The circuit with three bias points

A bias point is a solution found a diode’s operating point as the root of a transcendental equation, and the bias the gain does not set did the same for a transistor across a tenfold range of current gain. Both circuits had one root. Newton found it from any reasonable start, and the only question was how many steps it took.

That property is not a property of operating points. It is a property of circuits with negative feedback, or with none, and the moment a circuit feeds its output back to its input with the sign that reinforces it, the equations that define its operating point can have more than one solution. Every one of them satisfies Kirchhoff’s laws to the last digit. The circuit can sit at some of them and not at others, and nothing in the solution itself says which.

Two transistors, each biased from the other #

The circuit is the oldest memory in electronics. Two transistors each have 1 kΩ to a 5 V supply in the collector and a 10 kΩ resistor from base to ground; each base is fed from the other transistor’s collector through a coupling resistor $R_B$. If the first transistor conducts, its collector is low, the second’s base is starved, the second is off, its collector is high, and that high collector keeps the first transistor’s base fed. The same argument works with the roles exchanged.

That argument finds two operating points and misses the third. With $R_B$ = 10 kΩ, the solve finds:

The symmetric point is as much a solution as the other two. Both transistors conduct, equally; each collector sits at 1.702 V; each base is fed through $R_B$ by exactly the current it draws. Kirchhoff’s current law closes at every node to a part in 10⁹, checked by rebuilding it from the Ebers–Moll equations directly rather than from anything the iteration produced. By every test an operating point can be put to, it is one.

The answer a simulator finds first #

Newton’s method needs a starting guess, and a circuit simulator’s default is the obvious one: every junction voltage at nought, which is the state of a circuit before its supply is connected. For this circuit that start is exactly symmetric — both transistors at the same guess — and Newton’s iteration preserves symmetry exactly. The linearised network at a symmetric guess is symmetric, its solution is symmetric, the next guess is symmetric, and the iteration converges, in fifteen steps, to the only root the symmetric subspace contains.

So the cold start returns 1.702 V on both collectors. It is the answer a DC operating-point analysis prints for this circuit, and it is the one answer that describes no state the circuit is ever in.

That is not a numerical accident and no tolerance cures it. The iteration converged; the residual is below 10⁻⁹; the root is real. What is wrong is the question. The circuit’s DC equations have three solutions, and “the operating point” is a phrase that assumed one.

The basin map says how much of the space of starting guesses belongs to each answer. The diagonal, where the two guesses are equal, belongs entirely to the symmetric point, for the reason above. Near the diagonal the symmetric point still captures most starts — a small asymmetry in the guess is removed by the first few steps faster than it grows — and only well off the diagonal does the iteration fall towards whichever transistor started more strongly on. The boundaries are not straight, and along them there are fourteen starts from which the damped iteration wanders for four hundred steps without arriving anywhere. Those fourteen are caught by their residual, which stays near one; they are not answers, and they are drawn as their own colour rather than assigned to the nearest root.

Which roots the circuit can hold #

A real circuit has capacitance at every node — the junctions have it, the wiring has it — and at a DC operating point it is the capacitances that decide what happens to a small disturbance. With a positive capacitance $C_k$ at each node, the circuit near an operating point obeys $C\,\dot v = -J\,\delta v$, where $J$ is the Jacobian of the nodal currents at the root: the same matrix a Newton step solves.

That gives a test that needs no capacitance values at all. If $\det J < 0$, then $J$ has a real negative eigenvalue — an odd number of them — and $C^{-1}J$ has one too for any positive diagonal $C$, so $-C^{-1}J$ has a growing mode. The root cannot be held, whatever the capacitances are. At the symmetric point with 10 kΩ of coupling the determinant is negative. At both latched points it is positive.

The Jacobian here is written out by hand — four nodal equations from the Ebers–Moll laws and their sixteen derivatives — and its determinant is taken by elimination with pivoting. It is the second route in the figure: the netlist Newton finds the roots, and the hand-written equations both confirm each root and classify it.

