The pole that crosses with the resistance

The resistor that is not made of the resistors reduced a bootstrap to a Thévenin pair and found the resistance that no combination of its resistors can make. A source drives node $p$ through $R_1$; $p$ reaches node $q$ through $R_2$; and $q$ is held at $A$ times $p$ by a controlled source. The current $R_2$ takes from $p$ is then $v_p(1 - A)/R_2$, so $R_2$ behaves as $R_2/(1 - A)$ — larger than itself below unity gain, infinite at unity, negative above it — and the resistance looking into $p$ is $R_1$ in parallel with that. With two ten-kilohm resistors it runs from five kilohms at no gain, through infinity at $A = 2$, into negative values beyond.

That essay asked what a negative Thévenin resistance means and answered carefully: the equivalent is exact, and its resistor is not a resistor. It left one question for later, and it is the question a circuit designer asks first. Is the network stable? A reduction at direct current contains no time, so it cannot answer; the answer needs a capacitor, and a capacitor is what every real node has. This essay adds one, finds the network’s natural frequency, and then asks how far the obvious guess — that the network is unstable exactly when its resistance is negative — survives contact with a load, a lag, a resonator and a source of the other sign.

One capacitor, one pole #

Put a nanofarad from node $p$ to ground. The network now has one natural frequency, found as the root of its determinant at each gain, and it is a plain number: the rate at which a disturbance at the node grows or decays.

The natural frequency is −G/C to a part in a million at every gain drawn, where $G$ is the Thévenin conductance the reduction gave: −200 thousand per second with no gain, rising linearly through zero at $A = 2$, where the resistance passes through infinity, to +100 thousand at $A = 3$, where it is minus ten kilohms. Below the crossing a disturbance at the node decays; above it, it grows, and the node runs until something not in the linear model — a supply rail, a saturating amplifier — stops it.

So, for this network, the guess holds exactly. The node’s equation is $C\,dv/dt = -Gv$, the Thévenin conductance is the only conductance the capacitor sees, and its sign is the sign of the decay rate. The capacitance does nothing to the crossing: a tenth of a nanofarad makes the pole ten times faster on both sides of $A = 2$ and leaves the crossing where it was. That is a useful fact to have measured, because it is the reason the direct-current reduction is worth anything for stability at all. But the network has been chosen so that it is true, and the rest of the essay is the list of ways it stops being true.

A load decides whether a negative resistance is stable #

The first is the most important. A Thévenin resistance is what the network looks like from outside; what the capacitor at the node sees is the network and whatever is connected to it. Hang a load resistor across the node and the conductance the capacitor sees is the network’s plus the load’s.

Loaded by $R_L$, the node goes unstable at $A = 1 + R_2(1/R_1 + 1/R_L)$: 7 for a two-kilohm load, 4 for five kilohms, 3 for ten, 2.5 for twenty, 2.2 for fifty, each bisected on the natural frequency and agreeing with the expression to every printed figure. Between $A = 2$ and the loaded crossing the network’s resistance is negative and the node is stable, because the load’s positive conductance is larger than the network’s negative one. A negative resistance is stable across a load smaller than itself and unstable across a larger one, and with no load at all — an open circuit, an infinite resistance — it is unstable at any negative value.

That reverses the usual intuition about loads. Lightly loading a negative resistance destabilises it; heavily loading it — shorting it, nearly — makes it safe. A tunnel diode biased into its negative-resistance region is stable when driven from a low-impedance source and oscillates when driven from a high one, and the resistance that is below zero found an emitter follower’s negative output resistance doing exactly that against a capacitive load. Stability is a property of the network and its load together, and a Thévenin resistance is a property of the network alone; the direct-current reduction can only say where the boundary lies once the load is named.

A lag moves the speed, not the crossing #

The controlled source here is instantaneous. A real one — an amplifier — has a bandwidth, and a lag in a feedback loop is the classic way to destabilise it.

Give the source a first-order lag of a microsecond, then ten, and the node has two natural frequencies instead of one. The rightmost still crosses zero at exactly $A = 2$, for every lag, and above it grows more slowly: at $A = 3$, 100.0 thousand per second with an instant source, 78.2 with a one-microsecond lag, 30.3 with a ten. The lag has moved the speed of the runaway and not the gain at which it starts.

