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The mean noise voltage for Johnson (Thermal) Noise is given by the formula:

\$ v_n = \sqrt{4kTR\Delta{f}} \$.

The bigger the resistance of a resistor or the higher the frequency of a voltage supply, the higher the thermal noise the resistor will have. Does the formula also indicate if the voltage is supplied by a battery (DC, or a voltage of 0Hz frequency), any noise you measured on the resistor cannot be due to thermal noise because:

\$ v_n = \sqrt{4kTR \cdot 0} = 0V \$ ?

[Edit]
The question should be - what is \$ \Delta f? \$

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    \$\begingroup\$ It's not mean noise but RMS noise. The noise is intrinsic to the resistor itself. If shunted by a battery, that noise drops significantly because the effective resistance becomes that of the battery (low). \$\endgroup\$ – Andy aka Jan 11 at 14:40
  • \$\begingroup\$ @Andyaka I thought the internal resistance of a battery \$ R_b \$ is modeled to be in series with the battery itself, \$ R_b \$ can't shunt the load resistor \$ R_l \$ unless \$ R_b || R_l \$ but they are in series. How does this shunting happen? \$\endgroup\$ – KMC Jan 11 at 15:08
  • \$\begingroup\$ It is in series with the battery itself but the battery itself has zero AC impedance so, in effect, the series resistance is in parallel with the johnson noise resistor as far as AC (and noise) is concerned. \$\endgroup\$ – Andy aka Jan 11 at 15:17
  • \$\begingroup\$ The bigger the resistance of a resistor or the wider the bandwidth of the sample measurement, the higher the RMS thermal noise of the sample measurement. Measuring samples for a longer time duration, effectively increases the sample measurement bandwidth and also increases the noise in the measured sample. Measuring for less time gives less sample bandwidth and thus less RMS noise in the sample. \$\endgroup\$ – MarkU Jan 12 at 4:46
  • \$\begingroup\$ @MarkU I guess I don't understand what bandwidth really is in this context. The "bandwidth" are supposed to be the range of frequencies that superimpose and make up the noise, and for a random waveform like noise, its bandwidth should be infinitely large (there are infinite number of Fourier series to represent the noise). Why would noise increase for longer duration of time? If the signal noise bounce around in between ±0.1uV, regardless of how long the measurement takes, it'll stay that way. But what you're suggesting is if I keep my probe on a resistor, the noise will grow over time... \$\endgroup\$ – KMC Jan 12 at 7:43
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yes. Even a resistor with no power source will generate Johnson noise. This is thermal noise due to random movement of electrons in the resistor itself.

f in the formula is the bandwidth across which you wish to calculate the noise, it has nothing to do with the signal applied (or not) to the resistor.

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    \$\begingroup\$ Noise by definition is AC, and is always calculated across some bandwidth. For instance, in measuring noise in an audio system, we typically measure with a low pass filter at 20kHz, and a high pass at 20Hz. Widening the bandwidth will increase the reading, and vice versa. In this case, the thermal (Johnson) noise in a resistor lying on your desk (unconnected) is white noise. Connect it to a sensitive enough AC meter and you will measure it. \$\endgroup\$ – danmcb Jan 11 at 14:58
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    \$\begingroup\$ Noise is generally distributed across a wide bandwidth. You have white noise (linearly distributed) and pink (log) but noise in general is AC, the spectrum of which is always varying randomly, instant to instant (but predictable over a long time period). So when we make the measurement, we choose which part of the spectrum we will take into account (depending on the application). \$\endgroup\$ – danmcb Jan 11 at 16:08
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    \$\begingroup\$ There's always an upper and lower frequency. You might define say 339 to 441 and indeed noise would probably be so low it can't ever be measured then - but it does give you a number. But when (as in the case of audio) you want a lot of bandwidth and very small signals, this can become very significant. \$\endgroup\$ – danmcb Jan 11 at 17:49
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    \$\begingroup\$ I can't explain it any more clearly. Where are you getting a 1V some wave of noise from? That is neither correct not implied by the equation. \$\endgroup\$ – danmcb Jan 12 at 5:20
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    \$\begingroup\$ @KMC lose the battery. Just connect the resistor to the measurement device. \$\endgroup\$ – pjc50 Jan 12 at 9:50
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The bigger the resistance of a resistor or the wider the bandwidth of the sample measurement, the higher the RMS thermal noise of the sample measurement. Measuring samples for a longer time duration, effectively increases the sample measurement bandwidth and also increases the noise in the measured sample. Measuring for less time gives less sample bandwidth and thus less RMS noise in the sample. This effect is due to the act of sampling the noise. If you could take a measurement that ran for the entire duration of the universe, then the measurement would capture everything -- but you'd only get the measurement result at the end of the universe. Taking a smaller sample gives a less accurate result because it omits detail, but the measurement result is achieved faster. The bandwidth (or delta-f) term in Johnson's equation reflects the effect of how long an uncorrelated noise source is measured.

