The compensation that costs nothing

Where the trouble is at the input built a photodiode amplifier and found its difficulty at the summing junction: the diode’s own capacitance, which is the price of its area, makes the noise gain climb and the loop ring. The factor the expression leaves out compensated it the classical way, with a capacitor across the feedback resistor, and found that one capacitor moves bandwidth, peaking and total noise together. The tee that charges for its own compensation built the same transimpedance from a feedback tee instead and found that the tee raises the noise gain from the start — by one plus the tap’s ratio — and that this raises the phase margin as a side effect, from 1.8° to 15.9°, while multiplying the amplifier’s voltage noise.

So there are two ways of buying margin at a fixed transimpedance, and that essay ended by asking the question that decides between them: drawn on one axis, margin against signal-to-noise ratio, is the tee dominated everywhere, or is there a region where it wins? It guessed that the answer might depend on whether the amplifier’s voltage noise or the resistors’ own dominated. This page draws the axis.

The two compensations, for reference #

Both mechanisms work on the same quantity. The loop’s instability comes from the noise gain climbing from one towards $1 + C_d/C_f$ as frequency rises, where the detector’s capacitance $C_d$ takes over from the feedback. The capacitor $C_f$ stops the climb early by giving the feedback its own capacitance. The tee shortens the climb from the other end, by starting it high: with a noise gain of nine at direct current, the climb to $1 + C_d/C_f$ is shorter, and the loop crosses over where the phase is less bad. The first acts only where the climb happens. The second acts at every frequency, including all the frequencies where nothing was wrong.

The axis #

The comparison is set up to be fair to both. Both designs have the same transimpedance, 500 kΩ, on the same 33 pF detector, behind the same amplifier: a ten-megahertz gain-bandwidth product and 4 nV per root hertz of voltage noise. The amplifier’s own finite gain is kept in the netlist rather than assumed away, since how much of the amplifier gets through is exactly what a transimpedance depends on at its edges. The capacitor’s family sweeps $C_f$ from nothing to 4 pF across a single 500 kΩ resistor; the tee’s family sweeps the tee factor from 3 to 64, with the resistors chosen so the transimpedance stays 500 kΩ. For each design the phase margin is solved by cutting the loop at the inverting input, the closed-loop bandwidth is bisected on the solved transimpedance, and the output noise — every resistor’s Johnson noise and the amplifier’s voltage noise, each as its own source — is integrated from one hertz to that design’s own bandwidth.

That last choice matters. A design that bought margin by narrowing its band should not be credited with the noise it no longer passes, and a design that kept its band should not be charged for noise beyond it. Each is judged on the band it actually delivers.

The capacitor’s curve lies below the tee’s at every margin the tee can reach. At about fifty degrees, the margin most designs aim for, the capacitor needs 0.97 pF and its output carries 85.2 µV across its 396 kHz band; the tee needs a factor of 32 and carries 347 µV across 401 kHz — four times the noise for the same margin and the same band. Across all the margins compared the tee carries between 1.3 and 7.5 times the capacitor’s noise, never less.

The shapes of the curves say why. As the capacitor’s family buys margin its noise falls, steeply: adding capacitance cuts the noise gain’s peak and narrows the band at once, and both remove noise. As the tee’s family buys margin its noise barely moves, because the tee’s noise gain is already high at every frequency and raising the tap’s ratio raises it further everywhere while shortening the peak. The two families run in opposite directions on the noise axis as margin rises, and they never cross.

Equal margin, equal band #

The one thing the tee might be buying with its noise is bandwidth: a design whose noise gain is high everywhere crosses over differently, and perhaps keeps more of the band at a given margin.

It is not. At equal margin the two families keep the same band to within 1.3 per cent — 351 kHz against 347 at sixty degrees. There is a reason for that, and it is general. A phase margin fixes the shape of the closed-loop response, its peaking and its damping; a transimpedance and a detector capacitance fix its scale, since they set where the loop crosses over. Two designs matched in both have almost no freedom left in their band. So the tee does not trade noise for bandwidth. At a matched margin it simply carries more noise, in the same band, for the same signal.

Where the extra noise lives #

The density figure locates the difference. The capacitor’s design has a noise gain of exactly one at low frequency — the detector is a capacitor, so at direct current there is nothing for the feedback resistor to divide against — and its amplifier noise at the output is the amplifier’s own 4 nV per root hertz until the detector’s capacitance starts to count. The tee’s noise gain at direct current is its factor less one, 30.7, and its amplifier noise at the output is 123 nV per root hertz from the lowest frequency drawn. The tee’s penalty starts at direct current and runs across the whole band. The capacitor’s design pays only across the top of its band, where every design pays.

