Johnson noise thought experiment

[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.    

Are there other volt-nuts out there that think about this sort of thing?   

[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]

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!! :-)

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]

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.

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 .

Cheers

Alex

[Zeranin]

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.

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 .

Cheers

Alex

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. 

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. 

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. 

[Alex Nikitin]

Oh, if you prefer to play games here, the power created by this current is completely used up to generate that very current so no increase in temperature is possible .

Cheers

Alex

P.S. Wikipedia is a great reference if you know the answer before looking it up there.

[Zeranin]

Oh, if you prefer to play games here, the power created by this current is completely used up to generate that very current so no increase in temperature is possible .

Cheers

Alex

P.S. Wikipedia is a great reference if you know the answer before looking it up there.

I am not playing games at all. I do not know the answer to this apparent paradox.

I don't believe the 'thermal runaway' argument is valid though. The noise power at room temperature is on the order of a picowatt, which won't heat the resistor enough to cause thermal runaway. Put another way, the gain of the positive feedback loop is way too low to 'take off'.

That said, even the miniscule temperature rise from a pW of I^2R Johnson dissipation breaks the second law of thermodynamics.

Cheers

[RIS]

Oh, if you prefer to play games here, the power created by this current is completely used up to generate that very current so no increase in temperature is possible .

Cheers

Alex

P.S. Wikipedia is a great reference if you know the answer before looking it up there.

hooray we just proved that the closed box is always closed box that is really greatest achievement of energy conservation

[Kleinstein]

The power that causes the thermal noise of a resistor comes from the heat in the environment. So the resistor connected to the other resistor in the fridge gets colder. So the transfer of electrical power in the form of noise is just a kind of heat flow going from the hot to the cold side. There is nothing special about that.

Only because the energy is in the form of a small AC current does not meat it is "useful" energy, it is more a form of heat. So to make it useful one would need a kind of rectifier, but the efficiency is limited just the that of the Carnot maschine.  Efficiency could in theory be very good at low temperature, but not perfect.

Much of thermodynamics is about about free energy, but that is Gibbs free energy, not energy for free.

[RIS]

"Much of thermodynamics is about about free energy, but that is Gibbs free energy, not energy for free."

that is bloody ridiculous   that was like sun has to ask permission to shining

[Zeranin]

Oh, if you prefer to play games here, the power created by this current is completely used up to generate that very current so no increase in temperature is possible .

Cheers

Alex

P.S. Wikipedia is a great reference if you know the answer before looking it up there.

hooray we just proved that the closed box is always closed box that is really greatest achievement of energy conservation

Energy conservation, which is demanded by the first law of thermodynamics, is not the issue here. The issue is with the second law, which can be stated in may ways. However, one of the implications of the second law is that in a closed system at constant temperature, temperature differences (such as the increase in temperature of a resistor above it's surroundings) cannot spontaneously happen, but there is nothing in the first law to prevent this happening. The issue is not with energy conservation which is apparently what you are discussing. There is plenty of thermal energy to be had from the surroundings.

We can very easily say that the temperature of the shorted resistor cannot increase because the 2nd law forbids it, but that does not explain the paradox, it creates it.

The fact is that before the resistor is shorted, there is no I^2R dissipation, and the resistor is at the same temperature as it's surroundings. We then short the resistor, and it now dissipates Johnson noise power as given by In^2 R. If one accepts the 2nd law, then how does one explain that this sudden I^2R dissipation does not cause a small rise in temperature? Normally it is universally accepted that an I^2R dissipation leads to an increase in temperature. Can you explain?

Alex suggested tongue in cheek while 'playing games' that :-
...  the power created by this current is completely used up to generate that very current so no increase in temperature is possible .

I'm not sure if Alex meant his explanation to be taken seriously or not, but in my view it's probably about as good as we are going to get. My suspicion is that a detailed and fully satisfactory answer will be very complicated.

 

[RIS]

hooray we just proved that the closed box is always closed box that is really greatest achievement of energy conservation            I was being sarcastic on comment of his
how far are you from Papamoa Bay of Plenty -sory man wrong flag-I mean not wrong only has a few more stars.

[Zeranin]

The power that causes the thermal noise of a resistor comes from the heat in the environment. So the resistor connected to the other resistor in the fridge gets colder. So the transfer of electrical power in the form of noise is just a kind of heat flow going from the hot to the cold side. There is nothing special about that.

Previously I wrote :-

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.

So we are saying similar things, though personally I find the spontaneous flow of electrical energy to be more novel and unfamiliar that the spontaneous flow of heat energy that everyone knows about.

Only because the energy is in the form of a small AC current does not meat it is "useful" energy, it is more a form of heat. So to make it useful one would need a kind of rectifier, but the efficiency is limited just the that of the Carnot maschine.  Efficiency could in theory be very good at low temperature, but not perfect.

