The cure that becomes a different circuit

The current the instrument draws put an amplifier’s own input currents into the netlist and separated the three errors they make, and it ended with an instruction that every textbook gives in one sentence: make the two resistances the amplifier’s inputs look back into equal, and the bias current cancels, leaving the offset current.

That essay noticed, in passing, that the instruction has an asymmetry in it. If the feedback network is the larger resistance, balancing is a resistor in series with the source. If the source is the larger — which is the only case in which any of this mattered — there is no resistor to add and the feedback network has to be scaled up to meet it instead. It said so and moved on. It never asked where the boundary between the two is, what the scaled network costs, or whether the scaling leaves anything for a designer to choose.

All three are one solve away, and the answers are sharper than the sentence suggests.

What is being solved #

A non-inverting stage of gain eleven, as a netlist, with the amplifier’s two input currents as current sources at its two input nodes and its offset voltage as a source in series with the non-inverting one. Nothing is quoted: the reading is the solved output divided by the gain, which is what the instrument is wrong by, and each of the three contributions is measured by a solve with the other two set to zero — the arrangement what a network answers makes routine.

Two resistances decide everything. $R_p$ is the total from the non-inverting input to ground, which is the source and whatever is added in series with it. $R_m$ is what the feedback network presents to the inverting one, which is the bottom resistor in parallel with the feedback resistor. Balanced means those two are equal, and the interesting question is what it costs to make them so.

The reproduction, which comes before the departure #

Splitting one instruction into two circuits is the kind of claim that can be made to come out of a solve by mis-stating the solve, so the first thing asked of the netlist is a question whose answer is already on this ladder.

Every number in it comes back unchanged, which it must, because the netlist has not changed — only the questions asked of it. The two crossings are still $V_\mathrm{os}/I_B$ and $V_\mathrm{os}/I_\mathrm{os}$, at 1.77 kΩ and 10.0 kΩ for a part with fifty nanoamps of bias and five of offset. Below about a kilohm balancing still makes the reading worse rather than better, and that fact is the first sign of the boundary this essay is about: at a hundred ohms the feedback network is already the larger resistance, so equalising means adding 809 Ω to a source that had a hundred, and the amplifier’s bias current now flows in nine times what it flowed in before.

That is the same knee seen from the error rather than from the resistances. It is not a separate phenomenon and it is not a caveat: below 909 Ω the cure works by making the good input worse, and above it by making the bad one worse still. Only the second is what anybody means by balancing, and the instruction as usually stated covers both.

The knee, and it belongs to the feedback network #

Before the cure, $R_m = R_1 R_f/(R_1+R_f)$, and with a gain of $G$ that is $R_1(G-1)/G$. For a kilohm bottom resistor at a gain of eleven it is 909.09 Ω, and it is 909.09 Ω whatever the source is — the feedback network does not know what is on the other input.

So the instruction splits at exactly that number. Below it the source is the smaller resistance and the cure is a resistor of $R_m - R_\mathrm{source}$ in series with it: at a hundred ohms, an 809-ohm resistor. Above it there is nothing to add, because adding to the source makes it larger still, and the only way to make the two equal is to raise $R_m$ — which means scaling the whole divider, keeping its ratio so that the gain is unchanged.

$R_1(G-1)/G$ is the bottom resistor for any gain above about ten, to within a few per cent: 500 Ω at a gain of two, 909 at eleven, 990 at a hundred and one, 999 at a thousand and one. The resistance at which a stated instruction becomes a different circuit is the feedback divider’s bottom resistor and essentially nothing else — not the amplifier, not the source, not the gain.

Above the knee the network has left the answer #

Scaling by $R_\mathrm{source}/R_m$ makes the bottom resistor $R_\mathrm{source}\,G/(G-1)$ and the feedback resistor $G$ times the source, exactly, at every resistance drawn. A megohm source at a gain of eleven therefore asks for 11 MΩ in the feedback path and a hundred kilohms for 1.1 MΩ, and the size of the network the design started with has disappeared from both.

That is not a small observation dressed up. It means the balanced circuit is the same circuit whatever the designer chose, and the consequence is measurable in the reading.

