The resistor that is not made of the resistors

Exact outside and wrong within takes six elements — two sources and four resistors — and reduces them to one source and one resistor that no load can distinguish from them. It finds the resistor by two routes that share no arithmetic: looking in with the sources dead, and dividing the open-circuit voltage by the short-circuit current. They agree to twelve figures, and the reduction is exact about every voltage and every current outside the two terminals.

Everything in that essay is a network of resistors and independent sources, and for such a network there is a third route that everybody actually uses: combine the resistors. With the sources dead, a voltage source is a short and a current source is an open, and what is left is a network of resistors whose resistance between two nodes can be worked out by series and parallel reduction.

That third route stops working the moment a dependent source is in the network, and so does part of the second. This essay measures what is left.

The smallest network that shows it #

The network is a bootstrap and it has three elements besides the source. A five-volt source drives a node through a ten-kilohm resistor; the node also reaches a second ten-kilohm resistor; and the far end of that second resistor is held at A times the node’s own voltage by a controlled source.

The current the second resistor takes from the node is $(v - Av)/r_2$, which is $v(1 - A)/r_2$. So the resistor behaves as $r_2/(1 - A)$ — larger than itself below unity gain, infinite at unity where the two ends move together and the resistor carries no current at all, and negative above unity where the far end moves further than the near one and pushes current back in.

The resistance looking into the node is $r_1$ in parallel with that, and the solve returns it to a part in a billion without being told the expression. At no gain it is five kilohms, the parallel combination. At unity gain it is ten kilohms, which is $r_1$ alone. At $A = 1 + r_2/r_1 = 2$ the two conductances cancel exactly and there is no resistance at all — the network at that gain does not determine its own node voltage, and the solve refuses it by name rather than returning a large plausible number. Above two it is negative.

That is the arrangement the current that does not reach the input uses to make a leakage resistance disappear: a guard ring held at the input’s own potential is exactly this bootstrap with A near one, and the ten-to-the-eighteenth ohms that essay measures is the $r_2/(1 - A)$ of an amplifier whose gain is not quite one. Here the gain is taken past one deliberately.

Four numbers the resistors can make, and a curve that is none of them #

Two resistors can make four resistances between them: either alone, their sum, and their parallel combination. For ten kilohms and ten kilohms those are 10 k, 10 k, 20 k and 5 k.

Swept across the gain, the network’s Thévenin resistance crosses each of those at exactly one gain and is something else the rest of the time. It is not that the combination rule gives the wrong answer; it is that the answer is not in the set the rule can produce. Above the pole it is negative, and no arrangement of two positive resistors is negative.

The reason is the one sentence that does all the work in this essay. “Setting the sources dead” means setting the independent sources dead. A dependent source is not a source in that sense; it is part of the network’s own law, an element whose current or voltage is determined by something else inside the network, and killing it would be changing the network rather than de-energising it. So what is left when the independent sources are zeroed is not a network of resistors, and there is nothing to combine.

The route that does survive is the one that puts a test current in and reads the voltage that appears. That is what a bench measurement does, and it makes no assumption about what the network is made of.

The short-circuit current does not move at all #

One of the two surviving routes is $V_\mathrm{oc}$/$I_\mathrm{sc}$, and on this network it has a property worth stating separately because it is exact and it is not obvious.

The short-circuit current is 500 microamperes at every gain. With the node shorted to ground its voltage is zero, so the dependent source — which is controlled by that voltage — produces zero, so the second resistor has both ends at ground and carries nothing. The whole of the short-circuit current comes through $r_1$ from the five-volt source, and it is 5 V over 10 kΩ whatever A is.

So the open-circuit voltage is simply the short-circuit current times whatever the resistance is, and it follows the resistance everywhere: 2.5 volts at no gain, 5 at unity, 10 at A = 1.5, 500 at A = 1.99, and negative above the pole. A five-volt source in a network of two resistors produces an open-circuit voltage of five hundred volts, which is a perfectly ordinary consequence of positive feedback and is the number that says the reduction is describing something other than a divider.

