The reading that does care which way round

The reading that does not care which way round it is put one transconductance into a five-element ladder, injected a current at one port, read the voltage at the other, and then exchanged the two. With no controlled source the two readings were one number to $4.9\times10^{-14}$ of themselves. With one, they parted, in exact proportion to the transconductance and with no threshold under it — and the essay recorded the proportionality as a constant: 62.8 per siemens, for this network at this frequency.

That is the loose end. A constant with a network and a frequency attached to it is not a constant; it is a measurement that has not been identified yet. And 62.8 ohms at ten kilohertz is $2\pi f L$ for a millihenry — 62.832 ohms, to be exact — which is the reactance of the one inductor in that ladder, the one sitting between the controlled source’s control node and its output node.

The whole of the asymmetry belongs to that single branch. Nothing else in the network appears in it.

What is being solved #

A controlled source stamped into a nodal matrix is a rank-one addition: it puts a single entry at one off-diagonal position and nothing at its mirror, which is the whole of what what a network answers has to say about it. A rank-one addition to a matrix has a closed-form effect on the inverse, so the two transfer impedances of the active network can be written in terms of the passive one’s — four of its entries, and a denominator.

Do that for a source whose control terminals are $b$ and ground and whose output terminals are $c$ and ground, with the ports at $a$ and $d$, and the difference between the two readings is

$$z_{21} - z_{12} = -\frac{g_m\,(z_{dc}z_{ba} - z_{ac}z_{bd})}{1 + g_m z_{bc}}$$

where every $z$ on the right is a transfer impedance of the network with the source removed. That expression already says something the rung below could not: the departure is computed entirely from the passive network, and the controlled source enters only as its own transconductance and as a denominator that is one plus a loop gain.

The second step is where the network disappears. In a ladder — a chain, with the ports at its two ends — the four passive transfer impedances are not independent. Let $P$ be the impedance to ground looking left from $b$, $Q$ the impedance to ground looking right from $c$, and $Z$ the branch between them. Then $z_{bb}z_{cc} - z_{bc}^2$ is $PQZ(Z+P+Q)$ over $(P+Z+Q)^2$, and $z_{bc}$ is $PQ$ over $P+Z+Q$, so their quotient is $Z$ exactly — the branch, with $P$ and $Q$ cancelled out of it. Feed that back and the ratio of the two readings is

$$\frac{z_{21}}{z_{12}} = 1 - g_m Z.$$

Everything the network is made of has gone. What is left is the transconductance, and one impedance.

The reproduction, which comes before the departure #

An expression that predicts a small difference between two large numbers is exactly the kind that returns something plausible for a mis-stated problem, so the first thing asked of it is the case whose answer is already known.

With $g_m = 0$ the expression returns exactly 1, and so does the solver. That is not a coincidence to be admired, it is the calibration: the two routes must agree there or the comparison everywhere else is unattributable, which is the habit the digits the arithmetic did not have applies to any answer that looks fine. The floor it agrees at is the number rung one’s whole boundary was computed from.

Three branches, three slopes #

Now put something else between the two nodes. Same ladder, same ports, same transconductance; only the one component the earlier essay never named is changed.

At one millisiemens the inductive branch gives 0.0628319 at ten kilohertz, the 120 Ω resistor gives 0.120000 at every frequency there is, and 220 nF gives 0.0723432 at ten kilohertz. Those are $g_m|Z|$ three times, and the agreement with two full solves of the five-unknown matrix — four node potentials and the inductor’s own current — is at the arithmetic’s floor.

The flat line is the one worth pausing on. A resistive branch makes the departure from reciprocity frequency-independent — the same 12 per cent at a hundred hertz as at a megahertz — while the network around it changes its own transfer impedance from 37,887 ohms to 1.9032 — more than four decades — over the same span. A quantity that stays put while everything it is made of moves by four decades is a quantity that belongs to something small.

There is a practical reading of the three slopes, and it is about which asymmetry a designer is actually stuck with. An inductive branch makes reverse transmission worst at the top of the band, which is where a stage is hardest to keep stable; a capacitive one makes it worst at the bottom, where there is usually gain to spare; and a resistive one makes it the same everywhere, which is the only one of the three that can be cancelled once and stay cancelled.

What rung one’s boundary actually was #

The earlier essay’s headline number is 0.78 femtosiemens: the transconductance at which the two readings differ by more than the arithmetic can resolve. It is that floor, $4.910\times10^{-14}$, divided by 62.832 ohms.

So the boundary is not a property of the network, and it is not a constant of the subject either. It is the arithmetic’s floor divided by one branch’s impedance, and that impedance is a reactance. Hold the floor where the earlier essay measured it and the same boundary is 78 femtosiemens at a hundred hertz and 0.0078 femtosiemens at a megahertz — four decades of movement, produced by nothing except a millihenry being asked at different frequencies. Put a capacitor there instead and it moves the other way.

