a question about resistors in an arc experiment

S Claus · 2009-01-15T03:40:28+00:00

Hi allI was browsing through the book "Analog Circuit Design: Art, Scienceand Personalities" and checked on Chapter 8, titled "True AnalogCircuit Design". On page 59 in this chapter, Tom Hornak writes abouthow he got involved in electronics when he was younger. He explainshis and his friends first hands-on experiment which dealth withelectric arcs as thus:We used a 1kW smoothing iron as a series resistor and large iron nailsas electrodes.I had a dumb question about this, which is that what is the reason forusing a resistor in this type of experiement? I mean, would there notbe more current and therefore a larger arc if no resistor was used?How is the limiting of the current flow (by the resistor) desireable?

M

Mike Monett 17 years ago

John Larkin wrote: [...]

Jagadis Chandra Bose used mm wavelengths up to 60GHz in his experiments in the 1890's. He demonstrated many modern microwave techniques prior to 1897, including polarization, the parabolic reflector, dielectric lens, prisms, microwave absorbers, the cavity radiator, diffraction gratings, the radiating iris, twisted dielectric waveguides to rotate polarization, double-prism attenuator, and the pyramidal electromagnetic horn. He used round, square and rectangular waveguides several years before Rayleigh's

1896 theoretical solution for waveguide modes. He demonstrated an iron point contact semiconducting junction and plotted the V-I curves as a function of contact pressure. Much of his apparatus is still in existence.

Some good articles are:

formatting link

Mike

Vote

J

Jim Thompson 17 years ago

Wasn't he that nasty guy in "Batman" ?:-)

...Jim Thompson

formatting link

| 1962 | I love to cook with wine Sometimes I even put it in the food

Vote

J

John Larkin 17 years ago

Obviously not; there's no such word.

You must really enjoy being wrong, since you do it all the time.

John

Vote

J

John Larkin 17 years ago

Superman.

John

Vote

J

Jasen Betts 17 years ago

Martin Luther ?

Lex Luthor mainly against Superman. but yeah, a DC comics bad guy.

Vote

J

Jim Thompson 17 years ago

Actually not seen either one, just the TV ads ;-)

...Jim Thompson

Vote

R

Richard The Dreaded Libertaria 17 years ago

Out of _whose_ control, exactly?

Thanks, Rich

Vote

R

Richard The Dreaded Libertaria 17 years ago

Well, aren't we little miss prissy-pants.

Sorry we can't all be just like you, John.

Thank, Rich

Vote

J

John Larkin 17 years ago

I'm not prissy. But neither am I impressed by a popular culture that admires "daring, edgy" things like bravely using Those Words, or fart/scat/sexual organ jokes, or finding groups to insult, as if those things took insight or courage. Nowadays, anybody can do that, and too many people do.

Watch the comedy shows on TV, for pete's sake. This is Lenny's legacy.

John

Vote

J

John Larkin 17 years ago

His own, of course. He did kill himself, alone. Do you think he planned that? And if he did, why?

Control, to me, means being to *think* your way to the right things to do, and having the discipline to control your passions enough to follow through. That's exactly what engineering is about.

ODing on morphine is a calibration error.

John

Vote

J

John Larkin 17 years ago

The first Batman movie was excellent, especially the dark Gothic visuals. They went downhill from there.

John

Vote

J

John Larkin 17 years ago

Obviously a shill for the sunscreen industry.

John

Vote

M

Michael A. Terrell 17 years ago

No wonder the 'ware' is soft.

http://improve-usenet.org/index.html aioe.org, Goggle Groups, and Web TV users must request to be white listed, or I will not see your messages. If you have broadband, your ISP may have a NNTP news server included in your account: http://www.usenettools.net/ISP.htm There are two kinds of people on this earth: The crazy, and the insane. The first sign of insanity is denying that you\'re crazy.

Vote

S

Sylvia Else 17 years ago

A negative resistors is no more unstable at its operating point than a positive resistor is. Apply a particular potential to it, and you get a defined current. The only difference between a positive and negative resistor is the direction of the current.

A circuit that doesn't behave that way cannot be called a negative resistor under any circumstance.

Sylvia.

Vote

J

John Larkin 17 years ago

I suppose we could debate "at its operating point." But, connected to nothing, a negative resistor will generate a huge, theoretically infinite [1] voltage across its terminals, whereas a regular resistor just lays there and generates a tiny amount of Johnson noise.

The first circuit at the top is wrong. The second circuit is correct and behaves like a -1 ohm resistor, within its swing and bandwidth limitations. And it will rail if not connected to an external resistor of +1 ohm or less.

John

[1] or "unlimited voltage" to the pedants

John

Vote

S

Sylvia Else 17 years ago

I agree that your version is not stable in that case. But that's not the end of the story. You have to consider what would happen if you connected a high impedance input to a real negative resistor.

With a high impedance current source, any upwards deviation in the current coming out of it is countered by a reduction in the voltage. Connect such a current source to an ordinary resistor, and the result is stable.

But connect it to a negative resistor, and the situation changes. Since the current is coming out of the current source, the voltage on the side of the resistor connected to the current source is lower than the voltage on the side connected to ground. An upwards deviation in the current results in a reduction of the voltage, that is, it becomes even more negative, resulting in a further increase in the current.

In other words a high impedance current source connected to a real negative resistor would also be unstable.

So a purported negative resistor circuit that is stable when connected to a high impedance current source is not a correct emulation of a real negative resistor.

Now, I know you'll want to argue with me ;), but I still think your version of the circuit is correct, and the other version is wrong.

Sylvia.

Vote