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

Johnson noise — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz.

    The floor a resistor sets

    A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.

    part 1 · noise
  2. The same three stages, in two orders. computed by solving, not by drawing by Friis's cascade. With the low-noise amplifier first the chain's noise figure is 3.31 dB; with the mixer first it is 12.01 dB. The gain is identical either way — 48.0 dB — so the 8.70 dB is bought with nothing but an ordering. Everything after the first stage contributes 26.0% of the total. The 2.0 dB of cable in front has gain below one, so it multiplies every later stage's contribution rather than dividing it.

    The loss in front, counted twice

    A decibel of cable before an amplifier costs a decibel of signal, which everybody expects, and a decibel of noise figure, which is a separate decibel arriving from a separate place. Measured across the slider it is exact: 1.31 dB of chain noise figure becomes 2.31 with one decibel of cable and 9.31 with eight, every time, while the 8.70 dB that stage ordering is worth does not move at all. A lossless reactive network in the same position costs nothing, because only the real part of an impedance is warm.

    part 2 · noise
  3. Which resistor the noise of a Chebyshev 5 actually comes from. computed by solving, not by drawing, one solve per resistor. Each bar is that resistor's share of the noise power at the output, found by splitting its node, putting a source of √(4kTR) in series with it and re-solving the whole network — so what is drawn is not how much noise each resistor makes but how much of it arrives. The largest contributor is F2R at 64.1 per cent, the smallest F0R at 9.0, and the shares add to 1.000000000000 because noise powers add. A resistor's share of the noise is not its share of the resistance: the largest departure between the two is 3.9 percentage points.

    The resistor the noise comes from

    Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.

    part 3 · noise
  4. What is warm in a capacitor, by its two loss models. computed by solving, not by drawing. One 100 nF capacitor of loss tangent 0.02, written as a 3.183 Ω resistance in series with it and as a 7.958 kΩ resistance across it — the pair that converts exactly at 10.0 kHz and nowhere else. The noise at the terminals is 4kT times the real part of the impedance, so the two models give the same density at 10.0 kHz and are 33.0 dB apart at 100 Hz and 40.0 dB apart at a megahertz. The dots are the same quantity computed the other way — the resistor split out of the netlist, a source put in its place and the network re-solved — agreeing to 3.3e-16. The reactance itself contributes nothing at either end: a lossless capacitor has no real part and is not warm.

    Only the real part is warm

    Johnson's 4kTR is the special case of a statement about impedances: the noise across any passive two-terminal in equilibrium is 4kT·Re{Z}, so a reactance contributes nothing however large it is. That turns a modelling convenience into a noise figure. A 100 nF capacitor of loss tangent 0.02 written as 3.183 Ω in series and as 7.958 kΩ across it — the pair that converts exactly at 10 kHz — gives 0.226 and 10.10 nV/√Hz at 100 Hz, 33 dB apart, and 40 dB apart the other way at a megahertz.

    part 4 · noise
  5. The floor a biased resistor is not standing on. computed by solving, not by drawing. A 100 kΩ resistor with 10 V across it. Its Johnson noise is 40.02 nV/√Hz and does not depend on the voltage; its excess noise is a 1/f density proportional to that voltage, 0.1 µV per volt per decade here, and the two are equal at 271 Hz. Below that the resistor is dominated by a generator that is a property of how it was made rather than of its resistance, and the crossover moves with the SQUARE of the voltage. Splitting the same total resistance into eight parts in series leaves the Johnson noise exactly where it was and divides the excess by √8 = 2.828, taking the total at 1 Hz from 660.2 to 236.4 nV/√Hz. With no voltage across it the second generator is absent rather than small.

    The floor that is only a floor while nothing flows

    Johnson noise depends on a resistance and a temperature and on nothing else, which is what makes it a floor. A real resistor has a second generator that depends on how it was made and on the voltage across it: 0.1 µV per volt per decade for a metal film, rising as 1/√f. On 100 kΩ with 10 V across it the two are equal at 271 Hz, and the crossover moves with the SQUARE of the voltage — 0.678 Hz at half a volt, 2.44 kHz at thirty. Splitting the same total resistance into eight parts in series divides the excess by exactly √8 and leaves the Johnson noise where it was.

    part 5 · noise
  6. The net noise power between two resistors at two temperatures. computed by solving, not by drawing. A 1 kΩ resistor at 400 K joined to a second one at 290 K, whose resistance runs over six decades. Each delivers 4kTR₁R₂/(R₁+R₂)² per hertz to the other; the net is 1.5187 zW/Hz at the match, which is exactly kΔT and contains no resistance — a gigohm pair at the same two temperatures exchanges the same. Away from the match the exchange falls, to 0.0596 zW/Hz at a ratio of a hundred, so the maximum-power argument applies to noise as it does to a signal. At one temperature the net is zero at every ratio, to 7.5e-37 W/Hz, which is the statement a wrong noise model would break.

    Which way the noise goes

    Two warm resistors joined together each drive the other, and the net flow is 4kΔT·R₁R₂/(R₁+R₂)² per hertz. At the match that is kΔT exactly — 1.5187 zeptowatts per hertz between 400 K and 290 K — and a kilohm pair and a gigohm pair at the same two temperatures exchange the same, which is why noise is quoted as a temperature. At one temperature the net is zero at every ratio to a part in 10³⁷, and that zero is the second law rather than a tolerance.

    part 6 · noise

All series