The error a bigger resistor cannot help

Two earlier essays here have been about capacitance. Where the trouble is at the input found that a photodiode amplifier is the one arrangement in this field whose difficulty is at its input rather than at its output, and that the capacitance causing it is the diode’s junction — the price of the area the diode was bought for. The factor the expression leaves out found that the classical compensation for that capacitance is neither the forty-five degree choice nor the flat one, and that the noise the compensation capacitor is supposed to trade does not move inside the signal band at all.

Both essays take the feedback resistor as given, and both note in passing that a larger one is better. It is worth saying why, because the reason is the same for every quantity except one.

The signal is $I_{ph}R_f$, linear in the resistance. The resistor’s own thermal noise is $\sqrt{4kTR_f}$, which goes as the square root. So the signal-to-noise ratio goes as $\sqrt{R_f}$ and a resistor a hundred times larger is ten times better. The amplifier’s voltage noise is multiplied by a noise gain that depends on the capacitances and not on $R_f$, so it does not degrade. Every argument points the same way, and the only limits are the bandwidth the resistor’s own parasitic capacitance allows and the difficulty of buying a good gigohm.

The exception is the direct-current error, and it is an exception of a specific and useful kind: not that a bigger resistor makes it worse, but that a bigger resistor does nothing to it at all.

The ratio with the resistance cancelled #

The amplifier’s input bias current flows into the summing junction. It has nowhere else to go but through the feedback resistor, which is exactly where the photocurrent goes, so it produces an output of $I_bR_f$ against a signal of $I_{ph}R_f$.

$$\frac{\text{error}}{\text{signal}} = \frac{I_bR_f}{I_{ph}R_f} = \frac{I_b}{I_{ph}}$$

The resistance cancels. That is a one-line derivation and the figure does not use it: it changes the resistor and re-solves, at a megohm, ten megohms, a hundred and a gigohm, and gets the same 0.1000 per cent at every one of them to a part in a million. Testing a cancellation by cancelling it is not a test; testing it by varying the thing that is supposed to cancel is.

A picoamp of bias current against a nanoamp of photocurrent is a thousandth, and it does not move. Three decades of feedback resistance change the signal by three decades, the noise by one and a half, the bandwidth by three, and this by nothing.

Which makes it the binding constraint on how small a photocurrent can be measured. Everything else about the stage can be improved by spending money on the resistor; this can be improved only by buying a different amplifier or by measuring a larger photocurrent. A gigohm feedback resistor does not help a picoamp of bias current, and the one-line derivation of that is the most useful thing in this essay.

The other term behaves the other way #

The amplifier also has an offset voltage, and it reaches the output multiplied by the stage’s direct-current noise gain — one plus the ratio of the feedback resistance to whatever shunts the summing junction.

What shunts the summing junction is the photodiode. At zero bias a photodiode is a resistance, and it is the one the junction that is a resistor at zero volts computes: a gigohm for a small detector at room temperature, falling fast with temperature. So the offset’s contribution is

$$V_{os}\left(1 + \frac{R_f}{R_{sh}}\right)$$

which grows with the feedback resistance. The figure measures it by changing the resistor, the same way: 100.1 µV at a megohm and 200.0 µV at a gigohm, a factor of two, being $1 + R_f/R_{sh}$ evaluated at the two ends.

So the two terms are opposites. One is invariant under the choice of resistor and one is proportional to it, which means their sum has a minimum in $R_f$ — and the minimum is at a resistance so small that the signal-to-noise ratio is ruined, which is why nobody ever designs there and why the resistor is chosen by the noise argument in practice. The direct-current error is a consequence of that choice rather than an input to it.

The minimum that exists and is never used #

The two terms have opposite dependences, so their sum has an interior minimum, and it is worth locating once in order to see why nobody is there.

The sum is $I_b/I_{ph} + V_{os}(1 + R_f/R_{sh})/(I_{ph}R_f)$, and differentiating with respect to $R_f$ gives a minimum at $R_f = \sqrt{V_{os}R_{sh}/(I_{ph}\cdot 0)}$ — which does not exist, because the second term falls monotonically as $1/R_f$ towards the asymptote $V_{os}/(I_{ph}R_{sh})$. The sum therefore has no interior minimum at all: it falls with $R_f$ and flattens.

