Exact outside and wrong within

Every other model on this site is approximately right over a range and wrong outside it, and the work of the essay is finding the number where the change happens. This one is different, and it is worth spending an essay on precisely because it does not fit: a Thévenin equivalent is not an approximation at all. For a linear one-port it reproduces every measurement made at the terminals, exactly, for every load, at every frequency, and there is no number on any axis at which it starts to fail.

It still has an edge. The edge is not a frequency, an amplitude or a size — it is a question.

The network, and why it is not a divider #

Six elements. A twelve-volt source at one node and a five-volt source at another; a kilohm from the first to the port, 2.2 kΩ from the second to the port, 4.7 kΩ from the port to ground, and 3.3 kΩ bridging the two sources directly. Nothing exotic — it is the shape a bias network takes when two rails are available and somebody wanted a particular voltage — and it is deliberately not a divider, because a divider’s equivalent is one line of arithmetic and would make the theorem look like a convenience.

Solved with nothing on the port, the node sits at 8.56032482599 V. That is the open-circuit voltage, and it is the first of the two numbers the equivalent needs.

The resistance, obtained twice #

The second number is the resistance seen looking back into the port, and there are two ways to get it that share nothing but the netlist.

Set the sources to zero and measure. A voltage source with no voltage in it is a short circuit, so the sources stay in the netlist — the topology is not allowed to change — and their values become zero. Then drive one amp into the port and read the voltage. That gives 599.767981438515 Ω.

Short the port and divide. Put a zero-volt source across the port, solve the original network, and read the current through it: 14.2727272727 mA. The open-circuit voltage divided by the short-circuit current is 599.767981439 Ω.

The two agree to twelve figures, which is what the site’s habit of measuring twice is for. They are not the same computation dressed differently: the first inverts a matrix with a modified source vector, the second inverts a matrix with an extra unknown in it and reads a branch quantity out.

There is a third thing to notice, and it is the seed of everything below. That resistance is exactly the parallel combination of the kilohm, the 2.2 kΩ and the 4.7 kΩ — 599.767981438515 Ω, to every digit a double holds. The 3.3 kΩ bridging the two sources does not appear. It cannot: it runs between two ideal sources, both of which become shorts when they are zeroed, so it is a resistor across a short and contributes no conductance to the port at all.

A third route, and where the bridge went #

Superposition gives a third way to the open-circuit voltage, and it is worth taking because of what it exposes rather than because a third number is needed.

Solve the network with the five-volt source set to zero: the port sits at 7.19721577726218 V. Solve it again with the twelve-volt source zeroed instead: 1.36310904872390 V. Their sum is 8.56032482598608 V against the 8.56032482598608 V from the single solve — two units in the last place apart, which is two additions rounding differently and nothing else.

Look at what each contribution is. The first is $12 \times 599.768/1000$; the second is $5 \times 599.768/2200$. Each source’s share of the open-circuit voltage is its own value scaled by the equivalent resistance over the resistor through which it reaches the port — which is to say the two numbers the equivalent needs are not independent of one another at all. Compute the resistance and the open-circuit voltage follows from the source values by inspection.

And once more the 3.3 kΩ is absent. In the twelve-volt solve it is a resistor from the twelve-volt node to ground, drawing 3.6 mA that goes nowhere near the port; in the five-volt solve it is a resistor from the five-volt node to ground doing the same. It carries current in both, dissipates in both, and appears in neither answer.

What no load can tell apart #

The claim to be tested is that a source of 8.56032482599 V behind 599.767981438515 Ω is indistinguishable from the six elements. Testing it means attaching the same load to both and comparing, and the range of loads matters: an equivalence checked at one load is an equivalence checked nowhere.

Seventy-three loads, logarithmically spaced from 10 Ω to 10 MΩ. At each, both netlists are solved independently and the port voltage is compared.

The worst disagreement across all seventy-three is $2.2\times10^{-16}$ of the voltage, and the power into the load agrees to $7.8\times10^{-16}$. Those are not tolerances. A double holds about sixteen significant figures, so a relative difference of $2\times10^{-16}$ is one unit in the last place — the two computations produced the same number, and where they differ at all it is because they rounded a different intermediate quantity in the last digit available.

This is a stronger statement than the site usually gets to make. Elsewhere “the two routes agree” means to a part in $10^{12}$ and the remaining gap is a measurement of something. Here there is nothing left to measure.

