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Test jig I need to transfer power from ground level and from a normal 230 V outlet to up to a test jig which is at high potential. I need about 50-100 mW after all losses to power my circuit. Power is too low and voltage too high for a normal transformer to be viable. Power consumption isn't an issue. No production run, just a test jig. I need to run it for a long time so batteries will eventually be depleted.

Your normal mono-, polycrystalline or even CIGS PV panel is around 13-18 % efficient under 1000 W/m^2 sunlight, but what about you shine it with a COTS warm or cold white LED lamp? I didn't find anything useful online so far. Does it drop down to ridiculous levels or can I expect at least 5 % efficiency out of the PV panel?

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  • \$\begingroup\$ The power for my mobile phone charger travels through half a dozen transformers from the 380kV line at the other end of the district, so I wonder what voltages you have that a transformer is not viable, but at the same time you can power LEDs with it \$\endgroup\$ – PlasmaHH Oct 13 '16 at 7:35
  • \$\begingroup\$ @PlasmaHH Sure, but that's a 10 milion USD transformer, the size of a truck. Going down in VA raing will still leave you with bushings and creepage distances the size of the same truck at that voltage. \$\endgroup\$ – winny Oct 13 '16 at 7:57
  • \$\begingroup\$ But how are you going to power an LED with 380kV? \$\endgroup\$ – PlasmaHH Oct 13 '16 at 8:04
  • \$\begingroup\$ @PlasmaHH I'm not. I'm powering it from 230 Vac from the ground and shining the light up onto the test jig which is at potential. \$\endgroup\$ – winny Oct 13 '16 at 8:12
  • \$\begingroup\$ it might be possible to harvest the electrical field, its pretty steep surrounding a 380kV line \$\endgroup\$ – PlasmaHH Oct 13 '16 at 8:14
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A transformer is the way to go. Your claim of "Power is too low and voltage too high for a normal transformer to be viable" makes no sense. Power being low is a good thing. That means the transformer can be smaller, all else held equal. High voltage does need to be considered, but this is done routinely.

Since this is a one-off and you're already looking at very low efficiency solutions, you can make this transformer yourself. Get a long ferrite rod and wind magnet wire around a section at each end, leaving a long gap between the two coils. That won't be very efficient, but will still be better than LEDs and a solar cell.

Figure air is good for about 1 kV per mm. Derate that by half, so 500 V/mm for the gap between the two coils. A 6 inch gap is good for 76 kV, for example. I consider that "high" voltage. Since you haven't given any other spec for voltage other than that it is "high", this meets your requirements.

Note that you can drive the transformer from much higher frequency than line voltage. A few 100 kHz will help transfer more power across the same gap. Or conversely, allow for smaller coils at each end.

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(EDIT: Ah. I think I see better with your updated question. Removed my earlier confusion. So you are trying to provide some power to a circuit that is "far away" in the sense of its high voltage potential and you don't want any galvanic connection. Can you use a fiber optic cable?)

Okay. So the efficiency of photopic lumens is pretty high in these days. I think we are getting somewhere around \$80-90\$ lumens per watt in COTS LED lamps, though the theoretical limit with phosphorescence color mixing is up around 250. In the best case, a \$9\:\textrm{W}\$ LED light will deliver about 12% of the electrical energy into radiant flux. If most of that is near the peak efficiency of human vision, this means about \$650-700\$ lumens. But a solar panel doesn't care in the least little bit about lumens. It cares about radiant flux. And you will get only about \$1\:\textrm{W}\$ after spending \$9\:\textrm{W}\$ of electrical energy. You say you don't care about efficiency. So, that is fine here, I suppose.

Now you want to convert that radiant flux back into electrical energy. Well, the light bulb radiates in all directions, not just at the panel. To get it to radiate totally at the panel itself you will need a parabolic reflector with the lamp at the focal point (which, if the light were an actual point source, would radiate the light out in parallel lines outward.) So you need one of those to get the light onto the panel. (Or you could just build an integrating sphere out of your PV panels, I suppose, and place the light(s) inside that.)

Now, you have a lamp and a parabolic reflector (or integrating sphere) and you have converted each watt of electrical energy into \$\frac{1}{12}\$th of a watt of flux, which is now reaching your panel(s). Let's say your panels are 16% efficient with the wavelengths emitted by the LED lamp (not likely, but let's say it is, just the same.) This means about 2% of your electrical energy gets converted back to electrical energy. And this discounts the converter efficiency after that to make the energy useful to you.

Assuming you have the reflectors for all this, I would hope I could get 1% efficiency. So to get \$100\:\textrm{mW}\$, I'd expect to spend 100 times that, or \$10\:\textrm{W}\$. Which is about one of those COTS bulbs.

Sure. Give it a whirl.

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  • \$\begingroup\$ Sorry about not being clear. Hope the image helps. Fiber is not possible unless you use a bunch of them since your aperture in both ends are large. I don't need meters of air so a normal spot reflector LED would be focused enough on the solar panel. \$\endgroup\$ – winny Oct 13 '16 at 10:38
  • \$\begingroup\$ @winny Yeah. I only just noticed the difficulty of the application. The image did help a bit. Re: fiber and aperatures... sapphire has twice the angle of quartz. So that might be a way to widen things. I was thinking more about a laser or flash lamp to pump down optical power to a specialized high efficiency solar cell. But I take your points. It may be good enough to just use a reflector and LED. And I didn't know about the short distance, either. \$\endgroup\$ – jonk Oct 13 '16 at 10:58
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You may find some other LED (cool white, or blue or even red) has a higher overall efficiency.

And you probably need to arrange multiple LEDs for maximum evenness of illumination to avoid energy loss in the "shaded" part of the solar cell, and possibly reflectors to minimise spill outside the cell area.

But once you factor in the LED's own efficiency, even the best cheap solar cells (currently about 22%, back contact cells like Sunpower Maxeon) won't give you 5%. Jonk's 1% may be realistic, but with care and experimentation you may be able to reach 2% or so.

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  • \$\begingroup\$ I think 1%. The reflector absorbs some and won't direct all the light, anyway. So that will reduce efficiency. The solar panel will probably reflect some of the light where it doesn't get used. The air will have tiny particles that diffuse the parabolic beam. I hadn't accounted for the converter. I was being generous about the 16% the panel would do with the LED's obtuse set of wavelengths. I could go on and on. A good thing is that he won't be overheating his panel, anyway. But if he gets better than 1%, I'd love to hear about it. \$\endgroup\$ – jonk Oct 13 '16 at 10:20
  • \$\begingroup\$ @jonk Yes, I'm curious too, especially if he experiments with different spectral output LEDs. It's potentially a nice experiment with some good basic physics and practical engineering behind it, even if the practical applications are rather ... niche. (Could make a seriously cool contactless phone charger) \$\endgroup\$ – Brian Drummond Oct 13 '16 at 10:32
  • \$\begingroup\$ I hadn't noticed, until just now, that he's trying to power something where he cannot accept any chance at galvanic connection. I had only one case, years ago, where I had to power a circuit via fiber. (Didn't need 100mW, though.) I'd just forgotten about those oddball circumstances that do come up from time to time. There are some 'experimental' solar cells that achieve near 40%. I think Perovskite-on-silicon was used in one case I recall, with a pyramidal surface texture for light trapping and antireflection. \$\endgroup\$ – jonk Oct 13 '16 at 10:54
  • \$\begingroup\$ The efficiency of the LED is a separate issue. I can calculate the radiate power from the luminous efficiency and color temperature but the solar panel is the unknown parameter. \$\endgroup\$ – winny Oct 13 '16 at 13:05

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