The conclusion the earlier essay drew survives all of this intact, and is if anything stronger: every transistor in this collection is nine or more orders above the largest of those figures. What does not survive is the idea that the number is telling anyone something about the network it was measured on.

A transconductance that cancels a branch #

The expression has a zero in it, and the zero is reachable. With a resistor in the branch, $1 - g_m R$ vanishes at $g_m = 1/R$, and $z_{21}$ vanishes with it.

That is a network which transmits exactly nothing in one direction and 206.13 ohms in the other, built from seven resistors and capacitors and one transistor’s worth of transconductance. The zero is a hard floating-point zero at a hundred hertz, one kilohertz, ten kilohertz, a hundred kilohertz and a megahertz alike, and the reason is in the expression: the cancelling quantity is $g_m R$, which has no frequency in it. It is not a resonance, nothing is being tuned, and there is no bandwidth over which the cancellation holds because there is no mechanism by which it could fail.

The reverse reading at those five frequencies is 215.86, 215.76, 206.13, 56.692 and 1.0839 ohms — a factor of two hundred across the band, while the forward reading is zero at every one of them. Perfect unilaterality and a wildly frequency-dependent reverse path, in the same object.

A directional path is the whole point of an active circuit, and this is where its size comes from. What gets through from the rail measures two regulators whose loop gains agree to a millionth of a degree and which differ by sixty decibels in one direction and not in the other; that sixty decibels is a transconductance times an impedance, computed the same way as everything on this page, and it is large because the impedance the pass device bridges is its own output resistance rather than a millihenry. The asymmetry is not a defect that happens to an amplifier. It is the quantity being bought, and it is priced by one branch.

This is also what a designer is doing, without the arithmetic, when a stage is neutralised: a feedback element’s transmission is cancelled by a controlled source arranged against it. The cancellation is exact when the two are matched and the branch is a resistor, and it degrades into a phase error the moment the branch is not — which is why neutralising the collector-base capacitance in the frequency a device sets for itself is a narrowband trick and cancelling a resistor is not.

The same two sources, arranged three ways #

The expression takes two transconductances, one pointing each way, and the general form is $(1 - g_1 Z)/(1 - g_2 Z)$. Setting $g_2 = 0$ gives everything above. The other two settings are the interesting ones, and they are both reachable with the same two parts.

With $g_1 = g_2$ the expression is 1, and the network is reciprocal. Two controlled sources, and the two readings are one number to fourteen figures.

That is the sentence the rung below could not have written, and it corrects the shape of the claim rather than a number in it. “A controlled source breaks reciprocity” is false. What breaks reciprocity is an asymmetric matrix, and a controlled source is merely the commonest way to get one; put a second source in the mirror position and the matrix is symmetric again, gain and all. The property was never about passivity or about which elements are in the netlist — it is about where their entries land, which is what every derivative, and the one that is zero depends on when it transposes the matrix and calls the result the same network.

The pair reversed, which is a gyrator #

Setting $g_2 = -g_1$ makes the two stamps antisymmetric: the same number at one off-diagonal position and its negative at the mirror. That is a gyrator, the element the inductor that is an amplifier is built from, and the ratio of the two readings is $(1 - gZ)/(1 + gZ)$.

For a reactance that quotient has modulus one identically, because the numerator and denominator are complex conjugates.

So a network with a gyrator in it gives the same size of reading in both directions and a different phase. An instrument that reads magnitude cannot tell which way round it has been connected; one that reads phase can, and the angle it sees is exactly twice what a single source of the same size would have produced. That is as sharp a statement of non-reciprocity as this collection has: the failure is total, and it is confined to one of the two numbers a complex reading has.

The equality needs the branch to be a reactance. Over a resistor it goes at once — 0.53488 at fifteen millisiemens with 220 Ω — and the reason is that $|1-gR|$ and $|1+gR|$ are two different real numbers while $|1-jx|$ and $|1+jx|$ are one.

Two facts about the same arrangement #

The reversed pair has a second property and it is worth keeping separate from the first, because they are usually taught together and only one of them depends on the branch.

Solve the network, take each controlled source’s own voltage and its own law’s current, and add the real power: for the reversed pair it is $2.4\times10^{-16}$ of what the drive delivers, over every transconductance and over both branches, the lossy one included. For one source alone it is 0.168 of the drive’s power and for the mirrored pair 0.0819. The reversed pair dissipates nothing, and the arithmetic of why is one line: its two contributions are $g\,v_c v_b^*$ and $-g\,v_b v_c^*$, which are conjugates with opposite signs, so their real parts cancel identically for any pair of voltages whatever — including the pair a resistive branch produces, which is why this half of the result does not depend on the branch at all.

The imaginary parts do not cancel, and that difference is the element’s whole use. The same pair exchanges up to 146.55 reactive volt-amperes across the sweep, and the sign of it changes with frequency: at one millisiemens the reversed pair takes 204.88 reactive volt-amperes at one kilohertz, 20.982 at ten, and supplies 7.7438 at a hundred. An element that moves stored energy while dissipating none of it is precisely an element that can present a capacitance at one port as an inductance at the other, which is what the inductor that is an amplifier builds — and it is why the negative resistance that essay measures belongs to the amplifiers’ gain-bandwidth product rather than to the gyration, which has no resistance to give.

