Optikon

Ok so I'm currently studying techniques to model, analyze and reduce electronic circuit noise. I'm still very much an amateur in this area and I have a fundamental question. Can someone please clear up the following argument in noise modeling?

Consider a resistor component and the thermal noise component.

The noise model can be a noiseless resistor in series with a noise voltage source or in parallel with a noise current source. In the case of noise voltage, Et = SQRT(4*k*T*R*B) where B is bandwidth of interest, T is temp, k is boltz const and Et is the mean noise voltage.

Now the question. If one considers a very large R and a very large bandwidth, the noise voltage will be accordingly high. If 2 such resistors are connected in parallel (with no other source present) an intuitive thought would be that a noise current will circulate around them(assume uncorrelated).

This would be incorrect! according to the literature and common sense. This situation does not represent a real power source. The thermal noise is physically caused by thermal agitation of the electrons and that acts against the resistance to generate the noise voltage at the terminals but a NET current flow will not exist into and out of the component.

This I understand..

Now think about what happens to a floating input gate of a high impedance circuit. We can take a high inp impedance circuit like say an electrometer type amplifier, and place our same resistor in series with an input and power the amplifier but leave the input circuit floating. Invariably we all know that noise can cause the amplifier to saturate to one of the rails quite easily or even jostle back and forth between rails.

But for this to be the case, a REAL current (albeit VERY VERY small) must have flowed into the gate of the input circuit. I mean charge can accumulate on the high impedance but, what switches the input will actually be the voltage due to charge and a small current(moving charge) and we say "well the amplifier turned on because of noise on the floating input."

So is a current flowing or not!? Can the thermal energy of the electrons in the resistor actually get bumped out of the component and into the gate in such a quantity that it can turn it on? Or is that charge not part of the charge that enters the gate and these two are somehow distinguishable?

Anyone shed some light???

THANKS!!!!!!!!

williB

I have never studied Noise to this extent..but
my thinking is , it is a very high resistance, which would eliminate any significant current flowing anyway..
example :: when i was working on my stepper motors and put my hand anywhere near the gate inputs of the Mosfets the motors would vibrate indicating that the Fets were turning on from the Noise from my hand .

(later i tied the inputs to ground with a high resistance)

Russlk

As you stated in the beginning, there is a noise voltage associated with all resistors. Voltage is the force that drives current, so there will be a current flow when the resistor is attached to a load. If you connect a resistor Rn across the input of an amplifier which has an input impedance of Rin, the net resistance generating noise is Rn*Rin/(Rn+Rin). The amplifier amplifies this noise and if current is reguired to charge the input capacitance, it flows. But the input capacitance reduces the bandwidth and thereby reduces the noise amplitude.
BTW, you can find a model for pink noise on my website.

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bmcculla

To add to what Russlk said:

I think the problem is really when you have an amplifier with really high input impedance and high gain. So tiny currents, from a high impedance voltage source like your noise resistor, get amplified.

Optikon

But in the case of ONLY one resistor paralleled with another (no voltage source present), a current does NOT flow due to the noise voltage of either resistor. If the two resistors are in thermal equilibrium, no noise current actually flows but a noise _voltage_ exists! so how can this same resistor turn on an amplifier if no actual power can be delivered from it.?

Thanks for the comments!

Optikon

Right of course, but there needs to be an actual tiny current (very small but not ZERO)

So you are saying that a real noise current will flow which is in conflict with the literature.? According to the literature, if one tries to take a large value resistor and try and rectify the noise voltage to charge up say an ideal capacitor, this experiment will fail. It will fail because no net current can flow from the resistor due to the noise voltage only. The noise voltage can exist due to thermal agitation but no current can flow (not even TINY TINY TINY amounts.)

This is what doesnt make sense to me and I'm missing something. I feel that if you can measure this voltage with a high impedance amplifier, then some (albeit very very small maybe Fempto amps whatever) current must flow through the resistor and into the input gate of the Hi-Z measuring device.

