a question about resistors in an arc experiment

gative

Yes, over a band of frequencies they are. At too low of a frequency the winding resistance wins.

Looks like the Thompson retard is getting senile on top of his utter stupidity as well. Maybe it is just excessive lard on the brain.

negative

it? You're a goddamned retard FooseMEAT.

I don't know, Phil... her brain box looks a lot bigger than your does, and she doesn't need Lithium infusions to keep a modicum of civility in public.

Got any of her "troll" posts?

Those are "yet to be programmed in *features*".

Sure; software is a hierarchy of artificial constructs. Electronics is physics. You can't fool hardware.

John

LT Spice lets you place a resistor and assign it a value of -10 ohms. Works great.

Try a negative resistor in parallel with an inductor and a cap. You may have to nudge it a bit to get it going.

John

Why would I need to use LTspice? It's a simple enough circuit, after the polarity issue has been addressed, as pointed out by MooseFET.

A real negative resistance would be absorbing energy from its environment - that is, it would run cold. The proposed circuit wouldn't do that.

Sylvia.

Cool. Tell us something interesting about VLF.

John

Unless, as has been pointed out here, you simulate it.

Uh, stick to software. Electronics and thermodynamics don't seem to work for you.

John

No, Dimmie, I don't admit to "get" you. I'm not AlwaysWrong.

It's the difference between saying that Iraq has "Weapons of Mass Destruction" vs "Weapons of Mass Destruction Related Activities", a rhetorical technique that everyone should be familiar with.

power.

Well, I have several books that address VLF arc transmitters... Moorecroft, Ghirardi, Henney, Drake's. Cool stuff.

Spark gaps have been used in microwave and picosecond pulse applications, too, with some amazing results. The first lidar transmitters were spark gaps.

Most of you guys would rather bitch and guess than get off your butts and actually learn something.

John

Actually, it's not. Feed-forward is generally used to improve line regulation, not load regulation.

Wrong again.

John

My original: . 1 ohm . ___ . .---|___|-. . | | . -1 ohm --> | |\\| | . o--------o--|-\\ | . | >---| . .--|+/ | . | |/| | . | ___ | . o---|___|-' . | 1 ohm . .-. . | |1 ohm . | | . '-' . | . === . GND

(MooseFET version redone since google groups mangled it for me):

. 1 ohm . ___ . .---|___|-. . | | . -1 ohm --> | |\\| | . o--------o--|+\\ | . | >---| . .--|-/ | . | |/| | . | ___ | . o---|___|-' . | 10k . .-. . | |10k . | | . '-' . | . === . GND

I think yours is much more intuitive to grasp, thanks! Also it's interesting that the circuit works the same even with the inputs swapped. That's unusual :)

(For some reason I couldn't quite remember exactly how the "negative resistor" configuration went. I was annoyed at this, so I fiddled around for a bit, on paper, and my one was what I came up with!)

Does it? The rule of thumb that the inputs must be equal might seem to give that result, but consider the effect of a step rise in the input (connected to the -ve input of the op-amp). The voltage on the +ve op-amp input needs to rise to the same value, but because the step rise is to the -ve input, the output of the op-amp will drop, taking the +ve input with it. It seems to me that that configuration is unstable at its operating point, and the op-amp output will head towards one or other of the supply voltages.

Sylvia.