The input that pushes back

The buffer that is not a buffer measured what an emitter follower presents: nineteen ohms at low frequency from a kilohm of source, sixty-seven at 29 megahertz, with an interior maximum in the middle of the source-resistance range that cannot be designed around by making the driving impedance smaller or larger.

This is the other end of the same device, and it is worse in kind rather than in degree. The quantity looking into the base does not merely grow. It changes sign.

Where the negative resistance comes from #

Looking into the base of a follower, the impedance is the base-emitter network in series with the load seen through the current gain:

$$Z_{\text{in}} = \left(r_\pi \parallel \frac{1}{sC_\pi}\right) + \left(1 + g_m\!\left(r_\pi \parallel \frac{1}{sC_\pi}\right)\right) Z_E .$$

At low frequency the parallel combination is $r_\pi$, real, the multiplier is $1+\beta$, real, and the whole thing is a real resistance times a load — which is the ordinary story and has no surprises in it.

At higher frequency the parallel combination becomes capacitive, so the multiplier acquires a negative imaginary part. Multiply that by a load that is also capacitive — $Z_E = 1/sC_L$, another negative imaginary — and the product has a negative real part, because two negative imaginaries multiply to a negative real.

That is the whole mechanism, and the striking thing about it is how little it needs. No second device, no feedback loop, no unusual model. A small-signal transistor with its ordinary base-emitter capacitance, and a capacitor on the emitter.

The figure solves it — one ampere injected into the base of the netlist, and the voltage that appears — and with a nanofarad of load the real part is negative from 1.25 MHz upward and reaches −1182 Ω at 3.40 MHz.

A negative resistance is not yet an oscillator #

It is easy to over-read a negative real part. A one-port with negative resistance does not oscillate on its own; it oscillates when it is put in a loop whose total resistance is negative at a frequency where the total reactance is zero. That is the series-resonance condition, and stating it that way is what makes the boundary computable.

So the figure asks the question in two parts.

Where does the reactance cancel? The base sees whatever inductance is between it and its source — a lead, a track, a via, a ferrite bead — plus its own $\mathrm{Im}(Z_{\text{in}})$, which is negative over the region of interest. The sum passes through zero at a frequency the inductance chooses.

Is the total resistance negative there? That is $R_s + \mathrm{Re}(Z_{\text{in}})$ at that frequency, and it is the criterion.

Bisecting the inductance at which the answer flips, with a nanofarad of load:

Seventy-five nanohenries is about seven centimetres of wire, or a short length of track with a poor return path. This is not an exotic condition; it is a follower driven from a low-impedance source through an ordinary lead, which is exactly how a follower is normally driven.

The repair is backwards #

The table above says something that reads as a mistake and is not. A larger resistance in the base makes the stage more stable.

Every other stability repair in this collection goes the other way. A stiffer source is better for a divider, for a filter’s termination, for an amplifier’s noise gain. Here the source resistance is in series with a negative resistance, and the criterion is a sum, so more of it is straightforwardly better: ten ohms tolerates 74.9 nH and a hundred ohms tolerates 913.

This is the arithmetic under a well-known piece of received wisdom — the small resistor in the base of a follower, a few tens of ohms, which appears in circuits without explanation and is sometimes described as “damping”. It is not damping in any oscillatory sense. It is a positive resistance added to a loop whose resistance would otherwise be negative, and its required value is computable from the device’s own parameters and the load.

The cost is what a base resistor always costs: it forms a low-pass with the input capacitance, so the stage’s bandwidth falls, and it adds its own noise. Neither is usually significant at the tens of ohms required.

The dangerous capacitance is the small one #

Sweeping the load capacitance gives a result that inverts the usual advice about capacitive loads.

The most negative resistance has an interior maximum at a few hundred picofarads — the same shape the previous rung found in the source resistance, for the same reason: two frequency dependences moving in opposite directions with a product that peaks between them.

But the column that matters is the last one, and it is monotone the other way. A hundred picofarads is the dangerous load, not three nanofarads. The threshold inductance is 7.8 nanohenries — a few millimetres of track — against 731 nanohenries for a load thirty-three times larger.

The reason is that the reactance null moves. A smaller load capacitance puts the negative-resistance region at a higher frequency, and a higher frequency needs less inductance to resonate with. So the small capacitance is dangerous because it is fast, not because it is large.

