Concept

Negative feedback — where it appears

Returning part of an output to subtract from the input, so that the loop opposes any change it sees. It divides the forward element's variations by one plus the loop gain, trading gain for accuracy, linearity and a steadier operating point.

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

With Rₜₕ = 10 kΩ, a tenfold range of β moves the collector current by 17.3%. computed by solving, not by drawing. The collector current of a transistor biased from a 2 V Thévenin source through 1, 10 and 100 kΩ, with 4.7 kΩ in the collector, 1 kΩ in the emitter and a 12 V supply, against its current gain from 30 to 1000, each point a Newton solve of the netlist with both junctions in the device. Dotted: one resistor from the supply to the base, 876 kΩ, sized for the same current at β = 100; it is proportional to β until it saturates. With Rₜₕ = 10 kΩ the current moves by 17.3 per cent from β = 50 to 500.

The bias the gain does not set

A transistor's current gain varies tenfold across the parts one type number covers, and a bias network is judged by how little of that reaches the collector current. One resistor from the supply to the base passes all of it — 0.61 mA at β = 50, 1.16 at 100, and saturation from 200. The four-resistor network passes 98.5 per cent of a tenfold change with a 100 kΩ divider, 17.3 with 10 kΩ and 3.4 with 1 kΩ, and never less than the 1.78 per cent the ratio β/(β + 1) keeps however stiff the divider. Each tenfold cut in that spread costs a tenfold rise in the divider's standing current, so the design is a point on a hyperbola rather than a rule.

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

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

A terminated ladder network's output noise three ways: one solve per resistor, one solve from the output, and 4kT·Re{Z}. computed by solving, not by drawing, at 290 K. The line is 4kT·Re{Zout}. The dots sum each resistor's 4kT/R through its own transfer to the output, one solve per resistor. The rings get the same transfers from a single solve with an ampere injected at the output — reciprocity — and agree to 4.4e-16; the sum agrees with Nyquist's expression to 4.4e-16. 4.60 nV/√Hz at 100 Hz, 3.82 at 1 MHz.

The resistance colder than the room

In a passive network the noise at a port is 4kT times the real part of its impedance, and reciprocity is why: the transfer from each resistor to the port equals the transfer from the port to that resistor, so one solve from the output gives every resistor's share and the shares add to Nyquist's expression, here to 4.4 × 10⁻¹⁶. With gain in the network both fail. A 50-ohm termination made by feedback around an amplifier of gain A presents 50.000 Ω and has the noise of a 50-ohm resistor at 290/(1 + A) kelvin — 2.90 K at a gain of 99 — but only where the amplifier still has its gain.

networks · Reciprocity

Named alongside it

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

Controlled source Current gain Design tradeoff Device mismatch Differential pair Emitter degeneration Even harmonic cancellation Gain–bandwidth product Input impedance Johnson noise Newton raphson Noise temperature

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