Electronics > Metrology

Johnson noise thought experiment

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Zeranin:
We all know about Johnson noise in resistors. It’s real. We can even measure it with a basic low-noise circuit. For a 50 ohm resistor, the voltage noise is ~0.9 nV/rootHz at room temperature. The voltage scales as the root(R), so a 5K resistor has ~9nV/rootHz, and so on. That said, the noise power is the same for every resistor, because as resistance and noise voltage increase, the noise current correspondingly decreases, resulting in the same noise power for any resistance.

Now for the thought experiment. We have a 5K resistor (any resistance will do) at room temperature (RT), and another 5K resistor in a fridge. For simplicity, we’ll assume it’s a very good fridge, at 0 Kelvin inside temperature. In practice it could easily be at the temperature of liquid helium, for example, and the thought experiment would not really be altered, but for simplicity we’ll imaging a fridge producing  very close to 0 Kelvin.

We then electrically connect our RT resistor to our 0 Kelvin resistor. What happens? Presumably, the total noise voltage is 9nV/rootHz, from the 5K RT resistor. Let’s assume a bandwidth of 100 MHz, so the actual noise voltage is 9E-9 x root(1E8) = 90 uV RMS.

The total circuit resistance is 10K, so the current is 90uV/10K = 9nA.

The curious thing (to me) is that apparently electrical power is being delivered from the warm resistor to the cold resistor. The power being delivered is (9nA)^2 x 5K = 4E-13W = 0.4pW. Alternatively, we could note that the 2 resistors form a voltage divider, so half of the noise voltage ends up across the cold resistor, so the power delivered is (45uV)^2/5000 = 4E-13W, the same answer as it must be. This is the maximum power that can be transferred, because we have matched the load resistance (5K) to the source resistance (5K) We would get exactly the same amount of power delivered to the cold resistor independently of the actual resistances. They could both be 5 ohms, 5Kohms or whatever.

The exact amount of power delivered is not important, it’s small, but is real. OK, so I’m sitting inside this fridge and getting cold. That free 0.4 pW is useful, but I would prefer more. What is the limit? A million such resistors would certainly be possible, remembering that there is nothing in the Johnson noise math that says anything about the physical size of each resistor. With sufficient miniaturization, even a million million (1E12) such micro-resistors should be possible, so our 0.4pW becomes 0.4W. At such miniaturization, bandwidth could easily be 10GHz rather than 100Mhz, gaining another factor of x10 in power, for 4 Watts. Presumably, when this experiment is performed, heat has to constantly flow from the hot environment, into the hot-resistor ‘power station’, and is then delivered as useful electrical power to the person living in the fridge.

Is there a theoretical limit to how much electrical power can be harvested via Johnson noise in this way? Or is there some reason that I have missed why this doesn’t work in the first place?

In principle we can apparently harvest significant electrical power in this way But can such power ever be useful for anything other than driving millions of tiny electric heaters, that are themselves at lower temperature than the hot ‘generator’ resistors? Can we generate ‘useful’ electrical power in this way, for example, can we obtain our 4W as 4V at 1A, by series/parallel connection of the millions of wires coming out of the ‘hot-resistor-power-station’? I’m sure that John Heath (or Zeranin) would patiently and happily solder up the 1E12 wires if the result was ‘free power’. Will this work?

Here it seems that Mother Nature is most unkind for when multiple hot resistors are wired in series or parallel, the noise power available from the multiple resistors is no greater than the power available from a single resistor, essentially because of the random nature of the noise voltages, that add in RMS fashion rather than algebraically. So unless we have some method of converting the output of each hot resistor to DC, in which case we can series/parallel successfully, we can never produce a useful amount of thermally generated Johnson electrical power. Bugger! Mother Nature thwarts us yet again. If we were ever successful, we would violate the second Law of Thermodynamics, because we would be able to extract useable electrical power from a constant-temperature hot environment, and that is prohibited by Law. I’m not a Law Breaker myself, but I know that JH would like to be. :)

montemcguire:
Generally, these sorts of problems are easiest to view using some basic thermodynamic concepts. If you imagine this system in some sort of completely isolated "bottle" within which the total energy is constant, a basic idea is that, if one waits long enough, the system will achieve isothermy. What's more important however is that the total contained "heat" or entropy in the system will be constant, since the system is a sealed bottle, and aside form the energy gradients, is in a net energy "stasis" - no net energy appears or is removed, perhaps by chemical reactions etc.

So, when you say that energy is being delivered from the warm resistor to the cold resistor, this is simply the process by which the "system inside of the bottle" achieves isothermy. The net amount of energy in the bottle is constant, and the energy is merely being re-distributed because energy gradients encourage energy flow, and given enough time, these gradients will eventually disappear. However, keeping in mind that a "bottle" encloses our system, there is no net gain or loss of energy, just a simple redistribution of the fixed quantity of energy in the system.

So, no free energy. Thermodynamics: it's the law!! :-)

Zeranin:

--- Quote from: montemcguire on May 24, 2016, 04:59:40 am ---Generally, these sorts of problems are easiest to view using some basic thermodynamic concepts. If you imagine this system in some sort of completely isolated "bottle" within which the total energy is constant, a basic idea is that, if one waits long enough, the system will achieve isothermy. What's more important however is that the total contained "heat" or entropy in the system will be constant, since the system is a sealed bottle, and aside form the energy gradients, is in a net energy "stasis" - no net energy appears or is removed, perhaps by chemical reactions etc.

