Talk:Johnson–Nyquist noise

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From this paper:

"3) The radiation resistance of free space is √(μ0 /ε0). What is the rms noise voltage at the terminals of an antenna which will compete, say, with an FM radio signal?"

Is that supposed to imply that free space generates thermal noise??? - Omegatron 18:36, Apr 15, 2005 (UTC)

- You are confusing radiation resistance with the impedance of free space. Radiation resistance is a property of the antenna geometry, and is a measure of the power lost from the transmitter because it is radiated as electromagnetic waves. The impedance of free space describes the ratio of electrostatic to magnetic field in the propagating wave.

I thought noise power was 4kTBR where B is bandwith and R is the source resistance. Am I wrong again??Light current 01:46, 1 September 2005 (UTC)
No its Ok. I noticed they are quoting power so resistance drops out of the equation. Its noise voltage/current that depends on resistance Light current 12:29, 1 September 2005 (UTC
Yes, but the 4 also drops out. Noise power to a matched load resistance is kTB, not 4kTB! 10Jan2006
The page is rong. You can only state that the transfered johnson noise power to a matchin resistence is kTB, so this is the maximum transfered power. The noise can be modeled by a voltage source in series with the resistence with rms value of sqrt(4 K T B R) or by the norton equivalent of a current source in pararel with the resistence of sqrt(4 K T B / R). If no one corrects me I am editing the page.

I see a number of problems with the discussion of radio communications on this page. First of all, it is NOT true that the noise received by a receiver listening to a radio channel is kTB. It can be much higher OR lower than this. If we exclude external sources of noise (like lightning and the sun), then the dominant noise source is the receiver itself. Internal receiver noise is usually rated by Noise_figure, which, for a receiver, is the ratio of Pout/G*Pin, where Pin is the thermal noise from a resistor at 290K attached to the input of the receiver and Pout is the noise output power of the receiver (usually can consider just the first few stages) and G is the receiver gain. So, a perfect receiver would have a noise figure of 1 (0 dB). A receiver which introduced internal noise equal to thermal noise at 290K (which is what is implied in this article as being always true) would be 2 (3 dB).

UHF and microwave receivers regularly achieve noise figures of less than 1 dB (1.25).

In real use, the receiver is not attached to a resistor as its source but to an antenna with the proper feedpoint impedance. Because the resistive part of the feedpoint impedance is not a power loss but a power transformation (from an EM wave propagating in free space to a signal in the wires going to the receiver), it does not produce Johnson noise.

So, in summary, if you attach an antenna to a receiver, point it away from all radio sources (free space), then you will see simply the receiver internal noise (plus the cosmic background radiation, but that is almost surely much less than the receiver internal noise), and this internal noise may be much less than kTB. If no-one responds to this, I will change the article to reflect this, along with links to other relevant articles such as Noise_figure Sbreheny (talk) 06:10, 24 December 2007 (UTC)

It seems to me the Preceivernoise equation should represent a closed system at equilibrium. If true, then the antennas temperature is equal to its surrounding temperature. Therefore, the amount of thermal noise received by the antenna due to blackbody radiation is relative to the antennas temperature. Correct me if I'm wrong, but power noise (KTB) is applicable to antennas; e.g., a 10MHz-BW 50-ohm (radiation resistance) antenna at 300K should receive K*300*10MHz = 4E-14 watts. Paul Lowrance, April 13 2008. —Preceding unsigned comment added by 72.25.73.160 (talk) 16:27, 13 April 2008 (UTC)

Hi! Thanks for your interest in this topic. An antenna may not "see" surroundings as you expect, though. For example, a directive antenna pointed up into the sky is almost unaffected by the ground. The blackbody radiation it sees is just the cosmic background at 3K. Therefore, you cannot assume that the noise temperature of the antenna is the same as its physical temperature. There will be a contribution from its physical temperature, but that is determined by the amount of ohmic resistance in the antenna, NOT its entire feedpoint impedance. If the antenna is resonant (impedance is all real), for example, and has a 50 ohm radiation resistance and a 1 ohm ohmic resistance, then the noise power at the output will be attenuated because you have a 1 ohm noise source driving an unmatched load, namely 100 ohms (antenna plus receiver)) Sbreheny (talk) 06:04, 20 May 2008 (UTC)

