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

Input bias current — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · instruments
  2. 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.

    part 2 · instruments
  3. 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.

    part 3 · instruments
  4. 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.

    part 4 · instruments

All series