The shelf a capacitor makes

Three essays before this one have measured the impedance of one node — a current driven in, a voltage read, the input sources zeroed — and all three found the same three-part shape: a small value at direct current, a rise of a decade per decade, and a plateau at the two feedback resistors in parallel.

The plateau was attributed to the amplifier: above the loop’s crossover it has stopped participating, and what is left is the passive network. That is true and it is incomplete, because the passive network was two resistors. The whole shape belongs as much to the feedback element as to the amplifier, and changing that element for the other component in the analogue vocabulary changes all three parts of it.

The loop gain with no frequency in it #

An integrator’s loop gain is the one quantity in this collection that is genuinely constant over a wide band, and it is worth deriving because the derivation is the essay.

The loop gain is the amplifier’s gain times the fraction of its output that returns to the node. Above the amplifier’s own first pole the gain falls as $1/f$: it is $\mathrm{GBW}/f$. The feedback factor is $R_\mathrm{in}$ over $R_\mathrm{in}$ plus the capacitor’s reactance, and while that reactance is large it is $2\pi f C R_\mathrm{in}$ — proportional to $f$. Multiply:

$$T = \frac{\mathrm{GBW}}{f} \cdot 2\pi f C R_\mathrm{in} = 2\pi \cdot \mathrm{GBW} \cdot C \cdot R_\mathrm{in}$$

with no $f$ anywhere in it. For a megahertz part, ten nanofarads and a kilohm that is 62.83, and it is 62.83 at a hundred hertz and at five kilohertz alike.

A constant loop gain means a constant everything the loop divides. The node’s open-loop impedance is $R_\mathrm{in}$ in parallel with the capacitor’s reactance, which is $R_\mathrm{in}$ while the reactance is large, so the node is

$$Z = \frac{R_\mathrm{in}}{1 + 2\pi\,\mathrm{GBW}\,C R_\mathrm{in}}$$

— 15.666 ohms for these values, measured 15.67, and flat to half a per cent from a hundred hertz to five kilohertz. Nothing in these essays have produced a flat impedance from a loop before, because nothing else in it has had a loop gain that did not fall — what is left at crossover is where the falling kind is measured, and every result on this sequence of essays has been a division by it.

Which gives the design rule directly, and it is not the rule for a resistive stage. A bigger integrating capacitor is a stiffer summing node, in proportion, at every frequency in the band. That is available and it is not free: the same capacitor sets the integrator’s time constant, so a node ten times stiffer is an integrator ten times slower, and the exchange is exact.

Both ends the other way round #

The shelf is the middle. Both edges of it are set by something other than the capacitor, and both invert the resistive case.

Below the amplifier’s own first pole the gain stops falling — it is $A_0$, flat — while the feedback factor is still rising with frequency. So the loop gain rises with frequency there instead of being constant, and at low enough frequency it is gone entirely. At direct current the capacitor is an open circuit and there is no loop whatever: the node is $R_\mathrm{in}$ in parallel with an open circuit, which is $R_\mathrm{in}$.

Measured at a tenth of a hertz it is 843.0 ohms with ten nanofarads, and 997.4 with one. The resistive case at the same frequency is 0.100 49 ohms. An integrator’s summing junction is four orders of magnitude worse a virtual earth at direct current than an inverting amplifier’s — which is the same fact as an integrator having no defined direct-current gain, seen at the node rather than at the output, and is why every practical integrator has a resistor across its capacitor to give the loop something to hold with. The offset it would otherwise integrate is the quantity how much of the amplifier gets through is about, and a loop with no gain at direct current does not reject it at all.

Above the amplifier’s crossover the capacitor’s reactance has fallen below $R_\mathrm{in}$, the feedback factor stops rising, and the loop gain falls again — but so does the open-loop impedance, because the capacitor is becoming a short circuit. The node ends at 47.6 ohms where the resistive case ends at 909.5: nineteen times better, and for the reason the resistive case cannot share, namely that its feedback element still has ten kilohms in it at every frequency.

