# If you're good at breadboarding & simulations, please, join the discussion!

• posted

You have that backwards, which surprises nobody.

Absurd.

John

• posted

I hate it when that happens. :-)

• posted

If you use a gyrator, you can get negative resistance thus a frosty resistor.

• posted

No. The term "negative resistance" is usually applied in either one tow cases:

(1) The voltage decreases with increasing current. This is the case for parts like tunnel diodes.

(2) There is a source of power bringing energy into the system. This would be the case for an induction generator.

Neither of these are mysterious.

That is not a negative resistance. That is only a case like a zener where the current suddenly rises.

That is the sort of use of the term "negative resistance" that still says that power is being lost in the device.

temperature.http://ocw.mit.edu/NR/rdonlyres/Nuclear-Engineering/22-616Fall2003/CC...

Simple, yes. Correct, no.

Once ignited, a plasma has less resistance than air. This is no big deal because air has a really high resistance. Look carefully at the other path the current could take. It is through air not wires.

• posted

This is not quite right. You can have negative resistances if you have some other energy input. Consider a shunt connected DC machine. If a steam engine is spinning it, it has a negative resistance.

• posted

Sure you can, Mark ;-)

...Jim Thompson

```--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |```
• posted

Busted!

• posted

I read into carbon nano-tubes. I guess there's some convention of negative resistance I am unaware of. But, that's all it meant to me.

A zener diode then expresses a form of negative resistance when reverse biased.

• posted

Bzzt. Physics 142 at Berkeley.

Hah. You can calculate the resistivity of a plasma, sure. Try the Spitzer resistivity, for example. It's proportional to Z and inversely proportional to T^(3/2). At 100eV, a H plasma has a resistivity comparable to stainless steel. Time dependence gets more complicated.... Bottom line is there are charge carriers that need to diffuse and collide for current to move in a plasma. It's nowhere near a superconductor, otherwise the accelerator guys would be using it in their magnets, rather than liquid He.

To get low resistivity, you need to keep the plasma hot. Thermodynamics dictates that the distribution will relax into a Maxwellian. Even if you get a plasma with low resistivity, how does that help you with this whole energy scheme? What exactly does it have to do with TECs? If you're trying to cheat thermodynamics, you're bound to lose.

• posted

Here, let me help:

1. Electricity. a. of, pertaining to, or characterized by negative electricity. b. indicating a point in a circuit that has a lower potential than that of another point, the current flowing from the point of higher potential to the point of lower potential.

1. Electricity. a. Also called ohmic resistance. a property of a conductor by virtue of which the passage of current is opposed, causing electric energy to be transformed into heat: equal to the voltage across the conductor divided by the current flowing in the conductor: usually measured in ohms. Abbreviation: R

Note all the interesting and strange things a real negative resistor would do (including repeal the laws of thermodynamics).

Diacs have a negative derivative (for a portion of their I-V curve), or "negative INCREMENTAL resistance". Zener (usually avalanche) diodes do not. ...at least until they fail.

```--
Keith```
• posted

What you don't seem to realize is that the more conductive your plasma gets, the less energy gets dumped into it. P = I*V = I*I*R. This is precisely why superconductors are used for channeling amazing amounts of current for power transmission, electromagnets, and the like: they don't heat up (much). So your claims that the plasma becomes a "near superconductor" 1) are inaccurate 2) aren't even conducive to your scheme.

Why don't you put your plasma tube model into SPICE and see what sort of power you get out of it? A 0.0001 ohm resistor might do the trick.

Nobody here will bother to breadboard your circuit. You're relying on some serious misconceptions. Besides, we've all built resonant circuits and analyzed them to death, and now have more pressing concerns than some poor misguided fellow's over-unity boondoggle.

• posted

Someday we will read of your arrest for duping investors. Enjoy your prison stay.

PLONK

Gawwwd! I wish there was a way to killfile on google-groups ;-)

...Jim Thompson

```--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |```
• posted

Jim Thomps>PLONK

...then you'd miss the opportunity to yank Win's chain now that he's posting from there.

• posted

I guess he'll just have to grow up.

...Jim Thompson

```--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |```
• posted

nicely put. OTOH I did get a good laugh when I looked at his circuit.

I saw an even funnier one on JLNaudinlabs.org - a supposedly over-unity smps that was a modified push-pull converter using a center-tapped battery. To be fair, JLN never claimed it was only over-unity, but the "inventor" did.

Cheers Terry

• posted

Hey, where is the feedback loop? If the over-unity crowd sucked back in some of the hot air from the turbine the efficiency will skyrocket!

Dave.

• posted

```--