That inverts what a designer coming from the feedback-amplifier problem expects. There, a bigger capacitive load is unambiguously worse, and the standard question is “how much capacitance can this drive”. Here the answer is a band with the worst case in the middle, and a hundred picofarads of oscilloscope probe is closer to it than a nanofarad of cable.

What the oscillation looks like when it happens #

The criterion says whether a loop will grow and not what it grows into, and the difference is worth a paragraph because it decides how the fault is recognised on a bench.

A small-signal negative resistance is a statement about a linearisation, and a linearisation is about what happens near the operating point. As the amplitude grows, the device leaves the small-signal region — the transfer curve compresses, the effective transconductance falls, and $C_\pi$ moves with the current. So the negative resistance shrinks with amplitude and the oscillation settles at whatever amplitude makes the loop’s total resistance exactly zero, which is the same amplitude-stabilisation argument the oscillator field makes on purpose.

What that means in practice is that the symptom is a clean sinusoid of modest amplitude, at a frequency in the megahertz to hundreds of megahertz, sitting on top of whatever the stage was supposed to be doing. It does not look like instability in the way a feedback amplifier’s ringing does. It looks like interference, and it is routinely blamed on one — which is why the diagnostic that identifies it is not a measurement of the waveform but a touch on the base lead: anything that adds series resistance there stops it, and nothing that adds shielding does.

The other signature is that the frequency moves with the load capacitance and barely at all with anything else, since it is the reactance null of the base inductance against $\mathrm{Im}(Z_{\text{in}})$.

Why it does not appear in the usual model #

The follower’s small-signal model without capacitances gives $Z_{\text{in}} = r_\pi + (1+\beta)Z_E$, which for a capacitive $Z_E$ is a resistance in series with a capacitance divided by $1+\beta$. No negative real part appears anywhere, at any frequency, and the model is perfectly self-consistent.

What produces the effect is $C_\pi$, and $C_\pi$ is the capacitance that sets the device’s transition frequency: $f_T = g_m/2\pi(C_\pi + C_\mu)$, which here is 962 MHz. So a device with an infinite transition frequency has no negative input resistance at all, and the effect scales with how close the operating frequency is to $f_T$ — which is to say, with how good the transistor is not.

That is worth stating because it explains a piece of engineering folklore. Followers built from slow devices are well behaved; the trouble started when fast devices arrived, because a fast device has a high $f_T$ and therefore holds its current gain up to a frequency where the lead inductances of an ordinary layout are significant. The problem was created by the improvement.

The four repairs, and what each one moves #

The criterion is $R_s + \mathrm{Re}(Z_{\text{in}}) < 0$ at the frequency where $\omega L_b + \mathrm{Im}(Z_{\text{in}}) = 0$, and every fix in circulation moves one of those four terms.

A resistor in the base raises $R_s$. Measured above: ten ohms to a hundred moves the threshold from 74.9 to 913 nanohenries, a factor of twelve for a component that costs nothing. It is the standard repair and it is the one the criterion points at most directly.

Less inductance in the base lead removes the reactance null altogether if it goes far enough — the sum $\omega L_b + \mathrm{Im}(Z_{\text{in}})$ has no zero in the negative-resistance band if $L_b$ is small enough. That is a layout fix, and it is the one that fails silently when a circuit is moved from one board to another.

A resistor in series with the load raises $\mathrm{Re}(Z_E)$, which propagates to $\mathrm{Re}(Z_{\text{in}})$ through the $1+\beta$ multiplier and is therefore extremely effective — but it is also directly in the signal path to the load, so it costs exactly what the isolation resistor of the feedback essay costs: the loop, such as it is, no longer holds the load’s node.

A ferrite bead in the base looks like the second repair and is the first. A bead is an inductance at low frequency and a resistance at high frequency, and the frequencies involved here are usually above the bead’s transition — so it adds tens of ohms of $R_s$ exactly where the criterion wants it, while adding nothing at direct current. That is why it works, and it is not because it “blocks high frequencies”.

The two routes, and what each one cannot do #

The measurement above is a one-port statement: inject a current into the base, read the voltage, look at the sign of the real part. The stability conclusion is then read off the series-resonance condition by hand.

That is one route and it is worth saying why the obvious second one is harder here. Recovering the closed-loop poles of the whole circuit — source, lead, device, load — by sampling the determinant and rooting the polynomial is the route this site uses everywhere else, and on this network it is badly conditioned: the pole cluster spans from the emitter resistor’s kilohertz to the device’s gigahertz, which is seven decades, and a polynomial with roots spread that far has coefficients spanning $10^{7n}$. The trim that keeps that honest is the one the value theorems found had been silently dropping an order, and even repaired it is asking a great deal.