So the direct-current error argues for a large resistor too, and the shape is the useful part. At a megohm the offset contributes 10.0 per cent of the signal and the bias current 0.1; at ten megohms, 1.01 and 0.1; at a hundred megohms, 0.110 and 0.100; at a gigohm, 0.020 and 0.100. The offset term falls a decade per decade until it hits its asymptote, at which point the bias term has become the whole answer and nothing more is to be had.

That is the crossing worth knowing: the feedback resistor is worth increasing until the offset term reaches the bias term, and after that it buys noise and bandwidth and nothing else. That is the same shape of answer as the ammeter that is a resistor gives for a shunt — two errors moving oppositely in one variable — with the difference that here one of them flattens instead of continuing, so the answer is a threshold rather than a geometric mean. Here it is at about a hundred megohms, which is where the figure is drawn, and it is a coincidence of these particular numbers rather than a general rule — but the rule that produces it is general.

Both double every ten kelvin, for two unrelated reasons #

The temperature behaviour is where the two terms stop looking like separate problems.

A field-effect input amplifier’s bias current is a reverse-biased junction’s leakage, and a junction’s leakage doubles about every ten kelvin. A photodiode’s shunt resistance is set by its own junction’s leakage and halves about every ten kelvin. So over sixty kelvin the bias term rises by $2^6 = 64$ and the offset term’s multiplier rises by nearly as much, and the two curves in the figure climb together.

The last row is not a rounding artefact. At a hundred and twenty-five degrees a one-picoamp amplifier has a gigapicoamp — a nanoamp — of bias current, which is the entire signal, and the instrument reads approximately twice the light that is there. The bias current specification on the front of the data sheet is at twenty-five degrees, and the doubling is in a graph on page nine.

That is worth stating as a design rule rather than as a curiosity. A picoamp-class measurement has a temperature specification before it has an electrical one, and the useful number is not the bias current but the temperature at which it reaches some fraction of the signal — 50 °C for one per cent here, and 25 °C would be the answer for a tenth of the photocurrent.

The third current, which is in the diode rather than in the amplifier #

A photodiode has a dark current of its own — the reverse leakage of the same junction whose shunt resistance appeared above — and it arrives at the summing junction indistinguishably from the photocurrent.

It is not in the figure, because it is not an amplifier property and the figure is about what the amplifier does. It belongs here anyway, because it has the same shape as the bias current: it is a current into the summing junction, so its contribution is $I_{dark}/I_{ph}$ with the resistance cancelled, and it doubles every ten kelvin for the same physical reason.

For a small silicon detector at zero bias the dark current is femtoamps and is below the amplifier’s bias current; for a reverse-biased one it is picoamps to nanoamps, because reverse bias is what a designer applies to reduce the junction capacitance — which is where the trouble is at the input’s whole subject. So reverse-biasing the diode to buy bandwidth buys dark current at the same time, and the exchange rate is a property of the detector.

Three currents at one node, two of them errors, all three with the resistance cancelled out of their effect. That is the clean statement of what a transimpedance stage’s direct-current accuracy is: a ratio of currents, and nothing about the circuit appears in it.

What can actually be done about it #

Four things, in order of how much they cost.

Choose the amplifier for its bias current over temperature rather than at twenty-five degrees. A CMOS-input part might specify 1 pA typical at 25 °C and 500 pA maximum at 85; a JFET-input part with a larger room-temperature figure may be an order better hot. The specification that matters is the one at the top of the operating range and it is rarely on the front page.

Keep the junction cool. The doubling is in junction temperature, so the amplifier’s own dissipation counts. A part running at 5 mA from ±15 V dissipates 150 mW and sits 15 to 30 K above its board, which is one and a half to three doublings of bias current given away for nothing.

Cancel it. A second, matched amplifier input or a deliberate current injected into the summing junction removes the bias current’s contribution at one temperature and leaves the difference of two leakages, which is smaller and still doubles. That is worth a factor of ten and not a factor of a thousand, and it is the same kind of first-order cancellation as the path that buys the error back.

Chop it. Modulating the light and measuring the amplitude at the modulation frequency moves the measurement away from direct current entirely, at which point neither the bias current nor the offset appears — both are direct-current quantities — and the limit is the noise from the essay before it. This is the answer that actually works and it is why almost every picoamp-class optical measurement is made with a chopper or a lock-in.

The fourth is worth pausing on. It does not improve any of the numbers in the table; it makes the table irrelevant, by measuring something the errors are not present in. That is a different kind of solution from the first three and it is the one to reach for when a table like this one has a row in it reading 113 per cent.