The question it gets wrong #

Now ask the two networks something that is not a terminal measurement: how much power is being dissipated inside them.

At a load of 5.62 kΩ, the six-element network is turning 49.168 mW into heat in its own resistors. The equivalent, carrying the same current into the same load and putting the same voltage across it, dissipates 1.1349 mW in its 600 Ω. A factor of 43.3.

With the port open — nothing connected, the case in which the equivalent is most obviously “just a voltage” — the equivalent dissipates exactly nothing, because no current flows in it at all. The network is dissipating 48.033 mW. The two rails are twelve volts and five volts, they are joined through 3.3 kΩ and also through the 1 kΩ and 2.2 kΩ in series to the port node, and current runs round those loops whether or not anything is connected. None of it reaches the port. All of it comes out as heat.

The pattern is the one that matters in practice. At a heavy load nearly all the dissipation in either network is the current squared through the source resistance, so the two nearly agree — 1.5 times apart at 100 Ω. As the load lightens the equivalent’s internal dissipation falls towards zero and the network’s does not, so the disagreement grows without limit. The equivalent is least trustworthy about heat exactly where it is most often used, which is at loads much larger than the source resistance.

The slider, which was not planned #

The bridging resistor is on the slider, and what it does is the sharpest form of the whole argument.

Moving it from 1 kΩ to 150 kΩ changes the open-port dissipation from 82.184 mW to 33.511 mW — a factor of two and a half. It changes the open-circuit voltage by nothing: 8.5603248259860809 V at both ends, all seventeen digits. It changes the equivalent resistance by nothing, for the reason given above.

So there are seven different circuits here, dissipating between 33 and 82 milliwatts with nothing connected, and they have one equivalent between them, identical to the last bit. A reader who has the equivalent has everything a load can ever discover and no way at all to tell which of the seven is on the bench.

That is not a defect in the theorem. It is what the theorem says, stated in the direction nobody states it in: the equivalence is exact because it throws away everything that does not reach the terminals, and what it throws away is not small.

Where a designer meets this #

Three places, and none of them is contrived.

Sizing a heat sink. A regulator’s input network reduced to an equivalent gives the right input voltage under load and says nothing about the dissipation in the bleed resistors. The number above — 48 mW with nothing connected — is small; scale the same topology to a rail-splitter across a forty-volt supply and it is watts.

Battery life on a standby current. The quantity that empties a coin cell is not the load current; it is the load current plus everything the biasing network is doing on its own. An equivalent has divided that out by construction.

Explaining a measurement that does not add up. A supply is delivering more current than the sum of the loads. Every terminal voltage is exactly where the equivalent says it should be. The two facts are consistent, and reconciling them requires the netlist rather than its reduction.

What is linear, and what happens when it is not #

Every word above depends on the network being linear. Superposition is what makes the open-circuit voltage a well-defined thing to have, and it is what makes the source-zeroing route to the resistance legitimate.

Put one diode in the network and none of it survives in that form. There is still a curve of terminal voltage against load current — a one-port always has one — but it is not a straight line, so it is not a source behind a resistance. The best that can be said is local: about any operating point there is a slope, and the slope is a small-signal equivalent valid over the range in which the curvature has not yet cost more than whatever accuracy is wanted. That range is what the semiconductors field measures, and the number it returns is a few millivolts rather than a few volts.

What this essay does not claim #

That the equivalent is unique among reductions. A Norton source — the short-circuit current across the same resistance — is the same object read the other way, and every number above is unchanged by using it. There is nothing to choose between them beyond which of the two measurements is easier to make.

That the internal dissipation is hard to get. It is not. The netlist is right there and the solve takes microseconds. The point is that it is not in the equivalent, and a reader who has reduced a network and thrown the netlist away has thrown that away with it.

That this generalises to a two-port. Everything here is one port. A network with an input and an output has more that a terminal measurement can reach, and the magnetics field’s two-port essay is where that is measured.

Where the reduction is used, and where it is refused #

The equivalent this page measures is used everywhere in the field and refused in two places. The source that is not a source is the same object one rung down, where the internal resistance is the whole model rather than a reduction of six elements — and it has the same property, that no load can distinguish it from what it stands for. What a network answers, and how the answer is checked is where the refusal lives: a reduction is a claim about a port, and the two checks that run on every solve in this collection ask about branch currents and dissipated energy, neither of which the reduction contains.

