The peak the amplifier moves

The sign the null leaves behind put a twin-T in the feedback path of an amplifier of gain A and turned its null into a peak. The closed loop is $A/(1 + A\,T)$, and where the twin-T’s transfer $T$ is zero the loop’s gain is the whole of A. With exact parts the peak’s quality factor is $(A + 1)/4$. With the shunt elements out by a fraction $t$ the twin-T leaves a residue at the null, and that residue is a real number, $-t/4$. Multiplied by the gain it moves the Q to $(A + 1)/4 \div (1 - A t/4)$ — down towards $1/|t|$ one way, and up without limit the other, until the loop oscillates at $t = 4/A$.

Every amplifier in that essay had a gain of A at every frequency. A real one does not. A selective amplifier’s gain stage is an operational amplifier with local feedback, and a stage of gain A on an amplifier with gain–bandwidth GBW has a bandwidth of its own, GBW/A, and lags $\varphi = \arctan(f_0 A/\mathrm{GBW})$ at the null frequency $f_0$. That essay ended by expecting the lag to rotate the real residue: the stability boundary would then depend on both, and there would be a gain–bandwidth product below which even an exact twin-T never reaches its Q or oscillates. This essay measures it, and the expectation is wrong in an instructive way.

What the lag was expected to do #

The argument is short and plausible. The loop’s denominator is $1 + A\,T$. Near the null an exact twin-T’s transfer is a small imaginary number growing with the detuning, and the peak’s width is set by how fast $A\,T$ grows against the one. If the gain is $|A|e^{-j\varphi}$ rather than $|A|$, only its in-phase part $|A|\cos\varphi$ is set against the one, so the peak should be narrower by $\cos\varphi$: the Q should rise by $1/\cos\varphi$, and at 60 degrees of lag it should double. By the same reasoning the wall should move in, to $4\cos\varphi/A$, because only the gain’s in-phase part multiplies the real residue.

The twin-T is the one of the earlier essays, ten kilohms and ten nanofarads, with its null at 1591.5 Hz, in a loop of gain 100.

On a 300 kHz amplifier the stage lags 27.95 degrees at the null, and the argument says the Q should rise by thirteen per cent. It rises by 0.8 per cent: 25.44 against 25.25. The peak’s height barely moves either, 99.7 against 100. What does move is visible at a glance and was not in the argument at all: the whole peak has slid 1.02 per cent down in frequency, so that the null the parts were chosen for is no longer where the circuit selects. The slider on the figure at the head of the page takes the amplifier from 10 MHz to 100 kHz, and the picture stays the same — the Q fixed, the peak walking downward.

Why the Q stays #

The argument’s mistake is in treating the lag as a rotation of a gain of fixed size. A stage with a single pole of its own does not do that. Its gain is $A/(1 + j f A/\mathrm{GBW})$, and at the null that is $A\cos\varphi\, e^{-j\varphi}$: the same pole that turns the gain by $\varphi$ shrinks it by $\cos\varphi$.

Put that into the denominator. Near the null the exact twin-T’s transfer is $j s\,\delta$, with $\delta$ the fractional detuning and $s$ its slope, a half. The denominator is $1 + A\cos\varphi\,e^{-j\varphi}\, j s \delta$. Its magnitude squared is a quadratic in $\delta$ whose minimum is $\cos^2\varphi$ and whose curvature is $(A s \cos\varphi)^2$, so the half-power points are $\cos\varphi/(A s\cos\varphi)$ either side of the minimum: $1/(A s)$, with the lag cancelled out. The peak’s width, and so its Q, does not depend on $\varphi$ at all to first order. Its height is the gain there over the minimum, $A\cos\varphi/\cos\varphi = A$, which does not depend on it either.

The figure checks that across three gains and nine lags. Up to sixty degrees the Q stays within about one part in the gain of its ideal-stage value — 3.2 per cent at a gain of 30, 0.97 at 100, 0.10 at 1000 — where $1/\cos\varphi$ would have doubled it. The remainder is the second order: the stage’s gain is not constant across the peak but rolls off over it, and a peak whose width is a larger fraction of the stage’s own bandwidth sees more of that. At eighty degrees, where the stage’s gain changes appreciably across the peak, the Q falls, to 0.69 of its ideal value at a gain of 30.

Why the wall stays #

The same cancellation keeps the loop’s stability boundary where it was.

