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

Offset voltage — where it appears

The input voltage a real amplifier needs in order to produce zero output, standing in for every asymmetry between its two input devices. It adds to the signal and cannot be told from it, so it is answered by matching or by trimming rather than by feedback, and what is left drifts with temperature.

Named by 8 essays across 3 fields — each of them below, with the objects they name alongside it.

Where an amplifier's reading comes from, against the source it is reading. computed by solving, not by drawing. Three errors with three different dependences on the source, each measured by a solve with the other two set to zero. The offset voltage is flat — 50 microvolts wherever the source is. The bias current times the imbalance is linear in the source and is what balancing removes. The offset current times the source is linear too and is what balancing leaves. Unbalanced, the current overtakes the voltage at 1.77 kΩ; balanced, at 10.0 kΩ, which is the offset voltage divided by the OFFSET current and is the ratio of the two currents further along. Below about a kilohm, balancing makes the reading worse — the feedback network is already the larger resistance, and equalising means adding to the source.

The current the instrument draws

Every amplifier in this collection has had inputs that take no current, and that is not an idealisation of a small quantity — it is an idealisation of one whose size is decided by something outside the part. Fifty nanoamps is nothing until it flows in a megohm, and then it is fifty millivolts. The classical cure balances the two resistances and removes the bias current, leaving the offset current: worth a factor of ten, not a thousand, and it costs forty per cent of the noise density to get.

instruments · Input bias current 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 A 100 kΩ resistor across the capacitor: the node rises a decade per decade and stops at the shelf. computed by solving, not by drawing. The summing-junction impedance of the integrator (1 kΩ in, 10 nF feedback, 1 MHz amplifier) with 100 kΩ across its capacitor (solid), beside the ideal integrator (dashed) and the 10 kΩ inverting stage (dotted). At the lowest frequencies the resistor closes a direct-current loop and the node is 0.999 Ω; it rises a decade per decade, as a resistive stage's does, and stops on the integrator's shelf of 15.9 Ω at the feedback's own corner, 1/(2π·Rp·Cf) = 159 Hz. Its highest point below a tenth of the amplifier's bandwidth is 16.3 Ω.

The resistor that holds the bottom of the shelf

An integrator's summing junction sits on a flat shelf, 1/(2π·GBW·Cf), over its working band, and at direct current it is nearly its whole input resistor, because a capacitor closes no loop there. A resistor across the capacitor closes one. Below the feedback's own corner the node is then an inverting stage's, rising a decade per decade from Rp/A₀, and the rise meets the shelf exactly at that corner, 1/(2π·Rp·Cf), for any resistor and any amplifier — so the node never rises above its shelf at all. That holds for every resistor below 1/(2π·f₁·Cf), the one that puts the feedback's corner on the amplifier's own pole: 1.59 MΩ here. Above it the bottom climbs past the shelf. The same 1.59 MΩ is where a millivolt of offset becomes 1.58 V at the output, and the price below it is a flat gain of Rp/Rin and arctan(1/(2π·f·Rp·Cf)) of phase.

feedback · Virtual earth A tee of factor 10 turns 1 µV/K of offset drift into 65.1 pA of error at −40 °C, 10 times the single resistor's. computed by solving, not by drawing. The direct-current error of a 10 MΩ photodiode amplifier calibrated at 25 °C, as the input current that would give the same output, against temperature: the part from the amplifier's offset drift of 1 µV/K (solid) and the part from its input bias current, 1 pA at 25 °C and doubling every 10 K (dashed), for a tee of factor 10 (heavy) and a single resistor (light). The offset reaches the output multiplied by the direct-current noise gain — one for the single resistor, since the source is a capacitor, and 10 for the tee — while the bias current sees the same transimpedance through both. The bias term overtakes the offset term at 85 °C with the tee and 40 °C with one resistor; below that the tee's error is the offset's, 10 times as large.

The drift a tee multiplies

A feedback tee reaches a large transimpedance with small resistors, and on a photodiode its price has been counted in noise: at direct current its noise gain is its own factor, where a single resistor's is one. The same noise gain multiplies the amplifier's offset. Calibrated at 25 °C and read at −40, a 10 MΩ amplifier with 1 µV/K of offset drift is 6.56 pA wrong through one resistor and 65.1 pA through a tee of factor ten — ten times, exactly, while the input bias current's error is identical through both. So the tee's drift penalty is hidden wherever the bias current dominates, which for a field-effect input doubling every 10 K is the hot end, and fully exposed at the cold end. The worst error over −40 to 85 °C comes within a factor of two of the single resistor's only above 10 MΩ for a tee of ten, and above 100 MΩ for a tee of a hundred.

feedback · Transimpedance Degenerated to gₘRₑ = 1, a pair with 1 mV of offset peaks at 0.41% of second harmonic, against 0.74% bare. computed by solving, not by drawing, every harmonic read from a solved waveform. The second (solid) and third (dotted) harmonics, as fractions of the fundamental, of a differential pair with 1 mV of input offset and emitter degeneration gₘRₑ = 1, against the peak drive; the same pair undegenerated in grey (dashed: second; dot-dash: third). Degeneration lowers both and moves both to larger drives. At small drive the second harmonic follows (u/4)·g″/g′² at the offset's operating point, which for a small offset is (u/2)·(Vos/2Vₜ)/(1 + gₘRₑ)³. Its largest value over all drives is 0.406 per cent at 177 mV, against 0.736 per cent at 76 mV undegenerated.