Twelve figures is not agreement, it is identity: above its own knee each of the three has had its feedback network forced to the source’s own size, and there is nothing of the original choice left in the netlist. The balanced error is the offset current times the source, plus the offset voltage, and both of those belong to the part and to the thing being measured.

So the cure removes the designer’s one lever along with the error. Below the knee, the size of the feedback network is a real choice with real consequences — a factor of seven in the reading across the three drawn here. Above it, the choice has been made by the source.

What it buys and what it costs, both of which stop moving #

Two quantities are worth having as functions of the source rather than as sentences, and both settle onto constants.

What the cure buys is the ratio of the unbalanced reading to the balanced one. It is 1.08 at a kilohm, 5.32 at ten kilohms, 9.56 at a hundred, 10.40 at a megohm and 10.49 at ten megohms — walking up to the ratio of the two currents, which for this part is ten. That is the number the rung below quoted as a factor of ten rather than a thousand, seen arriving.

What it costs is at the floor, and it settles too.

Below the knee the cost is there and the benefit is not, which is the third face of the same boundary. At a hundred ohms the cure adds 809 Ω in series with the source, raising the density from 5.67 to 6.72 nV/√Hz — 1.18 times, most of the penalty already paid — while the reading goes from 12.1 µV to 54.5 µV, four and a half times worse. At the knee itself the two circuits are the same circuit: 6.72 against 6.72, and 54.5 µV against 54.5 µV, because the resistor to add has become zero. The cure has a range, and its lower end is not a place where it is merely less useful but a place where it is the wrong move.

The two limits above the knee are also worth separating by what kind of number each is. The benefit’s limit, 10.49, is a property of the part — the ratio of two currents on one data sheet, which for a bias-compensated amplifier can be two rather than ten and for a field-effect input can be anything at all, since its two currents are junction leakages that do not track. The cost’s limit, √2, is not a property of anything: it is what happens when one uncorrelated noise source becomes two of the same size, and no part, temperature or process changes it. So a designer reading a data sheet to decide whether to balance is reading exactly one number — the ratio of bias to offset current — against a constant.

The two limits together are the whole trade in two numbers: above the knee and well above it, balancing is worth a factor of ten in a constant and costs a factor of √2 in a density, whatever the design. Neither number can be improved by anything the designer of the feedback network does, because by then the feedback network is the source.

Which leaves the trade where the rung below left it, and for a reason that is now measured rather than argued. The offset removed is a constant that can be measured once and subtracted — by a trim, by a calibration cycle, by a digital correction — and what remains of it is its drift. The noise added by a resistance cannot be subtracted at all, and neither can the shot noise the bias current itself carries, which passes the source resistor’s own thermal noise beyond 1.00 MΩ for a fifty-nanoamp part. That boundary is $2kT/qI$, which is a resistance only because a current was divided into a voltage: the voltage is 49.98 millivolts at 290 kelvin and has no resistance in it at all.

Where the knee falls in circuits people build #

909 ohms sounds like a boundary a design would rarely be near, and the opposite is true: almost every circuit in which input current matters is above it, and almost every circuit in which it does not is below.

A feedback divider is sized by two pressures that both push it down. The bottom resistor and the feedback resistor together load the output, so a stage driving them at full swing wants them large; but every one of them is a noise source and forms a pole with the summing node’s own capacitance, so a stage that has to be quiet or fast wants them small. The result across this collection is a kilohm to a few tens of kilohms almost everywhere, which puts the knee between a few hundred ohms and ten kilohms.

And the sources that make input current a problem are not near that at all. A pH electrode is hundreds of megohms. A photodiode with no shunt is a gigohm. A thermocouple through a broken-guarded terminal block is whatever the flux residue decided. A strain-gauge bridge is 350 Ω and is genuinely below the knee, which is why bridges never need any of this. A high-impedance divider in front of a voltmeter, at ten megohms, is four orders above it.

So the case the textbook draws — a resistor in series with the source, the cheap version — is the case in which the cure was not needed, and the case in which it is needed is the one the drawing does not show. That inversion is worth carrying on its own, because it is why the instruction reads as cheap: it is usually illustrated in the regime where it is.

