Equipartition_Voltage-On-Capacitor

Steps

  • Derive the value of the square of the voltage on the capacitor
  • Assume the noise voltage on the resistor can be faked up and simulated by an equivalent voltage source driving the circuit with a noiseless resistor.  The ideal source is assumed to have uniform power spectral density.
  • Equate the value of the voltage on the capacitor with that given by Boltzmann's equipartition theorem and try not to get hanged in the mean time.

Equipartition Theorem: Each orthogonal degree of freedom gets 1/2KT of energy.  This can be thought of as maximum entropy using uniform distribution over each axis. Look for the quadratic energy terms.  With a capacitor all you have is:  E_Capacitor={1/2}C{V^2}

{1/2}KT={1/2}C{V^2}

{KT}/C={V^2}

Now the circuit analysis

P.S.D. == Power Spectral Density

PSD(Capacitor Voltage) = PSD(Ideal Voltage Source) * ( Frequency response of the RC circuit )    …….this is in the frequency domain

{V_C}^2(f)={{V^2}/{{Delta}{f}}}delim{|}{{1/{{j}{omega}{C}}}/{R+1/{{j}{omega}{C}}}}{|}^2

{V_C}^2(f)={{V^2}/{{Delta}{f}}}delim{|}{{1}/{{j}{omega}{R}{C}+1}}{|}^2

{V_C}^2(f)={{V^2}/{{Delta}{f}}}{{1}/{({2}{pi}{f}{R}{C})^2+1}}

Total energy on the capacitor is the integral [0,infinity]

int{0}{+infty}{{V_C}^2(f){df}}={{V^2}/{{Delta}{f}}}int{0}{+infty}{{{1}/{({2}{pi}{f}{R}{C})^2+1}}{df}}      The integral of this is arctan()      

{V^2}=int{0}{+infty}{{V_C}^2(f){df}}={{V^2}/{{Delta}{f}}}*{{pi}/{2}}*{{1}/{{2}{pi}{R}{C}}}

The value on the left hand side is the previous value we deduced using Boltzmann:

{KT}/C=int{0}{+infty}{{V_C}^2(f){df}}={{V^2}/{{Delta}{f}}}*{{pi}/{2}}*{{1}/{{2}{pi}{R}{C}}}

{KT}={{V^2}/{{Delta}{f}}}*{1/{2}}*{{1}/{{2}{R}}}

Which yields the relation between noise voltage, bandwidth and resistance.

{4}{KT}{R}{{Delta}{f}}={V^2}

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Categories: ElectronicsPhysics

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