I agree. Although I believe my example applies since I made reference to "a closed system at equilibrium." In your example the cosmic background is 2.7K, but if the antenna is 293K then we have an example that is not at equilibrium. BTW, one would be hard pressed to find a 3K night sky on Earth due to atmosphere. I'd hate to step outside with a 3K night sky. Kind of like being in outer space without an insulated spacesuit-- a bit chilly. :) When considering the cosmic background we must use the universe as the closed system. If both the antenna and surrounding ambient temperature is 3K then the Johnson noise from the antennas radiation resistance is Vn=sqrt(4 k 3Kelvin R BW). Although, if the antennas temperature is different than the surrounding ambient temperature then the problem is more complex. This is an interesting topic and reminds me of the "Dew effect," which is essentially all objects pointing up toward a clear night sky will drop in temperature at a faster rate than other objects, and therefore reaching the dew point temperature before other objects thereby forming dew. For example, the top part of a car will form dew before the sides of a car. A flat surface at 300K radiates 459 watts/m^2, and at 200K radiates 91W/m^2. So if such a surface is facing a 200K night sky the object is losing 368 watts per square meter.--PaulLowrance (talk) 14:42, 20 May 2008 (UTC)

I think that you a mixing things up a bit. Radiation is, as I'm sure you know, not the only heat transfer method. Therefore, you can have an antenna which is neither getting hotter or colder but yet "sees" a much colder temperature via radiation. The rest of the thermal input is coming from conduction and convection. The reason why you don't freeze instantly when you step outside is because of the warm blanket of air (as you say) but it is NOT because of radiation from the air, but from convection and conduction. I agree that an antenna pointing at the sky is rarely going to see anything close to 3K, but that is not because of the atmosphere - it is because of cosmic RF sources. The atmosphere's attenuation to some frequencies (for example, around 400MHz) is very low. Since the attenuation is low, this necessarily also means that the noise radiation at that frequency is also low. This is precisely why the "Dew effect" you mention happens. Objects which "see" the ground stay a bit warmer than those which "see" the sky. So, it really is not all that complicated to determine the noise output of an antenna which is pointed at a quiet portion of the sky, in a frequency range where atmospheric attenuation is negligible. It would be equal to the component received from space plus the contribution from the ohmic losses in the antenna. This is often much lower than a noise temperature of 300K (otherwise it wouldn't make sense to make preamps with very low noise figure).Sbreheny (talk) 04:35, 26 May 2008 (UTC)

I think you're misunderstanding what I am saying. I never said you can't have an antenna that is or is not getting hotter or colder. If the antenna is at a different temperature than its surroundings such as the cosmic sky then the system is not in equilibrium. You're using examples that are not at equilibrium. As stated, my reference to the Thermal noise equation was for a closed system at equilibrium, which means if the antenna is at the same temperature as its surrounding then the amount of Thermal noise received from the antenna is equal to the well known equation Vn = sqrt(4 k T R BW). As to why people don't freeze outside in the night sky, we probably don't agree. The cosmic background is 2.7K, but the night sky temperature seen from Earth is significantly higher. I just measured it a few minutes ago where I live and it's 247K. The reason it's not 2.7K on Earth is due to the atmosphere. The thick atmosphere absorbs an appreciable amount of blackbody radiation, and therefore the atmosphere radiates back to Earth an appreciable amount of blackbody radiation. I would agree convection and conduction play a significant role as to what temperature a body would drop to and reach thermal equilibrium. So going back to a real measurement of 247K (-26.2C, -15.1F) night sky, that's not all that cold. Trust me, a person would quickly freeze to death if the night sky from Earth was 2.7K (-270.5C, -454.8F).PaulLowrance (talk) 06:49, 26 May 2008 (UTC)

Hi again Paul. The sky's temperature is frequency dependent. I'm sure that in the IR spectrum, it is not 3K. However, at RF frequencies, it is. Check out [1] top of page 5. Since antennas do not receive thermal noise in the IR spectrum, I was focusing on the RF sky noise temperature. As for the first part, I honestly do not know the technical definition of equilibrium (I had thought it meant "not changing temperature" but perhaps I am wrong). My main concern is that antennas are very rarely surrounded by materials which are true blackbodies. Therefore, antennas will usually NOT output noise equal to the equation you give. You may be strictly speaking correct for thermal equilibrium, but since that situation is very rare with real antennas, I don't think it should be given as a typical example in this article. Sbreheny (talk) 05:33, 28 May 2008 (UTC)