The ideal amplifier is the limit, and the limit is attained #

There is one more element in this site’s library with a direction, and it is usually introduced by analogy: a nullor is like a controlled source with a very large gain. It is not like one. It is the limit of one, and the limit is reached exactly.

The forward readings agree to ten significant figures, which is what “the limit is attained” means on a machine. The reverse reading is the sharper half: $g_m z_{12}$ is 0.98545842 across seven decades of transconductance, to nine parts in a million, so the reverse path closes as $1/g_m$ and is never actually shut. The nullor’s is a hard zero, and for a structural reason rather than a numerical one — holding its input pair at the same potential leaves the far port as two resistors to ground with nothing driving them.

Which is the honest way to read what the ideal amplifier and its bandwidth trades away. An ideal amplifier is not an approximation with a large number inside it; it is a different element, with an exactly unidirectional transfer, and the finite-gain stage approaches it along a path this figure measures rather than assumes.

What it does not say #

The identity needs the control node and the output node joined by one branch. That is a condition on the netlist rather than on the physics, and it is checked here by breaking it: a 3.3 kΩ resistor from one end of the ladder to the other gives those two nodes a second path, and the prediction then runs from 0.008 to 4.7 times what the two solves actually return. Not a percentage — a factor of a hundred and twenty at the far end. The two nodes are still the same two nodes; what has gone is the cancellation of $P$ and $Q$.

And the ratio is silent about whether the network has an answer at all. At 80,875.37 hertz the passive ladder’s transfer impedance $z_{bc}$ is purely real and equal to $-20{,}630.60$ ohms, so $1 + g_m z_{bc}$ vanishes at $g_m = 48.4717$ microsiemens and the matrix is singular — the kind of refusal the answer that is perfect and absurd collects. A bipolar transistor reaches that transconductance at 1.2531 microamps of collector current, which is not an unusual bias. Just below it both readings are $2.03\times10^{8}$ volts per ampere and climbing, and their ratio is still $1 - j0.024629$, which is $1 - g_m j\omega L$ to eight figures. The quotient is exact and local and does not know that the network it belongs to has stopped having a solution.

The element that keeps reciprocity, and the boundary it has instead #

A coupled pair of windings is the element that looks as though it should have a direction. The rung below measured it keeping reciprocity to the last bit and left it there. It is worth asking what boundary it has instead, because it does have one, and it is of a kind nothing else on this page has.

The stored energy of a coupled pair is $\tfrac12(L_1i_1^2 + 2Mi_1i_2 + L_2i_2^2)$, and that form is positive for every pair of currents only while $M^2 < L_1L_2$. At $k = 1$ its smaller eigenvalue is zero, the two winding currents stop being independent unknowns, and above it the pair stores negative energy at some pair of currents, which no component does. The solver refuses $k = 1$, $1.0000001$ and $1.5$ on that argument rather than clamping them.

So the two kinds of boundary sit side by side. A reciprocal element’s limit is an energy condition on its own constants. A non-reciprocal one has no such limit — the transconductance can be anything at all — and what it runs into instead is the network ceasing to be solvable, which is a property of the whole circuit and not of the part. The coupled-pair result also settles what two ports from two one-ports was relying on when it recovered four impedance parameters from four solves and found two of them equal.

The number worth carrying #

$z_{21}/z_{12} = 1 - g_m Z$, and $Z$ is one branch.

The habit that goes with it is about where to look. A measured asymmetry in a network invites a search through the network, and the search is nearly always wasted: the asymmetry is produced by the controlled element and the one impedance it bridges, and every other component in the circuit cancels out of the answer exactly. That is why 62.8 per siemens read as a fact about a ladder and was a fact about a millihenry, and why the femtosiemens boundary derived from it moves four decades across the band while the ladder stays exactly where it is.

The corollary is worth stating because it decides what is worth measuring. Two readings of a network that differ tell nothing about the network — they are a statement about one component and the branch it sits across, and the same two readings would come back from any other ladder built around the same pair. What identifies the network is the readings themselves, not their quotient; the quotient identifies the source. That is why a swept measurement of reverse transmission is a good instrument for finding an unintended feedback path and a poor one for characterising anything else, and it is the same separation two solves that add makes between what superposition can be asked and what it cannot.

It also says which of the four arrangements to reach for. A single source gives a departure with a magnitude and a phase, and dissipates 0.168 of the drive’s power doing it. A mirrored pair gives no departure at all. A reversed pair gives a pure phase and dissipates nothing, which is the cheapest directionality on the list — the device that never sees the swing buys its isolation with a cascode at two volts out of a five-volt supply. And a nullor gives the whole of it — a reverse transmission that is not small but absent, which is the one thing here that a finite gain cannot buy at any price, only approach as $1/g_m$.