The residue that has only one sign

The sign the null leaves behind found that a twin-T’s null, filled by an error in its parts, leaves a residue that is a real number, $-t/4$, and that in a loop of gain A this residue sets the quality factor through one law: $(A + 1)/4$ divided by $1 - A t/4$. The sign of the error decides the side. One way the Q is damped towards $1/|t|$; the other way it grows until the loop oscillates. The peak the amplifier moves then found that a real amplifier’s bandwidth moves the peak without touching that law.

That essay’s own question was whether the law belongs to the twin-T or to every resistor-capacitor null. The obvious second network is the bridged-T, which nulls with five parts instead of six: two capacitors in series, a resistor $R_1$ bridging them from input to output, and a resistor $R_2$ from their junction to ground. It is the notch in older tone-control and oscillator circuits, and it is usually described as a twin-T with less symmetry. The measurement says it is something different in kind, and the difference is a sign.

A floor instead of a null #

The bridged-T here uses the twin-T’s ten-nanofarad capacitors and puts its minimum at the same 1591.5 Hz, which fixes the product of its two resistors, $\sqrt{R_1 R_2} = 10\text{ kΩ}$, and leaves their ratio $n = R_1/R_2$ free.

With a ratio of a hundred the bridged-T’s transfer does not reach zero. It bottoms at 0.0196, 34.2 dB down, and stays there whatever else is done to it. Around a gain of a thousand it makes a peak with a Q of 4.76. The twin-T around a tenth of that gain makes 25. The slider on the figure at the head of the page steps the ratio from ten to a thousand: the floor deepens from 15.6 to 54.0 dB and the loop’s Q rises from 1.53 to 10.5, always far below what the gain would buy with a twin-T.

The floor is not a tolerance effect. With every part exact, the bridged-T’s transfer at the null frequency is, from its two node equations,

$$T(j\omega_0) = \frac{2 j\omega_0 C G_1}{j\omega_0 C (2 G_1 + G_2)} = \frac{2 R_2}{2 R_2 + R_1} = \frac{2}{2 + n},$$

where the frequency is the one at which $\omega_0^2 C^2 = G_1 G_2$ removes the real parts. The ratio of two conductances, both positive, cannot be zero for any finite parts and cannot be negative for any parts at all.

Why one fewer part costs the null #

The algebra says where the null went, and it is worth reading because it is the same argument for any network. A transfer of this kind is a ratio of two quadratics in $s = j\omega$, and on the $j\omega$ axis the numerator’s even terms are real and its odd term imaginary. A true null needs both to vanish at one frequency: the two even terms cancel each other at $\omega_0$, and the odd term must be zero in its own right. That is what a transmission zero on the axis is, the kind the zeros that buy an order places deliberately in an elliptic filter’s stopband.

In the twin-T the odd term is a difference: the numerator’s middle coefficient is the difference of what the network’s two parallel paths contribute, one a resistive tee and one a capacitive tee, and with the right proportions the difference is zero. That is why its null is exact with exact parts, and why an error in either direction leaves a residue of either sign — the difference simply stops being zero, one way or the other. What actually fills a null and the null that is stationary in nothing measured the two ways that difference can be spoiled.

In the bridged-T the odd term of the numerator is $2\omega C G_1$, a single product of positive quantities. There is no second path to subtract it from: the bridging resistor and the capacitive tee meet at the output, but the tee’s contribution to the odd term has the same sign as the resistor’s. So the even terms can be made to cancel, which fixes the frequency, and the odd term cannot, which fixes the floor. Removing one part from the twin-T removed the subtraction, and with it the possibility of zero. The floor is then $2G_1$ over the denominator’s corresponding term, $2G_1 + G_2$, and it shrinks only as $G_1$ is made small against $G_2$ — a large bridging resistor against a small shunt one.

Real, like the twin-T’s, and of one sign #

The solve agrees with $2/(2 + n)$ to a part in a million across three decades of ratio, and the residue’s imaginary part is below a millionth of its real part. So the bridged-T shares the property that made the twin-T’s loop tractable: at the frequency that matters its residue is a real number, a small in-phase copy of the input, with no quadrature to rotate it. It does not share the other property. Every corner of ±1 per cent on its three parts leaves the residue between 0.0191 and 0.0201 at a ratio of a hundred: moved a little, and positive.

The dotted levels give the comparison its scale. A twin-T with one per cent parts leaves a residue of 0.0025; with a tenth of a per cent, 0.00025. A bridged-T needs a ratio of about 800 to match the first and 8000 to match the second, and with a ratio of a hundred it leaves eight times the residue of a one-per-cent twin-T. The bridged-T’s null is shallower by construction, not by error.

