The tee that charges for its own compensation

A large feedback resistor is an awkward part. Above about ten megohms the choice narrows, the tolerance loosens, the temperature coefficient worsens, and the resistor’s own end-to-end capacitance — a few tenths of a picofarad — starts to set the bandwidth before anything else does.

The standard answer is a tee. Replace the single resistor with two modest ones from the summing junction to a tap and from the tap to the output, and a third from the tap to ground. The tap divides the output down before it reaches the feedback path, so a given output voltage drives less current back into the junction, and the transimpedance is multiplied:

$$Z = R_1\left(1 + \frac{R_2}{R_3}\right) + R_2$$

Two fifty-kilohm resistors and a 6.25 kΩ tap give five hundred kilohms. Two megohms and a small tap give a hundred. The arithmetic is right, the transimpedance is real, and the figure solves it on the netlist rather than quoting it.

What the expression has no term for is the noise gain, and it turns out that the noise gain does everything else the tee does.

The noise gain a single resistor does not have #

A stage’s noise gain is what multiplies everything referred to the amplifier’s input: its voltage noise, its offset, its drift. For an inverting amplifier it is one plus the ratio of the feedback impedance to the impedance from the summing junction to ground.

For a transimpedance stage at direct current, the impedance from the summing junction to ground is a capacitor. A photodiode at zero bias is a capacitance and a very large shunt resistance, so the noise gain at direct current is one plus a ratio of a resistance to something enormous, which is one. Exactly one, at every feedback resistance, whatever it is. The figure measures it as ×1.000 at a single resistor of any value on the slider.

A tee is different because the tap is a resistance. The feedback network’s impedance seen from the junction is $R_1$ in series with $R_2 \parallel R_3$; the fraction of the output that arrives at the junction is divided by the tap; and what comes out is a direct-current noise gain of

$$1 + \frac{R_1}{R_3}$$

which for a factor of ten is nine, and for a factor of a hundred is ninety-nine. It is not approximately the tee factor; it is the tee factor less one, and the figure claims it as such at every tap on the slider.

So the tee has done something the single resistor does not do at all. It has taken a stage whose direct-current noise gain was exactly one and given it a noise gain of nine.

Three consequences, and the first makes the trick pointless #

The signal-to-noise ratio does not improve. This is the one that decides whether the tee is worth building and it is the one least often stated.

The signal at the output is $I_{ph}Z$, so multiplying the transimpedance by $k$ multiplies the signal by $k$. The dominant resistor noise is $R_1$’s, and $R_1$’s noise reaches the output through the same tap that the signal does — so it is multiplied by $k$ as well. The ratio between them is unchanged.

A single resistor of the same transimpedance $Z$ contributes $\sqrt{4kTZ}$, which is $\sqrt k$ times $\sqrt{4kTR_1}$ and therefore less than the tee’s $k\sqrt{4kTR_1}$ by a factor of $\sqrt k$. Measured at a nanoamp in one hertz: the tee gives 1846 and a single resistor of the same 500 kΩ gives 5582, a factor of 3.02, which is $\sqrt 9$ — the square root of the noise gain.

Read the two curves in the figure together and the point is unmissable. The single resistor’s signal-to-noise ratio climbs as $\sqrt Z$ across the slider; the tee’s is flat. A tee built from 50 kΩ resistors has the signal-to-noise ratio of 50 kΩ whatever the tap does, because the tap multiplies the signal and $R_1$’s noise by the same number.

The amplifier’s own contributions are multiplied where they were not before. A single feedback resistor gives a direct-current noise gain of one, so the amplifier’s offset appears at the output as its offset and its voltage noise appears as its voltage noise. Through a tee both are multiplied by nine at a factor of ten. The essay before this one measured a direct-current offset budget in which the amplifier’s hundred microvolts was 0.11 per cent of the signal; put a tee of factor ten in and the same hundred microvolts is 0.99 per cent.

And the phase margin rises. This is the one that is a benefit, and it is worth the most attention because it explains why tees survive in practice despite the first two.

Why the margin improves, and what that means the tee is #

The factor the expression leaves out measured what makes a transimpedance stage unstable, and it is not a pole. It is the climb of the noise gain: the summing-junction capacitance and the feedback resistance form a zero in the noise gain at $1/2\pi R_f(C_d + C_f)$, above which the noise gain rises from one towards $1 + C_d/C_f$, and the loop closes where that rising curve meets the amplifier’s falling open-loop gain. The steeper the climb at the crossing, the worse the margin, which is the whole of the compensation problem.

