# Is the so called kTC noise, that is inherent in an RC filter, dependent on bandwidth?

An RC filter has intrinsic noise called the kTC noise which is given by the equation $$v_\text{noise} = \sqrt{\frac{kT}{C}}.$$ Does this equation calculate a noise density with units of volts per root hertz? Or does this equation calculate an RMS noise voltage that is not bandwidth dependent?

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the ktC noise formula simply happens when you insert the noise bandwidth of an RC filter, B=1/(4RC), into the formula for voltage square over a resistor. The R cancels out.

So, no, there's correctly no bandwidth in that formula, because the bandwidth is "hidden" in the C.

The wikipedia article on Johnson-Nyquist Noise actually containst this explanation, and a lot more stuff that you might want to know about kTC noise, considering you seem to be coming at this from a non-intuitive, formularistic side. It's usually easier to understand what's happening than to have all the right formulas exactly right :)

• Thanks Marcus for referring me to the wikipedia article. It does say that the units of Vnoise calculated in the equation given in my original question are Volts RMS. I mostly wanted to confirm that the units of the calculated Vnoise were NOT Volts per root hertz. I finally convinced myself that Vnoise is measured in Volts RMS when I calculated the equivalent noise BW of an example RC filter and plugged that ENBW into the thermal noise formula for the resistor and got the same result for Vnoise as when using the kTC noise equation given in my original question. Your answer was most helpful. – natesrc 2 days ago

The KTC noise is actually the total RMS noise power at the output of the low pass RC filter. The noise power is the product of the noise power spectral density (PSD) and the bandwidth (more accurately "effective bandwidth") of the system. So, actually the total noise power does depend on the bandwidth.
Interestingly, for the case of RC filter although the resistor is the only noise source present in the system, the total noise power is independent of the resistor value, as seen from the noise power expression, $$v_n^2 = \frac{kT}{C}$$ This is because although increasing the resistor value increases its noise PSD (= 4kTR) but the effective bandwidth ($$\\frac{1}{4RC}\$$) goes down proportionally. $$v_{n}^2 = 4kTR \cdot \frac{1}{4RC}$$ This can also be understood from the principles of thermodynamics. Since the origin of Johnson noise is due to random motions of electrons inside the resistor, the kinetic theory of gases is applicable for it. And the equipartition theorem says that in thermal equilibrium, energy is shared equally between each state of the system and the shared energy is given by $$\E = \frac{1}{2}kT\$$.
For the RC system, the only energy state is the energy stored in the capacitor which is given by: $$E = \frac{1}{2}CV^2$$ Equating the two terms gives: $$\frac{1}{2}CV^2 = \frac{1}{2}kT$$ $$\implies V^2 = \frac{kT}{C}$$ This is exactly the noise power derived from circuit equations but is derived much easily from laws of thermodynamics.

Ultimately it is because the noise (and signal) is counted in electrons, and the electrons jiggle about a bit (technical term) and their speed of jiggling depends on temperature. So the count of the number of electrons in the capacitor changes constantly.

The noise (rms) on the number counted is simply the square root of the number (same as a repeated random coin toss).

The magic value 'k' is the conversion factor. The bandwidth is in cycles (in & out) per second, and the Temperature gives the 'how far/how many' move in and out of the capacitance.

The square root part annoys every one because it doesn't follow the normal electrical rules. Rather its because electrons are quanta (quantum mechanics).

As explained by others the formulas can be swapped around to pretend that the resistors are noisy, while I find it easier to simply count the electrons in the capacitor. The answer is the same. The Physics is the same. Very useful to know if you do photonics.