One law, two sides of it #

If the law of the earlier essay belongs to every resistor-capacitor null, the bridged-T’s loop should follow it with the residue $2/(2 + n)$ in place of $-t/4$ — and since that residue is positive, it should sit permanently on the damped side.

Both networks lie on one line. With the residue written $\rho$ — the twin-T’s $-t/4$, the bridged-T’s $2/(2 + n)$ — the loop’s Q is its residue-free value divided by $1 + A\rho$, to five per cent for every point measured. The law is a property of any loop whose feedback network passes through a small real value at the frequency of interest; the twin-T was simply the first place it was found.

The difference is where each network may sit. The twin-T’s squares straddle the line where $1 + A\rho = 1$, because its residue is zero with exact parts and takes the sign of whatever error the parts have: its loop can be damped or sharpened, and past $A\rho = -1$ it oscillates. The bridged-T’s dots all lie to the right. More gain moves them further right, deeper into damping, and there is no error in the parts that can move one to the left.

A ceiling instead of a wall #

The residue-free value of the Q is itself proportional to the gain — the gain times the network’s quadrature slope at the null, over two — so the law has a consequence the twin-T never showed. For the bridged-T,

$$Q = \frac{A\,s}{2\,(1 + A\rho)} \longrightarrow \frac{s}{2\rho} \quad (A\rho \gg 1),$$

and a Q that rises in proportion to the gain while $A\rho$ is small stops rising once it is large.

The measurement shows the ceiling plainly. With a ratio of a hundred the Q climbs to about 3.3 at a gain of a hundred, 4.8 at a thousand, and 5.00 at a hundred thousand, where it has stopped. With a ratio of a thousand the ceiling is 15.73; with ten, 1.54. For the two larger ratios these are $s/(2\rho)$ to three per cent, and they approach $\sqrt{n}/2$ as the ratio grows, because the bridged-T’s slope at its null falls as $2/\sqrt{n}$ while its residue falls as $2/n$. For a ratio of ten the peak is too broad for a slope taken at the null to describe, and the law reads a sixth low.

How close to its ceiling a loop gets is the law again. Dividing $A s/(2(1 + A\rho))$ by its limit $s/(2\rho)$ leaves $A\rho/(1 + A\rho)$, so a loop reaches ninety per cent of its ceiling when $A\rho = 9$ and ninety-five when $A\rho = 19$. With a ratio of a hundred the residue is 0.0196 and a gain of a thousand gives $A\rho = 19.6$: the 4.76 measured at that gain is 95.2 per cent of the ceiling, as the fraction says. The gain a bridged-T needs is therefore about $4.5\,n$ for ninety per cent, and since $n$ is about $4Q^2$, about $18\,Q^2$ — a demand that grows with the square of the Q, where a twin-T’s grows with the Q itself.

No gain up to a million makes any of the three oscillate. That is the other half of the one-sign residue: a loop whose feedback network never passes through zero at the frequency where the gain is largest has nothing to cancel the one in $1 + A\,T$ with. The twin-T has a wall at $A t/4 = 1$ and no ceiling; the bridged-T has a ceiling at $\sqrt{n}/2$ and no wall. To make a Q of 25 with a bridged-T takes a ratio of about 2500, a bridging resistor of 500 kΩ against a shunt resistor of 200 Ω, and an amplifier gain large against the 1250 that makes $A\rho$ equal to one.

What the ceiling buys #

A ceiling is a limitation, and it is also a kind of immunity. At the ceiling the Q is a ratio of two of the network’s own properties and the gain has dropped out of it — and the network’s properties are ratios of its parts.

Built for the same Q of five, the two networks respond to their parts very differently. The twin-T needs only a gain of 19, and its Q runs from 4.770 to 5.222 over the corners of one-per-cent parts, a spread of 9.1 per cent, because its residue is its parts’ error and the gain multiplies it. The bridged-T at its ceiling runs from 4.946 to 5.046, two per cent, because a one-per-cent change in either resistor changes the ratio $n$ by one per cent and the ceiling by only half of that. It buys that steadiness with a gain five thousand times larger and a Q it can never exceed.

So the choice between the two is not the component count it is usually made on. A twin-T is the network for a high Q from modest gain, with a null that has to be trimmed and a wall that has to be respected. A bridged-T is the network for a low, stable Q that no amount of gain and no error in the parts can push into oscillation — which is exactly the property an oscillator’s amplitude-control loop or a tone control wants, and exactly the wrong property for a narrow selective amplifier.

