a question about resistors in an arc experiment

So what? The current through R4 is not the current being input into the circuit, because some goes into the V+ pin of the Op-amp.

Sylvia.

Crimnany, he was trying to help you out. You were the one that was unfamiliar with simulation software, remember?

The issues are already stated in the text you've quoted above.

Sylvia.

Like anybody here gives a fat flying f*ck who you decide to 'write off'.

No, he does not particularly like me.

JF has pretty much sat idly by for the last couple years in the groups. Maybe a hundred posts, if we do not count all these reply posts once you get him going when you won't admit an error.

He is probably the calmest guy in here. He strikes me as a well read, well spoken old stoner engineer that probably has more on the ball than most folks in here do. Certainly more than your reactionary ass does. You seem to like jumping on bandwagons. Right now it is the krw name calling bullshit bandwagon... you even came up with your own names. I do not think you even know what an open mind is, and that is what has your mind throttled.

And that is dubious ;-)

...Jim Thompson

He's just reading the sign of the R4 current backwards. It starts at

The opamp is railed low, and does nothing.

ftp://jjlarkin.lmi.net/JF.jpg

Look, for example, when V3=10 volts. The R4-R3 junction is at 5 volts. The opamp output is as close to ground as doesn't matter. Current is flowing INTO R4.

John

Analyze JFs "negative resistance" circuit for us.

John

Muzak is uniformly irritating and banal, Phil is frequently amusing and never banal.

Best regards, Spehro Pefhany

Not if you apply Larkin's Power GAIN Theorem ;-)

...Jim Thompson

I never looked specifically at JF's schematic, only one of the original sketches... which definitely had a negative resistance region.

...Jim Thompson

JF's updated proposal was meant to address my objection that the previous one requires a power supply. It attempts to use the input to power the Op-Amp, while purporting to draw a negative current, and thus have a negative total resistance. It doesn't though, which is hardly surprising, because if it did, one could construct a real life version that would provide free energy.

Sylvia.

Quit weaseling and look at the circuit. Can't admit that a "girl" might be right?

John

Damn, you *can't* admit that Sylvia is right.

Funny.

John

you've got R4 installed backwards (which effects the orientation of the current probe) that's why you see a negative trace,

what really happens is, as the voltage decreases so does the current flowing from the supply through the resistor.

To get it to behave as a negative resistor you need to have more voltage on the OP-amp's supply than on the input.

I think I saw this one in AOE.

O | | +------. | | |--' | | | | | .-->|--. | | | | | | | | | | | `--|-->' | | | | | .--| | | `------+ | | O

Phil is almost always right (when on topic) There are other regular participants here with a worse signal to noise ratio.

if there is any capacitance parallel to the negative resistor while it is disconnected, it quickly becomes unstable.

you can enter a negative value for a resistor in ltspice, it makes a handy comparison for the circuits everyone has been posting.