The bandwidth the product still gets right

The ideal amplifier, and where it stops being one derived the gain–bandwidth product in one line. Above its dominant pole an amplifier’s gain falls as one over frequency, so gain times frequency is a constant; a loop closed for a gain G reaches unity loop gain where the open-loop gain has fallen to G, at GBW/G. The one line depends on the single-pole shape, and that essay said as much: an amplifier with two poles close together would not have a constant product and would be unstable at low gains, which is why it is not built that way.

Every real amplifier is built with a second pole anyway; it is only built far enough out. The question that leaves is the useful one. How far out is far enough for the product to predict a stage’s bandwidth, and which bandwidth? A stage has two that matter: the half-power frequency a data sheet quotes, and the noise bandwidth that sets how much of an amplifier’s own noise reaches the output. The measurement here finds that the second pole moves the first by as much as forty per cent and does not move the second at all.

The stage, and the one number that describes the second pole #

The amplifier is a netlist model with two poles: a dominant one set by its direct-current gain of a hundred thousand and its gain–bandwidth product of 1 MHz, and a second one at a stated frequency, with buffers between so each pole is what it says. Two resistors close it as a non-inverting stage of gain G. Its response is solved at each frequency and divided by its value at direct current, and the half-power frequency and the noise bandwidth are read from that.

With the dominant pole far below the band, the amplifier is $\omega_t/s$ times $1/(1 + s/p_2)$, and the closed stage is a second-order low-pass. Its natural frequency is $\sqrt{\beta\omega_t p_2}$ and its damping ratio $\zeta = \sqrt r/2$, where $r = p_2/(\beta\,\mathrm{GBW})$ is the second pole measured in units of the bandwidth the product predicts. That single ratio is all that describes the second pole’s effect: a 1 MHz part at a gain of ten with its second pole at 2 MHz has $r = 20$; the same part as a follower has $r = 2$. The half-power bandwidth in units of the prediction is then $\sqrt r\,g(\sqrt r/2)$, with $g(\zeta) = \sqrt{1 - 2\zeta^2 + \sqrt{4\zeta^4 - 4\zeta^2 + 2}}$ the textbook second-order bandwidth factor. The netlist agrees with it to two parts in a thousand at every placement, which is the netlist’s own resolution in a bandwidth found by bisection.

Wider, not narrower #

The product’s prediction is low. A second pole adds lag, and the intuition from every loop essay is that lag costs bandwidth; here it buys it. At $r = 100$ the half-power bandwidth is 1.010 times GBW/G, at ten 1.110, at five 1.231 and at two 1.414 — forty-one per cent wider than the product says. Below two it narrows again: 1.272 at one, and exactly the product’s prediction at a half, though by then the response is peaking.

The mechanism is the loop’s damping. With one pole the closed loop is first-order and its half-power point is where the loop gain is one. With a second pole the closed loop is second-order, and a second-order response with a damping ratio near 0.7 holds its gain flat further than a first-order one of the same pole product would, then falls faster. The damping ratio $\sqrt r/2$ is 0.707 at $r = 2$, the maximally flat Butterworth case, which is where the extra bandwidth is largest: exactly $\sqrt 2$ of the prediction, the closed form’s own value there.

What the extra bandwidth costs #

The extra bandwidth comes with a smaller phase margin, and the figure lines the two up. At $r = 2$ the loop has 65.5 degrees of margin, no peaking, and the widest half-power bandwidth. Below two it peaks — 1.25 decibels at $r = 1$, 3.59 at a half — and its margin falls through fifty degrees toward forty. Above two the peaking is gone and the margin rises: 76.3 degrees at four, 84.3 at ten. The product’s prediction is approached only as the margin approaches ninety, which is the single-pole loop it was derived for.

So “the product holds” means something narrow: the margin is above about eighty-five degrees. What is left at crossover measured a margin as the phase remaining at the loop’s crossover; a designer who wants forty-five or sixty degrees has, by choosing them, chosen a stage whose half-power bandwidth is ten to forty per cent wider than GBW/G. The gain added to buy a margin met the same arithmetic from the other side, raising a decompensated amplifier’s gain until its margin was acceptable; the margin it bought was paid for in bandwidth closer to the product’s.

One part, ten gains #

For a real part the second pole is fixed and the gain chooses $r$: a 1 MHz amplifier with its second pole at 2.5 MHz has $r = 2.5G$. Plotted against the frequency the product predicts, its responses at gains from one to a hundred would all pass three decibels down at one. They spread from 1.005 at a gain of a hundred to 1.397 at a gain of one. The follower, whose second pole is only two and a half times its predicted bandwidth, runs forty per cent wide; a gain of two, twenty-three per cent.

