# Thermal Noise in Electrical Circuits

1. Feb 20, 2013

### JeroenvG

Dear people,

it might be a simple question, but I have been on this for over 2 days now and can't find a proper explanation of this anywhere:

https://dl.dropbox.com/u/2264519/circuit.jpg [Broken]

Picture a RCL circuit which is kept at room temperature. Now even when there is no current going through the circuit, the resistor is still creating some thermal noise $S_{R}(f) = 4k_{B}TR$ due to the motioin of individual electrons. This is just simple white (/Johnson) noise, which is the same for all frequencies. I can work with Laplace transformations and stuff, calculating the resonance frequency of the circuit. Now my question is this:

Wouldn't the resistor also create a small portion of noise at the frequency that just so happens to be the resonance frequency of the RCL circuit? At all other frequencies the noise woudn't resonante through the circuit but decay, but what happens on this frequency?

Will this circuit just start to resonate out of nothing? Isn't that strange? Also, can you calculate what would be the current that will be resonating at this frequency? (I guess it would have to be T dependent, since a higher T causes more noise). What will the nosie spectrum of this circuit look like? What would I see when I measure the current through certain parts of the circuit?
A lot of questions, but I'm hoping to get a more clear picture of noise in electrical circuits ^^.

If anyone knows a nice essay that deals with noise in electrical circuits that would be great! I do know some mathematics, but as a physics student I don't work with electronics very often.

Last edited by a moderator: May 6, 2017
2. Feb 20, 2013

### marcusl

3. Feb 20, 2013

### Staff: Mentor

Your circuit is not perfect (every induced oscillation dies out) - and the noise energy at a perfect resonance frequency would be zero.

4. Feb 20, 2013

### JeroenvG

Dear mfb,
yes, only a circuit with zero resistance would act as a perfect oscillator. So when I put an AC current with a certain frequency $\omega$ through this RLC circuit the current will oscillate a little bit, but will of course it will decay as soon as I stop applying any voltage to the circuit. In that you are right, when I don't apply any voltage the current through the circuit will decay and go to zero.
But my point is that - taking the thermal current produced by the resistor into consideration - there will always be a current at each frequency with a corresponding power $4k_{B}TR [V^{2}/Hz]$. I am under the impression that this circuit will always be driven by noise and therefore the current moving through the circuit will never go to zero.

Marcusl,
I have read the paper you recommended (Thermal Agitation of Electric Charge in Conductors, H. Nyquist, 1928). For people interested in reading it: it is more of a thermodynamical deriviation of Johnson noise. However, page 4 and equation 5 and 6 gave me some new insights. Here Nyquist explains how you can calculate the noise in a certain filter or circuit.
For simple circuits you can say that the thermal noise is beeing created in the resistors. When you can calculate the admittance $Y(\omega)$ of your circuit (which might be frequency dependent) you can see which frequencies will be easily conducted by your circuit and which frequencies might have a hard time going through. Since thermal noise is constant over frequency, calculating the frequency dependent noise that will move through the circuit can be done by simply multiplying the thermal noise by the square of $|Y(\omega)|$. This way the frequencies which have a low admittance will be less dominant in the result.
$$I^{2}d\nu = E^{2}_{\nu}|Y(\omega)|^{2}d\nu = (2/\pi)k_{B}TR(\omega)|Y(\omega)|^{2}d\omega$$
We can find the total noise power by integrating over all frequencies
$$I^{2}=(2/\pi)k_{B}T\int^{\infty}_{0}R(\omega)|Y(\omega)|^{2}d\omega$$

Ok, but now back to the RLC circuit. When we are wondering what would be noise current we would measure over the resistor in this circuit we can also look at it in the following way. In this case our circuit will work as a band pass filter, only letting frequencies around the resonance frequency go through easily. Other frequencies will decay faster and will thus have a harder time passing through. Since thermal noise (or white noise) is constant over all frequencies I believe that when we multiply the square of the admittance with our original noise spectrum we will get the noise spectrum of this circuit.
$$Y(\omega) = \frac{s}{L(s^{2}+\frac{R}{L} s + \frac{1}{LC})}$$
Multiplying this function with the constant white noise spectrum we will look something like this:
http://dl.dropbox.com/u/2264519/lrc.JPG [Broken]
It is important to use $|Y(\omega)|^{2}$ and not $|Y|$ since we are working with a power spectrum (power is proportional to the square of the voltage).