Hi there. Aren't you mixing two of our main discussions together? My reference to the atmosphere was in reference to our discussion of a person freezing if the night sky was 2.7K from Earth perspective, which it is not-- note to my previous mention of the clear night sky measurement out here at 247K (-26.2C, -15.1F). That discussion is relevant to blackbody radiation, which is THz radiation near 30THz-- the atmosphere radiates an significant amount of such radiation. Our other discussion was regarding antennas receiving thermal noise due to blackbody radiation. I was pointing out that your example was not at equilibrium, and therefore the problem is far more complex *if* one wants an accurate answer. Although your example of an antenna on Earth does indeed receive an appreciable amount of thermal noise from the ground, and *can* receive up to the full amount, sqrt(4 k T R BW). One needs to consider ground type and surroundings. Highly reflective grounds drastically decrease the received thermal noise. Now if one wishes to see an antenna receive a lot of thermal noise then place a 3+ element yagi antenna inside a deep canyon. If an antenna is placed deep enough inside a cave then the antenna will receive nearly sqrt(4 K T R BW) of thermal noise. Another example of an antenna receiving near full thermal noise is say a 10 GHz antenna inside a thick walled building. Going back to antennas facing the night sky, one method of drastically decreasing the received thermal noise is a good ground reflective system. Amateurs radio hobbyist often place grids of wires in the ground under the antenna. Such a ground reflective system can also increases the antennas effective gain. Anyhow, yes, it would be nice to add to the article that antennas receive an appreciable amount of thermal noise and can receive the full Vn = sqrt(4 k T R BW) thermal noise. Would you mind if I added a few simple sentences to the article regarding antennas receiving thermal noise? I probably qualify since in college I wrote an antenna simulation program that specialized in Quad antennas.--PaulLowrance (talk) 15:57, 28 May 2008 (UTC)
Hi Paul. Why don't you give the specific example that you want to include here and then I (and others) can comment? I think you are right that we substantially agree and may not be talking about the same thing. Making it more concrete by looking at exactly what you want to add would probably be best.Sbreheny (talk) 15:27, 31 May 2008 (UTC)

Contents

[edit] Noise voltage and power

There are two statements in this section

"The root mean square (rms) of the voltage, \bar v_{n}, is given by..."

"The root mean square (rms) of the voltage, vn, is given by..."

I understand the difference in the formulas (the former is per sqrt(Hz) ), but the terminology is confusing. --agr 11:31, 17 October 2006 (UTC)

The first one is incorrect. Hanspi's diff of today added the formula for volts per root hertz, and his diff comment calls it power spectral density, which is the square root of actually. But he didn't fix the text. Dicklyon 15:22, 17 October 2006 (UTC)

"Although the rms value for thermal noise is well defined, the instantaneous value can only be defined in terms of probability. The instantaneous amplitude of thermal noise has a Gaussian, or normal, distribution." (p. 203)

"The crest factor of a waveform is defined as the ratio of the peak to the rms value." ". . . a crest value of approximately 4 is used for thermal noise." (p. 204) [2]—Preceding unsigned comment added by 70.18.4.75 (talk • contribs)

That would imply a peak that is four standard deviations from the mean, which is not how a Gaussian process works. In fact, the peak factor is essentially infinite for a Gaussian process, or is a random variable with an unboundedly increasing expectation the longer you wait. But stepping back, can you say what question your quotes here were intended to address? Dicklyon 06:43, 21 October 2006 (UTC)

Is the formula for the Johnson-Nyquist noise on capacitors valid for both RC circuits in series and in parallel? 66.31.1.215 18:21, 16 February 2007 (UTC)

Yes, if you use the effective R in both cases, they are equivalent (Norton vs Thevenin models). But if you just use the resistor value and you're hooked up to something with its own impedance that you don't take into account, you'll get the wrong answer. Dicklyon 19:00, 16 February 2007 (UTC)

Thanks for the explanation. I'm sorry though, but I do not understand your answer completely. If I understand you correctly, you say that the effective resistance is the same for both RC circuits. So that would mean you should use the square root of (4kTRf), with that effective resistor as R, right? But is this then, the same as the square root of (kT/C)? Also, I thought Norton and Thevenin only work for resistor circuits, not for RC circuits.

