Re: A collection of monographs on high accuracy electronics

On 2024-06-10 15:14, Phil Hobbs wrote:

On 2024-06-09 21:43, Phil Hobbs wrote:

<snip>

>

Bill was kind enough to send me a copy (thanks again, Bill), and right there on P. 374, the author says,
 Pn = 4kTB
 which is a factor of four too high.
 Twenty years ago I posted a brief derivation of the Johnson noise formula in the thread "thermal noise in resistors - Baffled!"....

<snip> >
And again the following year, with more discussion...this qualifies as a well-aged FAQ. ;)
Cheers
Phil Hobbs

Subject: Capacitor-feedback for low noise
Phil Hobbs
Aug 23, 2005, 11:16:25 AM
 Zigoteau wrote:

>
If you want to calculate the noise you get from an arbitrary circuit,
then you need a model for the noise behavior. The thermal noise of an
impedance Z(f) can be modeled by a Thevenin equivalent circuit, where
the voltage source in series with Z(f) is random with a spectral
density of 4kTRe(Z(f)) V2/Hz. Equivalently, its thermal noise can be
modeled by a Norton equivalent circuit, where the current source in
parallel is random with a spectral density of 4kTRe(1/Z(f)) A2/Hz.

 Yes, the physics behind it is summarized in the fluctuation-dissipation
theorem of statistical mechanics, which says that any mechanism that can
dissipate energy has associated fluctuations at finite temperature. If
this weren't so, you could make heat flow spontaneously from cold to hot.
 The usual way to derive the Johnson noise formula for a resistor is to
use classical equipartition of energy, which predicts that any single
degree of freedom, e.g. the charge on a capacitor, has an RMS energy of
kT/2. Classical equipartition is a very general consequence of
statistical mechanics, and even in a quantum treatment, it can be shown
to hold for frequencies << kT/h, about 6 THz at room temperature. (The
high-frequency correction is due to the Planck function rolloff.) Since
E=CV**2/2, kT/2 of energy corresponds to voltage Vrms = sqrt(kT/C), and
charge Qrms = CV = sqrt(kTC).
 If you have a parallel RC, isolated from the rest of the universe,
this fluctuation must be maintained in equilibrium by the resistor
noise--otherwise, the initial sqrt(kTC) would just discharge through the
resistor. This must be true regardless of the values of R and C.
Therefore, the open-circuit thermal fluctuations of the resistor, in the
bandwidth of the RC, must equal sqrt(kT/C) volts; since the noise BW is
1/(4RC) (noise BW = pi/2* 3 dB BW), the open-circuit resistor noise
voltage density is sqrt[(4RC)*(kT/C)] = sqrt(4kTR), which we all know
and love.
 You have to work a little harder to make this demonstration completely
rigorous, e.g. by showing that the fluctuations have to be flat with
frequency, but this is the idea. It can also be shown directly from
statistical mechanics applied to a semiclassical electron gas model of
metallic conduction, but I don't know how that derivation goes.

 

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510
http://electrooptical.net
http://hobbs-eo.com