That also settles which of the two surviving routes is the safer one. $V_\mathrm{oc}$/$I_\mathrm{sc}$ needs the network to have both an open-circuit voltage and a short-circuit current, and at the pole it has neither: the open-circuit voltage is unbounded and the ratio is an infinity over a finite number. The test-current route reports a conductance of exactly zero there, which is a true statement that the ratio cannot make.

The equivalent is exact, and its resistor is not a resistor #

The reduction’s own promise is that no load can tell the difference, and that promise survives everything above.

Swept across five decades of load on both sides of the pole, the whole network and its two-element equivalent agree to about one part in ten to the fourteenth — the solver’s own arithmetic. At a gain of 1.5 the equivalent is a ten-volt source behind twenty kilohms. At a gain of 2.5 it is a source of minus ten volts behind minus twenty kilohms, and it is exactly as good.

But the netlist will not accept it. The solver refuses a negative resistance by name, with the same message it gives for a zero one: a resistance is a positive number, and a negative one is not a resistor. It is a controlled source — a voltage-controlled current source whose transconductance is negative — and writing it as one is what makes the equivalent buildable. That refusal is correct and it is the sharpest statement of the essay: the reduction is exact, and its result is not made of the kind of part the reduction promised.

One load makes even the equivalent singular. With a Thévenin resistance of minus twenty kilohms, a load of exactly twenty kilohms cancels it, and both the network and its equivalent have no answer — refused by name, in both. Nineteen kilohms gives 190 volts at the node and twenty-one gives minus 210, which is what an unbounded response either side of a pole looks like when it is sampled rather than plotted.

What a negative resistance means here, and what it does not #

A negative Thévenin resistance is a statement about a direct-current solve, and it is worth saying precisely what it does and does not claim.

What it claims is that pushing current into the node lowers its voltage: the network supplies power to whatever is connected rather than absorbing it. That is a real and measurable property, and it is what a negative-impedance converter is for.

What it does not claim is that the circuit works. A solve at s = 0 finds the operating point a network would sit at, and says nothing about whether it will stay there. A negative resistance loaded by anything with the wrong sign of slope has an operating point that is a balance rather than an equilibrium, and the smallest disturbance takes it away — which is why this same arrangement, with a capacitor on the node, is an oscillator rather than a source. Nothing in a resistive equivalent can express that, because the equivalent has no time in it.

The same distinction turns up in a different place. The answer that is perfect and absurd is about a network whose solve is numerically flawless and physically meaningless, and this is its companion: a network whose solve is correct, whose reduction is exact, and whose reduced form is a part that does not exist. Neither is a failure of the arithmetic.

The heat, which is now worse than a factor #

Exact outside and wrong within found the reduction understating the network’s dissipation by a factor of forty-three, because two sources inside it push current round a loop the load never sees. Here the same comparison has no factor at all.

At no gain the network burns 1.944 milliwatts into a four-kilohm load and its equivalent says 0.694 — the familiar kind of error. At a gain of 1.5 the network burns 1.863 milliwatts, of which the dependent source supplies 0.188: it is delivering power into the network rather than dissipating it, which is what a controlled source is for and what makes the whole bootstrap possible.

Past the pole the equivalent reports the network dissipating minus 5.55 milliwatts while 4.01 milliwatts are actually being burned. A negative dissipation is not a large error or a small one; it is not a quantity of the same kind. The equivalent’s internal power is $(V_\mathrm{oc} - V_\mathrm{load})^2/R_\mathrm{th}$, and with $R_\mathrm{th}$ negative that expression is negative, and nothing about the reduction ever promised it would not be.

Every power in the real network does add up, at every gain, to a part in a billion: what the independent source delivers plus what the dependent source delivers equals what the three resistors burn. That check is what says the accounting is complete rather than the model being convenient.

A test source is a measurement, and it has a procedure #

The route that survives is worth setting out as a procedure rather than a formula, because it is the one a reader will have to use on a network they cannot reduce.

Zero every independent source — a voltage source becomes a short, a current source becomes an open — and leave every dependent source exactly as it is, still controlled by whatever it was controlled by. Then connect a source of one ampere between the two terminals and solve. The voltage that appears across those terminals, in volts, is the Thévenin resistance in ohms.