The bandwidth the instruction warns about, given a test #

The rung below asserted a third cost and did not measure it: a megohm in the feedback path forms a pole with the amplifier’s own input capacitance and with the board’s stray, and makes the circuit’s bandwidth depend on a capacitance nobody specified. It is exactly the kind of claim this collection is supposed to give a test it could fail, so here is one — a capacitance at the summing node, a real amplifier of a megahertz, and the closed-loop response solved before and after the cure.

At fifteen picofarads the warning is not merely absent, it points the wrong way: the scaled network is forty-three per cent wider, because the summing node’s capacitance now works against a resistance two orders larger and lifts the feedback factor’s phase before the loop has run out of gain. The unbalanced circuit does not move at any capacitance drawn, since the resistance its inverting input looks back into is 909 Ω whatever the source is.

The warning is right, though, and what it is right about is a product rather than a resistance. Peaking arrives when the corner at $1/2\pi R_m C$ falls below the frequency at which the amplifier has no gain left to give away, which is $\mathrm{GBW}/G$. Setting the two equal names a capacitance, $G/(2\pi\,\mathrm{GBW}\,R_m)$, and at a thirty-kilohm source that is 58.4 pF. Bisected on the solved response, three decibels of peaking arrives at 55.3 pF — five per cent apart, from an arithmetic that never solves a network and a network that never mentions a phase margin.

So the instruction’s third clause is a statement with a boundary in it, and the boundary contains the source only through $R_m$. Fifteen picofarads of amplifier and board at thirty kilohms is nowhere near it. Sixty picofarads — an amplifier in a socket, a length of coaxial screen brought to the summing node, a guard ring driven from the wrong place — is past it, and the same cure that cost nothing at all now costs 3.49 decibels of peaking and the phase margin that goes with it. The repair for that is a capacitor across the feedback resistor, which is the resistor that buys the margin back arriving from the other side of the same trade, and it costs the bandwidth the scaling had just given.

What it does not say #

It does not say the scaled network is always buildable. At a gain of eleven and a megohm of source it asks for 11 MΩ, which is an ordinary part with an extraordinary temperature coefficient. At a gain of a thousand and one it asks for 1001 MΩ, which is not a resistor anybody stocks, and the honest reading of that is not that the cure is expensive but that it is unavailable — the design has to change the gain or the architecture rather than the resistor.

It does not say the offset itself is always defined. With no direct-current path at the input — a coupling capacitor, a transformer, a photodiode with no shunt — the node’s potential is not determined by the network and the nodal matrix is singular, which is the refusal that essay is built on rather than a large number.

And it does not say the amplifier is the one modelled. The response above is a one-pole macro-model with a hundred decibels of open-loop gain, which is the amplifier the ideal one stops being at a stated frequency. A faster amplifier reaches the peaking boundary at a proportionally smaller capacitance — ten megahertz puts it at 5.8 pF, which is the amplifier’s own input capacitance and nothing else — so the clause the instruction attaches to a megohm is really attached to a gain–bandwidth product.

Where this sits beside the rest of the field #

The current the amplifier’s own inputs demand is one of two currents at that node, and the other one does not come from the amplifier at all: it comes across the board, and the current that never reaches the input puts the laminate into the netlist and drives a guard ring with the same loop gain this stage is using for its own feedback. That current has no cure of this shape, because there is no second resistance to make equal to it.

The other neighbour is the instrument this field spends four rungs on, where two of four errors are divided by the gain and two are not and the design question is again which quantity a chosen number is allowed to move. The shapes rhyme: a lever that looks like a free choice turns out to be pinned by something outside the part, and finding where it is pinned is more useful than optimising it.

The number worth carrying #

909 ohms, and √2.

The first is where one instruction becomes two circuits, and it is the feedback divider’s bottom resistor times $(G-1)/G$ — a number a designer chose, usually for reasons that had nothing to do with input current.

The second is what the cure costs at the floor once the source is large, and it cannot be moved by anything, because above the knee the balanced circuit is the source’s circuit and not the designer’s.

The habit that goes with it is the one every model has an edge states generally. An instruction phrased as an equality — make these two equal — hides which of the two is being moved, and the two answers are different circuits with different costs. Solving it once is cheap, and what comes back is not a correction to the instruction but a boundary inside it.