Circuit Response
every model has an edge, and the figure carries it
Concept

Gain–bandwidth product — where it appears

An amplifier's open-loop gain times the frequency it is measured at, constant above the first pole, so it fixes the bandwidth a closed-loop gain can have. It is a constant of the amplifier and not of the circuit, so a gain of a hundred and a bandwidth of a hundred kilohertz are one choice rather than two.

Named by 18 essays across 6 fields — each of them below, with the objects they name alongside it.

Where four of this site's models stop being true. In order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is.

Every model has an edge

Four assumptions this collection runs on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does.

limits · Model edges A gain of 100 asked of an amplifier with 1.00 MHz of gain–bandwidth. The ideal amplifier — a nullor, so the two golden rules exactly — holds 100 at every frequency. The real one is 0.10% low at direct current, 1% low by 1.35 kHz, and 3 dB down at 10.0 kHz. Above 10.0 kHz there is no loop gain left and the ideal answer is not an approximation to anything.

The ideal amplifier, and where it stops being one

An ideal operational amplifier's closed-loop gain is set by two resistors and nothing else — a horizontal line at every frequency. The real one is already a tenth of a per cent low at direct current, one per cent low by 1.35 kHz, and above 10 kHz has no loop gain left, at which point the ideal answer is not an approximation to anything.

feedback · Gain-bandwidth A common-emitter stage with 2.0 pF from collector to base. computed by solving, not by drawing. The stage's midband gain is 144.7 and its −3 dB point is at 504 kHz. The Miller approximation lumps 311 pF at the input and predicts 643 kHz — 21.6% high. The network's second pole is at 336 MHz and its right-half-plane zero at 3.08 GHz, both of which the approximation has no room for.

The frequency a device sets for itself

A common-emitter stage's bandwidth is decided by two picofarads between its collector and its base. The Miller approximation says how — lump it at the input, multiplied by one plus the gain — and predicts 643 kHz where the solved network gives 504 kHz. Twenty-two per cent optimistic, and it has no room at all for the second pole or for the zero in the right half-plane that the network also has.

semiconductors · Device bandwidth Two reasons the frequency is not 1/2πRC. The measured oscillation frequency sits below the network's own zero-phase frequency, and by two separate amounts. The amplifier's share falls as 1/ρ — the product of shift and ratio is constant to 6.7% over two decades — and the limiter's share is flat at 0.7713%. They are equal at ρ = 578, and above that a faster amplifier moves the frequency by nothing that matters.

The frequency that is not the formula

The Wien network's zero-phase frequency is one over two pi RC to every digit the arithmetic has. The circuit does not run there. With a perfect amplifier it runs 0.771 per cent low, because the limiter's harmonics are part of the waveform whose period is being measured; with a real one it runs lower still, by an amount inversely proportional to the gain-bandwidth product. The two are equal at a ratio of 578, and above that a faster amplifier buys nothing.

applied · Oscillator The ideal amplifier is good to 1% over a region, and its corner is 21% inside the specifications. computed by solving, not by drawing. The 1 per cent contour of the ideal-amplifier model for a non-inverting stage of gain 2 built from a 10 MHz part, drawn over frequency and output amplitude at once. Each point is bisected on a marched circuit: the error is the root-mean-square difference between the marched output and 2 times the input, which counts the gain that is low, the phase that is late and the peak that is flat. Three mechanisms bound the region — finite gain–bandwidth on the left, the input pair's slew rate on the diagonal, and the rails at 12.19 V along the top. The two dashed lines are the numbers a data sheet gives: a small-signal edge at 48.8 kHz with no amplitude in it, and a full-power bandwidth of slew rate over 2πV̂ with no gain–bandwidth in it. They cross at 10.60 V and 48.8 kHz; the measured contour passes 38.5 kHz at that amplitude, which is 0.790 of it.

