Couple quick comments/questions...
[EMAIL PROTECTED] wrote:
> Hi,
>
> This email will describe the simplest (as far as I know) method of
> capturing and storing ambient temperature energy. Hopefully those
> wanting to reply could first read the entire email since I'll
> address various possible questions later in this email.
>
> I was hoping at least someone would have answered my previously
> posted question to nail down their stance if they believe it's
> possible to capture and store energy taken from ambient
> temperature. Since nobody posted his or her stance I'll just go
> ahead and post the proof. This could be a fun ride, as debating
> experience shows most people won't be nailed, which allows them to
> weasel out of any situation, which is probably one reason there are
> so many formulations of the 2nd law. There's a well-taken 2nd law
> quote in the physics community by physicist P.W. Bridgman, "There
> are almost as many formulations of the second law as there have been
> discussions of it."
>
> Personally it's not my present goal or interest to focus on the 2nd
> law. Truthfully, there are too many 2nd law formulations, as one
> physicist may adhere to a stricter interpretation than another. My
> only assertion is that energy can be captured from ambient
> temperature, and here is how.
>
> Here is a clear-cut method to demonstrate the assertion. Using a
> low noise high gain amp and oscilloscope view a resistors thermal
> noise. This is an extremely simple task. I would be more than happy
> to provide anyone legitimately interested individual with a simple
> circuits to view such noise. You will see the thermal noise voltage
> fluctuating in a random unpredictable fashion. Guess what, you are
> witnessing a direct conversion from ambient temperature energy to
> battery storage. A capacitor stores energy in the form of electric
> potential. So where's the capacitor you ask. All measuring devices
> from common amps to oscilloscopes have input capacitance.
10x scope probes run around 10-15 pF, IIRC. I think a 100x probe is
rather lower.
Opamp inputs tend to be a lot lower, tho, but still definitely finite
and large enough to have a macroscopic impact on a circuit.
So, yeah, you're seeing a cap charge and discharge, alright...
> If you want more capacitance than simply place a small capacitor
> across the resistor. You will still see the thermal noise voltage,
> but the average rms voltage amplitude will decrease. There's now a
> total of 4 pF if your amp has 2 pF input and you add a 2pF across
> the resistor. Lets say at a given moment you see 10 mV across the
> capacitor. At that moment you could unplug the capacitor to claim
> your energy. LOL, indeed it's a small amount of energy, but it is
> true that you actually captured energy from ambient temperature. If
> you want more energy then simply make more devices.
>
> Please note I am not stating this is your "smoking gun!" This is
> ***MERELY*** to demonstrate the possibility, to let people know it
> is indeed possible!! If you have the money and technology such as
> IBM then it's possible to make trillions of such devices in a small
> area. One device could be a nanometer. One hundred trillion 2 pF
> capacitors at 10 mV each contains 10 mJ's of energy. If memory
> holds true, the human eye in complete darkness can see a flash of
> red focused light of less than 1 nJ. One 780 nm red light photon
> contains just 2.5E-19 J's!
>
> Ten mJ's may not sound like much, but it merely demonstrates that
> you can capture energy from ambient temperature. This is not the
> best method of capturing ambient temperature energy, but again it
> merely proves the assertion.
>
> Again, in the nutshell, a resistor generates thermal voltage
> noise. All measuring devices from common amps to oscilloscopes to
> multimeters always have a certain amount of capacitance. When you
> measured that thermal noise voltage that capacitor in the measuring
> device is charged to that value. You can also add your own capacitor
> across the resistor. Your capacitor would be completely discharged
> before you add it, but at any given moment once the capacitor is
> connected to the resistor their will be a certain charged voltage on
> the capacitor. At any given moment you could unplug the capacitor to
> retain such energy. You could perform the same experiment with an
> inductor since all measuring devices have inductance.
>
> What you do with such energy is your choice. One hundred 2 pF
> capacitors charged to 10 mV is very usable. That's equal to a 200
> farad capacitor charged to 10 mV. You could discharge the cap
> energy to an inductor followed by a quick field collapse to generate
> appreciable amount of voltage across a smaller cap. Or you could
> place a percentage of the caps in series to increase the voltage,
> etc. etc.
>
> Skeptics may wonder just how much energy is required to "unplug" the
> capacitor. There is no theoretical limit.
