a question about resistors in an arc experiment

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.

Why don't you try that and get back to us?

...Jim Thompson

OF COURSE it's unstable at its operating point. It's a negative resistor!

John

Oops, you're right. Sorry for my confusion :(

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A friend lent me a book written by some guy who did his work before tubes were common. He was doing UHF/microwaves with a spark gap. He didn't fully understand what exactly he was doing but he knew that something was resonating with the dimensions of the brass structures he was trying. He thought of it as organ pipes and the like to try to get a grasp on the effects he was seeing. Some of his structures had dimensions around 1/2 inch.

I think she's right Jim...

(If this keeps up I'm going to have to build the ****ing thing.)

It needn't get cold. Yes, I have complained about frost forming on the heatsink of the DC-DC converter design I was working on but that is not the only way the energy can be taken out of the environment. Consider the Dark Emitting Diode. It doesn't get cold because it takes in electromagnetic energy.

In real life, a dynamic microphone connected to a backwards diode based rectifier would also take in energy without getting cold.

to

Unless there's quantum physics involved. :)

The two circuits are quite different in how they act. Yours is only stable when the input impedance is high and mine is only stable when the input impedance is low. For yours, consider a 1mA current source on the input.

With 1mA in the op-amps output must be 1mV lower than the inputs.

Because of the divider, the inputs are at 1/2 the output.

The output is at -2mV and the inputs are at -1mV meaning we do have our minus ohms.

power.

Hertz demonstrated the predicted properties of Maxwell's waves using spark gap transmitters and receivers, in the hundreds of MHz I think.

John

Consider the infinite source impedance case as I did in the reply near here.

My version is not stable with a high impedance input.

All of my switchers that run much above 100% efficiency have that problem.

John

Above version, apply +1V voltage SOURCE to input... UNSTABLE

Above version, apply +1V voltage SOURCE to input... STABLE

See my annotations above.

You're basically playing with a Howland current source/pump. "Stability" is source impedance dependent.

...Jim Thompson

Were you looking for 'lathargic', you retarded twit?

The Cycles Of Time will prove you to be correct.

Only the second circuit is correct, in that it presents a linear negative resistance to the world.

I'd modify the circuit to

to make it more practical and increase headroom/swing. -1 ohm would only be usable over maybe a +-20 mV signal span with a typical opamp, and offset errors would be relatively huge.

John

Lotharian-- followers of German reformer Lex Lothar.

Best regards, Spehro Pefhany