Yes, I mentioned in my original posting that the ballgame changes if we can convert the AC Johnson noise to DC. As you say, that would require a rectifier. I'm not sure if you realize that if you could build such a rectifier, then you could break the 2nd Law, because then you could sum the noise power of 2 (or any number) of resistors algebraically, literally converting waste heat at constant temperature into useful electrical power. I don't believe the 2nd Law is about to be broken any time soon, so there must be fundamental (as opposed to practical) reasons why Johnson noise power cannot be efficiently converted to DC.

[CatalinaWOW]

I guess I am having trouble seeing the paradox.  The noise in the warm resistor is a reflection of the random thermal motions of electrons in the resistor.  When you short a warm resistor you are not changing those random thermal motions, you are just eliminating your ability to observe them.  When you connect the warm resistor to a cold one the electrons which happen to exit the warm resistor can now pass through the cold one, and as they bounce through the cold resistor transfer some of their energy.  There is very little difference between this and normal thermal conduction. 

[tszaboo]

If you put a pair of wire in a box, which is warmer, and bring the cables to a box, which is cooler, and place some imaginary resistors in both boxes, the colder box will get warmer. Because the wire will conduct the heat.

Food for thought: Most* materials conduct electricity and heat equally good. If it is a good electricity conductor like copper, it will conduct heat also well. If it is an insulator like glass? Well the oven at home has glass door doesnt it?
^{*please dont start giving examples where this doent apply!}

[RIS]

Zeranin technical side of this is horrible too much effort for very little but intellectual side is stronger than the atomic bomb
In other words Physics is a very good tool if you know how to use it,if you do not know how to use it then you become a citation of what is already verified." you can not get more out "
so If you expand your idea to other more productive fields
then you may become

[AF6LJ]

Now I have a headache, and it seems that my initial assumptions were correct to begin with.

1. you cannot gather power from a random source.
2. the bottle is closed and you only have a finite amount of energy to work with in the first. place.
3. There is no Free lunch.

I'll take my headache and leave now.

[T3sl4co1l]

I believe it has to do with degrees of freedom.

One wire (I'll assume an absolute ground in this case, so strictly speaking there are N+1 wires, one of which is ground) has two degrees of freedom: magnitude and phase (or I and Q phase, or..).  Or analogously in the time domain.  The amount of power delivered by those degrees of freedom is where your noise constant comes from.

If you connect resistors in parallel, you don't double the power, in fact you short it out, because you restrict the degrees of freedom.

If you have a million independent connections between the thermal reservoirs, you could indeed transfer uW instead of pW.

It does seem strange that more connections should make possible a phenomenon that simply using wider connections cannot, but this is well below our regime for intuition; electronic heat capacity is so small under ordinary conditions that it can be completely ignored.

(Note that an EM field of larger size has proportionally more modes available, and therefore black body radiation /does/ scale with size, as we expect.  Maybe you should try putting antennas inside the cold chamber, instead of resistors? )

Tim

[mswhin63]

There are two things that have not been considered. Shorting a resistor produces a lot of current but no power although power would be removed from the resistor into a wire.

A resistor terminating into another resistor of the same value maybe more deliverable of real power.

Second issue and could be the clincher is the AWGN (Additive White Gaussian Noise) response mean value is in the most part is zero. I haven't thought of the issue just a brief glance but this also may impact the paradox.

[John Heath]

I find it interesting that no where in a Johnson noise calculation is the element in the resistor irrelevant? Temperature K , band width but no distinction between tungsten resister , carbon resister or lead resistor.  A generalization could be made that atom properties are secondary to Johnson noise. The one variable that carbon , tungsten and lead would have in common would be the vacuum that permeates the three different elements. Possibly the true nature of Johnson noise is the sound of a vacuum. Virtual particles of different flavors and charges popping in and out somewhat like the hiss heard on a radio when not on a station.

[Zeranin]

If you connect resistors in parallel, you don't double the power, in fact you short it out, because you restrict the degrees of freedom.

As I'm sure you are aware, if you parallel 2 resistors, then the noise power is the same, but you get it at root(2) less voltage, and root(2) more current.

Two physically separate resistors definitely have double the noise power of of one resistor, yet no matter what you with those 2 resistors, you can't harvest more noise power than from one of them.   

[T3sl4co1l]

If you connect resistors in parallel, you don't double the power, in fact you short it out, because you restrict the degrees of freedom.

As I'm sure you are aware, if you parallel 2 resistors, then the noise power is the same, but you get it at root(2) less voltage, and root(2) more current.

Two physically separate resistors definitely have double the noise power of of one resistor, yet no matter what you with those 2 resistors, you can't harvest more noise power than from one of them. 

Er... so you're denying your initial premise?