A tee starts the noise gain at nine instead of at one. The climb to the same ceiling is therefore nine times shorter in ratio, the crossing happens at a lower frequency with a gentler rate of closure, and the margin improves: 1.8° for a single 500 kΩ resistor, 15.9° for the tee. At a factor of a hundred it is 50.8° against 1.8.

So the tee is a compensation. It is the same compensation a capacitor across the feedback resistor performs — both work by making the noise gain’s excursion smaller — and the figure states that the tee’s margin is never the worse of the two at any tap on the slider. That the margin is a property of the excursion rather than of a pole is what two measurements of one margin establishes by measuring the same margin twice, and it is why the resistor that buys the margin back works at all in the capacitive-load case: a stage is stabilised by changing what the loop closes against, not only by moving poles.

That reframes the whole comparison. The question is not “is a tee a good way to get transimpedance” but “is a tee a good way to buy phase margin”, and against the alternative it is expensive: a capacitor across the feedback resistor costs one part and no signal-to-noise ratio, while a tee costs one part, a factor of $\sqrt k$ in signal-to-noise ratio, and a factor of $k-1$ on the amplifier’s offset and drift.

A tee buys the transimpedance for free and the margin for a great deal. Which is exactly backwards from how it is usually described.

What it does to the bandwidth, which is why it was reached for #

The premise of the whole technique is that a large resistor has a large parasitic capacitance across it, and that capacitance sets the bandwidth before anything else does. It is worth checking that the tee actually helps with that, because the argument is not quite the one usually given.

A resistor’s end-to-end capacitance is set by its geometry rather than by its value — a few tenths of a picofarad for a surface-mount part, roughly the same whether it is ten kilohms or ten gigohms. A single five-megohm resistor with 0.2 pF across it has a corner at $1/2\pi R C$ = 159 kHz. Two fifty-kilohm resistors have 0.2 pF each, but they are in series with the tap between them, and what appears across the feedback path is the series combination of the two capacitances modified by the tap — so the pole is at a frequency set by 50 kΩ rather than by 5 MΩ, a hundred times higher.

That is real and it is the reason tees exist. A tee of factor a hundred built from fifty-kilohm parts has a feedback network whose own parasitic pole is two decades above where a five-megohm resistor’s would be.

What it does not do is give the stage that bandwidth. The closed-loop bandwidth of a transimpedance stage is set by where the rising noise gain meets the falling open-loop gain, which is the gain the loop closes against’s subject, and the tee has raised the noise gain — so the crossing happens earlier and the closed-loop bandwidth is lower, not higher. The tee removes one bandwidth limit and imposes another.

The net is usually favourable, because a five-megohm resistor’s own pole at 159 kHz is inside the band a photodiode amplifier wants and is uncompensatable, while the noise-gain limit is at least a designer’s choice. But “a tee gives more bandwidth” is not what the netlist says, and a design that assumed it will come out slow.

The tap resistor’s own noise, and why it is not the problem #

A reasonable objection at this point: the tap resistor $R_3$ is small — 6.25 kΩ for a factor of ten, 505 Ω for a factor of a hundred — and small resistors are quiet, so surely its own noise is negligible.

It is, and it is not where the noise comes from. $R_3$’s thermal noise is $\sqrt{4kTR_3}$, which at 505 Ω is 2.9 nV per root hertz, and it appears at the output multiplied by the noise gain — so a small tap produces little noise and multiplies everything else by a lot. The two effects are governed by the same resistance and they do not cancel: the noise gain is $1 + R_1/R_3$ and $R_3$’s own contribution goes as $\sqrt{R_3}$, so shrinking $R_3$ raises one as $1/R_3$ and lowers the other as $\sqrt{R_3}$.

The figure computes every resistor’s contribution separately — each one split out of the netlist, a source put in its place and the whole network re-solved, which is the construction the resistor the noise comes from uses — so the total is a sum over parts rather than an estimate. The result is that $R_1$ dominates at every tap, by construction: it is the largest resistor in the network and it is the one in series with the summing junction.

Which is the useful summary. The tee’s noise is $R_1$’s noise, amplified, and $R_1$ is a fraction of the transimpedance being claimed. The window the resistors own is the same accounting done for a filter: the parts that set the response are the parts that set the floor, and which of them dominates is a property of where they sit rather than of how large they are. The tap contributes almost nothing itself and is responsible for all of it.