The determinant’s sign is a sufficient condition for instability and not a necessary one. A positive determinant can still hide a pair of unstable eigenvalues, and whether a latched state is truly stable depends on the capacitances. Here the latched states have one transistor fully off and the loop broken by it, so no pair of eigenvalues has anywhere to come from.

The loop gain the determinant stands for #

The determinant has a more familiar name. Around the symmetric point the circuit is two inverting amplifiers in a ring, and a ring of two inversions is positive feedback. Its loop gain is the product of the two stages’ incremental gains: each is the transconductance at the solved current into the collector’s load — $R_C$ in parallel with the next stage’s base network — divided down by the coupling resistor into that base.

At 5 kΩ the loop gain is 380: a disturbance of a microvolt at one collector returns, a loop later, as 380 microvolts of the same sign. The symmetric point is balanced on that gain like a pencil on its point. As the coupling weakens the gain falls, and at 56.6 kΩ it reaches one. Past that, a disturbance comes back smaller than it left, the symmetric point becomes stable, and the two latched states have gone.

The loop-gain computation is a hand estimate from the solved slopes and it lands within five per cent of one at the fork; the determinant is exact and it changes sign at the fork by construction, since that is how the fork is located. The two are one statement. A determinant that crosses zero as a parameter moves is a pitchfork: the symmetric root persists on both sides and the two asymmetric roots are born from it at the crossing.

That is the same pairing of a loop’s gain with its stability that stable, and unstable with less gain draws for a frequency response — with the difference that here the loop is at DC and its instability is not an oscillation but a choice.

The fork, and what the circuit is either side of it #

Below the fork the circuit is a memory: two stable states and an unstable one between them, and which stable state it occupies depends on which it was last pushed into. Above the fork it is an amplifier — a ring of two stages with loop gain below one, one stable operating point, and an incremental gain from any input to any node that is large near the fork and finite everywhere.

The failures above the fork are worth a sentence of their own, because they reverse the intuition that one root is an easy problem. At 100 kΩ the coupling is weak and each base node is held mainly by its own 10 kΩ. The failing starts lie where a guess begins with a junction far into conduction, and the likely mechanism is the step limit’s: such a guess is pulled back by at most 0.15 V a step, overshoots, and the damped iteration circles rather than settles. The mechanism is inferred from where the failures sit, not measured step by step. Every one of those starts is caught by the residual, so nothing wrong is reported. What is lost is only the answer, and a simulator in that position reports a convergence failure rather than a number — which is the refusal this collection’s solves are built to make.

A root that is perfect and absurd #

The symmetric point has a relative in another network, and the comparison sharpens what kind of wrong it is. The answer that is perfect and absurd is a network whose equations have a solution that satisfies every law exactly and describes nothing physical, because the network was posed so that the arithmetic could not see what was missing. The symmetric point is the opposite case: the network is posed correctly, the arithmetic sees everything, and the solution describes a state that exists — for an instant, balanced, the way a coin can land on its edge. It is not absurd. It is unheld.

The distinction matters for what a solver should say. An absurd answer should be refused, and the network’s solver refuses singular networks for exactly that reason. An unheld answer should be reported and labelled, because it is real and it is useful: the symmetric point is where a latch’s decision is made, the metastable state every synchroniser is designed around, and the coupling at which it stops being unheld is the boundary between a memory and an amplifier.

The same trap waits in a hand analysis. The standard technique for a symmetric circuit is to cut it in half along its axis of symmetry and solve one half — which is how a differential pair is analysed, and correctly, because a pair’s symmetric point is its operating point. Applied to this circuit the technique finds 1.702 V on both collectors and nothing else, because a half-circuit cannot represent a state in which the halves differ. It is the cold-start Newton iteration done on paper, and it fails for the same reason: symmetry is preserved by the method, so a method that starts symmetric can only reach the symmetric answer.

The loop gain is what distinguishes the two cases, and it is the thing a half-circuit analysis never computes. A differential pair’s two halves are coupled through a shared tail with negative feedback on their sum; this circuit’s halves are coupled through each other with positive feedback on their difference. The gain that is exactly one finds an oscillator’s boundary where a loop’s gain reaches one at a frequency; here the loop reaches one at zero frequency, and what grows is not a sine wave but the difference between two collectors.