If I lay a 10GOhm resistor on my desk and use an electrometer to measure the noise voltage, is not a tiny current flowing? Is the electrometer sourcing this current or is the noise voltage or both?

Now if I lay this resistor down and short it out with another resistor, is a circulating current flowing due to the noise voltage? Never mind measuring this current, is it even flowing at all? I don't think it is (where is the energy coming from to move this current, or develop the noise voltage??) Looks like a violation if a current does flow.

So it seems there is some arguments both ways but we cant have it both ways..I'm missing something! :cry:

Thanks for the comments!!

TheOne

Don't overlook the fact that it is random

Statistics of Thermal Noise
". . . any linearly filtered or amplified thermal noise wave has a Gaussian distribution of instantaneous amplitude."

"Usually the Gaussian amplitude distribution of thermal noise is developed on the basis of a model source containing a very large number of independent generators each of which produces an infinitesimal contribution to the resultant amplitude. For example, in a resistor, each conduction band electron as it is buffeted about produces a random current wave. The total current is then shown to have a Gaussian distribution by the Central Limit Theorem."

Oliver, B. 1965. Thermal and Quantum Noise. Proc. IRE. 53: 436-454.
Also reprinted in: Electrical Noise: Fundamentals and Sources. 1977. M. Gupta, Ed. IEEE Press. 129-148.

Roff

There will be no net current, because, as TheOne said, the noise is random (and has no DC component). Nevertheless, it does have a real rms value, given by the equation, so, with a closed circuit, current (but no net current) must flow.

Optikon

I see....

So this current is also an RMS noise current and how then does this not deliver a small power in the RMS resistor heating sense?

Or does it actually deliver power? The fact that the resistor is not at absolute zero says that the resistor does contain energy(thermal) that might be able to be used.???

bmcculla

I'm sure there is some reason that you can't rectify the current from the resistor noise - the laws of thermodynamics say you can't get energy for free. You could transfer energy through heat flow but that would require a temperature differential - Like a peltier device.

This sounds like Maxwell's Demon. This is a thought problem that claimed to break the laws of thermodynamics. Apparently there is some quantum mechanical reason it doesn't work.

Russlk

I think bmcculla has hit on the answer: current drawn from the resistor will tend to reduce its temperature, but it can regain its temperature from its surroundings.

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Juglenaut

I cought wind of a effect called pyrolytic.

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Your website has been blocked

Oznog

I am intrigued. Indeed laws of thermodynamics say you can only get energy out of a system when you have a hot and cold side. Well, technically it's a matter of us never seeing it happen (and it's never been observed as far as I know), even though "Maxwell's Demon" was a theoretically sound, though realistically impossible, case where it could. Yet resistor noise, being a voltage, sounds realistic.

Certainly this has been considered by far better men than myself, so I don't really expect any earth-shattering results, but what restriction shows up to stop it?

bmcculla

I think the circuit we should consider would be a diode with the two leads connected together. The whole circuit would be at that same temperature so that any power generated would break the laws of thermodynamics. There would be thermal noise in the diode itself in addition to the wire connecting the two leads. With an ideal diode it seems that current should flow around the circuit and create power from nothing. The diode would be Maxwell's demon in the circuit.

There has to be some effect in a real diode that prevents this from working. My intuition sais that it's probably related to why the bandgap voltage from a diode doesn't create current in a circuit - the same reason a thermocouple requires a temperature differential, in addtion to dissimilar metals, to create current. I can't seem to remember it though and google isn't being very helpful tonight.

Come to think of it the diode in our circuit would be just like a TE cooler. the band gap voltage is there but without the temperature diffential no current flows.

Optikon

Yes, this is similar to my original question. I think the most confusing point is undertsanding what mechanisms are responsible for the fact that a voltage can exist but no current flow due to that voltage. i.e. band gap volts but no delta-T and no current. Or NOISE voltage, no delta-T and no current.