Optimum source resistance — where it appears
Named by 10 essays across 2 fields — each of them below, with the objects they name alongside it.
The floor a circuit has
A resistor's noise is 4kTR and there is nothing to choose about it. An amplifier adds two generators that belong to the device — 4 nV/√Hz in series with its input and 0.6 pA/√Hz across it — and because one matters most into a small source and the other into a large one, there is a source resistance at which their sum is least. It is 6.67 kΩ, it is the ratio of the two, and it is not the resistance that transfers maximum power.
The bowl, and the bottom of it
An amplifier's noise figure has a minimum against source resistance and its input-referred noise has none: the 4 nV/√Hz part reads 1.138 dB into 6.67 kΩ and 10.41 dB into 100 Ω, and is 2.81 times noisier in volts at the first. The bowl is one shape scaled by its own depth, so the quieter the part the flatter it is — ±30.1 times for a decibel on the best of four, ±2.16 on the worst — and three parts of equal eₙiₙ share a floor of 0.3138 dB at optima 16 times apart.
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.
The two generators that are one current
An amplifier's noise is two generators, a voltage in series with its input and a current across it, and the essays on its noise figure treat them as independent numbers. In a bipolar input stage they are one current's shot noise divided two ways, by the collector and the base, and their product is 2kT/√β at every collector current: 0.8008 nV·pA per hertz at a current gain of a hundred, from a microampere to ten milliamps. The best noise figure, 0.4139 dB, does not depend on the bias. The bias decides only where the best source is, and 50 ohms of base resistance decides what the best actually is below 500.
The window a field-effect input wins in
A bipolar input's two noise generators are one current's shot noise split two ways, so their product, and with it the best noise figure, is fixed at 1 + 1/√β, 0.414 dB at a gain of a hundred, whatever the bias. A field-effect input's are two unrelated things. Its voltage noise is its channel's and its current noise is the shot noise of its gate leakage, so their product is a design variable: 0.0007 dB at a milliampere of drain and a picoampere of leakage. Against a bipolar input biased for each source, the field-effect input wins in a window of source resistance with two ends in closed form, from 2.83 kΩ, set by the two inputs' noise resistances, up to 5 GΩ, set by the gate leakage. A nanoampere of leakage brings the top down to 5 MΩ, and 438 nA closes the window.
The source resistance the flicker moves
An amplifier's optimum source resistance is the ratio of its voltage noise to its current noise, and both densities rise below their flicker corners — which are rarely the same frequency. With the voltage noise's corner at 10 Hz and the current noise's at 1 kHz, a part whose optimum is 6.67 kΩ at high frequency wants 2.11 kΩ at 100 Hz and 699 Ω at 1 Hz, tending to the white value times the square root of the ratio of the corners. Over the audio band the drift costs 0.012 dB; over a band from 1 to 100 Hz, choosing the white optimum costs 3.77 dB, and the spot optimum at the band's centre recovers all but a tenth of a decibel.
Two generators with one cause
The two-generator model of an amplifier assumes its voltage and current noise independent. Give them a correlation γ and the source sees a cross term, 2·Re(γ·e·i·Z*), linear in its resistance and its reactance. A real correlation leaves the quietest source resistance at e/i and moves the floor in proportion to 1 + γ: 0.83 dB instead of 1.14 at γ = −0.3. An imaginary one moves the optimum off the resistance axis — 5.77 kΩ with 3.33 kΩ of capacitive reactance at γ = 0.5j — and the noise figure's contours become circles about it; a resistive source cannot use it at all. If the two flicker components share one cause, the sign of the sharing decides the low-frequency floor: at 1 Hz, 3.9 dB or 18.0 instead of 15.1. And the correlation is measurable with three resistors, best with the middle one next to the optimum.
One turns ratio for a band
A 50 Ω source and an amplifier whose optimum is 6.67 kΩ want a step-up transformer, and the ratio that matches the white optimum, 11.5, is the one a design note computes. Over the audio band it is nearly right: 1.31 dB against 1.30 at the band's own best ratio, 10.7. Over 1 Hz to 100 Hz it is badly wrong — 9.2 dB where a ratio of 4.8 gives 5.4 — because the band's noise lives in its bottom decade, where the flicker has moved the optimum down. One ratio for a two-decade band costs at most 0.24 dB against a ratio that could follow the optimum. The transformer's winding resistance multiplies the whole noise factor by one plus the winding's resistance over the source's, not just its own share: a tenth of the source costs 0.41 dB where its own noise alone would cost 0.31. And 115 amplifiers in parallel are an ideal 10.7:1 transformer, noise for noise.
The window that closes from the top
A field-effect input beats a bipolar one biased for each source between 2.83 kΩ and 5.14 GΩ at 25 °C, and the top of that window is twice the thermal voltage over √β times the transistor's gate leakage: it falls as fast as the input junction leaks. Carried through the junction's own law, the ceiling is 39.5 GΩ at 0 °C, 913 MΩ at 50 and 5.79 MΩ at 150 — while the floor does not move from 2.83 kΩ until 200 °C. A 1 GΩ sensor leaves the window at 48.6 °C, a 10 GΩ one at 16.4, about 33 kelvin a decade. The leakage's doubling interval is not ten kelvin but 7.5 at 0 °C and 15.0 at 150, so the rule of thumb is right at one temperature only.
The resistors that sum a hundred amplifiers
A hundred amplifiers in parallel across a 50 Ω source are an ideal transformer of ratio ten, and the summing network between them and the output is where the comparison was left. Its resistors turn out to be the least of it. N resistors averaging N outputs make the noise of R/N, falling exactly as the amplifiers' voltage noise does, so the resistor equal to a 4 nV/√Hz amplifier is about a kilohm at every N, and at 133 parts 200 Ω costs only a tenth of a decibel. The stage that was forgotten is whatever reads the sum. Averaging leaves it alone, so a following amplifier as good as one of the parts costs 12.14 dB, and each part needs a gain of 26 to bring that under a tenth. Parallel amplifiers then match a transformer with one ohm of winding.
Named alongside it
The objects these essays reach for when they reach for this one.
Current noiseNoise figureVoltage noiseDesign tradeoffFlicker noiseJohnson noiseNoise floorModel rangeShot noiseImpedance matchingLeakage currentTransconductance