The product predicts this part’s half-power bandwidth to within ten per cent only from a gain of five up. That is the range most data sheets illustrate the product with, and it is exactly the range where the second pole has moved out of the way. Below a gain of five the rule underestimates, and the error is largest at the gain most often used — a follower.

The bandwidth that does not move #

The noise bandwidth is a different integral. It is the area under the squared response, $\int_0^\infty |H(f)/H(0)|^2\,df$, and for a second-order low-pass with denominator $1 + a_1 s + a_2 s^2$ that area has a closed form with a surprising property: it is $1/(4a_1)$, independent of $a_2$. The stage’s $a_1$ is $1/(\beta\omega_t)$ — the integrator’s coefficient, the one the product is made of — and the second pole adds only $a_2$. So the noise bandwidth of a stage round a two-pole amplifier is $(\pi/2)\,\beta\,\mathrm{GBW}$ wherever the second pole sits, which is exactly the one-pole stage’s noise bandwidth.

The hero figure measures it on the netlist at thirteen placements of the second pole and finds it within half a per cent of $(\pi/2)\,\mathrm{GBW}/G$ at every one — through the forty-one per cent excursion of the half-power bandwidth, through the peaking below $r = 2$, at margins from thirty-eight degrees to ninety. The one-part figure finds the same at every gain to six parts in a thousand. A noise calculation made with the product is right for this part at any gain; a bandwidth calculation made with it is not.

The bandwidth noise sees found a single pole passing $\pi/2$ times as much noise as a brick wall at its corner, and the ratio that does not walk to one found that ratio behaving differently in each filter family. Here the ratio of noise bandwidth to half-power bandwidth is not a property of a family but of the margin: $\pi/2$ at ninety degrees, $(\pi/2)/\sqrt 2 = 1.11$ at $r = 2$, and the noise bandwidth stays put while the half-power point moves under it.

The identity is of second order only #

The independence of $a_2$ is a property of second-order denominators and does not survive a third pole. With a third pole at three times the second, the noise bandwidth rises above the product’s: a third more at $r = 1$, fourteen per cent more at two, five per cent at five, two and a half at ten. The excess fits $4r/(4r - 1)$ to two parts in a thousand, which is what the third-order integral gives for that pole ratio, and it grows without limit as $r$ falls toward a quarter, where the three-pole loop goes unstable.

That is the edge of the identity, and it is the important caveat for a real amplifier. A part whose second pole is the only one within a decade of its crossover behaves as a two-pole part and its noise bandwidth is the product’s. A part with a second and third pole close together — which is what a decompensated or a very fast amplifier often has — passes more noise than the product says, by the factor the figure shows, and the extra noise is concentrated where the response peaks.

Two routes to the noise bandwidth #

The noise bandwidth has been reached two ways that share only the circuit. The netlist route solves the stage at four thousand frequencies over eight decades and integrates the squared response by the trapezoid in logarithmic frequency. The closed-form route is the second-order integral, $\pi/2$ over the $s$-coefficient, which contains no second pole at all. They agree to half a per cent at every placement, and the half a per cent is the trapezoid’s error on a curve that rises and falls steeply near the peak. For the third-pole curve there is no netlist; the loop is evaluated in closed form, and the agreement with $4r/(4r - 1)$ to two parts in a thousand is the check on the integration.

What a designer can take from it #

The two bandwidths answer different questions, and the product answers one of them reliably. For noise — the output noise of a stage whose amplifier’s input noise density is flat — use $(\pi/2)\,\mathrm{GBW}/G$ and trust it, at any gain and any second-pole position, as long as only one extra pole is near the crossover. For the band a signal passes through, the product is a lower bound when the margin is below ninety degrees: forty per cent low for a follower round a part with its second pole at two and a half times its unity-gain frequency, ten per cent low at a gain of five.

That asymmetry matters in a stage meant to filter. An anti-alias stage whose corner is set by the product sits forty per cent above its design corner if it is a follower, and its aliases are larger by that much; its noise is exactly what the design said. A design that checks its noise against the product and its corner by simulation gets both right. The reverse mistake is the commoner one: a designer who measures a follower’s bandwidth on the bench, finds it forty per cent above GBW/G, and scales the expected noise up by the same factor has overestimated the noise by exactly that factor, because the noise bandwidth never moved.

