a question about resistors in an arc experiment

Out of _whose_ control, exactly?

Thanks, Rich

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

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

Thank, Rich

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

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

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

John

Obviously a shill for the sunscreen industry.

John

No wonder the 'ware' is soft.

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.

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

John

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.

Seems highly improbable, because it would be a free-energy device.

What is your circuit? I don't have software capable of reading a net list.

Sylvia.

All this seems to do is latch up the opamp with its output railed to ground.

John

Scrawny broads shouldn't join nudist groups ;-)

Neeerp! Not enough "active area" ;-)

...Jim Thompson

Sure you do (any Spice program). You just don't know how to do it, Scrawny Sylvia ;-)

...Jim Thompson

Get LT Spice. It's free.

John

ts

of

I have considered the case. It depends on whether the negative resistance is greater than infinite or less than zero. Unlike positive resistance, there are two types of negative resistance.

That is not correct for this circuit:

----/\\/\\--- ! ! -+--!-\\ ! ! >-----+ -!+/ ! ! ! +---/\\/\\--- ! \\ / \\ ! GND

This circuit is stable with a high impedance input.

Actually yes it is. It follows exactly Ohms law and has a minus sign in it.

I like skinny women. You can have the tubbies.

John

It's not clear to me what it means to say that a resistor has a resistance that is greater than infinite. What is the defined behaviour of such a resistor.

Yes. Which means it's not a correct emulation of a negative resistor, because the latter is not stable with a high impedance source.

It has a characteristic of which Ohms law is a solution, but different stability from a real negative resistor.

Sylvia.