If you put an oscilloscope probe on a noise source and estimate its noise based on the measured vertical deflection, there will be more RMS and more peak-to-peak noise in the sample if the horizontal sweep rate is slower. If you check the datasheet of a reference such as the Maxim Integrated MAX6126, on page 13 there are typical operating characteristics graphs showing exactly that measurement, "Output Noise (0.1Hz TO 10Hz)". (I am an applications engineer at Maxim Integrated.) Whenever we provide noise measurement characteristic graphs like this, we have to provide the measurement bandwidth as one of the test conditions because it will change with different test conditions.

MAX6126 pg 13 toc14 5V output noise (0.1Hz to 10Hz)

An oscilloscope has limited bandwidth to what range of frequencies it can capture for a given sweep. For a horizontal sweep rate of 1 second per division, and a display width of 10 horizontal divisions, the lowest measurable frequency would be 0.1Hz (all 10 divisions). The highest measurable frequency is deemed to be 10Hz (1/10th of 1 division) based on the limit of the display's resolution. (At higher frequencies there would also be some additional bandwidth limits in the analog / data acquisition path, but they're negligible at 10Hz.)

For many types of precision analog and data acquisition systems (ADCs, DACs, precision op amps) there are noise specifications given in units of nV/sqrt(Hz) for Johnson noise and nA/sqrt(Hz) for shot noise. Both of these types of noise vary with the measurement bandwidth.

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  • \$\begingroup\$ What is "sample measurement"? Does "sample" refer to a resistor or a time duration of noise? \$\endgroup\$ – KMC Jan 12 at 10:24
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    \$\begingroup\$ I'm not sure that first paragraph is correct - a longer sample-and-hold period has lower bandwidth. \$\endgroup\$ – pjc50 Jan 12 at 11:15
  • \$\begingroup\$ @pjc50: Or at best equal max frequency, if you sample at the same sample-rate and just end up with more samples. (Longer .wav file for the example case of recording the input of a sound-card.) Or for a pure analog scope, same vertical bandwidth but slower horizontal sweep. For random noise it is true that you're likely to see a higher peak over a longer measurement interval (with constant bandwidth). My math is rusty on this, but the "more RMS power" seems suspect. But again, all of this is assuming constant sample rate, not constant number of samples over the interval. \$\endgroup\$ – Peter Cordes Jan 13 at 4:35
  • \$\begingroup\$ @pjc50: Err, my previous comment was only considering the upper end of the band, but I think Mark's point is about the other end of the band. Given constant sample rate (fixed Nyquist frequency), a longer interval will let you see lower frequencies, so yes larger band width. Pushing the min frequency down closer to zero doesn't open up the bandwidth by much, because what matters is linear f_max - f_min = Δf, not the ratio fmax/fmin, but it does increase it asymptotically towards the full nyquist frequency of your sample rate. \$\endgroup\$ – Peter Cordes Jan 13 at 4:46
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    \$\begingroup\$ I think I'm remembering this right but I googled to confirm, and a random FFT sample-window help page (for audio software) came up: support.ircam.fr/docs/AudioSculpt/3.0/co/Window%20Size.html suggests that the lowest detectable frequency (the lowest FFT bin) is F_0 = 5*(SampleRate/WindowSize). Having more FFT bins closer together over the interval is probably not significant if you only care about total energy though; any energy between the min and nyquist frequency should be distributed across the bins. (Or did that blurring happen as linear sum, not sum of squares for RMS?) \$\endgroup\$ – Peter Cordes Jan 13 at 4:51
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Johnson-Nyquist noise has nothing to do with the power supply. The \$f\$ in the formula is the bandwidth of the noise, not any applied frequency.

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Maybe this needs a link to the original Johnson paper. It's fairly well explained and doesn't involve any quantum mechanics, since this is the experimental measurement paper; Nyquist has a corresponding paper with the theory in.

