## Derive kT/C from Equipartition Theorem

It’s better to know(actually have to know) kT/C if you design ADCs.

First, let’s take a look at its inverse-derivation (for detailed explanation, please go for elsewhere, there are lots of…):

$kT/C = 4kTR \times BW_{noise} = 4kTR \times \frac{\pi}{2} \times f_{3dB}= 4kTR \times \frac{\pi}{2} \times \frac{1}{2\pi RC}$

Then, comes the dessert of today – Equipartition Theorem 😉

General Note 1:

The equipartition theorem states that energy is shared equally amongst all energetically accessible degrees of freedom of a system. That is also to say, a system will generally try to maximise its entropy (i.e. how ‘spread out’ the energy is in the system) by distributing the available energy evenly amongst all the accessible modes of motion.

General Note 2:

The equipartition theorem states that if a system is in thermal equilibrium with a reservoir of temperature T, then each quadratic degree of freedom of the system will possess an energy ½kT. A ‘quadratic degree of freedom’ is one for which the energy depends on the square of some property.

Specific Note here:

The thermal fluctuations in the voltage on a capacitor is the ultimate origin of circuit noise. Since the energy stored on the capacitor is $CV^2/2$, the voltage on the capacitor constitutes a degree of freedom. Thus,

$\frac{C\Delta V^2}{2} = \frac{kT}{2} \Rightarrow \Delta V^2 = \frac{kT}{C}$

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All credits go for: 1) an old paper published in 1993 – “White Noise in MOS Transistors and Resistors”; 2) the online document which explains the equipartition theorem; 3) Dr. Timmy’s PhD thesis, where introduces the interesting year-1993-paper.

Btw, sorry for the equations, which are really ugly!

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### 1 Response to Derive kT/C from Equipartition Theorem

1. Load Pull says:

Great useful post.