The edge that is a region

Every boundary this collection has drawn is a number on one axis, and the figure that gathers four of them admits in its own caption that the fifth is an amplitude and cannot go there. Drawn on both axes at once, the ideal amplifier's one per cent boundary is a region with three sides and a corner — and the corner sits at 38.5 kilohertz where the two numbers a data sheet quotes cross at 48.8, because the two mechanisms are lags on the same waveform and add as magnitudes rather than in quadrature.

limits · Model edges An inverting unity gain, and the 100 pF that only the loop can see. computed by solving, not by drawing. Two ten-kilohm resistors around a 10 MHz amplifier make a gain of 1.00, and a loop that closes against 2.01 — one plus the ratio, not the ratio. Adding 100 pF at the summing junction leaves the closed-loop gain at a kilohertz unchanged — 0.99998051 against 0.99997988, three parts in a million at the far end of the slider — and takes the phase margin from 90.0° to 14.4°, because the noise gain now rises a decade per decade and the loop closes at forty decibels per decade instead of twenty. Forty-five degrees is reached at 9.00 pF, bisected on the netlist. The capacitance is not part of the signal path and does not appear in any expression for the gain.

The gain the loop closes against

An inverting amplifier with two equal resistors has a gain of one and a loop that closes against two, so it has half the bandwidth of a follower built from the same part — 4.99 megahertz against ten. Nine picofarads at the summing junction, less than a scope probe, takes the phase margin from ninety degrees to forty-five, and a hundred picofarads puts twelve decibels of peaking on a response whose designed gain is nought decibels and whose measured gain at a kilohertz has not moved by three parts in a million.

feedback · Noise gain Linearising an exponential at 125 °C, and what it costs. The linear model understates the gain by 1% at 9.69 mV and by 10% at 30.2 mV. The thermal voltage at this temperature is 34.3 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong.

The edges that move with the room

Every boundary in this collection is quoted at one temperature and most of them are functions of it. The small-signal edge is proportional to the thermal voltage, so it runs from 5.67 millivolts at −40 degrees to 9.69 at +125 — a factor of 1.71, the ratio of the absolute temperatures exactly. A realised Q is 1.54 per cent high at one end of that range and 2.58 at the other. The numbers are right; the condition attached to them was left off, and it is the same condition every time.

limits · Model edges Five boundaries, one tolerance, and three exponents. computed by solving, not by drawing. Each of five model boundaries re-solved at forty-one tolerances from 0.1% to 30%, divided by its own value at 0.1% so that an amplitude in millivolts and four frequencies can share one axis — an exponent has no units. Fitted over the two decades to 10%: Kirchhoff's laws 1.000, the ideal amplifier 0.513, the small-signal model 0.497, the ideal capacitor 0.500, and the full-power bandwidth 0.000. A boundary set by a first-order departure moves in proportion to the tolerance, one set by a second-order departure moves as its square root, and a refusal does not move at all — so relaxing the tolerance from 0.1% to 10% buys a factor of 100 on the board and 10.0 on the capacitor.

A boundary is a model and a tolerance

Every edge in this collection is computed from a fraction of error nobody states, and the four on its opening axis use three different ones. Swept over two decades, each boundary moves as a power of that fraction — Kirchhoff's laws exactly as the first power, the amplifier and the capacitor and the small-signal model as its square root to within three per cent, and a full-power bandwidth not at all. The exponent identifies the mechanism, and it re-orders the axis twice: the board fails before the capacitor below 0.424 per cent, and the output before the amplifier above 21.7.

limits · Model edges A first stage of 100 buys 40.0 dB of rejection, and gives it back above 10.0 kHz. computed by solving, not by drawing. The rejection of a three-amplifier instrumentation amplifier against frequency, beside the one-amplifier difference stage it is built around. At low frequency the two differ by 39.99 dB against 20 log 100 = 40.00 dB, and the reason is that the input stage passes a common-mode voltage at exactly unity: the common-mode gain of the whole instrument is 1.998 mV/V, which is the difference stage's own. So the four resistors around the last amplifier decide the rejection and the two that set the gain do not — ten per cent between them moves it by less than a hundredth of a decibel. What ends it is bandwidth: above 10.0 kHz, which is the amplifier's gain–bandwidth divided by the gain that bought the rejection, the differential gain falls and the rejection falls with it at twenty decibels a decade.