The step the extra bandwidth comes with #

A half-power bandwidth forty per cent wide at $r = 2$ is not a free improvement, and the time domain says what it is. A second-order stage at a damping ratio of 0.707 rises faster than the one-pole stage of the same product — its rise time is set by its natural frequency, $\sqrt 2$ times the one-pole stage’s corner — and overshoots by 4.3 per cent before it settles. At $r = 1$, where the damping ratio is a half, it overshoots by 16 per cent; at a half, by 30. The one-pole stage the product describes never overshoots at all. One margin, and the gain under it found overshoot predicted better by a loop’s closed-loop peak than by its margin, and here both come from the same damping ratio, $\sqrt r/2$.

So the forty per cent is bandwidth bought with ringing, and a stage that must settle cleanly is designed at the larger $r$ that kills the overshoot — which brings the half-power bandwidth back toward the product. The noise bandwidth does not take part in that trade. Whatever $r$ the designer chooses for the step, the noise the stage passes is the same, because the integral that measures it has already been shown not to depend on the second pole.

Why one coefficient #

It is worth seeing why the second-order noise integral forgets its $s^2$ coefficient, because the reason generalises. The squared magnitude of $1/(1 + a_1 s + a_2 s^2)$ on the frequency axis is $1/((1 - a_2\omega^2)^2 + a_1^2\omega^2)$. Moving $a_2$ moves the resonance and raises or lowers a peak near $\omega = 1/\sqrt{a_2}$, and the integral of the extra area under the peak exactly balances the area lost or gained on its skirts: the total depends on $a_1$ alone. That is a balance of the same kind the area that feedback only moves found in Bode’s sensitivity integral — feedback can move a quantity’s distribution over frequency but not its total — applied to the stage’s own response. A third pole adds a term the balance does not cover, and the total grows.

Where the second pole comes from #

A real amplifier’s second pole is usually the one its own internal stages set — an output stage’s, a level shifter’s — and a unity-gain-stable amplifier is compensated so that it sits a little above its unity-gain frequency, at the margin the manufacturer chose. That puts a follower at an $r$ of one and a half to three, exactly the region where the half-power bandwidth is furthest from the product. The node the bandwidth belongs to found the same kind of fact at the level of a single transistor stage: the bandwidth belongs to whichever node is slowest, and an estimate made at one node misreports it when another is close. The amplifier’s second pole is that other node, seen from outside.

What the model leaves out #

The amplifier has exactly two poles, with buffers between them and an output resistance of a micro-ohm, and the feedback network is resistive. A real feedback network’s resistors and the amplifier’s input capacitance add a pole of their own inside the loop, which for large resistors can sit as low as the second pole and act as a third; the third-pole figure is then the one that applies. And the noise bandwidth assumes the amplifier’s noise is white and enters at its input, where the loop shapes it with the same response as the signal; a noise current through the feedback resistors, or flicker noise, has its own weighting and is outside this identity.

The number to keep #

With a second pole at $r$ times GBW/G, a stage’s half-power bandwidth is $\sqrt r\,g(\sqrt r/2)$ times the product’s prediction: up to $\sqrt 2$ at $r = 2$, within ten per cent only beyond $r = 10$, so a follower round a 1 MHz part with its second pole at 2.5 MHz is forty per cent wider than GBW/G. Its noise bandwidth is $(\pi/2)\,\mathrm{GBW}/G$ exactly, at every $r$ and every gain, because a second-order noise bandwidth depends on one coefficient only. A third pole at three times the second raises it by $4r/(4r - 1)$.

Still open: the step that settles, the inverting stage, and the third pole a resistor makes #

The step response. A wider half-power bandwidth at $r = 2$ comes with a step that rises faster and overshoots a little. Whether the stage settles to a tolerance sooner or later than the one-pole stage the product describes — and at which $r$ the settling time is least — is the time-domain reading of the same second pole, and it need not agree with either bandwidth.

The inverting stage. An inverting stage of gain −G closes its loop against G + 1, as the gain the loop closes against found, so its $r$ for the same part is larger and its half-power bandwidth closer to the product’s. Whether its noise bandwidth is still exactly $(\pi/2)\,\mathrm{GBW}/(G + 1)$ with the input resistor’s noise included is the question for a summing node.

The third pole a feedback network makes. Large feedback resistors and the amplifier’s input capacitance put a pole inside the loop. Locating it as a third pole and reading the noise bandwidth off the $4r/(4r - 1)$ family would say how large the resistors can be before the product stops predicting a stage’s noise.