[T3sl4co1l]

I find it interesting that no where in a Johnson noise calculation is the element in the resistor irrelevant? Temperature K , band width but no distinction between tungsten resister , carbon resister or lead resistor.  A generalization could be made that atom properties are secondary to Johnson noise. The one variable that carbon , tungsten and lead would have in common would be the vacuum that permeates the three different elements. Possibly the true nature of Johnson noise is the sound of a vacuum. Virtual particles of different flavors and charges popping in and out somewhat like the hiss heard on a radio when not on a station.

...

But yeah, Johnson noise arises simply and fundamentally from the presence of free moving electrons.  All metals, by definition, contain free unbound electrons, therefore they all exhibit Johnson noise.

Moreover, anything that acts like a metal (like an "on" MOSFET) will also, though there may be additional sources of noise from other phenomena.

Devices with energy barriers, like semiconductor junctions, or vacuum tube diodes, exhibit shot noise instead, because the electrons are not fully free to move, but must cross a barrier (or not).

Tim

[Zeranin]

If you connect resistors in parallel, you don't double the power, in fact you short it out, because you restrict the degrees of freedom.

As I'm sure you are aware, if you parallel 2 resistors, then the noise power is the same, but you get it at root(2) less voltage, and root(2) more current.

Two physically separate resistors definitely have double the noise power of of one resistor, yet no matter what you with those 2 resistors, you can't harvest more noise power than from one of them. 

Er... so you're denying your initial premise?

No, we are at cross purposes, and it's my fault for presenting two scenarios in the same posting. The original posting spoke of resistors at different temperatures, in which case we can definitely get more electrical power to flow from the hot resistors to the cold resistors just by adding more resistors and interconnecting wires, in theory without limit. That said, while the guy in the fridge can keep adding resistors to keep himself warm, there is nothing he can do AFAIK to harvest useful electrical power from his millions of cold resistors that are being electrically driven from warmer resistors. He will never be able to run his TV set, even with 1E999 resistors hooked up.

I later posed the question of what happens when you short a resistor, and also discussed ‘harvesting’ the Johnson noise power from 2 or more resistors sitting on your bench at room temperature.

As with the guy in the fridge, if you have millions of warm resistors, each with the same noise power, then no matter what we do, we apparently cannot harvest useful electrical noise power either from a single resistor, or any large number of resistors. If we could then the 2nd Law would be broken, but I’m sure that hasn’t stopped many people from trying! That said, as per previous discussion, if you can rectify the noise voltage of a single resistor, with any rectification efficiency greater than zero, then the 2nd Law is busted. In other words, if you can produce any rectifying device, solid-state, metal-oxide or anything else, and connect it either in parallel or series with a hot resistor so as to produce a net DC voltage no matter how small, then the 2nd Law is toast, because you could then connect large numbers of such resistor-rectifier modules in series/parallel combinations to produce useful electrical power from the thermal energy of the resistors and their surroundings.

[T3sl4co1l]

As with the guy in the fridge, if you have millions of warm resistors, each with the same noise power, then no matter what we do, we apparently cannot harvest useful electrical noise power either from a single resistor, or any large number of resistors. If we could then the 2nd Law would be broken, but I’m sure that hasn’t stopped many people from trying!

Well why not?  Evidently the guy stuck in the 'fridge is "harvesting useful power".  He could use it to power a conventional heat engine and generate electrical power (at fairly poor efficiency, but decidedly nonzero amount).

Presuming you could have a rectifying junction that worked on the magnitude of currents and voltages present in thermal noise of a given source resistance: what's stopping that from being unlimited?

There's no contradiction here.

That said, as per previous discussion, if you can rectify the noise voltage of a single resistor, with any rectification efficiency greater than zero, then the 2nd Law is busted. In other words, if you can produce any rectifying device, solid-state, metal-oxide or anything else, and connect it either in parallel or series with a hot resistor so as to produce a net DC voltage no matter how small, then the 2nd Law is toast, because you could then connect large numbers of such resistor-rectifier modules in series/parallel combinations to produce useful electrical power from the thermal energy of the resistors and their surroundings.

The trick, then, is that you can't.

Or, perhaps more importantly: you can't, without the diode junction being in the refrigerator.  Otherwise, guess what?  The diode will have some incremental resistance around its threshold voltage; which will therefore exhibit Johnson noise, of its own accord; it will dither about the threshold and not deliver net current unless the diode junction itself is cooler.

So, once again, we have a heat engine, and all is well with the universe.

It's a presumed semiconductor implementation of the thought experiment:
https://en.wikipedia.org/wiki/Brownian_ratchet
We like to think that, by some intuition, the ratchet mechanism has zero forward force and infinite reverse force required to turn it; but that's clearly not true, as the pawl is pushed up a ramp as the crank turns, and for the pawl to remain in position, it must be spring-loaded, and this exerts a lever force on the crank.  For thermal motion to be able to turn it, the spring must be so slight that, not only will the crank turn by thermal motion, but the pawl will jitter about as well, allowing the crank to skip backwards after all.  The combined effect is no net movement.



Tim