Where the tee is the right answer anyway #

That the transimpedance expression is wrong. It is right, it is solved on the netlist to two parts in a thousand at the largest tap, and the small departure at large taps is the amplifier’s own finite gain rather than the arithmetic — the tee divides the loop gain by the same factor it multiplies the transimpedance by, so the closed-form’s assumption of infinite loop gain gets worse as the tap gets larger. That the gap grows with the tap is the same effect this essay is about, showing up in the quantity the tap was bought for.

That a single large resistor is always available. It is not, above some value, which is the premise of the whole technique. Where the trouble is at the input sets out why a photodiode amplifier wants a large resistance in the first place. What this essay establishes is the exchange rate, not a recommendation: a tee is the right answer when the resistor cannot be bought, and it should be chosen knowing that it costs $\sqrt k$ of signal-to-noise ratio and $k$ on every input-referred quantity.

That the noise gain at direct current tells the whole stability story. It does not — the climb matters more than the starting point, and the ceiling $1 + C_d/C_f$ is set by capacitances the tee does not touch. What the tee changes is the ratio between the start and the ceiling, which is the excursion, and the margin follows that.

That a bipolar-input amplifier behaves the same way. Its bias current flows in the tee’s resistors and produces an offset multiplied by the same factor, so everything above is worse for one. A tee with a bipolar-input part is a poor combination and the arithmetic here says why in one number.

Every resistor summed separately, against one resistor of the same value #

The transimpedance is solved on the netlist and checked against $R_1(1+R_2/R_3)+R_2$ to three parts in a thousand at every tap.

The noise gain is checked against $1 + R_1/R_3$ — and separately required to be exactly one for a single resistor of the same transimpedance, which is the comparison the essay turns on and which would be easy to assume.

Every resistor’s noise contribution is summed part by part, each one split out and the network re-solved, so the total is a sum over the network rather than an approximation.

The single resistor is checked to be at least as good at every tap, and better by at least a clear margin above a factor of two, so that “the tee buys no signal-to-noise ratio” is a measurement across the whole slider.

And with no tap the three penalties are refused: the noise gain, the resistor noise and the margin are required to equal a single resistor’s to a part in ten thousand, which is what says the penalties belong to the tap and not to the topology.

An expression with a missing term, and the term doing all the work #

The pattern this essay belongs to is the one that returns most often here: a design expression that is correct and incomplete, where the missing quantity is not a second-order correction but the thing the component actually does.

$R_1(1 + R_2/R_3) + R_2$ is exactly the transimpedance. It is also the only thing about the tee that is usually computed, and the noise gain — which the expression has no room for — is what sets the noise, the offset, the drift and the stability. A designer working from the expression alone gets the one quantity it describes and is surprised by the other four — which is the situation the ideal amplifier, and where it stops being one describes for the amplifier itself, where an expression with no bandwidth in it is used up to and past the frequency where the bandwidth decides everything.

The closest relative here is the factor the expression leaves out, a companion essay, where the classical compensation formula is exactly right about a phase margin nobody wants and silent about the peaking. The general form is worth stating: when a component is introduced by an expression, ask what the expression’s variables do not appear in. Here the transimpedance expression contains no capacitance, and the stage’s whole behaviour is set by one.

And the honest verdict on the tee follows from that. It is not a bad component and it is not a trick; it is a compensation network that has been sold as a gain network, and every one of its costs is invisible from the equation it is sold with.

Still open: the tee against a capacitor, the tap that is a capacitor, and the drift #

The two compensations on one axis. A capacitor across the feedback resistor and a tee both reduce the noise gain’s excursion. Drawing margin against signal-to-noise ratio for the two, at equal transimpedance, would give a single picture in which the tee is dominated everywhere or is not — and the answer probably depends on whether the amplifier’s voltage noise or the resistor’s dominates, which is an axis already drawn here.

A tap made of a capacitor. If the tap is a capacitor rather than a resistor, the noise gain at direct current returns to one and rises only above the tap’s corner. That would give the transimpedance multiplication at high frequency and none at direct current, which is nearly useless — but the mirror arrangement, a resistive tap with a capacitor across $R_1$, might give the margin without the direct-current penalty. Solving it is one netlist.

And the drift, which is the cost nobody prices. The amplifier’s offset drift is multiplied by $k-1$ exactly as its offset is, so a tee of factor a hundred turns a 1 µV/K part into a 99 µV/K one referred to the input. Over an industrial range that is a millivolt of output drift on a stage whose signal is a hundred millivolts, and it is the number that would actually stop a design.