Three things about that procedure are worth noticing. It never asks what the network is made of, so it works on a network containing dependent sources, transformers, gyrators or anything else linear — including the reciprocal two-ports the reading that does not care which way round it is is about. It returns a signed number, so a negative resistance comes out negative rather than as an error. And it is one solve, where combining resistors is a sequence of steps each of which has to be right.

The procedure also has a failure mode, and it is the one at the pole. If the network’s conductance between the two terminals is exactly zero, driving it with a current source gives it no answer, and the solve refuses. Driving it with a voltage source instead and reading the current gives zero, which is a conductance of zero and a perfectly good statement — so the two duals fail at opposite points, and a network whose resistance is unbounded is best measured with a voltage and one whose resistance is zero with a current. That is the same pairing the bench uses for the same reason.

What is doing the work, said once more #

The whole of this essay follows from one distinction that the word “source” hides, and it is worth separating the two meanings explicitly because the theorem’s usual statement runs them together.

An independent source is an input. It is where the energy and the signal come from, and zeroing it is what turns a question about a circuit’s response into a question about its impedance. There is nothing conceptually difficult about killing one; it is what a small-signal analysis does to every bias in the netlist.

A dependent source is an element. Its current is a function of a voltage somewhere else in the network, exactly as a resistor’s current is a function of the voltage across itself, and the only difference is which voltage. Killing it would be deleting an element, which changes the network’s topology and therefore its impedance. A transconductance of a hundred millisiemens is no more an input than a ten-ohm resistor is.

Once the two are separated the rest is arithmetic, and the reason the shortcut fails becomes almost obvious. What is left after the independent sources are zeroed is a network of elements, and “combine the resistors” assumes all of those elements are resistors. On a network of resistors it is a theorem. On this one it is a guess.

Where the reduction still earns its keep #

Nothing above is an argument against Thévenin’s theorem, and it is worth being clear which of its claims survive.

The theorem itself is untouched. Any linear network with two terminals — dependent sources included, and solved the way what a network answers sets out — has a two-element equivalent that no load can distinguish from it, and the measurements here confirm that to the solver’s own precision on both sides of a pole and with the resistance negative. Linearity is what the theorem needs, and a dependent source is linear.

The recipe is what breaks. “Kill the sources and combine the resistors” is a shortcut that works on a network of resistors and independent sources and is not the theorem. Two routes survive — a test source, and $V_\mathrm{oc}$/$I_\mathrm{sc}$ where both exist — and of those the test source is the one that works everywhere.

And a resistance that varies is not a resistance that is wrong. That first reduction was exact outside and wrong within, and the source that is not a source measured the same reduction’s resistance from a slope. This one adds that the equivalent’s own parts may be things the reduction cannot name, and that both of those are properties of what a reduction is for rather than defects in it.

Still open: a measurement, the frequency, and the second reduction #

A measurement that could tell the two apart. Everything here is a solve. A bench measurement of the same network would put a small current in and read the voltage, which is the surviving route — and near the pole that measurement is ill-conditioned in a way the matrix that is ill, and the answer that is not distinguishes carefully from the solve being ill-conditioned, because the voltage it reads grows without bound while the current it drives is fixed. How close to the pole a measurement of stated precision can get, and whether driving with a voltage and reading a current is better there, is a question about instruments that this network is a clean test case for.

The same reduction with s in it. A Thévenin resistance that is negative at direct current is a Thévenin impedance at every other frequency, and whether the network is stable is a question about where that impedance’s zeros are rather than about its sign at one point. The answer that is perfect and absurd’s companion question — which of the operating points a network with several will be found in — needs the same extension. Adding one capacitor to the node turns this into a circuit whose poles can be computed, and the gain at which they cross into the right half-plane should be the same gain the resistance changes sign at.

Norton, and whether it fails in the same place. The dual reduction gives a current source in parallel with the same impedance, and it is usually said to be equally valid. Here the short-circuit current is well behaved at every gain and the open-circuit voltage is not, so the Norton form has a finite, gain-independent source and all of its pathology in the parallel element. Whether that makes it the better form to use near the pole, or merely moves the same difficulty, is a question the two solves would answer directly.