Where the two cross, and which stage has the better node #

The two curves cross once, and the crossing is the practical output of the figure: it is the frequency above which an integrator is the better virtual earth of the two and below which it is worse.

The crossing is where the resistive case’s rising curve passes through the integrator’s flat one, so it happens at the shelf — and since the resistive node is $R_f/A$ in the rising band, which the node that does not care how many derives, the crossing frequency is where $R_f/A(f)$ equals the shelf. That is $f = \mathrm{GBW}\cdot R_f/(R_\mathrm{in} \cdot (1 + T))$ near enough: 1.57 kilohertz for the default values, measured 1.56.

So the design statement is short. Below a couple of kilohertz on a megahertz part, an inverting amplifier’s summing node is the stiffer; above it, an integrator’s is — by a factor that grows with frequency and reaches nineteen. The stage that holds its node worst over the audio band holds it best above it, and vice versa, and neither fact is visible in either circuit’s gain expression.

Why this matters for what an integrator is used for #

An integrator is not usually chosen for its summing node, so it is worth saying which designs the shelf decides.

A charge amplifier. A piezoelectric or photodiode sensor is a current source with a capacitance across it, and the standard arrangement is an integrator whose feedback capacitor sets the charge-to-voltage gain. What the sensor’s own capacitance sees is the summing node, and if the node were the rising curve of an inverting stage, the sensor’s capacitance and the node’s rise would put a zero in the noise gain at a frequency inside the band. The shelf means it does not: a flat node and a capacitance across it give a single corner rather than a rise meeting a fall. Where the trouble is at the input measures the transimpedance case, where the feedback element is a resistor and that zero is the whole problem.

A sigma-delta loop filter. The first integrator of a modulator has the quantiser’s output current injected into its summing node, and what that current does to the node’s voltage is the shelf times the current. A flat node means the injected disturbance produces a flat voltage rather than one rising with frequency, which is the behaviour the loop’s own analysis assumes and which an inverting stage would not supply. One bit, and where the noise went is where the modulator’s own shaping is measured, and it takes the loop filter’s response as given.

And a state-variable filter. Every integrator in one has a summing node carrying several currents, so the summing-junction crosstalk of the multi-input measurement applies — and the shelf changes its frequency dependence completely. The leakage between two inputs is the node’s impedance over the input resistance, so with a resistive feedback element it rises twenty decibels a decade — which is exactly the figure the node that does not care how many reports for a multi-input node, and which it shows to be independent of the channel count — and with a capacitive one it is flat. A filter built from integrators has a channel-to-channel leakage that is a constant rather than a slope, which is a much easier thing to design against and is not, as far as anything here shows, usually noticed.

The gain error that does not vary with frequency #

The shelf has a consequence for what the integrator computes, and it is worth taking out of the “not measured” list because it follows from the numbers above by one division.

An integrator’s job is to put all of the input resistor’s current into the capacitor. What actually reaches the capacitor is reduced by the fraction of the input voltage that appears on the node rather than across the resistor, which is the node’s impedance over the input resistance. With a shelf of 15.666 ohms and a kilohm input resistor that is 1.567 per cent, and it is the same 1.567 per cent at every frequency in the band, because both quantities are constants there.

That is an unusual shape for an error in this collection. Nearly every departure measured here is a function of frequency — a gain that falls, a rejection that runs out, an impedance that rises — and the reason is always that a loop gain is falling. An integrator’s is not, so its finite-loop-gain error is a number:

$$\frac{\Delta}{1} = \frac{1}{1+T} = \frac{1}{1 + 2\pi\,\mathrm{GBW}\,C R_\mathrm{in}}$$

A one-nanofarad integrator on a megahertz part is fourteen per cent wrong, at every frequency, and a frequency sweep of it would show a perfectly clean one-over-f magnitude with the wrong constant in front. That is the failure worth naming, because the usual test for an integrator is exactly that sweep: a slope of minus one over five decades is taken as evidence that the integrator is working, and the slope is right while the constant is not.