Two decibels of matching for one of gain

An offset gives a differential pair back the second harmonic its symmetry removed. Put resistors in the emitters and that second harmonic falls as the cube of one plus gₘRₑ at a fixed input and as its square at a fixed output — measured slopes of −3.00 and −2.00 — so each decibel of gain given up buys two decibels of distortion, or two decibels more offset at the same distortion. A matching specification that allowed 0.535 mV for −60 dB at 10 mV of output allows 2.14 mV at gₘRₑ = 1. The worst case over all drives relaxes more slowly: the offset below which the second harmonic can never reach one per cent goes from 1.36 mV to 4.39 mV as gₘRₑ goes to three, because the peak it limits sits where the pair saturates, and degeneration only moves that further out.

limits · Small-signal An integrator of 100 kΩ and 1 µF with its input grounded: what its own input currents do to it. computed by solving, not by drawing, marched for ten seconds from rest. With 50 nA flowing out of each input and the + input grounded, the output ramps at 50.0 mV/s — the bias current over the capacitance, to six figures — and reaches 0.500 V. With 100 kΩ in the + input to balance the resistances, only the 5 nA of offset current is left and the ramp is 5.00 mV/s. Neither stops: an integrator has no gain to quote, because its gain is time.

The current a capacitor adds up

An amplifier's input current in a resistor is an offset; in a capacitor it is a rate, and time is the gain. An integrator of 100 kΩ and 1 µF with its input grounded ramps at 50.0 mV/s from 50 nA, and balancing the inputs leaves the 5.00 mV/s of the offset current, still unbounded. A leak resistor bounds it at the current times the leak and costs the lowest frequency the circuit integrates. A hold capacitor's droop times its acquisition time is fixed, so 50 nA through a 100 Ω switch needs 655 nF and 0.77 ms for half a 16-bit step. A charge amplifier's offset times its low corner is Ib/2πCf, whatever its bleed.

instruments · Input bias current What balancing the source resistances is worth, against how correlated the two input currents are, for a bias-cancelled input. computed by solving, not by drawing; 400 seeded parts at each correlation, each solved unbalanced and balanced at a 1 MΩ source, input currents of zero mean and 1 nA spread. The ratio of rms errors (dots) follows 1/√(2(1 − ρ)) (line): 0.72 for independent currents — balancing makes it 38% worse — one at ρ = ½, and 6.5 at 0.99. The cure removes what the two currents share, and is a cure only if they share more than half.

The cure that needs one current

Balancing the resistances an amplifier's inputs look into removes whatever its two input currents have in common and leaves their difference. On a plain bipolar input the two are one current at two pins, correlated at 0.991, and the cure is worth ten. On an input whose bias current has been cancelled they are two residues of either sign, correlated at 0.10, and over 400 seeded parts the cure makes the error 38 per cent worse. It is worth 1/√(2(1 − ρ)) for a zero-mean current, nothing at ρ = ½, and in general exactly the input current over the offset current.

instruments · Input bias current At a fixed output a 0.5 mV junction mismatch's second harmonic falls as (1 + gₘRₑ)⁻² and a 1% resistor mismatch's as (1 + gₘRₑ)⁻¹: past gₘRₑ = 1.93 the resistors set it. computed by solving, not by drawing, every waveform solved arm by arm in currents. A pair on a 1 mA tail, its output held at what the bare, balanced pair gives for 10 mV of peak drive; the second harmonic over the fundamental from a 0.5 mV VBE mismatch alone (dashed), from a 1% mismatch between the emitter resistors alone (dotted), and from both in the same sense (solid), against one plus the degeneration. Between gₘRₑ = 5 and 20 the junction's falls with slope −2.00 and the resistors' with −0.89, still approaching −1, since the local slope of gₘRₑ/(1 + gₘRₑ)² is 1/gₘRₑ − 1. The resistors' mismatch is an input offset of Vₜ·gₘRₑ·δ, which grows with the degeneration that divides it, so the two cross where gₘRₑ = ΔVBE/(Vₜδ) = 1.93; beyond it each decibel of gain given up buys one decibel of distortion, not two.

The resistor mismatch that is only an offset

Two emitter resistors that differ by δ unbalance a degenerated pair, and with an ideal tail they do it in the simplest way possible: the arm currents always sum to the tail, so the mismatch adds the constant (Iₜ/2)·R·δ = Vₜ·gₘRₑ·δ to the input and nothing else. Solved arm by arm, a 1% mismatch at gₘRₑ = 3 is the balanced pair moved by 0.776 mV to 4 × 10⁻¹⁶ of the tail, and a trim of that size leaves no second harmonic at any drive. What separates it from a junction's offset is that it grows with the degeneration meant to divide it: at a fixed output its second harmonic falls as gₘRₑ/(1 + gₘRₑ)², one decibel per decibel of gain rather than two, and it takes over from 0.5 mV of base–emitter mismatch at gₘRₑ = 1.93. For −80 dB at 10 mV of output the resistors must match to 0.823% at gₘRₑ = 1 even with perfect transistors.

limits · Small-signal

Named alongside it

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

Design tradeoffInput bias currentDevice mismatchJohnson noiseModel rangeSource impedanceDifferential pairEmitter degenerationEven harmonic cancellationIntegratorNoise gainShot noise

All concepts