Now, to speculate about the answers to some of my earlier questions:

Will this circuit just start to resonate out of nothing? You don't have to picture it like that. Originally the resistor made white noise (inserting all frequencies into the circuit), since only frequencies near the oscillation frequency can easily pass through this circuit I expect that only at this frequency a portion of the original noise spectrum will remain. So yes, you will see the circuit mainly oscillating at these frequencies. But not because the whole system magically started oscillating, but because these frequencies of the white noise spectrum are 'better preserved' while other frequencies are filtered away.

Can you calculate what would be the current that will be resonating at this frequency? (I guess it would have to be T dependent, since a higher T causes more noise).
The power of the noise at the ocillation frequency will be equal to a fraction of the original white noise spectrum at this frequency. The total noise power will therefore be less than the power of the white nosie of the resistor. This is because we filtered the noise away at certain frequencies. Increasing the temperature will increase the white noise in the resistor and thus also the noise we see through this RLC circuit.

I want to thank both of you for your time and I do have the feeling I am seeing things more clearly now. Is there maybe somebody who knows whether this is correct or are there still some flaws in my reasoning? Also, any good text on noise in electrical circuits would still be appreciated :).

Last edited by a moderator: May 6, 2017
5. Feb 20, 2013

### marcusl

I think your reasoning is correct--the power spectrum of the current will be |Y|^2. No power is dissipated in the LC resonator, although energy is stored there. Applying Nyquist's equipartition theorem arguments, the energies stored in the L and C are the same on resonance, $$\frac{1}{2}LI_{rms}^2=\frac{1}{2}CV_{rms}^2=\frac{1}{2}kT.$$
I am not familiar with texts on noise in electrical circuits. If I had to investigate this, I'd look back in time to the classics (late 40's - 60's?). Uhlenbeck was a big name in noise, and the book by van der Ziel is well-known (though I've never looked at it).

EDIT: Your comment about the power on resonance being a fraction of that with no LC does not seem correct to me. If |Y|=1 there, it should be the same.

6. Feb 20, 2013

### lolone

The best way to think of this is in terms of thermodynamics and in particular the laws of thermodynamics. The Energy must come from somewhere, therefore without an input the resonance of the circuit would go unused.

The OP's line of thinking is that the quantum effects would input energy and cause the circuit to ring. The problems are:

a. Quantum effects are not coherent, they do not produce a "frequency" and certainly none that would be suitable for the circuit.
b. Without an additional source of energy, the circuit is already at its maximum output.
c. Any net motion would be dissipated by the thermal losses in the circuit.

If this were not the case, the circuit would be the biggest weapon ever created...on par with the "Big Bang" if it could hold its integrity.

7. Feb 21, 2013

### the_emi_guy

Expanding on Marcusl,

The noise voltage sourced by resistor is given by: $$V_r=\sqrt{4KTRΔf}$$

Voltage across the capacitor in a bandwidth Δf as a function of quality factor is given by: $$V_c=QV_r$$

Quality factor of capacitor is given by: $$Q=\frac{1}{ωRC}$$

Integrated bandwidth of circuit is given by $$Δf=\frac{ω}{4Q}$$

Using the above we can solve for the capacitor voltage: $$V_c=\sqrt{\frac{KT}{C}}$$

And the energy within the capacitor will be: $$E_c=\frac{1}{2}CV_c^2=\frac{1}{2}KT$$

Which is what we expect from equipartition theorem for single degree of freedom (capacitor voltage).

So, yes, you can say that the circuit resonates all by itself, but it is probably more accurate to say that the capacitor's E-field and inductor's H-field each carry one degree of freedom's worth of thermal noise.

8. Feb 21, 2013

### f95toli

Yes, I think the description is OK>

I sometimes use circuits like this as "noise standards" when I test equipment (amplifiers ec). If you hook up just a resistor to an FFT analyzer you will (obviously) see a white noise level corresponding to the thermal noise, by adding caps and inductors you can quite easily filter that noise to get whatever spectral shape you want. Mostly, you can just think of the resistor as a broad-band noise generator.

In your case you would see a peak in the noise correponding to the resonance frequency.

Also, if you build an oscillator using an LC circuit and an amplifier (in a feedback configuration with the Barkhausen condition met)) that circuit WILL spontaneously start to oscillate thanks to the thermal noise; it won't need any external "kick" to get going (any energy lost is of course replaced by the power supply of the amplifier).

9. Feb 21, 2013

### lolone

If the energy is coming from the power supply, then the oscillation is not due to thermal noise in the LC circuit. The oscillation is coming from noise in the amplifier.

In any circuit, follow the thermodynamic chain to discover the input source.

10. Feb 21, 2013