The voltage(rms) created by thermal noise is:

What is the bandwidth(Δf) of the thermal voltage?

  • \$\begingroup\$ It's whatever you want it to be. \$\endgroup\$
    – Andy aka
    Oct 16, 2021 at 14:09
  • \$\begingroup\$ What do you mean by "whatever I want it to be"? \$\endgroup\$ Oct 16, 2021 at 14:10
  • 2
    \$\begingroup\$ It's a system specification, and you haven't told us what the system is. For example if it's an audio system, the bandwidth specification may be 3.4 kHz, or 15 kHz or 20 kHz. \$\endgroup\$
    – user16324
    Oct 16, 2021 at 14:29
  • 1
    \$\begingroup\$ Yes but it is up to you to define at what bandwidth you are interested in the noise. It is unlikely that your bandwidth would be 0 or infinity, but some practical value depending on what is important. \$\endgroup\$
    – Justme
    Oct 16, 2021 at 14:40
  • 1
    \$\begingroup\$ Let's say you are feeding the signal to an amplifier. In that case you would probably use the bandwidth of the amplifier. Let's say you are feeding the signal to a band pass filter. In that case, you would use the bandwidth of the filter. Etc. This is one of the reasons why you usually do want to filter out unwanted signals, to reject the out-of-band noise. \$\endgroup\$
    – user57037
    Oct 16, 2021 at 18:02

4 Answers 4


It's the frequency range that you are examining for noise. The normalisations are such that you'd get infinite results when looking at an infinite frequency range. Most noise will not even be relevant for the application you need the analysis for.


Thermal noise produces a flat power spectral density i.e. it produces "so-many" watts per hertz of bandwidth. A wider bandwidth measurement system will see more noise produced.

To convert watts per hertz into an RMS voltage we assume the resistance is 1 Ω and compute voltage using the following standard equation: -

$$P = \dfrac{V^2}{R}$$

$$V = \sqrt{P\cdot R}$$

The same applies to power as power spectral density hence, the RMS voltage across a warm resistor is proportional to the square root of the bandwidth of the measurement instrument.

  • \$\begingroup\$ I’ve seen the 1 ohm assumption innEMC testing - do you know what the basis of this is? Perhaps I should ask it as a separate question. \$\endgroup\$
    – Frog
    Oct 16, 2021 at 20:22

At first glance, the equation you show (probably comes from the Wikipedia article on Johnson noise) looks a bit confusing since \$ v_n\$ is on both sides of the equation.

\$ \bar v_n \$ is the resistor thermal noise spectral density in \$volts_{rms}/ \sqrt{Hz}\$.

\$v_n\$ is the total noise in a given bandwidth in \$volts_{rms}\$. This bandwidth, \$\Delta f\$, is the bandwidth of your circuit. Lets say there is a unity gain noiseless amplifier that has an bandwidth of 10 kHz with a 1k resistor connected to the input. The 1k resistor thermal noise spectral density is about 4 \$ nV/\sqrt{Hz} \approx 0.13\sqrt{R}\$ at room temperature. The total output noise voltage in the 10 kHz bandwidth is \$ \bar v_n \sqrt{\Delta f} = 4 nV/\sqrt{Hz} \times \sqrt{10 kHz} = 400 nV\$.

This is a cute concept and is easily observable in real electronics.

\$\Delta f\$ is actually the equivalent noise bandwidth (ENBW) which is different from the traditional -3 dB bandwidth. A simple RC low-pass filter will have a ENBW of 1.57 times the -3 dB bandwidth since there is energy beyond the -3 dB bandwidth frequency points. Higher order filters will have a smaller fudge factor since the filter skirts are steeper.

If you want to delve deeper in to noise, the traditional go-to book is "Low-Noise Electronic Design", C. D. Mochenbacher.

  • \$\begingroup\$ \$\bar v_n\$ is in units of V/√Hz, not V. \$\endgroup\$
    – Hearth
    Oct 16, 2021 at 20:03
  • \$\begingroup\$ @Hearth Corrected, thanks. \$\endgroup\$
    – qrk
    Oct 16, 2021 at 20:17

Your formula is derived from a simpler formula P=kTB. It presents how much any resistor, no matter how many ohms it is, can output thermal noise over bandwidth B (=your delta f) if there's a matched load, for ex. another resistor with the same resistance. T=temperature in kelvins and k is Boltzmann's constant.

Your formula shows what's the rms output voltage of the imagined noise voltage source if we think a practical resistor R is a series circuit of theoretical noiseless resistor R and a noise voltage source.

About the bandwidth: It's the bandwidth of the AC voltage or power meter which measures the voltage or power of the noise. That can look ridiculous - the noise voltage or power can be as high as wanted if we measure over a wide enough bandwidth. In theory a resistor seems to be able to produce an infinite noise power. But the formula is perfectly right if we assume the classical statistical theory of heat is valid.

There are few limitations:

  1. all circuits have parasitic capacitance and inductance which limit the bandwidth, the theoretical infinite power has no infinite bandwidth route out of the resistor and if it had, the resistor would cool down to zero kelvin immediately - assuming the formula is right and nothing feeds replacement energy to the resistor.

  2. the classical theory of heat is not valid. Quantum physics stops its validity range well below the infrared frequencies.

Actually physicist Max Planck found the quantum physics when he started to think possible explanations for the measurable fact: Formula P=kTB is valid up to microwaves, but it's not valid in infrared nor higher frequencies. The density of noise W/Hz drops drastically when the frequency is higher than a certain limit which depends on temperature.

Mr. Planck, of course didn't have in front of him electronic circuits and he didn't use formula P=kTB, but he had a possibility to observe the direct electromagnetic radiation from pieces of material to the environment and his problem was the validity of the classical heat theory from where formula P=kTB was quite straightforwardly derived by Harry Nyquist (=pioneering scientist of many phenomena in electronic circuits) after his colleague Johnson had become (by making measurements) convinced it's true.

How do we use the formula:

We calculate with it the amount of thermal noise in our circuit. The calculation is essential for radio communication, radars and sensitive microphone amplifiers. We know the bandwidth of our circuit by design.

If we measure DC voltage with a multimeter we can well assume that the bandwidth of the multimeter is only few Hz in DC ranges - see how many new readings it can take per second. The noise voltages in practical resistors cannot make the ordinary multimeter reading of a DC voltage fluctuating, because the maximum thermal noise voltage in ordinary circuits is max. few microvolts; that's far less than DMMs can show as changing DC reading.

  • \$\begingroup\$ @Heath Thanks for fixing badly written sentences. \$\endgroup\$
    – user136077
    Oct 16, 2021 at 20:19

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