I'm trying to understand Johnson noise and I already read Nyquists paper and did some other reading, but I'm pretty stuck. As I guess there is this missing piece that relates everything, I'll just write down my many points in the hope for an answer that relates everything more or less:

  1. What do we measure if we take a resistor $R$ and measure the thermal voltage with a perfect voltmeter?

  2. If we take two resistors $R_1$ and $R_2$ at random temperatures and random resistances, connect them and put them into vaccuum. Will their temperatures converge to some temperature $T$?

  3. If we take again two resistors $R_1,R_2$, but with different resistances. Now put two voltmeter directly besides of each of them and then connect them again. Will both of the voltmeters always show the same voltage? What if the temperature differs at the beginning?
  4. Is there a general equivalent circuit for a resistor to account for the thermal noise? What would that look like?
  5. Assuming that in #2 their temperatures converge. Eventually they are pretty close to being in thermal equilibrium and the power transfer in both direction would need to be the same. How would I calculate/show that for $R_1\ne R_2$?

Unfortunately my knowledge about transmission lines is nonexistent, and I didn't understand the part in Nyquist's paper, where he requires the impedance of the transmission line to be $(LC)^{\frac{1}{2}}=R$. How important is this detail?


2 Answers 2


Johnson noise, like Brownian motion, is an observable effect of temperature in a population of identical particles.
What you measure, with an ideal voltmeter across a resistor, is the small fluctuations in local charge density at the terminals of that resistor, basically the charge on a (very small) parasitic capacitor.

If we take two resistors R1 R 1 and R2 R 2 at random temperatures and random resistances, connect them and put them into vaccuum. Will their temperatures converge...

If we take again two resistors R1,R2, but with different resistances. Now put two voltmeter directly besides of each of them and then connect them again....

If you connect two resistors, they DO come to the same temperature, because electrons flow freely through an electrical connection, and electrons have temperature (and carry heat). This is why metals are usually good conductors of heat, the freely moving electrons carry heat. The sum of Johnson voltages (which are random fluctuations) is done in quadrature (i.e. the squared voltages add) because there is no phase synchronization possible. In-phase (zero degrees) voltages add, out-of-phase (180 degrees) voltages subtract, and random (orthogonal, 90 degrees average) voltages add in quadrature.

So the mean-square Johnson voltage is proportional to resistance, and the mean-square Johnson voltage of two resistors in series is proportional to the sum of the two resistors. There's no significance to temperature distributions in the resistor(s), it just means a messier calculation. EVERY resistor is a series-connected multiplicity of smaller resistors, each at a temperature.

  1. assuming that your resistor is connected to a lossless ideal flat bandpass filter of bandwidth $W$ then a perfect voltmeter will measure at the output of the filter an average 0, but a perfect rms meter will measure a voltage $v_{rms} = \sqrt{Rk_BTW}$
  2. if you connect to resistors at different temperatures then they equilibrate, i.e., then their temperatures will be equal after a while.
  3. see the formula above in 1
  4. The Thevenin equivalent would be a resistor connected in series with a random voltage source of $v_{rms}$ in 1. There is an obvious analog of this for the Norton equivalent current source.
  5. Use the Thevenin equivalents and voltage division.
  6. Nyquist assumes that the transmission line connecting the resistors at either end is such that it is terminated without reflection, thus whatever noise wave one resistor emits at one end is fully absorbed by the other.
  • $\begingroup$ 1. Not sure, how such a circuit would look like, I specifically meant to have the resistor open-circuited... 2. ok; 3. I did intentionally not speak about the rms value. I'm interested if two resistors are connected, and we measure the actual voltage at both of them, do we always get the exact same value...in principle if the connection between them is a perfect cable, I expect them to be the same..not sure if my question is kind of a "what is the voltage across a short-circuited perfect voltage source" question; 4. ok; $\endgroup$ Commented Jan 4, 2018 at 22:02
  • $\begingroup$ 5. If their resistances don't match how do I calculate the rms voltage of both them. Obviously they can't be the same, for if they would, thermal equilibrium wouldn't be preserved. 6. Can you please in some words explain reflection? $\endgroup$ Commented Jan 4, 2018 at 22:02
  • $\begingroup$ The ideal bandpass (or lowpass) filter just represents the finite effective input bandwidth of the meter (of very large input impedance) you are connecting to the resistor you are measuring. What you are measuring is noise whose average is 0; if you connect two resistors of the same temperature then the mean square voltages per unit bandwidth add just as the variances of two independent r.v.-s. Being noise there is no single value to the voltage, it is an r.v., whose distribution is normal (Gaussian) of average 0, and dispersion (standard deviation) is given in 1. $\endgroup$
    – hyportnex
    Commented Jan 4, 2018 at 22:23
  • $\begingroup$ RF waves on a transmission line reflect off a load impedance just as light reflects from a mirror. If the load impedance is the same as the wave impedance of the line then there is no reflection and you get perfect absorption of the traveling wave by the load. $\endgroup$
    – hyportnex
    Commented Jan 4, 2018 at 22:26
  • $\begingroup$ If the resistors are of the same temperature then the Thevenin for series or Norton for parallel equivalent circuits can be used to get a single noise generator by adding the variances of the noise voltages or noise currents using the formula in 1. $\endgroup$
    – hyportnex
    Commented Jan 4, 2018 at 22:28

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