The resistor that holds the bottom of the shelf

The shelf a capacitor makes found that an integrator’s summing junction is flat over its working band, because its loop gain has no frequency in it there, and poor at direct current, because a capacitor is an open circuit and closes no loop. With a kilohm in, ten nanofarads of feedback and a one-megahertz amplifier, the node sits at about 15.9 Ω across the band and climbs to nearly the whole kilohm below the amplifier’s own pole. The sensor an integrator does not see then found that shelf untouched by a sensor’s capacitance on the node.

Both essays described an integrator that nobody builds. An integrator with nothing across its capacitor integrates its own amplifier’s offset voltage and bias current along with its signal, and drifts to a rail within seconds. Every practical one has a resistor across the capacitor, which makes it a lossy integrator: an integrator above the corner the resistor and capacitor make together, and a flat gain below it. The shelf essay argued from the topology that the resistor would change only the bottom of the band. This essay solves it, and finds that the resistor changes the bottom in the one way that matters and in a closed form.

A rise that stops at the shelf #

The resistor goes directly across the ten-nanofarad feedback capacitor. Nothing else changes: the same amplifier, with a gain of $10^5$, a pole at 10 Hz and fifty ohms of output resistance; the same kilohm in.

With a hundred kilohms across the capacitor the node’s worst stretch is gone. At the bottom of the band it is 0.999 Ω, where the ideal integrator was nearly a kilohm. It rises a decade per decade, exactly as an inverting stage’s node does, and it stops rising where it meets the shelf, at 159 Hz. From there up it is the integrator’s node, flat at 15.9 Ω, and then the amplifier runs out. The highest the node reaches anywhere below a tenth of the amplifier’s bandwidth is 16.3 Ω — the shelf, plus the few per cent the amplifier’s output resistance adds near its top.

That is a shape none of the earlier essays drew: a rising virtual earth whose ceiling is not the passive network. The node that is an inductance found a node rising past its ceiling when the amplifier had a second pole; this one stops short of it. An inverting stage’s node rises to its resistors in parallel, 909 Ω for this amplifier’s companion stage, and stops there because the loop has run out. The lossy integrator’s rise stops at 15.9 Ω, long before the loop has run out, because at that frequency the capacitor takes over the feedback.

Where the two meet, and why the answer has no amplifier in it #

The frequency at which the rise meets the shelf can be written down, and it is simpler than it has any right to be.

Below the feedback’s corner the resistor dominates the feedback element and the circuit is an inverting stage with $R_p$ as its feedback resistor. The node that is at ground for a while found such a node to be the open-loop impedance divided by the loop gain. The open-loop impedance is $R_{in} \parallel R_p$; the loop gain is the amplifier’s $\mathrm{GBW}/f$ times the feedback factor $R_{in}/(R_{in} + R_p)$. Their quotient is

$$Z_{node} \approx \frac{R_{in} R_p}{R_{in} + R_p}\cdot\frac{R_{in} + R_p}{R_{in}}\cdot\frac{f}{\mathrm{GBW}} = \frac{R_p\, f}{\mathrm{GBW}},$$

with the input resistor gone from it. The shelf is $1/(2\pi\,\mathrm{GBW}\,C_f)$. Setting the two equal, the amplifier’s product cancels too, and they meet at

$$f = \frac{1}{2\pi R_p C_f},$$

which is the feedback network’s own corner. For 100 kΩ and 10 nF that is 159 Hz, where the solve puts the node at the shelf over $\sqrt2$ to within the ten per cent the figure allows; the slider on the figure at the head of the page moves the resistor to a megohm and the meeting point to 15.9 Hz.

So the node changes character exactly where the integrator does. Below $1/(2\pi R_p C_f)$ the circuit is an amplifier with a flat gain of $R_p/R_{in}$ and an inverting stage’s rising node; above it, an integrator on a shelf. The two descriptions of the circuit, as a transfer function and as a virtual earth, change over at the same frequency and for the same reason. Nothing about the amplifier decides where — only how good the shelf is.

The bottom, in closed form #

The claim that the resistor makes the bottom of the band an inverting stage’s node can be checked at the very bottom, where there is a closed form to check it against.

At a millihertz the capacitor is an open circuit and the lossy integrator is exactly a resistive inverting stage. Its node is $(R_{in} \parallel R_p)/(1 + A_0 R_{in}/(R_{in} + R_p))$, with the amplifier’s fifty ohms added to $R_p$ on the way back, and the dots follow that line to two per cent at every resistor from a kilohm to ten gigohms. For any resistor much smaller than $A_0 R_{in}$, a hundred megohms here, it simplifies to $R_p/A_0$: 0.999 Ω for 100 kΩ. For larger resistors the direct-current loop is too weak to hold the node, and it tends to the input resistor, as the ideal integrator’s did.