That is the structural difference between the two compensations, and it is why no margin rescues the tee. A compensation should act where the instability is, at the crossover. The capacitor does exactly that. The tee acts everywhere and gets its margin as a by-product of making the noise gain large throughout, and everywhere below the crossover that largeness is pure cost.

Whether the conjecture holds #

The earlier essay guessed that the comparison might depend on which noise dominates: the tee multiplies the amplifier’s voltage noise, so with a very quiet amplifier, whose noise is small against the resistors’, its penalty might disappear.

It does not. With the amplifier’s voltage noise cut tenfold, the tee is worse by comparison, not better: 5.6 times the capacitor’s noise at fifty degrees, where it was 4.1. The capacitor’s design lost nearly a third of its noise when the amplifier got quieter — 85.2 µV down to 59.5 — because a large share of what remained was the amplifier’s; the tee’s barely moved, 347 to 331, because most of its noise was never the amplifier’s in the first place.

The reason is the resistor inside the tee. The tee’s transimpedance comes from multiplying the voltage across $R_1$ by the tap’s ratio, and $R_1$’s own Johnson noise is multiplied with it — the earlier essay found that the tee buys no signal-to-noise ratio for exactly this reason. $R_1$ is small, a thirty-second of the transimpedance at a factor of 32, so its Johnson noise density is larger relative to the signal it carries, and the tap multiplies it all the way out. A quieter amplifier removes the tee’s amplifier penalty and leaves its resistor penalty, which the capacitor’s design never had. The conjecture had the dependence the wrong way round: the tee is most dominated when the amplifier is best.

A calculation the curves agree with #

The two solved numbers at the quiet amplifier can be reached on the back of an envelope, and doing so shows that the ratio is not an accident of the chosen values. A single 500 kΩ resistor at room temperature has a Johnson density of 91 nV per root hertz — the floor a resistor sets scaled from a kilohm’s 4 nV by the square root of five hundred — and across a capacitor-compensated stage it appears at the output essentially undivided until the band’s upper edge. Integrated flat over 396 kHz that is 91 nV times the square root of 396,000, or 57 µV — against the solved 59.5, the difference being the little that the quiet amplifier and the peaking near the corner add.

The tee’s resistors at a factor of 32 are two of 15.6 kΩ and a tap of 521 Ω. The first resistor has a density of 16.1 nV per root hertz, and the tap multiplies it by its noise gain of 31, to 499 nV per root hertz. Each resistor’s noise is shaped by what lies between it and the output, as the resistor the noise comes from found in a filter, and here the tap resistor’s own noise reaches the output through much the same gain and adds about 91 nV per root hertz; the second resistor adds its 16 nV almost undivided. In quadrature these come to about 507 nV per root hertz, and that is exactly $\sqrt{31}$ times the single resistor’s 91. Integrated over 401 kHz it is 321 µV, against the solved 331.

So the quiet-amplifier ratio is, to within the peaking, the square root of the tee’s noise gain: $\sqrt{31}$ is 5.57 and the solved ratio is 5.6. The same law gave the 3.02 of the earlier essay’s tee, which was $\sqrt 9$. It is a clean way to state the tee’s price. Building a transimpedance of $Z$ as a tee of factor $k$ costs the output a factor of about $\sqrt{k-1}$ in Johnson noise, whatever the amplifier, and that factor is paid before the amplifier’s own contribution is counted at all.

With the ordinary 4 nV per root hertz amplifier the tee adds 123 nV per root hertz of amplifier noise to its 507 of resistor noise, which in quadrature is 522 — the amplifier is the lesser share even there. The capacitor’s design is where the amplifier matters: its noise gain climbs from one to $1 + C_d/C_f$, which at 0.97 pF is about 35, so across the top of its band the amplifier’s 4 nV becomes 140 nV and rivals the resistor’s 91. That is why cutting the amplifier’s noise helped the capacitor’s design by nearly a third and the tee’s by five per cent.

The price in parts #

The matched tee has a practical cost that the noise figures do not show. At a factor of 32 the tap is 521 Ω between two resistors of 15.6 kΩ, and the transimpedance depends on it almost one for one: differentiating $R_1(1 + R_2/R_3) + R_2$ with respect to the tap — the normalised derivative every derivative, and the one that is zero computes for every part at once — gives a sensitivity of $(k-2)/k$, which is 0.94 at this factor. A one per cent tap moves the gain by 0.94 per cent, and its temperature coefficient sets the gain’s drift in the same proportion. The single resistor’s gain depends on the single resistor and on nothing else, at a sensitivity of exactly one, so the tee has not escaped the precision requirement. It has moved that requirement onto a small resistor where the other two resistors now matter too.