The loop oscillates where $A(f)\,T(f) = -1$. With a lagging gain that is $T = -(1 + j\tan\varphi)/A$ at the stage’s own nominal gain: a real part of $-1/A$ and an imaginary part of $-\tan\varphi/A$. The twin-T supplies the imaginary part itself, a little below its null, where its own quadrature has grown to meet it. The real part has to come from the residue, and it is $-1/A$, exactly as with an ideal stage. So the wall stays at the shunt error that gives a residue of $-1/A$: 4.126 per cent at a gain of 100 with an ideal stage — the first-order $4/A$ plus a second-order three per cent — and 4.139 at 30 degrees, 4.244 at 60. It moves only once the stage’s lag is steep enough that its gain changes across the frequency range involved, 4.64 at 75 degrees.

The argument’s $4\cos\varphi/A$, the dashed curve, would have put the wall at 2.0 per cent by sixty degrees. It is nowhere near it. For a one-pole stage, the null depth the sign the null leaves behind required for a stated Q is the requirement whatever the amplifier.

Where the peak goes #

The lag has to go somewhere, and it goes into frequency. The minimum of the denominator is not at the null but at the detuning where the twin-T’s quadrature cancels the stage’s: $\delta^{*} = -\tan\varphi/(A s)$. With $\tan\varphi = f_0 A/\mathrm{GBW}$ and $s = 1/2$ that is

$$\delta^{*} = -\frac{2 f_0}{\mathrm{GBW}},$$

and the gain is gone from it.

Across three gains spanning a factor of thirty, every point with less than 45 degrees of lag lies on the line: the peak sits $2f_0/\mathrm{GBW}$ below the null, less a correction of about two parts in the gain — 0.936 of the line at a gain of 30, 0.979 at 100, 0.997 at 1000. The larger gain lags more at the null but its peak is narrower in exactly proportion, and the two cancel. The amplifier’s gain–bandwidth product alone decides how far the circuit’s selected frequency is from the one its parts put there. A tenth of a per cent needs 3.18 MHz at this null, whatever the Q; a hundredth needs ten times that.

That is a different kind of requirement from the null depth. The null depth sets how precisely the Q is known; the gain–bandwidth sets how precisely the peak’s frequency is known. A selective amplifier is usually built to pick out one frequency, and its frequency error from this cause is systematic, predictable and the same sign every time — it can be designed out by moving the twin-T’s null up by $2f_0/\mathrm{GBW}$, which is the one thing the earlier essays’ analysis could never have asked for.

The selected frequency on its own skirt #

The shift is small in absolute terms. What makes it matter is its size against the peak’s width.

The peak’s half-width is $1/(A s)$ in the same units as the shift, so the shift is $\tan\varphi$ half-widths. At the design frequency the loop’s gain is therefore the stage’s own, $A\cos\varphi$, against a peak that has stayed at A: the fraction is cos φ, and the measurement follows it to three per cent up to sixty degrees. At 45 degrees of lag — a stage whose own bandwidth equals the null frequency, which for a gain of 100 is an amplifier of only 159 kHz — the frequency the circuit was built to select sits 2.93 dB down its own skirt, at the peak’s half-power edge.

Stated as a rule it is simple. A twin-T selective amplifier keeps its design frequency within its own passband while $f_0 A < \mathrm{GBW}$, the stage’s lag under 45 degrees; it keeps it within a tenth of a decibel of the top while $\cos\varphi > 0.989$, a lag under 8.5 degrees, which is $\mathrm{GBW} > 6.7\,A f_0$. For a Q of 25 at 1.6 kHz that is 1.07 MHz — an ordinary amplifier — and for a Q of 250 it is 10.7 MHz.

Two amplifiers, worked #

Take the circuit on this page as a tone detector for its own 1591.5 Hz, with a Q of about 25 from a stage of gain 100, and build it twice.

On a general-purpose amplifier of 1 MHz, the stage has 10 kHz of bandwidth and lags 9.04 degrees at the null. The Q is 25.32 against the ideal stage’s 25.25, and the peak sits 0.311 per cent low, 4.9 Hz. The design frequency gets 0.988 of the peak’s gain, a tenth of a decibel down. For a detector whose passband is 63 Hz wide at the half-power points that is nothing, and nothing about the design needs to change — which is the usual experience with selective amplifiers, and the reason the amplifier’s bandwidth is rarely suspected.

On a micropower amplifier of 100 kHz, the stage has a kilohertz of bandwidth, less than the null frequency, and lags 57.9 degrees. The Q is 25.43, still indistinguishable from the ideal. But the peak has moved 2.98 per cent down, 47 Hz, and the design frequency gets 0.544 of the peak’s gain — 5.3 dB down, well outside the passband. A bench measurement of the Q would find nothing wrong, since the peak is as sharp as it should be. A measurement at the design frequency would find a detector 5.3 dB down at the frequency it was built for, and no component whose value explains it.

