The floor the gains bring

The ceiling a gain gives back took a fifth-order Chebyshev whose first section’s capacitor node carries 4.537 times the input, and applied the textbook repair. Divide the signal in front of the loud section, restore it later, and every node peaks at the same level. Done exactly, with silent gains, it lowered the ceiling to the input and gave back 2.16 of the 13.14 decibels the internal peak had taken. It said clearly where that figure was flattered: the inserted gains were controlled sources, and a real gain stage has two feedback resistors and an amplifier of a few nanovolts per root hertz. Every one of the 2.16 decibels was therefore an upper bound on what scaling buys, and it suggested that with that floor in the budget there would be an optimum, some scaling less than equal maxima, trading a little ceiling for a lot less floor.

The essay’s suggestion is reasonable, and the arithmetic that follows from taking it seriously does not come out as it predicted. There is no such optimum. A real amplifier noise has to go in everywhere for the comparison to be fair, which raises a question that the silent budget could ignore: the filter’s own amplifiers are not silent either. Counting them changes what the floor is. Once that is done, the scaling that seemed to be a small sure gain is a decision with a very clear boundary, and the Chebyshev 5 is on the wrong side of it with an ordinary amplifier.

Three budgets in one figure #

The filter is the earlier essay’s, a half-decibel Chebyshev of order five at a kilohertz, three Sallen–Key sections with unity-gain followers. Its signal budget is unchanged: gains $d_k$, determined from the unscaled solve so that the largest node of each section reaches the input’s level, and with $\alpha$ applied as $d_k^\alpha$ so that the amount of scaling can be walked from none to all of it. The response stays exact at every $\alpha$, to a part in a billion, because the gains always multiply to one.

The noise budget now has three kinds of source in it. The resistors were always there. Their Johnson noise is carried to the output by each resistor’s own transfer. The amplifiers inside the sections are new here: each follower gets a voltage noise $e_n$ in series with its non-inverting input, carried out by its own transfer. And each gain stage, wherever one exists, gets the same amplifier and also its resistors: below a gain of one a divider with a kilohm source resistance into a follower, above it a non-inverting stage with a kilohm lower arm. All of it is integrated from 1 Hz to 1 MHz on the sixty-points-a-decade grid the earlier budgets use, and all of it is summed in power. An everything-silent run has to reproduce 2.16 dB, and it does, which is what connects this to the earlier measurement: 142.44 to 144.60.

With everything at 4 $\text{nV}/\sqrt{\text{Hz}}$, the range is 118.32 dB without stages. It is already 24 dB below the silent figure. That is the first thing a reader of the earlier numbers should note, because almost all of the silent budget’s 250 nV was the resistors’ noise in the band of the filter, and now the amplifiers supply about 4 µV. A followers’ noise at the filter’s output fills the whole 1 MHz of the measurement. The filter cannot filter what is behind its last resistor.

There is no optimum in the middle #

The suggested trade has a clear shape on the figure and it is not what was expected. Going from no stages to $\alpha = 0.2$ costs about 2.6 dB, with very little gain to show for it, since a real amplifier has been put in the signal path where before there was a wire. From there the curve rises monotonically: 116.27 at 0.4, 116.86 at 0.6, 117.50 at 0.8 and 118.18 at 1. So the largest $\alpha$ is the best $\alpha$ among those with stages. A little less than equal maxima is not better than equal maxima, and that holds at all the amplifier noises considered.

Why it rises is simple once the floor is known. The ceiling improves with $\alpha$ directly, since the loudest node is reduced by the divider in front of it. The floor is dominated by the last gain, whose gain also rises with $\alpha$. Both are exponents of $\alpha$ applied to fixed numbers, and the ceiling’s is the larger: 13.14 dB to gain against 9.04 dB of last gain. Going partway gets the same fraction of each, so in decibels the curve is close to a straight line, with no curvature for an optimum to sit at. The price of stages is paid once, at the step, and it is a fixed cost that any further scaling reduces.

So the design question is a yes or a no: either the full scaling, or none. An engineer considering a little scaling to be careful has found the worst of both.

Where the floor lands #

The source-by-source budget at equal maxima says why the yes costs so much.

The last stage is 14.5 µV by itself. It is a gain of 2.833 with an amplifier whose noise fills the whole band, multiplied at the output, with nothing following it. Its Johnson noise share is also multiplied by the same gain. And the sections’ amplifiers, which together contribute 4 µV to the unscaled filter, contribute 11.3 µV at equal maxima, since each one’s noise is now multiplied by the last gain as well.

The stages inside the chain are almost invisible by comparison: 797 nV from the input stage, 493 nV from the middle one. Those noise sources are filtered by the rest of the filter, and they contribute only in-band. Any noise source before a filter is filtered. Every noise source after it is not. So the expensive stage is the one outside the filter, and the expensive thing about equal maxima is that its gain sits there.