The capacitor’s design asks for a 0.97 pF part, which is the other side of the ledger. A picofarad is small enough that the resistor’s own stray and the board’s are a noticeable fraction of it, and a designer usually trims it, or makes it from two traces, or accepts a margin that wanders by ten degrees from one board to the next. None of that changes the comparison — a capacitor that is 0.8 pF instead of 0.97 moves along the lower curve, not onto the upper one — but it is the real difficulty with the better compensation, and it is why the tee gets chosen by people who have already fought a sub-picofarad capacitor once.

What a designer should take #

At a fixed transimpedance and detector, the capacitor across the feedback resistor is the compensation to use, at every margin and whichever noise dominates. The tee’s reason to exist is elsewhere: it makes a large transimpedance out of resistors that are practical to buy and lay out, and it keeps the feedback resistor’s own stray capacitance, which a large single resistor has, from setting the bandwidth. When a tee is used for those reasons, its effect on the margin is a side effect to be accounted for, not a compensation to be designed with, and the capacitor across the tee — or across its first resistor — is still the compensation that should set the margin.

The error a bigger resistor cannot help found the same stage’s direct-current errors indifferent to the size of the feedback resistor; this page finds its noise very much dependent on how the resistance is built. The gain the loop closes against introduced the noise gain as the quantity a loop’s stability is decided by, and the lesson here is its converse: the noise gain is also the quantity a compensation should disturb as little as possible, anywhere it is not needed.

How the numbers were obtained #

Each design is a netlist: the detector’s 33 pF, the feedback network, and a one-pole amplifier model of 10⁵ at direct current and 10 MHz of gain-bandwidth with 50 Ω of output resistance. The phase margin is found by breaking the loop at the inverting input, driving it, and bisecting for unity loop gain; the bandwidth is bisected on the solved transimpedance for a fall of three decibels. The noise is computed by splitting each resistor into a noiseless resistor and a unit source in series, solving for its gain to the output at each frequency, and adding 4kTR times that gain squared; the amplifier’s voltage noise is a source at its non-inverting input. The densities are integrated by trapezoids on 240 logarithmically spaced frequencies from 1 Hz to each design’s bandwidth. Matched pairs are found by bisecting the capacitor or the tee factor for the stated margin.

What it leaves out #

The feedback resistor’s own capacitance. A single 500 kΩ resistor has a fraction of a picofarad across it, which is itself a feedback capacitor and sets a minimum $C_f$; a tee of small resistors largely escapes it. For a very large transimpedance that stray can be the capacitor the design did not choose, and it is the tee’s legitimate advantage.

Current noise at the input. The amplifier here has no current noise, which is right for a field-effect input and flatters neither design: current noise at the summing junction is converted by the transimpedance identically in both.

The detector’s shunt resistance. A real photodiode is a capacitor with a large resistor across it, and at direct current that resistor sets the noise gain to one plus the feedback resistance over the shunt. For a good diode of hundreds of megohms against 500 kΩ the correction is a part in several hundred and changes neither curve; for a large-area diode of 5 MΩ it adds about the transimpedance over the shunt, a tenth, to the noise gain at low frequency in both designs — the capacitor’s from one to 1.1 and the tee’s from 30.7 to about 30.8 — so the comparison survives it unchanged.

And a combined design, a tee with a capacitor across it. That is how a tee is usually built when it is built, and whether a small capacitor across the tee recovers most of the capacitor’s advantage — or whether the tee’s direct-current noise gain, which no capacitor touches, leaves it dominated anyway — is the natural next measurement.

Still open: the tee with a capacitor across it, the stray that forces a tee, and the drift #

A tee with its own capacitor. The tee’s penalty is its noise gain at direct current, and a capacitor across the tee acts only at high frequency, so it should supply margin without touching the penalty. Sweeping both the tee factor and a capacitor across it would say whether any combination approaches the single resistor’s curve, and by the argument above it should not at any factor above two.

The stray that forces a tee. A single resistor’s own stray capacitance sets a floor on $C_f$, and so a ceiling on the band a single resistor can offer. There is a transimpedance above which that ceiling falls below the band a tee could deliver, and above it the tee’s noise is the price of the band rather than a cost for nothing. Finding that transimpedance for stated strays would say where the capacitor’s dominance ends.

The drift. The tee multiplies the amplifier’s offset by its factor less one, and so its offset drift. A tee of factor a hundred turns a microvolt a kelvin into a hundred, and over an industrial range that is a direct-current error nobody prices at the time the tee is chosen for its noise gain or its resistors.