So, when you say that energy is being delivered from the warm resistor to the cold resistor, this is simply the process by which the "system inside of the bottle" achieves isothermy. The net amount of energy in the bottle is constant, and the energy is merely being re-distributed because energy gradients encourage energy flow, and given enough time, these gradients will eventually disappear. However, keeping in mind that a "bottle" encloses our system, there is no net gain or loss of energy, just a simple redistribution of the fixed quantity of energy in the system.

So, no free energy. Thermodynamics: it's the law!! :-)

--- End quote ---

Yes, I agree with all you say. We are all familiar with the spontaneous flow of heat energy from hot to cold, but what I described was something much more novel and unfamiliar, namely the spontaneous flow of electrical energy from a hot resistor to a cold resistor. Don’t know about you, but I have never heard or read of this mechanism of energy transfer between hotter and cooler environments, so I find it curious and novel. As you say, there is no thermodynamic law being broken here, so no reason I can think of why what I described does not happen, and there must be countless billions of examples of it happening all the time, given the number of resistors in the world, that will often be at different temperatures, and connected in ways that permit flow of current from one to another.

Here is something to think about. We have an unconnected resistor in a constant temperature environment, with the resistor at the same temperature as it’s environment, as we would expect it to be. There is a Johnson noise voltage developed across the ends of this resistor, but no noise current, as the resistor ends are not connected to anything, so there is no I^2R heating.

Now short the resistor, so there will be a noise current of Enoise/R. This noise current is real, and it could be measured. For example, one end of R could be connected to circuit ground, and the other end to the virtual ground input of a trans-impedance amplifier, AKA current-to-voltage converter.  The TIA feedback resistor could be cooled to near zero Kelvin, so contribute negligible additional noise, so we really could measure the noise current flowing through our resistor.

You can guess what I’m going to ask.  Before we shorted it, the resistor was at the same temperature as its surroundings. Now we short it, and it has a current (Inoise) flowing through it, so presumably it now dissipates a power of Inoise^2R. Keep in mind, this is a real current, and we can even measure it if we want.  So we should expect the resistor to heat up slightly, on account of this (admittedly tiny) I^2R heating.

However, such an increase in temperature is specifically prohibited by the Second Law.

Enjoy. :)

Alex Nikitin:

--- Quote from: Zeranin on May 24, 2016, 06:26:59 am ---
You can guess what I’m going to ask.  Before we shorted it, the resistor was at the same temperature as its surroundings. Now we short it, and it has a current (Inoise) flowing through it, so presumably it now dissipates a power of Inoise^2R. Keep in mind, this is a real current, and we can even measure it if we want.  So we should expect the resistor to heat up slightly, on account of this (admittedly tiny) I^2R heating.

However, such an increase in temperature is specifically prohibited by the Second Law.

Enjoy. :)

--- End quote ---

Nah. There is no current through the short. And we can not measure that current (as it is not there  ;) ). An easy way to understand it is to remember that the noise power is the same for the same temperature independent of the resistor value. So there is no power transfer possible between the resistor and another resistor in parallel to it in a thermal equilibrium -  and the "short" is also a resistor, just of a much smaller value. Your trick here is to substitute a real "short" with a virtual one. In this particular case it is not equivalent enough 8) .

Cheers

Alex

Zeranin:

--- Quote from: Alex Nikitin on May 24, 2016, 08:20:21 am ---
--- Quote from: Zeranin on May 24, 2016, 06:26:59 am ---
You can guess what I’m going to ask.  Before we shorted it, the resistor was at the same temperature as its surroundings. Now we short it, and it has a current (Inoise) flowing through it, so presumably it now dissipates a power of Inoise^2R. Keep in mind, this is a real current, and we can even measure it if we want.  So we should expect the resistor to heat up slightly, on account of this (admittedly tiny) I^2R heating.

However, such an increase in temperature is specifically prohibited by the Second Law.

Enjoy. :)

--- End quote ---

Nah. There is no current through the short. And we can not measure that current (as it is not there  ;) ). An easy way to understand it is to remember that the noise power is the same for the same temperature independent of the resistor value. So there is no power transfer possible between the resistor and another resistor in parallel to it in a thermal equilibrium -  and the "short" is also a resistor, just of a much smaller value. Your trick here is to substitute a real "short" with a virtual one. In this particular case it is not equivalent enough 8) .

Cheers

Alex

--- End quote ---

I'm having trouble judging which parts of your posting are serious, and which not.  :-//

Therefore all I can do is reply as if it is all serious.  :-DD

I do agree that the entire paradox is solved if the noise current of a shorted resistor is zero, but I don't think you'll have much support for that proposition among the noise gurus here, and I don't agree either. Johnson noise can be expressed as a voltage or current, and the noise current (with resistor shorted) is given by Ohms's Law as Vnoise/R. Mr. Ohm isn't going to take a holiday just because you want him to.

I didn't look up any references before writing any of this, cos that would be boring and cheating, but Wikpedia is in full agreement with me :-

A resistor in a short circuit dissipates a noise power of P = Vn^2/R

So I have played no 'tricks'. The noise current that flows through the resistor when shorted is absolutely real in every way, and it must be dissipated in the resistor. For those that think I'm just playing games, I'm not. We have a paradox here that needs to be solved.  |O

But it gets worse. A mate of mine at work (who cursed me because now he won't sleep tonight) pointed out that we have a thermal runaway situation here. We short a resistor, so the noise power dissipation (Vn^2/R or In^2 R, same thing) heats it up a tiny amount, so the noise power increases slightly on account of the higher temperature, so it heats up more, and so on. This is of serious concern, so do not leave shorted resistors unattended on your workbench or in the drawer. Hmm. This could explain why sometimes resistors in my circuits suddenly overheat and explode in a puff of smoke for no apparent reason ...

Enjoy.  :)