The contrast with the parts is worth stating. The null that is stationary in nothing found that a twin-T’s tolerance fills its null rather than moving it; that is what made its residue real in the first place, and why a Q could be set by null depth. The amplifier does the opposite. It leaves the null and the Q alone and moves the frequency, which is the one quantity the parts’ tolerance does not move, so the two errors land on different axes and can be budgeted separately.

Why a gain with a pole is not a rotated gain #

The whole of this essay turns on one property of a single pole, and it is worth having in isolation. A gain $A/(1 + jx)$ has magnitude $A/\sqrt{1 + x^2}$ and phase $-\arctan x$. Its real part is $A/(1 + x^2)$ and its imaginary part $-Ax/(1 + x^2)$. What the loop’s denominator responds to is the gain’s real part set against one, and the twin-T’s own quadrature set against the gain’s imaginary part — and for a single pole the imaginary part is exactly $x$ times the real part, with no freedom between them. The ideal amplifier, and where it stops being one draws the same stage as a horizontal line that bends at GBW/A; the bend is where the real part starts to fall, and the lag is the same bend seen in the phase.

A drawing with the gain as an arrow of fixed length rotated by $\varphi$ is a gain with a pole and an all-pass section together, and it predicts a loop that does not exist. The lesson generalises to any argument that treats a phase as free of its magnitude, which in a minimum-phase network it never is — the phase the magnitude already knows recovers one from the other to a fraction of a degree.

What a designer should take #

A finite gain–bandwidth does not spoil a twin-T selective amplifier’s Q or its stability margin, whatever the intuition about a rotated gain says. It moves the peak down by $2f_0/\mathrm{GBW}$. Budget for that as a frequency error, correct it by trimming the twin-T’s null upward if the frequency matters, and keep $\mathrm{GBW} > A f_0$ at the very least so that the design frequency stays inside the peak.

The null-depth requirement of the earlier essay stands as it was: a null deeper than $20\log(44\,Q)$ for a Q held to ten per cent. The amplifier adds a second, independent requirement on frequency rather than on Q, and the two can be met separately.

The broader lesson is about arguing from a phasor diagram with the gain drawn as a fixed-length arrow. A real gain’s size and angle are not independent when they come from one pole, and a cancellation between them is the rule rather than the exception — the Q the amplifier decides met the active section’s version of the same question, where the amplifier’s finite bandwidth moved both the pole frequency and the Q, and the ratio decided which moved more.

How the numbers were obtained #

The twin-T is solved as a netlist at each frequency, with its shunt elements perturbed in opposite directions as in the earlier essays. The stage is a gain of A with one pole at GBW/A, so its lag at the null is $\arctan(f_0 A/\mathrm{GBW})$, and every figure sets the product from the lag wanted. Each loop’s stability is decided by the winding of $1 + A(f)\,T(f)$ about the origin, the criterion stable, and unstable with less gain uses where a single margin cannot decide, over twelve decades and a dense band around the null; the network is passive and the stage has no zeros, so any winding means poles in the right half plane. The peak is found by a coarse scan over half to one and a half times the null and refined by golden section, and both half-power edges by bisection. Each wall is bisected on stability over the shunt error, thirty-six iterations.

What it leaves out #

A stage with a second pole. Everything above rests on the stage’s gain and phase coming from one pole, which is what makes them cancel. An amplifier with a second pole below the stage’s bandwidth lags without shrinking in proportion, and then the argument in the first section is closer to right: the Q rises and the wall moves in. How far depends on where the second pole sits against $f_0$.

The amplifier’s output resistance. The twin-T is driven from an ideal source here. A few tens of ohms against the twin-T’s input impedance near the null, which falls to a few kilohms, is a small change in the network’s transfer and, since the residue at the null is what matters, a small change in the residue.

Non-inverting against inverting stages. The stage here is a gain with no sign. An inverting stage puts the twin-T’s residue on the other side of the law, and the whole of this essay’s arithmetic carries across with its sign reversed.

Still open: the second pole, the trim that corrects the frequency, and the bridged-T #

An amplifier with a second pole. The cancellation that protects the Q and the wall holds for one pole. A second, at a frequency comparable to the null, breaks it: the stage’s phase then outruns its magnitude’s fall. Solving the loop with a two-pole stage would find the product of gain and second-pole frequency below which the Q is no longer the parts’ and the wall moves.

Trimming the null upward. The shift is systematic, so the twin-T could be designed with its null $2f_0/\mathrm{GBW}$ high and the peak would land on the design frequency. Whether a trim that corrects frequency also disturbs the residue — and so the Q — is a question of which element is trimmed, and the answer decides whether frequency and Q can be adjusted independently.

The bridged-T in the same loop. A bridged-T nulls with one fewer component, and the residue that has only one sign asks whether its residue is real, signed and on the same law.