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Subject: Re: A collection of monographs on high accuracy electronics
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From: Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net>
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Date: Mon, 10 Jun 2024 15:20:31 -0400
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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