What the reduction is used for #

An equivalent that no load can distinguish is used everywhere in the field and refused in two places. The source that is not a source is the same object with the internal resistance as its whole content. The divider, and the thing it does not know about is the smallest circuit in which the reduction is worth making. What a network answers, and how the answer is checked is where the two checks live that the reduction cannot satisfy, because both are about branch currents and dissipated energy. Two solves that add, and the one that does not is the other place where a shortcut is exact about voltages and wrong about power, and The answer that is perfect and absurd is where a reduction to an ideal source is refused outright.

The gate #

The two routes to the resistance agree to twelve figures — 599.767981438515 Ω from a source-zeroed solve and 599.767981439 Ω from the open-circuit voltage over the short-circuit current — and the site’s gate requires them to.

The terminal agreement is the last bit and is checked as such. $2.2\times10^{-16}$ on the voltage and $7.8\times10^{-16}$ on the delivered power, across seventy-three loads spanning six decades. The gate asserts a bound of $10^{-14}$, which is loose enough to survive a different rounding order and tight enough that any real disagreement would fail it.

The disagreement about heat is asserted to grow monotonically with the load, not merely to exist. That is the claim worth making, because it is the one that says where the reduction is dangerous: at a heavy load the two nearly agree, and by the open port the equivalent has lost the whole 48 milliwatts.

And the equivalent is asserted to be bit-identical across the slider, at every one of the seven bridging resistances, while the internal dissipation moves by two and a half times.

The other quantity a reduction throws away #

Heat is the obvious casualty of a Thévenin reduction and it is not the only one. Two other essays in this field measure quantities that survive at the terminals and are destroyed inside, and together the three say what an equivalent is actually equivalent for.

Two solves that add, and the one that does not is the closest relative and its arithmetic is the same arithmetic: every node voltage and every branch current in a linear network is the sum of the per-source solves, to the last bit of a double at eighty-one settings, and the power is not — two equal sources in antiphase put nothing at all into a load while adding their powers gives 222 milliwatts. Both essays are about the same fact from opposite ends: voltage and current are linear in the sources and power is quadratic in them, so every technique built on linearity is exact about the first pair and silent about the second.

The reading that does not care which way round it is is the property a reduction preserves and that a reader might not expect it to: a passive network gives the same transfer in both directions to five parts in $10^{14}$, and a two-terminal equivalent is trivially reciprocal, so nothing is lost. What breaks reciprocity is a controlled source, at a transconductance of eight tenths of a femtosiemens — and a network containing one has no source-and-resistance equivalent in the first place.

So the rule that comes out of the three is narrow and usable. A Thévenin equivalent is exact for everything the terminals can be asked about and wrong about everything happening behind them, and the line between those two is whether the quantity is linear in the sources. Voltage, current and impedance are; power, dissipation and efficiency are not.

That last word is the one that makes the rule worth having rather than pedantic. The load that takes the most is a whole essay about a quantity this reduction cannot see: a load equal to the source resistance takes more power than any other and does so at exactly fifty per cent efficiency, with the source burning as much as the load receives. Every number in it is a statement about what is happening inside the equivalent — and a network reduced to one source and one resistance has an internal dissipation that is a property of the reduction rather than of the circuit, which this essay measures as a factor of forty-three.

The practical form of that is a rule about where a thermal question is allowed to be asked. Maximum power transfer is a correct statement about the terminals and a nonsense statement about the source’s temperature, so a design matched for maximum transfer has to be told what the real network behind the equivalent is dissipating before anybody sizes a heatsink — and that number is not recoverable from the equivalent at all. It has to come from the netlist the equivalent replaced, which is the one thing a reduction was performed in order to stop carrying.

Which is the small irony this essay ends on. A Thévenin reduction is done to avoid keeping the network around, and the one question it cannot answer is the one that requires keeping the network around. The reduction is not therefore a bad technique; it is a technique with a stated domain, and the domain is written on the terminals.

What makes it worth a figure rather than a footnote is the size. Forty-three times is not a subtlety about where heat appears; it is an answer of the wrong order, produced by a method that is exact to the last bit of a double about everything else it is asked. A technique that is either perfect or wrong by a factor of forty-three, with nothing in the output to say which, is exactly the kind this collection exists to put a boundary around.