The four resistors that decide, and the two that do not

A difference amplifier's rejection is decided by four resistors and one-tenth-per-cent parts give 54 decibels. Putting a two-amplifier stage in front adds exactly twenty times the log of its gain — 94 decibels at a gain of a hundred — and the reason is not that the input stage rejects anything. It passes common mode at exactly unity, so the common-mode gain of the whole instrument is 1.998 millivolts per volt at every gain tried, and the improvement is entirely the differential signal arriving larger. The two resistors that set that gain may be ten per cent apart without moving the answer a hundredth of a decibel.

instruments · Common-mode rejection Two of the three are one mechanism at 19.0°; the third arrives at ninety. computed by solving, not by drawing. The cosine between each pair of error waveforms at 30.0 kHz, against how much output the ideal model is asked for. Bandwidth and slewing sit at 0.9454 — 19.0 degrees — and close only slowly, reaching 0.8813 at 16 V. Clipping does not exist below 12.00 V, where its error is 4.0e-9 per cent of the signal and its direction is the direction of rounding; above it the mechanism is real — 18.1 per cent at the top of the sweep — and its cosine against both of the others stays under 0.0025. The bandwidth error is 0.615 per cent at every amplitude here, unchanged to 5.3e-15, because a linear stage's fractional error has no amplitude in it.

Where the mechanisms are one mechanism

An amplifier is said to run out of three separate things — bandwidth, slew rate and rails — and errors from separate mechanisms add in quadrature while errors from one mechanism add as magnitudes. Measured as waveforms rather than as numbers, two of the three sit 18.4349 degrees apart, which is exactly the angle between a sinusoid and its own cube, and the third sits at ninety: its cosine against both of the others stays under 0.0025 wherever it exists. So the arithmetic is neither of the two anybody reaches for, and a budget built the right way is within 2.9 per cent where quadrature is 17 per cent low and a straight sum 37 per cent high.

limits · Small-signal Matched parts cost nothing; a 2 dB difference between them sets a 112 dB ceiling. computed by solving, not by drawing. The common-mode rejection of a three-amplifier instrumentation amplifier against the gain of its input stage, with each amplifier's own rejection in the netlist as an input-referred error of the common-mode voltage over the rejection. The architecture's own figure rises decibel for decibel with the gain, because the difference stage sees a larger differential signal beside the same common-mode one. The parts' contribution does not rise with anything, and the part of it that matters is not their rejection but the difference between their rejections: two amplifiers of 98 dB that are identical cost 0.000 dB, while 100 dB against 98 dB leaves a ceiling of 111.7 dB with no gain in it. The two mechanisms cross: below a gain of 1903 the four resistors decide everything, and above it more gain buys no more rejection at all — 111.9 dB at a gain of 100000, where the arrangement alone would have been worth 148. The one place the instrument beats its own floor is a gain of 1000, where the two errors cancel; that is a coincidence of signs and not something a design can hold.

The rejection the parts have

Two essays measured the architecture: four resistors decide an instrumentation amplifier's rejection, two do not, and the answer is the one-amplifier figure plus twenty log of the first stage's gain — exactly, with amplifiers of infinite rejection. Give each amplifier its own and something unobvious happens: two matched but individually mediocre parts cost nothing at all, because their error is a common-mode signal at the difference stage and is rejected there. What costs is the difference between them, and it sets a ceiling with no gain in it.

instruments · Common-mode rejection A one-henry inductor with nothing magnetic in it, good for 3.6 decades. computed by solving, not by drawing. The impedance at the input of an Antoniou impedance converter, read as an inductance: a current source drives the node and the voltage is solved for, and the imaginary part divided by ω is what is plotted. Five components — four resistors of 10 kΩ and a 10 nF capacitor — behave as 1000 mH, which as a wound coil would be several henries of wire. It is that inductance to within one per cent from 1.00 Hz to 3.65 kHz, 3.56 decades, and the upper edge belongs to the amplifiers rather than to the arrangement: with ideal ones in the same netlist the inductance is exact everywhere drawn. Nothing in it stores energy in a magnetic field — the current lags because an amplifier is holding a capacitor's voltage somewhere else in the loop.