No, I don't mean they have the same effective R. I mean that IF they have the same effective R, then they have the same RC, same noise bandwidth, and same noise. The R comes from a Norton or Thevenin equivalent of the circuit connected to the C. So a series RC connected to a voltage source, or a parallel RC connected to a current source have the same noise. A series RC and a parallel RC connected both to the same source, of whatever impedance, can never have the same effective R if the two circuits have the same R, because the impedance of the source modifies them in different ways. Dicklyon 00:55, 18 February 2007 (UTC)

Ok, I understand it better now. So only if the Norton and/or Thevenin equivalence hold, are they the same and does the R of the RC not contribute to the noise, right?

[edit] Switched Capacitor Filters

This article seems to imply that noise from Caps is equivalent to noise from Resistors. I'm not sure that this is true, given the earlier teachings on the benefits (noise) of switched-capacitor filters. Maybe an explanation of the paradox is necessary. Petersk 16:41, 12 April 2007 (UTC)

Why do you think it's not true? Check any book on switch-cap filter noise analysis. Dicklyon 18:36, 12 April 2007 (UTC)

[edit] KTC noise (reset noise)

KTC noise is not fundamental noise. The source of Thermal noise across a capacitor comes from resistance. The source of KTC noise is easy to understand from .noise Spice analysis. Regardless of the parallel resistance across the cap (it could theoretically be infinite, or 1T ohm, 1p Ohm, etc.), the noise comes from parallel resistance across the capacitor. Kent H. Lundberg properly describes KTC noise, Please see page 10: http://web.mit.edu/klund/www/papers/UNP_noise.pdf —Preceding unsigned comment added by 72.25.73.160 (talk) 01:58, 18 April 2008 (UTC)

Also the KTC noise equation, Vn = sqrt(K T / C), is accurate for pure white noise with infinite bandwidth. For example, consider an amazing 220uF cap that has 100T ohm parallel internal resistance. According to the KTC equation the cap noise should be 4.3nV rms, but in reality such KTC noise would be 4.1nV rms due to the limitation of thermal noise bandwidth, which is ~ 7 THz. With a parallel resistance of 500T ohms the KTC noise is 3.4nV rms. —Preceding unsigned comment added by 72.25.73.160 (talk) 14:37, 18 April 2008 (UTC)