The right test is a ratio. An integrator’s gain error is one over one plus $2\pi\,\mathrm{GBW} C R_\mathrm{in}$, so it is fixed by three quantities and none of them is the frequency. Measure the magnitude at one frequency against $1/(2\pi f C R_\mathrm{in})$ and the discrepancy is that number, directly — which is a one-point measurement where the sweep is a five-decade one, and it is the only one of the two that can fail.

It also says which component to change, and the answer is not the amplifier’s speed alone. Ten times the gain–bandwidth divides the error by ten, and so does ten times the capacitor, and so does ten times the input resistor — and the last of those costs nothing in speed, because the product $CR_\mathrm{in}$ is the time constant and it is the product that sets it. A given integrator time constant can be built from a small capacitor and a large resistor or the reverse, the two are usually treated as equivalent, and they are not: the large-resistor version has the smaller gain error and the stiffer node, both by the same factor. The reason to prefer the other is noise, which the floor a resistor sets prices at 4.00 nanovolts per root hertz for a kilohm.

The resistor every real integrator has, and what it does to the shelf #

An integrator with nothing across its capacitor has no defined output: the amplifier’s own input offset is integrated without limit and the output arrives at a rail. Every practical one therefore has a resistor $R_p$ in parallel with the capacitor, chosen to put the resulting pole below the band of interest, and it changes the direct-current end of this essay’s curve and nothing else.

At direct current the feedback element is now $R_p$ rather than an open circuit, so the circuit is an inverting amplifier of gain $R_p/R_\mathrm{in}$ and its node is the resistive case’s tenth of an ohm. The 843 ohms is gone. Above $1/(2\pi R_p C)$ the capacitor takes over and the shelf is exactly as measured here, because the shelf never depended on what happens at direct current.

So the honest statement about the low end is conditional and worth putting precisely. An integrator’s summing node is $R_\mathrm{in}$ below the frequency at which its feedback element stops being resistive, and that frequency is a design choice rather than a property of the integrator. Put the pole at a millihertz and the node is poor from a millihertz down, which nothing cares about; put it at a kilohertz, as a lossy integrator in a filter does, and the node is poor across the band the filter works in.

Which is a boundary of the kind the habit here is to state as a number rather than as a caution. The shelf is $R_\mathrm{in}/(1+T)$ with $T$ constant; the low end is $R_\mathrm{in}$; and the frequency between them is $1/(2\pi R_p C)$, so the ratio of the two is $1+T$ and the transition is one decade per decade like everything else on this sequence of essays. The only thing the capacitor changed was to put a ceiling on that rise at the shelf instead of letting it run to the resistors.

Two symbols, and a difference that is not of degree #

It is worth stopping on what changed between that essay’s circuit and this one, because the answer is one component symbol and the consequence is not a shift in a number.

The netlists differ in exactly one element: a ten-kilohm resistor between the node and the output, or a ten-nanofarad capacitor. Everything else — the amplifier’s model, its gain–bandwidth, its fifty ohms of output resistance, the kilohm input resistor, the measurement — is identical, element for element and name for name. Both circuits are drawn in every textbook, side by side, as “the inverting amplifier” and “the integrator”, and the difference between them is presented as what they compute.

What this essay finds is that they differ in the kind of object their summing node is. One is a node whose impedance rises monotonically over five decades and stops at a value set by the passive network. The other is a node with a flat region in the middle, and the flatness comes from two frequency dependences cancelling rather than from either of them being absent. There is no value of the resistor that makes the first circuit flat and no value of the capacitor that makes the second rise: the shapes belong to the element types.

Which is these essays’ standing argument about schematics, in the place where it is easiest to check. Kirchhoff’s own frequency makes the strong form — that the drawing contains none of the variable which decides the answer, so no amount of care in the drawing can address it — and that is about geometry, which a schematic genuinely cannot carry. This is milder and more awkward: the schematic carries everything, both circuits are completely specified by it, and the two behave differently in a way that requires a solve rather than a reading.