The inductor that is an amplifier

Four resistors and a capacitor, arranged around two amplifiers, present one henry at a node — an inductance with nothing magnetic in it, which as a wound coil would be several henries of wire. It is that inductance to within one per cent over three and a half decades, and its series resistance goes negative at 63 hertz, which is well inside the band where it is still an excellent inductor. A resonator built around it there does not have a high quality factor; it has a negative loss, and starts on its own noise.

filters · Gyrator The step at which the output impedance stops being a number. computed by solving, not by drawing. The excursion divided by the step, against the step. The flat line is the linear model, and it is flat to 0.0 parts per million across four decades — which is what an impedance is. The rising curve is the same netlist with the differential pair's tanh in the transconductor, and it leaves at 10.6 mA: the input error there is 3.63 thermal voltages, so the boundary is an amplitude in the pair's own units rather than a current with the amplifier's name on it. At 300 mA the ratio is 54.5 Ω against the linear 23.6 — 131 per cent, and it is no longer a property of the circuit at all. The slew rate that decides it is 3.25 V/µs, which is twice the thermal voltage times the gain-bandwidth in radians, and contains no design choice.

The step too large to have an impedance

The rung below drove the load node with a current step and reported an impedance: a voltage divided by a current, which is a number only if the ratio does not depend on the current. Give the amplifier the differential pair's own tanh in place of a linear transconductor and it is a number up to 10.6 milliamps and not above — where the input error is 3.63 thermal voltages, and where the slew rate that decides it is twice the thermal voltage times the gain-bandwidth in radians, containing no design choice at all.

feedback · Capacitive load A cascode multiplies rₒ by β, not by gₘrₒ — and the two are 21× apart. computed by solving, not by drawing. The output resistance of a cascode stage, measured by driving the output node with a current source and reading the voltage, against the current gain of the upper device. The plain stage's is 80 kΩ — rₒ and nothing else. The cascode's is 11.5 MΩ at β = 150, which is βrₒ to within a tenth and is 21 times below the gₘrₒ² every reference gives. The reason is in the netlist rather than in the algebra: the upper device's base draws current, so its rπ sits from the lower device's collector to signal ground and shunts the node the feedback works through. What the arrangement buys therefore scales with β and stops when β does, and the curve is the two expressions drawn against the measurement.

The device that never sees the swing

A second transistor standing between the first and the load does two things that every reference gives one expression each for, and one of the two expressions has no ceiling in it. The output resistance is not gₘrₒ² — that is 248 megohms here, and the measurement is 11.5 — it is βrₒ, because the upper device's base draws current and shunts the very node the feedback works through. The bandwidth really is fourteen times better, and what it costs is two volts of a five-volt supply.

semiconductors · Cascode An amplifier a hundred times the corner moves the unity-gain section's Q by 2.0%. computed by solving, not by drawing. A unity-gain Sallen–Key section designed for Q = 2 at 1.00 kHz, built with a one-pole amplifier, and its poles recovered by rooting the determinant. The section as drawn has two poles; as built it has three, and the pair is not where it was put. At a gain-bandwidth of a hundred times the corner — which is the rule of thumb — the quality factor is 1.97 per cent high and the pole frequency is 1.97 per cent low. The two are the same number with opposite signs over the window where the amplifier is well clear of the section and its own pole is still resolvable, and the number is the designed quality factor, 2, divided by the ratio: 2.00 per cent at 100 times the corner.

The Q the amplifier decides

A second-order section's quality factor is set by a capacitance ratio and its pole frequency by a product of four passive values, and neither expression contains the amplifier. Build it with one that has a gain-bandwidth a hundred times the corner — the usual rule — and the Q comes out 1.97 per cent high while the pole comes out 1.97 per cent low, the same number in both directions, and the number is the designed Q divided by the ratio. A fifth-order half-decibel Chebyshev built that way has 2.1 decibels of ripple.

filters · Q enhancement A half-decibel filter that meets one decibel at 25 °C and does not at 89. computed by solving, not by drawing. The passband ripple of a fifth-order 0.5 dB Chebyshev built from three Sallen–Key sections, against temperature, with a part whose gain-bandwidth at 300 K is 300 kHz against a 1 kHz corner. Not one passive component has a temperature coefficient in this model. The ripple is 0.824 dB at −40 °C and 1.049 at +125, and it crosses a one-decibel specification at 88.6 °C — a boundary in temperature, which every other edge in this collection is not. The dashed line is the same filter with a tail current proportional to absolute temperature: 0.916 dB at both ends, and no crossing anywhere.

The ripple that is a temperature

The rung below found that an amplifier a hundred times the corner leaves a section's quality factor two per cent high. That two per cent has a temperature in it: with a tail current a resistor sets, the transconductance falls as one over absolute temperature, and a fifth-order half-decibel design whose passives have no temperature coefficient at all goes from 0.82 decibels of ripple at minus forty to 1.05 at a hundred and twenty-five — crossing a one-decibel specification at 89 °C. With the other bias it does not move at all.

filters · Q enhancement At Q = 2 and a hundred times the corner, one arrangement is 1.97% out and the other 2.83%. computed by solving, not by drawing. The same second-order section designed twice — a follower with a capacitance ratio, and equal passives with the Q supplied by a closed-loop gain of 2.5000 — built with the same one-pole amplifier and swept over its gain-bandwidth, with both pole pairs recovered by rooting the determinant. The unity-gain arrangement's error is Q over the ratio: 2.00 per cent at a hundred times the corner. The equal-component arrangement's is K²/2 over the ratio, 3.13 per cent, because its amplifier is a gain-of-K stage and therefore has K times less bandwidth to spend. Below about thirty times the corner both laws fail, and the second one changes sign.

Where the Q comes from

Two second-order sections with no component value in common have the same transfer function to 1.8×10⁻¹⁰ of a decibel, and are not the same circuit. Against a slow amplifier the follower's Q error is Q over the gain-bandwidth ratio and the gain stage's is K²/2 over it — so the second is worse below Q = 3.08 and better above it. Against component tolerance the follower's worst element carries ½ and the gain stage's carries 2Q − ½, nineteen times as much at Q = 5. Neither oscillates at any gain-bandwidth at all.

filters · Q enhancement The sign change follows the 0.51 power of the amplifier, not the inductor. computed by solving, not by drawing. Both of the arrangement's frequency boundaries against the gain–bandwidth of the two amplifiers in it, over three decades. The lower curve is the frequency at which the series resistance changes sign, bisected on the sign of the real part; the upper one is where the inductance leaves one per cent. The crossing grows as the 0.513 power of the gain–bandwidth, and the dashed prediction over it is ½√(f_c·f_p) — half the geometric mean of the arrangement's own corner r/2πL = 1.59 kHz and the amplifier's open-loop pole f_t/A₀ — which is inside one per cent while that pole is at least fifteen times below the corner and 8.7 per cent out at the top of the sweep, where it is not. The two boundaries stay between 1.61 and 2.30 per cent of one another throughout, so a faster amplifier moves the active region rather than removing it.

The boundary that improves when the part gets worse

A synthetic inductor's series resistance changes sign at 63.0 Hz with one-megahertz amplifiers, and that frequency is not a property of the inductor. It is half the geometric mean of the arrangement's own corner and the amplifier's open-loop pole — half the square root of their product, which the bisection confirms to a part in a thousand — and it therefore falls as the amplifier's direct-current gain rises, from 686 Hz at a gain of a thousand to 19.9 Hz at a million. The quantity that decides whether a resonator starts does not move at all: it is the transition frequency over four times the Q, 2.50 kHz for a tank of a hundred.

filters · Gyrator

Named alongside it

The objects these essays reach for when they reach for this one.

Model rangeDesign tradeoffQuality factorSlew rateLoop gainClosed-loop gainComponent toleranceModel refusalSallen-keyThermal voltageValidity regionCommon-mode rejection

All concepts