It's not clear what you mean by "not a fundamental noise"; your source says it's just thermal noise, which is true. But it does not say that it comes from a resistance across the capacitor (or if it says that, I can't find it). And it's not clear where the Lundberg paper is published, so not clear that you can even rely on it as a source. Please clarify and point out what you think he says in relation to your assertions. Dicklyon (talk) 15:57, 30 April 2008 (UTC)
Kent H. Lundberg pdf goes over the basic math how kTC noise is derived. Kent starts with the spectral density of the thermal noise of the resistor, which is V2=4kTR. Kent then shows the noise bandwidth of the resistors thermal noise, which is 1/2pi * pi/2RC = 1/4RC. Kent continues, "Thus, the total output noise voltage that is measurable across the capacitor, in volts, is simply \sqrt{V^2 df} = \sqrt{kT/C}" As you can see, the kTC equation is derive from the thermal noise equation and the resistors thermal noise bandwidth-- \sqrt{V^2 df} = \sqrt{4kTR * 1/4RC} = \sqrt{kT/C} The capacitor is acting as a low-pass filter for the caps parallel resistance thermal noise.--PaulLowrance (talk)
Yes, that's the conventional "hard way" to get to kTC, the long way around by assuming a white noise in a resistor. The only trouble is that when it's not clear what the resistance is, or there's an ideal switch instead of a resistor, it's not clear how it applies. One might think that using a 0-resistance switch, opening to infinite resistance, would avoid the noise, if the noise comes from resistance. But, the noise is NOT avoided in this case, since it's not really a consequence of resistance. It's a consequence of electrons in a state variable in a quantum system in thermodynamic equilibrium with an environment at temperature T; the kTC falls out easily from such an analysis, in which it becomes clear that Johnson noise need not be thought of as being caused by resistors. Dicklyon (talk) 05:57, 1 May 2008 (UTC)
You said, "It's a consequence of electrons in a state variable in a quantum system in thermodynamic equilibrium with an environment at temperature T" Yes, but such a system still has resistance. You have referred to an ideal switch across the cap, and such an ideal switch has infinite *resistance." Even a single ideal capacitor is properly modeled with infinite parallel resistance. I am unaware of how to derive the kTC noise voltage equation without using resistance.--PaulLowrance (talk) 17:35, 1 May 2008 (UTC)
It is impossible to make a switch with infinite resistance or a perfect capacitor with infinite parallel resistance. BTW, what do you mean by "that's the conventional 'hard way' to get to kTC"? Can you please show the easy way of deriving Vn = \sqrt{kT/C} ?--PaulLowrance (talk)
You mentioned "One might think that using a 0-resistance switch, opening to infinite resistance, would avoid the noise, if the noise comes from resistance." As you know this is theoretical since an infinite switch is not real, but mathematically there is no problem with this because the resistance cancels out from the noise bandwidth equation and the thermal noise equation. In non-mathematical terms this means the theoretical resistor may be infinite resistance, but the thermal noise is also infinite. What is nice about this conventional method is the use of fundamental equations that apply such as thermal noise and resistance and capacitor filtering work out mathematically on paper and in Spice simulations. In Spice we can take a real capacitor, apply realistic parallel resistance, and arrive at correct results matching the kTC equation. Not to sound repetitive, but it's important to note Spice clearly shows the noise comes from the capacitors parallel resistor.--PaulLowrance (talk)