The practical form of it is a question to ask of any circuit with a loop round it, and it is one question rather than two. Not what is the loop gain — the essay before it reduced everything to that — but what is the loop gain’s slope, and what is the open-loop quantity’s slope, and do they cancel. Everything on this sequence of essays is a division of the second by the first, and the three shapes the four essays have drawn are the three possible answers: the divisor falls faster (a rise), the two fall together (a shelf), or the divisor dips below one (a peak).

What is not measured #

One pole in the amplifier. The essay before this one shows what a second pole does to a resistive node — it peaks above its ceiling and becomes inductive — and the same must happen here, at the frequency where the loop’s phase runs out. The shelf’s upper edge is set by the capacitor’s reactance passing $R_\mathrm{in}$, which is below the crossover for every value on the slider, so the peak would sit on the falling part of the curve rather than at its top. Whether it is visible there is not measured.

No capacitance on the node itself. The netlist has the feedback capacitor and nothing else reactive. A sensor’s capacitance across the node is exactly the case the charge-amplifier paragraph is about, and it is the case that matters most, and it is not here.

And the gain error above is arithmetic rather than a solve. The section that derives it divides the measured shelf by the input resistance, which is right and is not the same as solving the circuit for its output and comparing that with an ideal integration. The two should agree; only one of them has been done.

Still open: the sensor’s capacitance, and the integrator’s flat gain error #

A capacitance across the node, swept. A charge amplifier’s sensor is a capacitance at the summing junction, and everything this essay establishes says it should behave differently there than at an inverting stage’s node: a flat impedance and a capacitance give one corner, where a rising impedance and a capacitance give a peak. Measuring both on one axis would say how much of the noise-gain trouble that the gain the loop closes against prices for an inverting stage simply does not arise for an integrator — which would make the choice between the two topologies a measurement rather than a convention.

The flat gain error, in a filter made of several. One over one plus a constant loop gain is a constant, and what that costs in a state-variable filter built from three such integrators — where the errors compound rather than add, and where each integrator’s own node carries the others’ currents — is the question such a design actually asks. The Q the amplifier decides measures the equivalent for an active section built from a resistor and a capacitor, and finds the quality factor two per cent high and the pole frequency two per cent low for an amplifier a hundred times the corner; the integrator version of that arithmetic should be flat where this one is not.

And the resistor that spoils it. The section above argues from the topology that the parallel resistor moves only the low end. It is an argument and not a solve, and the thing to check is the corner: a lossy integrator’s node should rise from a tenth of an ohm at a decade per decade and stop at the shelf, which is a shape nothing on this sequence of essays has drawn — a rise with a ceiling that is not the passive network.

What is checked #

The shelf is held against its closed form, $R_\mathrm{in}/(1 + 2\pi\,\mathrm{GBW}\,C R_\mathrm{in})$, to five parts in a thousand at every value of the capacitor. The closed form contains the amplifier’s gain–bandwidth product and two component values and no frequency, which is the claim.

The flatness is required over a band bounded at both ends by something else — from ten times the amplifier’s own first pole to a third of the capacitor’s corner — and required to hold to two per cent. The first version measured from three decades below the capacitor’s corner, which at these values is a decade below the amplifier’s pole, and reported 113 per cent of variation on a quantity that is flat to a half. A flatness measured across the edge of the flat band is not a flatness.

The resistive case is required to rise a decade per decade over the same band, on the same amplifier, which is what makes the flatness a statement about the feedback element rather than about the part they share. That slope is taken per decade over the decades actually spanned; taking the ratio without dividing by the span reported 0.72 for a slope that is 0.99, at the settings where the band is narrower than a decade.

The direct-current end is stated as a ratio and not as a level. It is 997 ohms with one nanofarad and 48 with three hundred and thirty, so a threshold like “more than half the input resistor” holds at one end of the slider and fails at the other — which is what the first version required. What holds at every setting is that the integrator’s node there is at least a hundred times the resistive case’s, which is the comparison the essay is about.

And the crossing is bisected on the two solved curves, not computed from the closed forms, so the 1.56 kilohertz in the caption is a measurement against the 1.57 the algebra predicts.