Clamp-on ammeters have a ferromagnetic "clamp" that encloses a wire in a closed magnetic circuit; the magnetic field induced in the ferromagnetic material is proportional to the total net current flux enclosed by the clamp, and you can measure this field precisely with a Hall-effect sensor, thus distinguishing wires carrying a load from wires that don't.
You could use the same approach to build a low-power electronic device that powers itself by leeching energy from the magnetic field around a current-carrying wire, without needing a direct electrical connection. This could enhance safety and reliability and ease installation, since you can "plug in" such a device without making a direct electrical connection, and if the device shorts out, it won't cause an electrical fire. Not only could it harvest power without ever coming in contact with the wire, it could harvest power without even coming near the wire; it need only enclose the wire with a loop of ferrite without also enclosing its return path. Effectively this is a clamp-on transformer, with the "primary winding" of the transformer having only one turn of wire. If that wire is normally carrying, say, 100 amps, then the secondary winding of the transformer with, say, 1000 turns, could draw up to 100 milliamps; but if the total available voltage to be dropped through that one turn is 240 volts, the secondary winding would need to handle 240kV, which is difficult. If we limit the secondary winding voltage to 10V, then the voltage drop on the primary will be an insignificantly small 10mV, and the total power being transmitted can be up to 10mV * 100 A = 1 watt; the secondary winding then can draw up to 1 watt / 10 volts = 100 mA still. That's enough power to allow the use of a very simple power supply (say, a diode, a small capacitor, and a 7805) and a relatively inefficient electronic device. If you don't enclose the wire in ferrite, but simply put a ferrite rod near the wire and perpendicular to it, you'll be able to harvest orders of magnitude less energy, but installation would be even easier. Inductive coupling is only one possible way to harvest energy from power-line fields. Capacitive coupling is also feasible. However, the 60Hz of typical line power presents an extremely difficult problem for capacitive-coupling energy harvesting. If you manage 10pF of capacitive coupling to a power line field, which is probably about all you can hope for, your 1/ωC capacitive reactance is 270 megohms at 60Hz. The other side of your power-supply input is then going to be your capacitive coupling to ground at around 100pF, I think. (Would it be better or worse if you had a wire to ground? Better, I assume.) You're going to need a power supply with an input impedance in the hundreds of megohms in order to be able to take advantage of that. Megohms or tens of megohms is probably feasible with CMOS; hundreds of megohms is probably not. However, all is not lost! Fluorescent lights and high-intensity discharge lights powered by AC produce much-higher-frequency harmonics, some one to two orders of magnitude higher in frequency. This brings the capacitive reactance down to the megohms to tens of megohms range you need. So you could parasitically capture a significant fraction of the line voltage, but only in the high-frequency harmonics produced by the discharge. Another approach, which would be more practical in areas without electrical lines, is to harvest atmospheric electricity, either from actual lightning strikes or directly from the atmospheric voltage gradient. During a lightning storm, atmospheric voltage gradients can reach 100kV/m, only an order of magnitude below air's ionization strength. In clear weather (the "fair weather condition"), it falls three orders of magnitude to some 100V/m. A lightning rod struck by lightning has some 30kA available, but only for tens to hundreds of microseconds. If you erect a ten-meter lightning rod, you could in theory harvest up to a megavolt of the lightning's voltage --- a few megajoules per lightning strike. If you put your lightning rod on a mountain top, you could perhaps get several lightning strikes per month, for a total average power on the order of a watt. Harvesting the energy of a lightning strike, however, seems like a really difficult problem. A 1000:1 step-up transformer as discussed above could reduce your 30kA to some 30A, at the cost of boosting your megavolt to a gigavolt. If you put a series of these transformers along the lightning rod's path to ground, you could drop only a tiny fraction of the voltage through each one, making the situation more manageable. If you can dump this massive amount of power through a low-resistance path into some kind of resonating circuit, you could then store it for milliseconds up to seconds in order to harvest it at a more reasonable pace. Harvesting the atmospheric voltage gradient directly seems much more feasible, and I've heard you can do it as simply as holding up a spent fluorescent light tube in one hand in a thunderstorm. In the absence of a separate source of ions, you need a corona discharge to couple your wire to the atmospheric charge, which means that you need points sharp enough that the electric field intensity at the point is above air's ionization strength. If you're working with, say, only 1000 volts, then you need micron-scale conductive sharp points to generate ionization, and preferably enough of them to support a substantial ionic current. By contrast, if you have 400kV --- say, a two-meter fluorescent tube plus a two-meter-tall person, in a thunderstorm with 100kV/m --- then any conductive point radius below around 40cm will produce a sufficient field, if I remember my electrodynamics correctly. The points of the prongs on the end of the fluorescent tube are on the order of 0.4 mm in radius, so they should work down to about 400 volts. The question, then, is how much current and thus power you can expect to draw at these voltages. It's observed that the fluorescent lamp in the thunderstorm experiment will simply flash periodically as the lamp's parasitic capacitance to ground charges sufficiently to ionize its contents and discharge the capacitance. If I SWAG this, we have a 1ms flash per 5 seconds at 40 watts with a 200V breakdown voltage, giving 40 microamps average charging current; or maybe charging a 1pF parasitic tube capacitance to 200V in 5 seconds, giving about 40 picoamps. This difference of six orders of magnitude suggests that I don't know enough about the problem even to guess. -- To unsubscribe: http://lists.canonical.org/mailman/listinfo/kragen-tol
