Lighting a LED with ambient RF (was candle)

Why not just use a long vertical antenna and tap into the ES field of the Earth?

Ken Gillmore

How does the transmitter know that a receiver is there, as opposed to the signal passing on into space?

Not really. A mile of wet mud under the line would extract a lot of power, and not be accountable. Line losses will vary with temperature, and they can't know the temperature of every point on the transmission line. And they can only meter to a reasonable fraction of a per cent accuracy, to detect line losses. A kilowatt out of a gigawatt would never be detected... 1 PPM.

Hey, this is cool...

There are huge potential transformers, too.

Yes!

In principle.

You might've shot off a reply before noticing the "impractical" part though.

They don't. I said "tuned", implying the antenna is a net absorber, not just a phase-shifting reflector (i.e., reactive, but unmatched loading).

It does, but the antenna either absorbs or re-radiates it. Actually, if the antenna is resonant, it does that anyway regardless -- this has applications in passive RADAR, of course.

This is a trivial consequence of the boundary conditions imposed by a hunk of metal. Fields don't just pass willy-nilly, they interact in an analytical and in-principle-detectable manner. Whether you can detect ppm or ppt in practice is for the engineers to figure out.

Absolutely!

If a heat source at 500K is radiating into a room at 500K, is it really radiating?

Physics tells us, yes, and the room is likewise radiating. The net balance is zero, so no heat transfer occurs (they are in equilibrium). But that doesn't mean there's nothing there.

What really amazes me is, John doesn't know this.

Tim

Ok, but unless that re-radiation is re-absorbed by the source, it can't tell the difference. The coupling coefficient is very small.

No, it's not. That's a closed system. The only way for the room to be in equilibrium at 500K and the radiator at 500K is if the net power radiated is zero.

On a sunny day (Tue, 09 Oct 2012 14:16:47 -0700) it happened mike wrote in :

If the fence is an open circuit not much power will be delivered. If you take the ends and connect a light bulb maybe some power is delivered. That *could* be the reasoning.

scope

I often find my own posts when I search for stuff. Not that I even post all that much. Must have very specialised interests... (or a specialised way of putting things).

Exactly! I *did* use words like "in principle", "it's up to engineering", and "10^-12" in there. :)

But I didn't say net power radiated. I said radiated, i.e. emitted alone.

All things radiate all the time!

Tim

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Then build it, and show us.

low

Not this month; I have a day job. The big hassle would be stringing up the antenna.

There's little doubt, from the math, that it would work, at least enough to get some visible light out.

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Cat whiskers & Germanium diodes have a very low forward drop.

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Low barrier schottkies are good, too, and have lower series resistance. But it might be better to let the LED rectify its own RF, and have zero diode loss.

The best RF detectors are germanium back diodes, as far as I know the only germanium devices fabbed using modern lithography. Except photodides.

LED capacitances are all over the place, 10 pF to hundreds. A low c part would be easier to use here.

line.

Ok, but where'd the differential power come from? Does it increase the line loss? Or does it merely reroute loss power through the light bulb as it moves into the ground and radiate it as light?

Ok, but so what? All things absorb all the time. In equilibrium, the absorption and radiation cancel. If you have a covered bucket of water, you have evaporation and condensation happening in equilibrium. But there's no LOSS of water.

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its basically an aircored transformer, a load on the secondary, your wire and bulb, will load the primary, the line

-Lasse

Exactly! And just as a sealed container, sitting in the refrigerator, collects condensation on one side (which is actually due to net work, due to a thermal gradient, but it needn't be much), so too, an antenna can detect the nature of the EM waves around it, and anything affecting them (the net power transfer, back and forth, likewise being potentially very small).

Tim

Yes, adding a load (the lightbulb) to the fence increases the power loss from the transmission line. But if the fence had a resistor at the end, such as a damp pole, it will do the same thing. I don't think anyone can claim you were then stealing power. If the damp pole is replaced with the lightbulb what is the difference? Intent, I suppose.

Rick

I get tired of you guys arguing about your personal differences. I know I'm not going to stop it, but maybe I can give it a little reality.

I want to learn about this and am willing to do the experimental stuff. Give me the info on what circuit to use and how long a wire to string, and I'll build it. I'll also need info on how close to the radio station to be and what transmission level is needed.

Rick

There is no damp pole. Your gain. Their loss. No compensation.

On a sunny day (Wed, 10 Oct 2012 13:15:25 -0700) it happened mike wrote in :

Depends on the fence. I assume we are talking about inductive coupling.

=========================== HV lines

------------------- | | | gate | | --------------------

If the fence wires are on poles, and go round like this, replacing 'gate' with 'bulb' will perhaps produce light. No blub no current no light.

If you close the 'gate' short, then current cause some dissipation in the fence wires, how much 'power' is then consumed depends on the coupling and the wire resistance + inductance. Inductance of a large loop of wire can be very big actually I tried that.

I would not expect a lot of power from this whole setup, but for sure a LED or small bulb would work. If you use a fence with many wires and put the turns in series...