Totalled, the floor is 18.5 µV against 4.01 µV, up 13.28 dB. The ceiling has gained 13.14 dB. The difference between them is the 0.14 dB of loss on the figure, and the two are close for a reason that can be read off. The last gain raises the sections’ amplifiers by its own 9.04 dB, and that alone leaves 4.10 dB of the ceiling’s gain in hand. The stage then brings a second source of about the same size as what it has just amplified, since its own amplifier is multiplied by the same gain, and two equal sources are 3 dB; its feedback resistors supply the rest. So the budget is 13.14 dB of ceiling against 9.04 dB for the gain and a little over 4 dB for being a real stage. The earlier essay’s 2.16 dB came from a budget in which that last 4 dB did not exist.

A noisier amplifier makes it look better #

The next figure is the counter-intuitive one. Moving the amplifier’s noise changes how good scaling looks, and the direction is opposite to what a reasonable reader expects.

At 0.5 $\text{nV}/\sqrt{\text{Hz}}$ equal maxima costs 11.38 dB. At 1 $\text{nV}/\sqrt{\text{Hz}}$ it costs 6.61, at 2 it costs 2.50, at 4 it costs 0.14, at 8 it gains 0.73, and at 16 it gains 0.99. A quieter amplifier does not want the scaling, and a very quiet amplifier is badly hurt by it. The reason is the stages’ resistors. The kilohm in each stage’s arms has its own noise, 4 $\text{nV}/\sqrt{\text{Hz}}$, which is fixed regardless of what the amplifier is doing. With a 0.5 $\text{nV}/\sqrt{\text{Hz}}$ amplifier, the unscaled filter’s floor is down near the amplifier, and adding a stage adds a kilohm’s noise that was not in the design at all. With a 16 $\text{nV}/\sqrt{\text{Hz}}$ amplifier the kilohm is negligible in comparison, and the stage then costs only the 3 dB of being a second source, which leaves the 0.99 dB.

So the reader who supposes that a careful low-noise design is the one that would most benefit from scaling has it backwards. The careful designer has already spent money on an amplifier, and scaling with a kilohm stage throws it away. For a low-noise filter, the impedance scaling that the earlier essay named as the other knob is the right tool, because it lowers the stages’ resistors along with everything else. Stage resistors of 100 Ω rather than a kilohm would move the crossing on this figure a good way to the left, at the cost of the stage’s current.

It also says something about the “2.16 dB” as a result. It was exact, it was an upper bound, and on the circuit it described it was right. In this budget it does not survive even as a rough guide. What survives is the idea that a filter’s internal peak costs it range, and the ceiling is not at the output stands. The repair is what depends on the amplifier.

The peak sets the prize #

The 0.14 dB is specific to this filter’s 4.537. A Chebyshev of order seven has a much louder inner node, so it has much more to give back.

At 4 $\text{nV}/\sqrt{\text{Hz}}$ the same arrangement gains 6.15 dB on the Chebyshev 7, and it gains nothing on the Chebyshev 5 or the Butterworth 5. So the question of whether to scale has a practical threshold in the peak’s size. With a peak under about five times the input, an ordinary amplifier’s filter gains nothing from it, and the search returns $\alpha = 0$. A peak of more than ten times the input is worth scaling at any ordinary amplifier noise, and with an amplifier as low as 1 $\text{nV}/\sqrt{\text{Hz}}$ it is still not. This is the same structure as which section goes first, which found that an ordering decision has much more to recover on a filter with a much higher peak.

The Butterworth 5 is worth a sentence. Its peak of 1.70 costs it only 4.6 dB of ceiling, its sections are nearly quiet relative to each other, and no stage on any budget here repays the amplifier it adds. Scaling a low-Q filter with a real amplifier is simply a loss.

A narrower band does not rescue it #

It would be natural to suppose that all this is an artefact of the 1 MHz measurement band. The last stage’s noise is out of band for a kilohertz filter, and a real system often has a following filter or a converter’s anti-alias filter that would remove most of it. With a narrower measuring band, it would seem, the last stage should stop dominating and the scaling should begin to pay.

It does not, and that is a useful correction to an instinct. The absolute range improves greatly as the band narrows: 139.40 dB measured to 3 kHz against 118.32 to 1 MHz. But the comparison does not change sign. Equal maxima loses 0.67 dB at 3 kHz, and at 1 MHz loses 0.14. In fact it loses a little more with the narrower band. The reason is that the unscaled filter has a last amplifier too, its own output follower, whose noise has the same bandwidth and is in the same position. Narrowing the band improves both floors by the same amount. What separates them is the gain of 2.833 and the extra stages, which are independent of bandwidth.