The impedance route asks for none of it. Every quantity in it is a solve at one frequency on the imaginary axis, which is the best-conditioned thing this solver does, and the criterion is a sign rather than a root. So the essay’s boundary is bisected on two solved quantities — the frequency where a reactance passes through zero, and the resistance there — and neither needs a polynomial.

The price is that it is a sufficient condition for oscillation rather than a complete stability analysis: it finds the series resonance and asks the question there, and a network with several reactance nulls would need each of them checked. The figure does check each of them and takes the worst, which is what makes the answer a boundary rather than an example.

The three things a follower gives back #

An emitter follower is reached for to avoid loading something, and it returns three problems in place of the one it solves. This page is the negative input resistance. The buffer that is not a buffer is the inductive output impedance, which peaks into a capacitive load. The resistance that is below zero is where the same negative resistance is measured with a source inductance in front of it and the poles are found. The frequency a device sets for itself is the bandwidth a follower is chosen to avoid spending, and The load that gets inside the loop is the same failure one field over, where the load is inside a feedback path rather than on an emitter. A bias point is a solution, not a choice is where every operating point on this page comes from.

What is checked #

Three assertions, and the third is the one that names the repair.

That a capacitive load makes the input resistance negative, with nothing added to the model — the most negative value and the frequency it occurs at, taken off the solved one-port. At zero load capacitance the same assertion is made in reverse and requires the real part to stay positive everywhere, so the claim rejects as well as accepts.

That the series loop’s resistance goes negative above a computable inductance, found by bisecting on two solved quantities: the frequency at which the total reactance passes through zero, and the sign of the total resistance there.

And that a larger source resistance raises that inductance by more than three times between ten ohms and a hundred — the assertion that makes the repair the opposite of the instinct, and the one a reader is most likely to disbelieve.

Seven centimetres, and where that puts this boundary #

The threshold being an inductance rather than a frequency or an amplitude puts this result in a small class, and the edges that are lengths is where the class is collected: boundaries nobody chooses at the schematic, set by whoever builds the thing, appearing in no netlist at all.

What distinguishes this one from the others on that list is that the length is not even a layout decision. A gap in a core is machined to a dimension; a track’s height above its plane is a stackup chosen once. Seven centimetres of wire between a source and a base is whatever the assembly happens to produce — a lead not trimmed, a connector, a wire looped rather than routed — and it varies between units of the same design built on the same day.

Which is why the remedy has to be a schematic change rather than a mechanical instruction. Raising the source resistance from ten ohms to a hundred moves the threshold from 74.9 nanohenries to 913, which is most of a metre and is outside what any assembly produces. That is not a smaller sensitivity; it is the same sensitivity with the threshold moved past every value the uncontrolled quantity can take, and it is the only form of robustness available against a boundary that is not on the drawing.

The buffer that is not a buffer is the rung below and shows the same mechanism before it changes sign — an emitter presenting 19.08 ohms at low frequency and 67 at 29 megahertz, with the peak worst in the middle of the source-resistance slider, so it cannot be avoided by making the source stiffer or softer. The repair that fails there is the one that works here, which is worth noticing: the two essays are about the same falling current gain and they disagree about what to do with it.

The disagreement is only apparent, and resolving it says what the source resistance is actually doing. It is not making the impedance better; it is damping a loop. The negative resistance looking into the base is a negative element in series with the source and the base lead, so what decides stability is the sign of the total — and adding ninety ohms of positive resistance to a loop containing a thousand ohms of negative one at resonance does nothing, while adding it to a loop where the negative part is a hundred and ten ohms is decisive. The mechanism scales with the reactance rather than with the source, which is why the threshold moves by more than an order for one decade of source resistance.

Which is also why the repair has a ceiling. A source resistance large enough to guarantee stability is a source resistance that has undone the reason for fitting a follower — the buffer that is not a buffer measures the emitter presenting 19.08 ohms at a kilohm of source, and the impedance a follower presents rises with the source resistance it is driven from, because the current gain divides one into the other.

So the repair is bounded above by the reason the stage exists, which is a shape this collection meets often enough to be worth naming: a fix whose cost is measured in the quantity the circuit was built to provide. Ten ohms is comfortably affordable and a hundred is usually affordable; a kilohm would make the follower pointless, and there are lead inductances a kilohm would not cover.