And kTC noise from opening a switch is DC, not infinite bandwidth. Dicklyon (talk) 15:57, 30 April 2008 (UTC)
Can you please explain this DC kTC noise? Thanks. As far as I know, kTC noise is not DC and comes from thermal noise.--PaulLowrance (talk)
It can be DC when considering the situation after a reset event. The amount of charge remaining on a capacitor when the switch is opened is a random variable that does not change with time after the event. Each time you do it, you get an independent sample; each one is a different DC value; the variance of the distribution is noise. It's a thermal noise, in that it depends on the thermal coming and going of electrons through the switch when the switch is closed; the variance is proportional to absolute temperature, based on the Boltzmann distribution of electron energies. Dicklyon (talk) 20:55, 30 April 2008 (UTC)
I fail to see why you believe such kTC noise is DC. The thermal noise caused by the caps parallel resistance causes the voltage across the cap to vary over time, which is not DC since it is random. If you analyze kTC Spice sims you will see the fluctuating charge across the capacitor is due to the capacitors parallel resistance. I see below you believe kTC is immediate and somehow DC. If that is what you believe then perhaps we should first focus on the follow thread before returning to this thread.--PaulLowrance (talk)
If there's a parallel resistor you'll get an AC noise. That's not the situation I was referring to. I'm talking about what's called kTC noise or "reset noise", which is a DC random variable that comes from opening a switch to a voltage source. Sorry if I got us off track, as I notice that "reset noise" is only one part of what the article section discusses. The point of discussing it, however, is to show that the thermal noise in an RC circuit can be attributed to the capacitor as well as to the resistor; it's just a different point of view; instead of resistance and bandwidth, the charge variance can be directly computed from just the capacitance. The noise is not caused by the resistor; it's caused by the electrons. Dicklyon (talk) 03:58, 1 May 2008 (UTC)
It appears we agree that reset noise is not DC noise since it is caused by non-DC noise, which randomly charges/discharges the capacitor over time.--PaulLowrance (talk) 16:35, 1 May 2008 (UTC)
You have said kTC noise is not due to resistance. Although it appears we agree conventional method derives the kTC equations from resistance and cap bandwidth. Can you show an alternative mathematical method of deriving the kTC noise voltage equation?--PaulLowrance (talk) 16:35, 1 May 2008 (UTC)
Regarding the 2nd equation in your article, charge variance noise. Can we derive this equation without resistance? In your article you derive Qn from the kT/C noise voltage equation, which as I have pointed out is derived from resistance. If true, then both equations are derived from resistance.--PaulLowrance (talk) 16:35, 1 May 2008 (UTC)
I believe the Sarpeshkar et al article has a pretty good explanation toward the end. Have you read it? Dicklyon (talk) 21:45, 1 May 2008 (UTC)
I didn't read the entire book, but read all the his section on kTC noise. BTW, I was asking for an alternative method of deriving kTC noise voltage equation, not an explanation. Perhaps we can end this thread since it's now similar to another thread below.--PaulLowrance (talk) 23:42, 1 May 2008 (UTC)
By the way, the beauty of the kTC approach is that it works the same for the RC circuit as for the reset noise. The variance of charge is kTC, which means the variance of voltage is kT/C. This is a whole lot easier than integrating the response band of a white noise with spectral density determined by a resistor. It's the same physics, just a different viewpoint on it. Dicklyon (talk) 04:10, 1 May 2008 (UTC)
What do you mean by the kTC approach? As I understood, you said Kent H. Lundberg approach was the conventional method, which includes resistance. So far I am unaware of your approach as to how you derive the kTC noise *voltage* equation. Furthermore, as shown in your article the charge variance noise equation is derived from the kTC noise voltage equation, which as I pointed out is derived from resistance and cap bandwidth.--PaulLowrance (talk) 16:44, 1 May 2008 (UTC)
I'm not sure what you're referring to by "your article". Do you mean the Sarpeshkar et al? It recalls that each degree of freedom has mean fluctuation energy kT/2, and from that directly gets the noise voltage on the capacitor, mean square voltage kT/C. The charge variance kTC can be equally easily derived that way; I probably misremembered exactly how they had done it, but in any case resistance is not relevant to the physical argument. Dicklyon (talk) 21:45, 1 May 2008 (UTC)
No, I'm referring to your Wikipedia article. I should have been more specific. You still have not shown mathematically where the noise voltage comes from and I don't see this in the book by Sarpeshkar. You mentioned kT/2, but that is in joules, not volts. The noise voltage I am referring to is shown in your wikipedia article on how to derive Qn. Your article says, "Using the formula for energy on a capacitor (E=1/2*C*V2), noise energy on a capacitor can be seen to also be 1/2*C*(k*T/C), or also kT/2." As you can see, you replace V with the KTC noise voltage equation, which is sqrt(kT/C). So 1/2CV2 = 1/2C(sqrt(kT/C))2 = 1/2C(kT/C) = kT/2. So you're using the kTC noise voltage equation, but you don't show how it's derived. On the other hand, Kent H. Lundberg shows how the KTC noise voltage is derived, from resistance and bandwidth.--PaulLowrance (talk) 23:42, 1 May 2008 (UTC)


When a resistor is involved, the kTC noise is in whatever the bandwidth is. But it's always a charge variance of kTC. Dicklyon (talk) 15:57, 30 April 2008 (UTC)
It's unclear what you're saying. As far as I know, there's always parallel resistance across a capacitor. Note that the transition from a real to perfect capacitor is an increase in parallel resistance. Therefore a theoretically perfect capacitor has infinite parallel resistance.--PaulLowrance (talk)
What I'm saying is that the variance of the charge of a capacitor, when it has been reset to a voltage by closing and opening a switch, is kTC. This is true immediately; it does not depend on waiting for a time constant having to do with the usually very huge parallel leakage resistance; it is true even if that resistance is infinite or missing. That leakage is too small to have any effect, and is not any part of the explanation of kTC noise. It's unclear why you think it's relevant. The more usual explanation is the switch resistance. But it's not necessary, since there are physical explanations that give the right result even if the switch goes ideally from 0 to infinite resistance. Dicklyon (talk) 20:51, 30 April 2008 (UTC)
I'm uncertain why you thought I was talking about cap leakage. Cap leakage is indeed caused by parallel resistance, but I am talking about thermal noise caused by the caps parallel resistance. Anyhow, I don't believe your above description is kTC noise. You said the caps charged voltage changes immediately upon closing or opening a switch. Can you provide a real example so we can analyze this to find the source. Thanks. BTW, I noticed you did not comment on the above math. Do you now see how the kTC noise equation is derived from the caps parallel resistance and thermal noise??"--PaulLowrance (talk)
What parallel cap do you mean? An R across a C would normally be caused a leakage resistance, would it not? For specific situations in which kTC reset noise is important, see these books. Dicklyon (talk) 03:58, 1 May 2008 (UTC)
Again, I was not talking about capacitor leakage. A capacitor has parallel resistance, which does indeed cause leakage, but such parallel resistance is also the cause of thermal noise. As seen in my statements, I was talking about the thermal noise caused by the caps parallel resistance. Can you please show where I even used the term "leakage?"--PaulLowrance (talk) 17:09, 1 May 2008 (UTC)