The simplification says where the bottom crosses the shelf: where $R_p/A_0 = 1/(2\pi\,\mathrm{GBW}\,C_f)$. Since $\mathrm{GBW}/A_0$ is the amplifier’s own pole $f_1$, that is

$$R_p^{*} = \frac{1}{2\pi f_1 C_f},$$

the resistor that puts the feedback’s corner exactly on the amplifier’s pole. With a 10 Hz pole and ten nanofarads it is 1.59 MΩ.

The largest resistor that keeps the node under its shelf #

The two closed forms together predict a threshold. Below $R_p^{*}$ the node rises from $R_p/A_0$, which is under the shelf, and meets the shelf at the feedback’s corner, so it is never above the shelf. Above $R_p^{*}$ the bottom starts above the shelf and the node falls to it, so the bottom is the worst point.

The sweep confirms it. For every resistor below a third of 1.59 MΩ the highest point of the node below a tenth of the amplifier’s bandwidth is the shelf, to five per cent. Above three times 1.59 MΩ it is more than twice the shelf, climbing with the resistor to the unloaded integrator’s thousand ohms. The transition between is the region where the feedback’s corner and the amplifier’s pole are close together and neither description is quite right.

This answers the shelf essay’s question in the form a designer can use. A lossy integrator’s summing junction is as good at every frequency below the amplifier’s bandwidth as it is on its shelf, provided the feedback’s corner lies above the amplifier’s own open-loop pole. For a general-purpose amplifier with its pole at a few hertz that is a mild condition. For a precision amplifier whose open-loop gain is $10^7$ and whose pole is a tenth of a hertz, the same capacitor allows a resistor a hundred times larger.

The same threshold, read as a loop gain #

The threshold has a second reading that needs no impedance at all, and it is the one that explains why it exists. A summing junction is as good as its loop is strong — the node is its open-loop impedance divided by one plus the loop gain, which is what what is left at crossover measures by cutting the loop. On the shelf the integrator’s loop gain is $2\pi\,\mathrm{GBW}\,C_f R_{in}$, 62.8 here, flat. At direct current, with the resistor across the capacitor, it is the amplifier’s full gain times the resistive divider, $A_0 R_{in}/(R_{in} + R_p)$, which for a hundred kilohms is about 990.

The node stays under its shelf if the loop is at least as strong at the bottom as on the shelf, and setting the two loop gains equal gives $R_p = A_0/(2\pi\,\mathrm{GBW}\,C_f)$, which is $R_p^{*}$ again. So the threshold says something simple: the resistor must leave the loop stronger at direct current than across the band. Below $R_p^{*}$ the loop gain falls monotonically from its direct-current value to the shelf’s and the node rises monotonically to meet the shelf. Above it the loop gain at direct current is the weaker, and the node is correspondingly worse there.

The same reading shows why the ideal integrator was worst at direct current. Its direct-current loop gain is zero: the capacitor closes no loop, and no amplifier gain can multiply a feedback factor of nothing. How much of the amplifier gets through measures what a loop gain buys at each frequency; an integrator with no resistor buys nothing at the one frequency its drift lives at.

A precision amplifier, worked #

The threshold depends on the amplifier’s open-loop pole, and precision amplifiers move it a long way. A chopper-stabilised or auto-zeroed amplifier with an open-loop gain of $10^7$ on the same megahertz of bandwidth has its pole at a tenth of a hertz, so with ten nanofarads $R_p^{*}$ is 159 MΩ, a hundred times the general-purpose amplifier’s. Its offset is also a few microvolts rather than a millivolt: 25 µV through a gain of $1 + R_p/R_{in}$ with ten megohms across the capacitor is 0.25 V at the output. So a precision integrator can carry a resistor that puts its corner at 1.6 Hz, integrate accurately down to a few tens of hertz, and keep both its output and its virtual earth where they belong.

The general-purpose amplifier cannot. With its 10 Hz pole the same ten megohms is six times its threshold, puts a node of 90.9 Ω at the bottom of the band where the shelf is 15.9, and gives 9.18 V of output from a millivolt of offset. The integrator’s accuracy at low frequency is limited by the amplifier’s open-loop gain twice over — through the drift, and through the virtual earth — and both limits are the same ratio.

What the resistor is there for #

The resistor was not put across the capacitor to improve the node. It is there to give the integrator a direct-current operating point, and the same threshold measures how well it does that.

At direct current the lossy integrator is an inverting amplifier of gain $R_p/R_{in}$, so its own offset voltage appears at the output multiplied by $1 + R_p/R_{in}$ and its bias current by $R_p$. With a millivolt and ten nanoamps, a hundred kilohms gives 0.102 V at the output. The threshold resistor, 1.59 MΩ, gives 1.58 V, and ten megohms 9.18 V, most of a twelve-volt supply, which leaves an integrator with almost no range for its signal. Without the resistor the output sits at the amplifier’s gain times the offset, a hundred volts, which is to say at a rail.