Where the doubling model stops #

That the bias current is exactly a junction leakage. In a CMOS-input amplifier it is, and it doubles. In a bipolar-input one it is a base current, which changes by tens of per cent over the range rather than by orders — so a bipolar part with a hundred times more bias current at room temperature can be the better choice at a hundred degrees, and the crossing is the design question. The model here is the field-effect one.

That ten kelvin is the exact doubling interval. It is between eight and eleven for silicon depending on the mechanism, and it is a fit to an exponential rather than a law. The constant that is a window is what usually gets said here about such fits, and the same caution applies: a doubling interval extracted over sixty kelvin is not the same number as one extracted over two hundred.

That the shunt resistance is the only thing at the summing junction. The board is, too. A gigohm-class measurement on ordinary FR4 has surface leakage comparable with the diode’s, which is what a guard ring exists to remove and what the sign of what the guard gives back measures. The figure models the diode and the board is the designer’s problem.

That the offset term’s growth is unbounded. It is $1 + R_f/R_{sh}$, and once $R_f$ exceeds $R_{sh}$ the term is dominated by the ratio, so a gigohm feedback resistor on a gigohm diode doubles the offset and a ten-gigohm one multiplies it by eleven. That is a real limit on how large the resistor can usefully be and it is a different limit from the bandwidth one.

The invariance tested by moving the resistor three decades #

The bias term’s independence of the resistance is measured rather than derived, by solving the stage at four feedback resistances spanning three decades and requiring the fractional error to agree to a part in a million.

And it is checked against $I_b/I_{ph}$, so that the invariant is the right invariant rather than merely a constant.

The offset term is checked to grow, by exactly $1 + R_f/R_{sh}$ evaluated at two resistances three decades apart, which is the statement that the two terms behave oppositely.

The doubling is checked over sixty kelvin as a factor of exactly $2^6$ on the bias term, and the offset term is checked to grow over the same interval by its own separate mechanism.

And the whole ratio-free term is refused at zero bias current, where the error is exactly zero at every resistance rather than small — which is what makes “the resistance cancels” a statement about a term that exists.

An invariance that is a limit rather than a convenience #

Quantities that do not depend on things turn up throughout these essays, and they are usually good news. The burden voltage that contains no current sizes every shunt with one number. The available noise power that contains no resistance makes noise figures add. The reuse of $\sqrt{kT/C}$ makes a sampled noise floor a property of one capacitor.

This one is the other kind. $I_b/I_{ph}$ contains no resistance, and what that means is that the designer’s main lever does not reach it. An invariance is a statement that a quantity is unmoved by something, and whether it is welcome depends entirely on whether that something is a lever or a nuisance.

The useful discipline is to ask, for each error in a budget, which of the design’s variables it responds to. In this stage: the resistor moves the noise and the bandwidth and the offset term and not the bias term; the diode’s area moves the capacitance and the shunt resistance and the dark current; the temperature moves everything exponentially; and the modulation frequency moves all the direct-current terms out of the measurement at once. Four variables, and the error that looks most alarming in the table responds to only one of them.

Which is why the fourth remedy is the one that gets used. Everything else on the list trades one of three variables against another inside a picture where the errors stay; chopping moves the measurement to a place where two of them are not defined. It is the same manoeuvre the corner where averaging stops working describes from the other side — a measurement moved away from direct current stops meeting the terms that live there, and starts meeting whatever lives where it went.

Still open: the bipolar crossing, the guard’s own leakage, and the chopped measurement’s floor #

Where a bipolar input becomes the better choice. A bipolar-input amplifier’s bias current is a base current — larger at room temperature, nearly flat with temperature — and a field-effect one’s doubles. There is a temperature at which they cross, it depends on both parts’ room-temperature figures, and solving for it would turn a rule of thumb into a number a designer could select against.

The board, solved rather than assumed away. A guard ring drives the region around the summing junction to the same potential, so the surface leakage sees no voltage and contributes nothing — until the guard’s own driver has an offset, at which point the leakage is proportional to that offset rather than to the signal. Adding the guard to this netlist would say how much of a gigohm-class measurement’s error the guard’s amplifier owns.

And the floor a chopped measurement actually has. Moving the measurement to a modulation frequency removes the bias current and the offset. What is left is the noise from the essay before it, the diode’s shot noise on its own dark current, and whatever the chopper adds — and the last of those is a switched-capacitor-like injection of charge that the apparatus here could compute and has not pointed at an optical front end.