Here is a book that talks about charge variance kTC being the same as Johnson noise, and from "the Brownian motion of charge carriers in a resistor"; he would have been even more correct to omit "in a resistor", since the same effect is found with an ideal switch. Also, read the cited article by Sarpeshkar et al. (the link works to get a copy), and note on p.27 where they explain how all thermal noises can be attributed to the capacitors and inductors, whose respective voltages and currents are state variable in thermodynamic equilibrium; they unify the concepts of Johnson noise and shot noise by looking carefully at the physics of quantum-mechanical systems. Dicklyon (talk) 05:12, 1 May 2008 (UTC)
Indeed, the book you mention above is correct and clarifies my point that kTC noise comes from resistance, not pure capacitance and/or inductance. As you can see he clearly shows the resistor in kTC noise model. So far I am unaware of how you are going to derive the kTC noise voltage equation without including resistance. BTW, it appears one point you are missing is that an ideal switch is modeled as infinite resistance. Perhaps an easier method of understanding what is occurring in theoretical infinite resistance is to begin with finite resistance and begin increasing the resistance toward infinity. In doing so you will note the thermal noise increases as resistance increases.--PaulLowrance (talk) 17:09, 1 May 2008 (UTC)
Yes, it's correct, but the interpretation that kTC noise "comes from resistance" is still just one possible interpretation. Sarpeshkar's is another, with a simpler tie to the underlying quantum physics, which works even when there is not relevant resistance instead of only in the limit. Dicklyon (talk) 21:45, 1 May 2008 (UTC)
I understand Sarpeshkar offers an "interpretation," but I don't see where he derived V2, the kTC noise voltage. At some level you'll need to include resistance.--PaulLowrance (talk) 23:42, 1 May 2008 (UTC)
The V2 of kT/C follows directly from the mean fluctuation energy kT/2 that any state variable in thermal equilibrium has; he states that; is it not sufficiently clear? No resistance is needed. Dicklyon (talk) 16:09, 4 May 2008 (UTC)

Mike Engelhardt, the creator of LTspice, gave permission to post his email, as follows.

Paul,

Yes, I fully agree with you that kTC noise is due to the resistance(even though the switch resistor specifics don't matter). This is basically lot and parcel that reactances generate no noise to the extent that they have no losses. That is why SPICE is correct not to include *any* noise whatsoever from a capacitor. There simply is none if the capacitor has no loss(resistance mechanism). BTW, LTspice will include noise from a capacitor because it supports series and parallel losses in the cap and those do generate noise. So while LTspice has noise coming from capacitors, its due to the losses in the capacitor, not the capacitance itself.

Now the problem with using SPICE or LTspice with kTC noise is that LTspice is doing a noise computation of the linear circuit. But you can't see any kTC noise in a linear circuit. It is the result of the resistance of the switch as it opens. It occurs during the non-linear operation. It isn't that LTspice is incorrect, it just that it doesn't do non-linear noise analysis.

Feel free to post the above as a quote of private communication with me.