The reason is the sign of the feedback. The bootstrap is positive feedback — the source pushes the node’s far end in the same direction as the node — and a positive-feedback loop with a low-pass lag in it reaches a loop gain of one first at direct current, where the lag is no lag at all. Its instability is a runaway rather than an oscillation, and a runaway is decided at zero frequency, which is where the Thévenin reduction lives. That is the deeper reason the reduction predicted the crossing correctly in the first figure, and it is a condition rather than a coincidence: it fails for feedback of the other sign, as the last section shows.

A band in which the network supplies power #

With a lag in the source, the network’s impedance looking in is a function of frequency, and it is worth looking at directly rather than through the poles.

The real part of the admittance is $G_0 - (A/R_2)/(1 + \omega^2\tau^2)$, where $G_0$ is the conductance with the source turned off. At $A = 2.5$ it is negative below 79.6 kHz, where the lagging source still keeps up with the node, and positive above, where it does not. A negative real part means the network delivers power to whatever is connected to it at that frequency; direct current is simply the bottom of the band in which it does so.

That changes what the reduction is for. A Thévenin resistance is the admittance’s value at one point; the question a designer needs answered is whether anything connected to the node can resonate inside the band where the real part is negative, because a resonance there is fed by the network. At $A = 1.5$ the band is empty — the real part is positive at every frequency — and nothing connected to the node can be driven into oscillation by it. At $A = 2.5$ it runs from zero to 79.6 kHz, and the next figure puts something in it.

A tank in the band is an oscillator #

Hang a parallel tank across the node — a millihenry coil with some series resistance, and 27 nanofarads — resonant at 30.6 kHz, inside the negative band.

Without the tank, the node at $A = 2.5$ has a real pole at +40 thousand per second and runs to a rail. The tank’s coil shorts the node at direct current, which removes the runaway outright: the conductance the node sees at zero frequency is now infinite. What is left is a complex pair near the tank’s own frequency, 30.5 kHz, and whether it grows is a contest between the network’s negative conductance in the band and the tank’s own loss. With twenty ohms in the coil the pair decays at 9.25 thousand per second; with ten, at 4.25; with five, at 1.75; with one ohm it grows at 0.26 thousand per second — an oscillator, the rate crossing zero near one and a half ohms.

That is how every negative-resistance oscillator is built, from a tunnel diode to the crystal oscillator inside a microcontroller: a network with a negative real part over a band, and a resonator inside the band whose loss is smaller than the negative part. The gain that is exactly one designed an oscillator as a loop with a gain of one; the same oscillator described from the node is a negative conductance cancelling a tank’s positive one, and the condition is the same equation read from a different place. The inductor that is an amplifier is a reminder that the tank need not be made of coils at all.

A positive resistance that does not promise stability #

The last case turns the source round. With an inverting controlled source — $A$ negative — the bootstrap’s resistance is $R_1 \parallel R_2/(1 - A)$, positive and falling at every gain. The direct-current reduction says, correctly, that nothing about it is negative anywhere.

Give that source three equal lags of a microsecond, as a real multi-stage amplifier has, and the node is stable only down to $A = -16$, where a complex pair crosses the axis at 276 kHz and the network oscillates. The Thévenin resistance there is 556 ohms, positive, and shrinking. The crossing is the classic three-lag result: the loop through the source and the divider has a gain of $|A|\,R_1/(R_1 + R_2)$, half of $|A|$ here, and three equal lags make 180 degrees at $\sqrt3/2\pi\tau$ with a gain of eight needed to reach one — so $|A| = 16$.

So the direct-current reduction is a reliable stability test only for the runaway kind of instability, the kind that crosses at zero frequency. A negative Thévenin resistance warns of a runaway; a positive one says nothing about an oscillation, because an oscillation crosses at a frequency the reduction never looked at. What is left at crossover is the tool for that kind: the loop gain cut and measured, its phase read where its magnitude is one. The two tests are complements, not alternatives, and the bootstrap needs both depending on the sign of its gain.

A pole at zero is a singular matrix #

There is a second way to see why the crossing at $A = 2$ is where the direct-current reduction breaks, and it connects this essay to the solver rather than to the circuit. A natural frequency is a value of $s$ at which the network’s matrix — conductances plus $s$ times capacitances — is singular: the network can sustain a voltage pattern with no source driving it. A natural frequency at zero is therefore a value of the gain at which the conductance matrix alone is singular, which is to say a gain at which a direct-current solve has no unique answer.

