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Johnson noise
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author Johnson noise
555x555
Help me understand...

So I know Johnson noise is proportional to resistance and temperature, and unrelated to other things like voltage across or current through the resistor. But surely there must be more to it than that? I mean...I don't think the noise profile would change if I soldered one end of a resistor to the signal path, and left the other unconnected, would it? Doesn't the air have a finite, but very large resistance? And so the air ought to contribute more significant Johnson noise than any resistor if all there was to it was that little equation...

Anyway, I'm clearly missing some piece of the puzzle here.
guest
if you lifted an input resistor, the noise would definitely go up. but, im pretty sure johnson noise is only valid for a conductor, so the air doesnt really count here. basically, as things get warmer, the electrons move around more, creating more random current. those electrons move through the resistor, and you have V=IR getting larger.
slow_riot
Johnson/Brownian noise is from the quantum waterfall effect, where electrons move erraticly in clumps, therefore there is noise in I, proportional to R which is then heard as V.
555x555
Well I realized one thing I wasn’t thinking of—it’s noise across (V) and through (I) the resistor, not currents magically coming out of one end or a voltage magically being injected. And yes I’d expect random unterminated resistors to add noise, but for other reasons...but maybe this *is* Johnson noise...idk seems like with one end unclamped the voltage has no reference and so that’s why it wouldn’t affect the signal. Like attaching one side of a battery doesn’t affect the voltage at all...

So basically there’s a noise current through anything that is proportional to temperature and not really resistance, which then becomes voltage noise through the IR drop. That’s helpful. So signals based on current level are not really affected...? like the current limiting resistor after a V to I converter isn’t going to add noise proportional to its resistivity. I often wonder how many problems we’d solve if we just used currents as signals everywhere, though I suppose we’d also cause a bunch of other problems we don’t currently have to worry about.
Graham Hinton
555x555 wrote:
So basically there’s a noise current through anything that is proportional to temperature and not really resistance, which then becomes voltage noise through the IR drop. That’s helpful. So signals based on current level are not really affected...?


No. Johnson noise may be modelled as a voltage source in series with an ideal noiseless resistor or a current source in parallel with an ideal noiseless resistor. It's just a model that allows us to predict a result.
neil.johnson
A page I am working on covers noise sources:
https://www.njohnson.co.uk/index.php?menu=2&submenu=2&subsubmenu=8

Neil
555x555
neil.johnson wrote:
A page I am working on covers noise sources:
https://www.njohnson.co.uk/index.php?menu=2&submenu=2&subsubmenu=8


This is great, thanks and let us all know if/when you update the WIP sections!
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