So a bandwidth limit does not change the verdict. What it would change is a design in which the output filter is placed between the last section’s output and a separate gain stage, that is, one where the gain stage’s noise does get filtered while the filter’s own output amplifier has the larger signal. There are several arrangements of that kind and they are what a design with a real range requirement would explore.

Where the gain ought to go #

The arithmetic implies its own remedy. The expensive stage is expensive because nothing follows it, and there is no law requiring it to sit after the last section. A gain placed before the last section’s filter has its noise filtered by that section, at the cost of the section then having a louder input, which is a ceiling decision. So the equal-maxima prescription, which optimises only the ceiling and places the last gain after the last section because that is where its restoring factor is needed, is doing the wrong thing with respect to this budget. A prescription that optimised the range would put the gains earlier and accept that the last section’s nodes are above the input. The ceiling a gain gives back noted that the silent gains could be moved without affecting anything. Real ones cannot, and this budget sets how to choose. That optimisation is not computed here. The fact that it exists is what the source-by-source figure shows.

The other remedy is a filter topology that has no inner peak in the first place. A doubly terminated ladder filter spreads the resonant energy among its nodes, and its internal peaks are known to be smaller. A floor and a ceiling measured the same quantity for a single section. Whether a ladder filter reaches the 155.57 dB the output reading promised is the third of the earlier essay’s questions, and now it looks more interesting than the first: it would avoid the gain stage’s floor altogether.

What a kilohm is doing here #

The choice of a kilohm for the stages’ resistors is a number that enters the result, and an honest report says so. With a 100 Ω arm the resistors’ contribution falls by 10 dB of noise power and the crossing on the amplifier figure moves left. With a 10 kΩ arm it moves well to the right. The kilohm was chosen because it is what the filter’s own sections use in this realisation, and for no deeper reason. A design that scales the impedance of its stages to its amplifier’s noise resistance would have stages that cost little, which is another way of saying that the question of this essay and the same filter a thousand times larger are one question about the relationship between impedance level and amplifier noise. The earlier measurement of how the bowl gives each part an optimum source resistance is what a stage’s arm resistance should be compared with, for a 4 $\text{nV}/\sqrt{\text{Hz}}$ amplifier about a kilohm, which is what makes this example honest and not a tilted one.

What the budget leaves out #

Only voltage noise is counted for the amplifiers. Their current noise flows in each section’s resistors and adds to the floor, and with flicker in both generators the corner where averaging stops working applies. Both are real, and neither would favour the stages. The sections’ followers are ideal in every other respect, and a follower with a finite gain–bandwidth moves the filter’s response as the Q the amplifier decides shows. The measurement band has a hard edge at its top. And the rails are taken as fixed at ±15 V, so that the range is what is measured, with no statement about the distortion at the ceiling, which is a real limit before the rail.

The numbers to keep #

Silent gains and silent amplifiers recover 2.16 dB on a Chebyshev 5. With every amplifier at 4 $\text{nV}/\sqrt{\text{Hz}}$ over a 1 MHz band, the unscaled filter reads 118.32 dB and the scaled one 118.18: the ceiling rises 13.14 dB and the floor 13.28, nearly all of it from a last gain of 2.833 with nothing after it. There is no optimum at partial scaling. A stage is a fixed cost of about 2.6 dB, and the curve rises from there to equal maxima. A quieter amplifier wants scaling less, −6.61 dB at 1 $\text{nV}/\sqrt{\text{Hz}}$, and a noisier one wants it more, +0.99 at 16, with the crossing near 4.4. A narrower band does not change the sign. A Chebyshev 7, with a peak of 13.26 times the input, gains 6.15 dB at 4 $\text{nV}/\sqrt{\text{Hz}}$. Scaling is worth building on a filter whose peak is large and whose amplifiers are ordinary.

Still open: a last gain inside the filter, impedance scaling per stage, and a ladder with no inner peak #

A last gain with a section after it. Moving the restoring gain in front of the last section filters its noise and raises that section’s nodes. The optimum placement is a function of each section’s Q and of the amplifier, and it is a search over three gains with the ceiling and floor both moving. Whether it can bring the Chebyshev 5 back above the unscaled filter at 4 $\text{nV}/\sqrt{\text{Hz}}$ is a direct measurement on this netlist.

Impedance scaling as the companion knob. Lowering the impedance of the loud sections, and of the stages, by a factor costs current and lowers exactly the contributions that rose. Combining equal maxima with stage resistors matched to the amplifier’s noise resistance would say how much of the remaining eight decibels a current budget can buy.

A realisation that does not hold the energy at one node. A doubly terminated ladder filter realises the same Chebyshev with its resonant energy shared among several nodes, and has no stage to pay for. Its range at real amplifier noise, with each of its amplifiers counted, is the comparison between a cascade and a ladder filter made on range rather than on sensitivity.