Regards,

--Mike

I understand that's the conventional view. And nobody is claiming that the noise is "generated" the reactance. The noise is generated by discrete electrons behaving thermodynamically. Dicklyon (talk) 16:09, 4 May 2008 (UTC)
You still haven't shown any method of deriving the kTC noise equations without resistance. I agree with Mike that kTC noise does indeed come from resistance. Your description of such noise being generated by discrete electrons behaving thermodynamically is a non-mathematical description of how the noise is generated, and resistance is included in such a process.
The method is as follows: invoke thermodynamics fundamental conclusion that in a system in thermodynamic equilibrium, every state variable has a mean fluctuation energy of kT/2. Equate that to the noise energy of a capacitor, CV2/2, for V being the rms voltage fluctuation. Solve for V. No resistance needs to be invoked. Dicklyon (talk) 03:09, 6 May 2008 (UTC)
Regarding your above text, "for V being the rms voltage fluctuation", that is what I was talking about in our previous discussion. Such rms Voltage is derived from resistance. As you describe, the cap energy is 0.5CV^2. The V is kTC noise voltage, sqrt(kT/C). Now again I'll show how such noise voltage is derived -->
A capacitor has parallel resistance. Even a perfect cap has parallel resistance, infinite. We know that a resistor generates thermal noise voltage. We know that a capacitor is a filter and the resulting bandwidth is 1/4RC. That is fundamental science. In continuing, we know that Thermal noise is Vn = sqrt(4kTRB), and thus the Thermal noise voltage caused by resistance in parallel with the capacitor is Vn = sqrt(4kTR(1/4RC)) = sqrt(kT(1/C)) = sqrt(kT/C). As you can see, we know from fundamental science that a resistor of *any* resistance across a capacitor will generate sqrt(kT/C) noise.
Note that Thermal noise begins to drop near 1 THz. The kTC noise voltage equation is not perfect in a real world. For example, consider an amazing 220uF capacitor that has 100T ohms of parallel resistance. According to the kTC noise voltage equation the noise should be 4.3nV rms, but in reality it is appreciably less due to the limitation of thermal noise bandwidth.--PaulLowrance (talk) 16:20, 6 May 2008 (UTC)
The important thing, however, is that resistance is not needed to capture some of that noise on a capacitor. At any time, the charge on a capacitor being held at voltage V has a variance kTC, and if you cut it off by opening an ideal switch, you're left that with that charge variance, and a corresponding voltage variance. No resistance is needed. Dicklyon (talk) 16:09, 4 May 2008 (UTC)
Again, a theoretical ideal switch has infinite resistance with infinite Thermal noise. Real switches have finite resistance. This issue is simple to model in Spice and one can see where the noise comes from, resistance.
Yes you can model it that way, in the case of finite R. But it's nice to have a more direct connection to the physics, independent of R. Dicklyon (talk) 03:09, 6 May 2008 (UTC)

[edit] Noise at very high frequencies

So resistors emit white light? What's the connection to blackbody radiation? — Omegatron (talk) 00:34, 4 May 2008 (UTC)

Resistors emit blackbody radiation; but that radiation has a completely different spectrum from the electrical noise that is due to discrete charges moving in thermal equilibrium. Dicklyon (talk) 16:03, 4 May 2008 (UTC)
Dicklyon, we are in sync. I clicked "Save page" to see someone else had already edited the page.
Anyhow, it depends what you mean by white light. All objects emit blackbody radiation, and in a sense the amount of such radiation in the visible spectrum would to some degree resemble "white light," but such blackbody radiation in the visible spectrum is so weak that you cannot sense it with your naked eyes. Here's a blackbody radiation calculator,
http://www.spectralcalc.com/blackbody_calculator/blackbody.php
As far as Johnson noise, the normalized power density begins to drop off at around 1THz. See the graph at the bottom of the following pdf,
http://www.claysturner.com/dsp/Johnson-Nyquist%20Noise.pdf
--PaulLowrance (talk) 16:09, 4 May 2008 (UTC)
The "white" spectrum of thermal noise does not resemble the spectrum of blackbody radiation, does it? Dicklyon (talk) 03:05, 6 May 2008 (UTC)
A year or so ago I worked out the comparison between Thermal noise and Blackbody radiation. Both are caused by Thermal energy, but we must remember that Blackbody radiation energy varies relative to frequency-- E = hf. For example, the Thermal noise current in material is appreciably equal at 2KHz and 1HHz, but such noise current @ 2KHz emits twice the blackbody radiation compared to 1KHz noise current. That's why the blackbody radiation spectrum is not flat like Thermal noise. Thermal noise is appreciably flat till near 1THz.--PaulLowrance (talk) 16:20, 6 May 2008 (UTC)