So the resistor that keeps the node under its shelf and the resistor that keeps the output off its rail are bounded by the same kind of number, a direct-current gain against the amplifier’s. A design choosing the largest resistor that keeps the output usable will usually land below $R_p^{*}$ anyway, and will get the flat virtual earth without asking for it.

What it costs the integrator #

The resistor spoils the integration, and the cost is in closed form too.

Below the corner the output is a flat gain of $R_p/R_{in}$, a hundred with 100 kΩ, rather than a rising one. Above it the circuit integrates, but not exactly: the phase falls short of ninety degrees by $\arctan(1/(2\pi f R_p C_f))$. At a kilohertz that is 8.853 degrees with 100 kΩ and 0.514 with the threshold resistor, and the solved transfers agree with the expression to a fifth of a degree, the remainder being the amplifier’s finite gain.

For an integrator whose job is to integrate — a ramp generator, a charge amplifier reading slow events — the corner has to sit well below the lowest frequency of interest, which pushes the resistor up towards the threshold. For an integrator whose job is a low-pass shape in a filter, the corner is part of the design and nothing is lost. Either way the node is fine as long as the corner stays above the amplifier’s pole, and that is a condition on the amplifier as much as on the resistor.

What a designer should take #

Put a resistor across every integrator’s capacitor, choose it for the output’s direct-current range and for where the integration must be accurate, and then check one inequality: the corner $1/(2\pi R_p C_f)$ should lie above the amplifier’s open-loop pole $\mathrm{GBW}/A_0$. If it does, the summing junction is never worse than its shelf, $1/(2\pi\,\mathrm{GBW}\,C_f)$, anywhere below the amplifier’s bandwidth. If it does not, the bottom of the band is worse than the shelf by the ratio of the two, and the integrator’s direct-current output is correspondingly close to a rail.

The shelf essay said an integrator’s virtual earth is poor at direct current and good across its band, the other way round from an inverting stage. With the resistor every practical integrator has, it is good at both ends, and the two regimes meet at exactly the frequency at which the circuit stops being one kind of amplifier and becomes the other.

How the numbers were obtained #

The amplifier is the one-pole model used throughout this sequence, with its gain, its bandwidth and its fifty ohms of output resistance as elements. The node impedance is measured by driving a unit current into the summing junction with every source zeroed and reading the voltage, at thirty points a decade from a hundredth of a hertz. The highest point below 100 kHz is taken over twenty points a decade for each of twenty-five resistors spaced a quarter-decade apart. The direct-current node is the same measurement at a millihertz, and the offset is the full netlist with a 1 mV source in series with the amplifier’s input and a 10 nA source into the node, solved at $10^{-7}$ Hz. The transfer is the integrator driven through its input resistor. Every closed form quoted is checked against the solves, and none is used to draw.

What it leaves out #

Bias-current cancellation. A resistor from the non-inverting input to ground, equal to $R_{in} \parallel R_p$, cancels the bias current’s contribution to first order and leaves the offset current’s. It changes the offset figure, not the node.

The resistor’s own noise. $R_p$ is a noise source across the feedback element, priced in the way the gain the loop closes against prices a feedback network’s contribution, and below the feedback’s corner its current noise appears at the output through a gain of $R_p$. A larger resistor has more voltage noise and less current noise, so the integrator’s low-frequency noise falls as $R_p$ rises — a reason to push the resistor up that runs against the offset.

A T network for the resistor. Very large effective feedback resistances are often built from a tee of three smaller resistors, which the tee that charges for its own compensation found costs noise gain in a transimpedance stage. What it does to the node here is not measured.

Still open: the tee across the capacitor, the reset switch, and the noise the resistor sets #

A tee network across the capacitor. A tee multiplies a resistor’s effective value without its size, and it also raises the noise gain at direct current by its multiplying factor. Whether the threshold $R_p^{*}$ applies to the tee’s effective value or to its physical resistors — and so whether a tee can buy a lower corner without giving up the flat node — is the measurement that decides whether it belongs across an integrator at all.

A reset switch instead of a resistor. Many integrators are held at zero by a switch across the capacitor between measurements, so there is no resistor during integration and the drift is bounded by time rather than by a gain. The node during the hold is the switch’s on-resistance across the capacitor, a very small $R_p$, and at the instant of release it becomes the ideal integrator’s; the transient between the two is the part a charge measurement sees.

The resistor’s noise, integrated. Below the corner, the resistor’s current noise and the amplifier’s voltage noise times the flat gain both appear at the output. The resistor that minimises their sum over a stated band, against the one that minimises the offset, would say whether the noise or the drift decides the choice, and whether either lands above $R_p^{*}$.