(Some nice experimental detail on performing small-value measurements in the vacuum tube era, too.)

So the original misconception in your question was thinking that the bandwidth related to the bandwidth of a supply. No supply to the resistor is required - we're discussing the bandwidth of the noise measurement. The noise itself is comprised of a series of point events, the collisions of individual electrons. However, if you feed a near-zero-duration point event through a measurement device with finite bandwidth, you don't get a point measurement, you get "wavelets" (sinc? gaussian? not actually sure) which do have frequency components. See Nyquist-Shannon's other work on sampling.

Conceptually, if you have a positive voltage spike and a negative voltage spike immediately afterwards, a low-bandwidth measurement will average them together and see low noise power. A higher-bandwidth measurement will be able to separate them out and count the power properly.

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  • \$\begingroup\$ It seems I won't get any of these unless I read through a proper textbook - just for the fundamentals and nothing advance. I couldn't find any relevant material without getting utterly lost or overwhelmed on the first page (e.g. SPECTRAL ANALYSIS OF SIGNALS), and my few linear circuit analysis textbooks barely touch on this subject other having some Fourier Analysis problems which appear remotely relevant. Could anyone suggest a book or tutorial that may bridge this gap? \$\endgroup\$ – KMC Jan 12 at 11:11
  • \$\begingroup\$ I don't think I have any specific advice because I don't know what level you're starting from, but did you read the Johnson paper? \$\endgroup\$ – pjc50 Jan 12 at 11:16
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    \$\begingroup\$ It's not so much the noise that has bandwidth as the input that's reading the noise has a bandwidth. \$\endgroup\$ – pjc50 Jan 12 at 13:07
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    \$\begingroup\$ "bandwidth of the noise measurement" = decomposing the noise into each constituent frequencies and amplitude? Seems I'm interpreting it wrong again. So the bandwidth is the bandwidth of my oscilloscope? \$\endgroup\$ – KMC Jan 12 at 13:22
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    \$\begingroup\$ Yes, the bandwidth is that of the measurement. If you're connecting a resistor to a scope, it's the bandwidth of the scope. If you're connecting it to an amplifier, it's the bandwidth of the amplifier. Note that most op-amp circuits have resistors on the input - whose Johnson noise will be amplified! An important part of amplifier design. \$\endgroup\$ – pjc50 Jan 12 at 16:26
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I struggled to understand the given answers and comments, and kept redirecting myself but all to wrong directions. Now that I think I have it figured out, the question really should be asking - what is \$ \Delta f \$?

A resistor by itself generates noise from random movement of electrons inside that resistor. Without noise, the resistor measures at 0V flat. With noise, it measures at some random voltage values fluctuating above and below 0V. Noise, like signals, is a waveform of voltage (or current). Like any other waveform, periodic or aperiodic, noise is a superposition of sinusoids varying in frequency and amplitude.

If we model the noise generated by a resistor as a voltage source \$ v_{noise} \$, and that we (unrealistically) assume the noise to be a single sinusoid that has frequency within the bandwidth of a passband filter, we can measure the power of that noise (or equivalently, the power of that single frequency). I think the graph (Power/Hz vs. Frequency) is what they call the Power Spectral Density (PSD).

deltaf_01

Johnson Noise (or Johnson-Nyquist, Nyquist, Thermal, White Noise), as it turns out, generates an average \$ k_bT \$ watt of power for each and every constituent frequency of that noise. Thermal noise, as a random waveform, is composed of an infinite number (hence, a continuum) of sinusoids, each having a constant power of \$ k_bT \$. That explains the straight flat line valued at \$ k_bT \$ across the entire spectrum to infinity.

deltaf_02

The passband filter let through the set of frequencies between 1k-10kHz and filters the rest of the noise frequencies to ground. \$ \Delta f \$ is the bandwidth of the filter, and the area is dimensionally the power of noise! The wider the bandwidth, the bigger the noise power.

If I remove the passband filter, the noise will now be filtered by the bandwidth of my oscilloscope and the intrinsic/parasitic serial or parallel capacitance or inductance (which also acts like filters) of the resistor. Thus in reality, there's always a bandwidth existed somewhere to prevent you from collecting an infinite amount of noise power, and that the finite power is a function of the bandwidth \$ \Delta f \$.

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