What the step limit does to the basins #

The basin map is a picture of the method as much as of the circuit, and changing the method changes it. Newton’s raw step on an exponential is enormous — a volt of error across a junction asks for $e^{40}$ — so every practical implementation limits it, and this one allows 0.15 V a step. The limit decides the paths, and the paths decide the basins.

With a step a third the size, the symmetric basin falls from 437 starts to 275 and the non-converging starts rise from 14 to 54. Neither answer is more correct; both are the same three roots. What moved is which of them the iteration reports for a given guess, and that is the plain statement of the problem: for a circuit with several operating points, the answer is a function of the solver as well as of the circuit. A simulator’s option that changes its damping can change the printed operating point of a latch, and the change is not a bug.

What a solver should do instead #

Knowing this, there are three honest ways to answer the question “what is this circuit’s operating point”, and one dishonest one, which is to report the first root found.

Find all of them. Starting from a spread of guesses and de-duplicating, as the figures here do, finds every root this circuit has. For larger circuits there are systematic methods — continuation from a circuit with a known single root, following each root as a parameter moves — and they are how a simulator reports that a circuit has several states.

Classify them. The Jacobian determinant is already computed by the last Newton step. Its sign costs nothing, and a root with a negative determinant should never be reported as the circuit’s state without saying so.

Or ask the history. A real latch is in the state its past put it in, so the physically meaningful operating point is the end of a transient from a stated initial condition — power applied, with whatever asymmetry the parts have. That is a march, not a Newton solve, and the amplitude nothing linear predicts is what marching a nonlinear circuit costs.

How the roots were found and checked #

Six starting guesses per coupling, spread over both latched states, the diagonal and two off-diagonal points, each run to a residual below 10⁻⁸ and de-duplicated to a microvolt. Below the fork six starts always find three roots and above it one, which the fork figure requires on every coupling drawn more than three per cent from the fork.

Every root is checked against Kirchhoff’s law written by hand from the Ebers–Moll equations — not from the netlist, not from the companion model — and required to close to a nanoampere.

The fork is located by bisection on the symmetric root’s determinant, stopped at four parts in a million. Bisecting further walks onto the fork itself, where the Jacobian at the root is singular — that is what a fork is — and the network solver, correctly, refuses a singular matrix rather than returning a number from it.

The basin maps are 1,681 solves each, and every start that does not land on a known root is required to have failed visibly, with a residual above 10⁻³, rather than to have converged to something the six-start search missed.

A perfectly symmetric, static pair #

The circuit is exactly symmetric. Real transistors differ, and a mismatch tilts the picture: one latched state’s basin grows, the symmetric root moves off the diagonal, and the cold start no longer lands on it exactly. What a pair cancels, and what it only halves is the same symmetry in a circuit that uses it rather than breaks it.

Nothing here is dynamic. The instability test uses capacitances only through their sign, and how fast the symmetric point is left — which is what decides how long a latch takes to resolve a near-balanced input — needs their values.

And the Early effect is absent, which would give each collector a finite output resistance and slightly lower both stages’ gains, moving the fork to a smaller coupling resistance.

Still open: resolving time, the tilted pair, and the latch as a comparator #

How long the symmetric point takes to leave. A latch released from exactly balanced inputs departs from the unstable point with a growth rate set by the loop gain and the node capacitances, $e^{t/\tau}$ with $\tau$ about $C/g_m$ divided by the loop gain less one. The time to reach a decision grows as the logarithm of how close to balance it started, and that logarithm is metastability — the reason a synchroniser has a failure rate rather than a guarantee.

A mismatched pair. With the two transistors’ saturation currents differing by a per cent, the fork unfolds: one latched branch connects smoothly to the symmetric one and the other appears abruptly at a coupling of its own. The asymmetry that selects a state at power-up is then a number, and so is the input offset a latch used as a comparator has.

The latch as the decision element of a comparator. Driving the two bases differentially and then closing the coupling turns an analogue difference into a digital state, and the hysteresis this introduces is the latch’s two states seen from the input. Two thresholds because there is a floor prices hysteresis from the noise side; the latch prices it from the circuit side.