A low-Q peak, worked #

The bridged-T’s natural territory is a Q of two to eight, and the numbers there are easy to read off the figures. For a gentle peak with a Q near 2.7 — a presence boost, a loop that must emphasise one band without ringing — a ratio of thirty does it: $R_1$ of 54.8 kΩ and $R_2$ of 1.83 kΩ with the ten-nanofarad capacitors, a floor of 24.1 dB, and a Q of 2.69 around a gain of a thousand that no further gain can raise past its ceiling. For a Q near 7.5 a ratio of three hundred gives 7.53 at the same gain, with a floor of 43.6 dB.

The twin-T would reach either Q with a gain of ten or thirty, and it would reach it with a Q that moves by the parts’ error times the gain, and a wall a few per cent away. Where the Q comes from found the same trade inside active sections, where two circuits with the same transfer function differ in which of their parts the Q depends on; the bridged-T is the passive network in which the Q depends on a ratio of two resistors and on nothing else once the gain is large. The price, which that essay would recognise, is an amplifier whose gain is spent on holding the ceiling rather than on the Q itself — the Q the amplifier decides prices exactly that kind of gain in an active section.

What a designer should take #

The twin-T’s law, $Q = Q_0/(1 + A\rho)$, is general: find the feedback network’s residue at the frequency of interest, check that it is real, and read the loop from it. For a bridged-T the residue is $2R_2/(2R_2 + R_1)$ and positive, so the loop is damped for any gain and its Q is capped near $\sqrt{R_1/R_2}/2$. Choose the ratio for the Q wanted — about $4Q^2$ — and give the amplifier enough gain to make $A\rho$ several times one, at which point the Q is set by the ratio and is as steady as the two resistors’ ratio is.

For a Q above about ten, the bridged-T’s ratio becomes awkward and its null shallow, and a twin-T or an active section is the better tool. Below about five, where the twin-T’s sensitivity to its parts is at its worst relative to what it delivers, the bridged-T’s ceiling is an advantage rather than a limit.

How the numbers were obtained #

The bridged-T is solved as a five-element netlist at each frequency, with a teraohm load on its output, and its residue is its transfer at $1/(2\pi\sqrt{R_1R_2}\,C)$. The quadrature slope is a central difference of the transfer’s imaginary part at a hundredth of a per cent either side. Each loop is an ideal gain around the network; its stability is decided by the winding of $1 + A\,T$ over twelve decades and a dense band near the null, its peak by a scan and golden section, and its half-power edges by bisection over a range wide enough for a Q of one. The twin-T’s points are the earlier essay’s loops, located the same way. The tolerance corners are every combination of ±1 per cent on the three parts that matter in each network.

What it leaves out #

The other bridged-T. The dual arrangement — two resistors in series, bridged by a capacitor, with a capacitor to ground from their junction — should, by the same argument with the roles of resistors and capacitors exchanged, have a positive residue of its own. It is not solved here, and the argument is not a measurement.

A bridged-T with a negative element. A bridged-T whose shunt resistor is partly cancelled by a negative resistance, or whose bridging element includes a gain, can have its residue pushed through zero, and then the one-sign argument no longer holds. A bridged-T that reaches a high Q does it this way, as an active network in which the amplifier supplies what the passive five parts cannot.

The amplifier. Every loop here has an ideal gain. The peak the amplifier moves found a one-pole stage moving the twin-T loop’s peak by $2f_0/\mathrm{GBW}$ without touching its Q; the same argument should apply to the bridged-T’s, with the network’s own slope in place of the twin-T’s half.

Still open: the dual network, a trimmed bridged-T, and the loop that needs a ceiling #

The dual bridged-T. Two resistors bridged by a capacitor, with the second capacitor to ground, should have a positive residue and a ceiling of its own. Whether its ceiling is the same function of its ratio, or whether its slope at the null falls differently with the ratio, decides which of the two is the better low-Q network.

A bridged-T trimmed towards zero. Adding a small resistance in series with the shunt capacitor path, or a fraction of the output fed back to the junction, can move the residue towards zero from the positive side. How close it can come before the parts’ tolerances make its sign uncertain — at which point it has become a twin-T’s problem — is the measurement that would say whether a high-Q bridged-T is ever a good idea.

The loop that needs a ceiling. A Wien-bridge oscillator needs a loop that sits exactly on the boundary; an automatic gain control needs one that cannot be pushed past it. A bridged-T in the gain-control path, whose Q cannot grow whatever the control amplifier does, might be the passive part of such a loop, and marching one would say whether the ceiling survives the nonlinearity.