Right -- if there's a fatal flaw in the scheme, the energy to unplug
the capacitor is _not_ that flaw!
> How much energy does it require to move a nanometer filament a
> fraction of a nanometer? History demonstrates that the amount of
> energy required from an electrical switch has drastically
> decreased. Consider the FET, which on average has roughly 1E+12 ohms
> DC resistance. Sure, the FET has capacitance, but that in itself is
> stored energy. This is akin to how much energy is require to stop an
> object. One might think it requires a lot pressure to stop the
> object. Consider a spinning wheel next to a table. On the table is a
> hollow metal tube welded to the table. To stop the spinning wheel
> one merely needs to slide a metal bar in the hollow tube extending
> out the other end of the hollow tube, which jams in the wheels
> spokes, which abruptly stops the spinning wheel. The only amount of
> energy required to stop the wheel merely depends how much energy was
> required to slide the metal bar to jam the spokes.
>
> On many occasions I've described a device that has far higher
> potential for "free energy" than the aforementioned example. The
> above is to provide a simple undeniable clear-cut example. Of course
> there will always be those who will deny anything that goes against
> their beliefs. A more practical device that requires ***NO***
> energy such as from a switch would be my resistor and LED device.
> The thermal voltage noise from the resistor will generate thermal
> current in the LED. All LED's emit photos at any applied voltage. It
> just turns out the LED is exponentially more efficient above the
> forward voltage level. In such a device the LED would emit more
> photons when connected to a resistor of high resistance.
I still have some problems with this one.
First, an LED typically has a large forward drop (or at least they
used to, I assume that hasn't changed in the last decade or so). If
there's any effect at all, most of it's going to get cut off due to
that big drop.
If rectifying noise is to work, I should think you'd want to use
something (like a Schottky diode) with a very low forward drop.
Second, nearly all your noise is very close to zero volts, and close
to zero volts, diodes are close to linear. They conduct as something
like I = I_s * (exp(x*v) - 1) where "x" is a constant I don't feel
like writing out. This is from a book but the general form is easy
enough to verify in a lab (though tedious). Very close to zero volts,
this formula is very close to
I = I_s * x * v
or, in other words, the diode is (nearly) linear at zero volts. That
suggests that it might leak really badly in an application where
the signal strength is totally minute.
These may just be practical concerns but it's not clear how to get
past them.
I don't know enough about LEDs to answer this additional question: Can
an LED operate "backwards", as a solar cell? This may be a concern at
very low emission rates.
>
> Lets consider photovoltaic cells. Even at room temperature in
> complete darkness (no solar) there are visible light photons
> striking the cell. I calculate a 10 cm x 10 cm common solar cell
> would generate roughly 1E-30 volts. Not much voltage, lol, but
> still something nonetheless. The amount of radiated blackbody
> energy is small in the visible region. Although the FIR region is
> another story. Both sides of a thin sheet of 1m x 1m material
> radiates roughly 920 watts continuously in complete darkness at room
> temperature. Technology is improving, thereby allowing photovoltaic
> cells to capture lower and lower frequencies. A Canadian university
> succeeded in creating a 1355 nm photovoltaic cell! That's only
> 1/11th the wavelength away from the peak 15000 nm 920 watts/m^2
> blackbody 300 K radiation. BTW, blackbody radiation at 1355 nm is
> 2E+18 times greater than visible region of 600 nm. To calculate this
> I compared the radiation from 16667 to 16677 cm^-1, which is
> 3.907E-29 watts to 7380 to 7390 cm^-1, which is 7.499e-11 watts.
>
> University of Toronto in Canada achieves 1355 nm photovoltaic cell:
> http://nanotechweb.org/articles/news/4/1/7/1
>
> Eventually technology will reach the peak 15000 nm region where a
> thin double sided 1m x 1m sheet receives ~920 watts. It's difficult
> for a person to believe they are surrounded by a source "free
> energy" because we don't see such energy with our eyes.
This one still really bugs me. I don't understand solar cells well
enough to know if this could work, but it just seems /wrong/ to me
that a cell at the same temperature as a blackbody could generate
useful electricity from the blackbody's radiation! (Even a solar
cell made in Canada!) :-)
Whatever...
>
>
> Regards,
> Paul Lowrance
>
