error.

The band of load capacitance with quiet ground on both sides of it is the detail that makes this difficult to find on a bench. A fault that appears above a threshold is found by sweeping upwards; a fault that appears between 4.7 and 100 picofarads and not at 1 or 470 is found only by somebody who happened to fit a value inside it — and the natural response to an oscillation, which is to add capacitance, walks the design out of the band from one side or deeper into it from the other.

Part 3 on emitter follower

One argument about Emitter follower, 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: