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

Noise gain — where it appears

The gain an amplifier's own input noise sees on its way to the output, which is one over the feedback factor and is not the signal gain. A capacitance at the summing junction raises it with frequency; a capacitance in the feedback path lowers it.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.

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 The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range.

What the second path costs at the floor

The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

feedback · Capacitive load A photodiode's own capacitance sets the bandwidth, as its -0.50 power. computed by solving, not by drawing. The bandwidth of a 1.0 MΩ transimpedance stage against the capacitance of the diode driving it, with the feedback capacitor at each point bisected to give exactly forty-five degrees of phase margin on the solved loop. The classical expression √(GBW/2π·rf·cd) is drawn over it: the right shape, and conservative by about a fifth at every capacitance. The bandwidth falls as the -0.497 power of the capacitance — a square-root law, so a diode of four times the area costs half the bandwidth rather than three quarters of it. At 30 pF the compensation is 0.598 pF against the expression's 0.725, and the bandwidth 295 kHz against 220.

Where the trouble is at the input

Every other arrangement in this field has its difficulty at the output — a capacitive load, an isolation resistor, a load that draws current. A photodiode amplifier has it at the input, and the capacitance causing it is not a parasitic: it is the diode's junction, which is the price of its area, and area is what a photodiode is bought for. The feedback resistor's own noise is 127 nV/√Hz against the amplifier's 4, and the amplifier is still ninety-five per cent of the noise at a large diode.

feedback · Transimpedance The resistors own the floor between 2.91k Ω and 85.9k Ω, and the part owns it outside. computed by solving, not by drawing. The noise at the load of the two-path compensation against the impedance of its own feedback network, with the resistors scaled together and the compensation capacitor taken down in proportion so that Rf·Cf — the handover between the two feedback paths — does not move. Four contributions are integrated over 10 Hz to 100 MHz: the amplifier's 4 nV/√Hz, fitted as the 0.005 power of the impedance and so flat; the two resistors' √(4kTR), the 0.501 power; and the amplifier's 0.60 pA/√Hz flowing in the feedback resistor, the 0.997 power. Two different powers of one quantity cross twice. The resistors carry more than half the power only between 2.91k Ω and 85.9k Ω; outside that window, in both directions, the part does. The part's share is least at 15.8k Ω, which is not eₙ/iₙ — it is that ratio multiplied by the noise gain of 2.000 and again by 1.187, the square root of the ratio of the bandwidths the two generators actually see; there it carries 26.26 per cent. The model stops where the amplifier's output current does: at 100 Ω the feedback resistor alone draws 10 mA a volt.

The window the resistors own

Eight essays have priced one compensated stage, and the fourth of them left two of the amplifier's own generators named and uncounted. With the current generator put in the netlist the floor at ten kilohms goes from 28.88 microvolts to 30.13, and the resistors carry more than half the noise power only between 2.91 kΩ and 85.9 kΩ — outside that window, in both directions, the part does. The flicker corner turns out to be worth 1.00009 in this stage's own band, and 3.474 one band away.

feedback · Capacitive load The cure changes shape at 909 Ω, which is a property of the feedback network and of nothing else. computed by solving, not by drawing. What balancing actually does to the circuit, against the source resistance it is done for, at a gain of 11 with a 1.0 kΩ bottom resistor. The inverting input looks back into 909 Ω — the bottom resistor times (G−1)/G — and that number is the whole of the knee. Below it the cure is a resistor in series with the source and the feedback network is untouched. Above it there is no resistor to add, and the network is scaled up to meet the source instead: 1100× at 1.0 MΩ, which puts 11 MΩ in the feedback path. The scaled feedback resistor is the source resistance times the gain exactly, so the network's own size has left the answer — it decided where the knee was and nothing after it.

The cure that becomes a different circuit

The classical cure for an amplifier's input current is to make the two resistances its inputs look back into equal, and it reads as one instruction. Solved, it is two circuits meeting at 909 ohms — the feedback network's bottom resistor times (G−1)/G — and above that knee there is no resistor to add: the network is scaled to the source, which at a megohm means 11 megohms of feedback and at a gain of 1001 means 1001. Above the knee three different networks become one instrument to twelve figures, the noise penalty settles at 1.41420 against a √2 of 1.41421, and the benefit at 10.49 against two currents whose ratio is ten.

instruments · Input bias current One capacitor moves three quantities, and the expression's own answer peaks by 1.18 dB. computed by solving, not by drawing. The bandwidth, the peaking and the total output noise of a 1.0 MΩ transimpedance stage against its feedback capacitor, swept from 0.30 to 4.20 times what the classical expression asks for. Over that factor of fourteen the bandwidth falls from 334 to 55.4 kHz, the total noise from 230 to 65 µV, and the peaking from 10.0 dB to nothing. The expression's own answer sits at 1.18 dB of peaking, 0.82× is where the loop reaches forty-five degrees, and √2× is where the response is flat — so the choice usually quoted as maximally flat is neither of the two conditions it is quoted for. The faint families are the same three quantities at 3 pF and 300 pF of diode: normalised this way they are one curve, so the diode sets the scale and the multiple sets the shape.

The factor the expression leaves out

The classical compensation for a photodiode amplifier is quoted both as the forty-five degree choice and as the maximally flat one, and it is neither: it leaves 1.18 dB of peaking and 52.4° of margin. Flat is at exactly √2 times it — fitted at 1.4186 against 1.4142, at every detector from 3 pF to 1 nF. And the third quantity the capacitor is supposed to trade, the noise, does not move at all inside the signal band: three compensations spanning a factor of fourteen give 8.37 against 8.36 µV in a 4.36 kHz measurement and 230 against 65 µV over the whole plane.

feedback · Transimpedance What a feedback tee buys, and the single thing it charges. computed by solving, not by drawing. Two 50 kΩ resistors with a 6.250 kΩ tap give 500.0 kΩ of transimpedance, which is R₁(1 + R₂/R₃) + R₂ solved on the netlist. That expression has no term for the noise gain, and the tap sets it to 1 + R₁/R₃ = ×9.00 — against exactly one for a single resistor of any value, because at direct current the source is a capacitor. One quantity then does all three things at once. The signal and R₁'s own noise are multiplied together, so the tee buys no signal-to-noise ratio at all: 1846 against 5582 in a hertz at a nanoamp for one resistor of the same transimpedance, a factor of 3.02 which is the square root of the noise gain. The amplifier's own voltage noise and offset are multiplied where they were not before. And the phase margin RISES, from 1.8° to 15.9°, because starting the noise gain high shortens its climb to the peak — so the tee is a compensation that charges for itself in noise, and a capacitor across the feedback is the same compensation for nothing. With no tap the three effects are absent rather than small.

The tee that charges for its own compensation

A feedback tee makes a large transimpedance out of small resistors, and R₁(1 + R₂/R₃) + R₂ is the whole of what is usually said about it. The expression has no term for the noise gain, and the tap sets that to 1 + R₁/R₃ — nine, where a single feedback resistor of any value gives exactly one, because at direct current the source is a capacitor. One quantity then does everything: the signal and R₁'s own noise are multiplied together so the tee buys no signal-to-noise ratio at all, and the phase margin RISES from 1.8° to 15.9°.

feedback · Transimpedance

Named alongside it

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

Design tradeoffPhase marginJohnson noiseModel rangeTransimpedanceCapacitive loadInput capacitanceLoop gainSignal-to-noise ratioAnti-alias filterClosed formCompensation network

All concepts