Ohm's law

On Mon, 16 Nov 2015 07:21:28 -0800, John Larkin Gave us: snip

You just 'shot' the noise figure all to hell one over f times. :-)

It's cubic around zero, and has to have odd symmetry because otherwise if you flipped the resistor around, the nonlinearity would suddenly change sign, which is unphysical for a normal conductor.

The Taylor series expanded around some nonzero bias point will in general have terms of all orders, e.g. (x-1)**3 = x**3 - 3x**2 + 3x - 1.

Cheers

Phil Hobbs

Hmm, Phil I sorta get what you mean. But I don't see any problem with a quadratic term. R = R(1 - a*V) gives, for a*V

I suspect there's significant injection of charge from the end-wires. The high resistivity material (assume it's a metal) gets extra charge carriers, which fills the band, which means there's slightly less mobility of the charge. It equilibrates at some rate, which means that at small current (low terminal voltage) the resistivity is that of a material with the usual carrier concentration, and at higher current there's a different distribution of carrier concentration.

If the field is high enough, some carriers will collide and cause non-equilibrium charge concentration because they ionize (or otherwise promote) low-energy charges that aren't mobile until excited. That gives you Geiger tubes and avalanche diodes, neither of which is an Ohm's-law exemplar.

Yeah, lotsa theories are possible. This was my first hit on a search. (long wrap.?)

As JL said the granular stuff is worse, which sorta makes sense. A little more voltage opens up another little channel here or there. (I mean carbon comp's stink in every other way, so pile on!) (I still use some ~50 ohm carbon comps, cause they are non-magnetic.)

The Barth link says the voltage coef, in wire is "almost" un-measurable. films are different, but maybe a surface thing.

George H.

Carbon comps have a positive d2I/dV2, for sure, but AFAIK no metal has. IBM has used a nonlinear conductivity tester to look for "mouse bites" in MCM wiring for a good 30 years now, and the coefficient is always negative in that application.

Cheers

Phil Hobbs

It's convenient to use "Ohm", and those who know enough to be analysing circuits this way are unlikely to be mislead, so I don't see a problem with it. Still, this usage doesn't support the OP's position.

Sylvia.

I've seen the claim that thinfilm resistors, metal on a substrate, have a positive vc. But they are alloys, not pure metals.

Could be. It's really really small, anyway.

Cheers

Phil Hobbs

Yup, if it's hard to measure you can probably ignore it. (like current noise in FET opamps.... not to be confused with the voltage noise acting through the input capacitance)

George H.

WOW!

ent units.

Sure. But there's no reason to think that the "voltage coefficient" so defi ned is independent of bias. Any nonlinear differentiable function whatsoeve r will have a quadratic term if you Taylor-expand it about some random bias point, just because its second derivative is in general nonzero. That does n't affect the argument. Symmetry properties are a fundamental point in al l of physics, after all.

ut that's

Not unless you genuinely think the behaviour is nondifferentiable at zero. That would _really_ be news.

Cheers

Phil Hobbs

sideways ohms.

NT

Swohms.

Sometimes? The majority of research is bunk

NT

Circuits would be swarmed with swohms.

Sideohms are not sidereal.

NT

On Wed, 18 Nov 2015 13:01:36 -0800 (PST), snipped-for-privacy@gmail.com Gave us:

That's why there was never a hit song called siderealistic pillow.

Sideohms are illegal in Australia.

Slohmans? That's about as sideways as it comes.