At $A = 2$ that is exactly what happens. The reduction’s own solve, looking into the node with the sources dead, refuses: the conductance matrix has a zero pivot, and the network does not determine its own node voltage. The answer that is perfect and absurd met that refusal as a fault in a netlist — a wire written as a resistance, a node with nothing holding it. Here it is not a fault. It is the circuit announcing, through the arithmetic, that it has a pole at the origin; and the matrix that is ill, and the answer that is not is the general warning that a matrix approaching singularity says something about the problem, not only about the precision.

So the direct-current reduction can find the gain at which a runaway begins without ever adding a capacitor: it is the gain at which its own solve becomes singular. What it cannot find is a crossing at a frequency, because a direct-current solve is the matrix at one value of $s$, and an oscillation needs the matrix singular somewhere on the imaginary axis. That is the arithmetic form of the conclusion the last section reached by circuit reasoning.

Where real circuits sit on this map #

The bootstrap is the smallest network with this behaviour, and several familiar circuits are it with different values filled in.

A bootstrapped input. An amplifier’s input bias resistor, returned to the output of a follower rather than to ground, is a bootstrap with $A$ just below one: the resistor appears many times larger than itself, which is the point — a high input resistance from a modest resistor. It sits far from the crossing at $A = 1 + R_2/R_1$, and its pole is in the left half-plane by a wide margin; its risk is the lag of the follower, which the figures above say cannot destabilise a positive-feedback loop of this kind below the direct-current crossing.

A current source built to sit on the pole. The Howland current source is a bootstrap deliberately tuned to $A = 1 + R_2/R_1$, where the resistance looking in is infinite and the output current independent of the load. It is the second figure’s arrangement exactly: stable across any load smaller than its own resistance, which an infinite resistance makes every load — until resistor tolerances put its resistance a little on the negative side, and an open-circuited output then runs to a rail. The next essay measures how often.

A negative-impedance converter. A bootstrap run well past the crossing is used on purpose to cancel a positive resistance — the loss of a line, the resistance of a coil in a gyrator — and is stable only because it is always connected across something smaller than itself. The load figure is its design rule, and its failure mode is the connection being opened.

A cross-coupled pair. Two transistors each driving the other’s input are a bootstrap whose gain is the product of two stage gains, and above the crossing they latch rather than oscillate, because their feedback is positive and their crossing is at zero frequency. Where the behaviour is written down puts a circuit’s whole future in its poles; for the latch it is one real pole on the wrong side of the axis.

What the poles were computed from, and what they assume #

Every natural frequency is a root of the network’s determinant: the netlist assembled into its conductance and capacitance matrices, the determinant sampled on a circle and fitted as a polynomial, and the polynomial rooted. The controlled source’s lags are buffered RC stages in the netlist rather than a transfer function written down, so the lags are solved like everything else. The single-pole case is held against $-G/C$ to a part in a million at every gain; the loaded crossings are bisected on the rightmost root and held against $1 + R_2(1/R_1 + 1/R_L)$ to a part in a million; the inverting crossing is held against the three-lag gain of sixteen to a part in ten thousand. The admittance is solved at each frequency with a test source at the node and held against $G_0 - (A/R_2)/(1 + \omega^2\tau^2)$ to the solver’s precision.

All of it is linear. Above a crossing, the model says a node grows without bound; a real node grows until a rail or a saturating source stops it, and what it does then — latches, as the circuit with three bias points found for a pair of cross-coupled transistors, or settles into an oscillation of some amplitude — is a nonlinear question. The tank’s oscillation grows at 260 per second at one ohm, which says it starts; how large it becomes is set by whatever limits the source, as it was for the Wien oscillator’s diodes.

Still open: the amplitude, two nodes, and the reduction with s in it #

The amplitude the runaway stops at. Give the controlled source a saturating characteristic — a tanh, as a differential pair has — and the bootstrap above $A = 2$ has two stable states rather than a runaway, and with a tank it has a limit cycle whose amplitude is set by how far the source’s gain must fall for its average negative conductance to equal the tank’s loss. That is a describing-function calculation with a marched check, the same pair the oscillator essays made, asked of a node rather than a loop.

Two nodes with negative resistance between them. A negative resistance in series between two nodes, each with its own capacitance to ground, has two natural frequencies, and the stability condition is no longer a comparison of one resistance with one load. Whether the crossing is still at a gain the direct-current reduction names — and which reduction, since a two-port has several — is the first step from a one-port to a network.

The reduction carried to every frequency. The admittance figure is the Norton conductance as a function of frequency, and its zero crossings are where a resonance can be fed. The Norton form near the pole measures why that form, rather than the Thévenin one, is the one to carry: at $A = 2$ the conductance is zero and finite while the resistance is undefined.