How to invisibly provide power to an LED?
September 2, 2006 10:02 AM   Subscribe

How could I have LED's on a pane of glass light up with no visible contacts like This amazing table?

At an Ingo Maurer exhibit this weekend I saw a glass bench and table embedded with white LED's. This is the first thing I've ever seen with LED's that I would ever call elegant or beautiful. However the table and bench are essentially not for sale.

Is it possible for a non-international artist to recreate this effect? I looked very closely at the table and bench and couldn't find any obvious way that the electricity was being transmitted, not even thin traces like the ones you can barely see in transparent LCD displays.
posted by Ookseer to Technology (28 answers total) 13 users marked this as a favorite
"278 white LEDs per table top, emitting light on both sides. 230/125 volts, 48 volts DC. INVISIBLE LIVE PARTS. There are tiny transparent wires embedded between the 2 sheets of glass, connecting the LEDs. The lights can be dimmed and even turned off if you suddenly want your LED table to look like a regular glass table again."

posted by nthdegx at 10:06 AM on September 2, 2006

Response by poster: So, where can I buy myself some "INVISIBLE LIVE PARTS"?
posted by Ookseer at 10:22 AM on September 2, 2006

This looks promising. At least you could buy the translucent material there...I think. Very cool.
posted by alteredcarbon at 10:39 AM on September 2, 2006

Another conceivable way to do it would be to use water channels as your "wires". If you get all the air out, you shouldn't be able to see where the plexi ends and the water starts.
posted by smackfu at 10:51 AM on September 2, 2006

Beleive it or not, this is almost a double.
posted by Terminal Verbosity at 12:28 PM on September 2, 2006

You might do some digging on Indium Tin Oxide conductive glass coatings.
posted by Orb2069 at 1:12 PM on September 2, 2006

I don't think that kind of glass coating could carry the kind of current that would be required. I'm not even sure just how you'd connect the LED lead to such a coating anyway, but even if you did, the current flow in the last couple of millimeters before the LED lead through the coating would be enough to burn it off. (We might be talking as much as a hundred milliamps, maybe even more, and the coating isn't going to be very thick.

Those kinds of coatings work really well for LCDs, but that's because LCDs operate on microamps.
posted by Steven C. Den Beste at 5:20 PM on September 2, 2006

Best answer: orb's prolly right - it's probably some glass plates with patterned ITO and some surface mount LED's. here's what i imagine it looks like in cross-section:

you could probably make this yourself but expect it to put you back several grand.

the highest cost will come from the ITO glass which is expensive and even more so the larger the pieces get.

buy the LED's in bulk from somewhere like digikey and they'll cost you about 50c - a buck each. then you'd need to pattern the ITO to form electrodes. i'd do something like this:

where here the blue and pink areas are what's left of the ITO film (blue is the positive electrode, say, and pink is the negative electrode), and the white area is the clean glass that you've etched away. there's one of the surface mount diodes on there so you can see how they sit.

you can use HCl as an ITO etch (get some at a pool store) and for large-scale stuff like this, scotch tape works fine for patterning a surface if you apply it very cleanly. basically you want to leave exposed the areas that are white in the image above. you might have to experiment with the etch times so get some ITO slides to play with first. but i suspect it'd be very quick, on the order of a few seconds.

when you place the diodes you'll probably want to use some kind of conductive adhesive like silver epoxy (just a tiny, TINY bit) to stick the diodes to the surface. i'd mount them with the circuit live, so that you can see them light up when you know you've got good contact.

once they're all mounted and dried, pour some clear acrylic over the surface if you want an acrylic top, or even another piece of glass to make a sandwich. (might have an issue with bubbles under the glass, but i'm sure there are clever ways around this.)

anyway altogether you're probably looking at a few thousand for the ITO glass, a few hundred bucks for all the LEDs and a few hundred more for the miscellaneous supplies, power supply, and the "it's a table" stuff.

let us know how it turns out!
posted by sergeant sandwich at 5:35 PM on September 2, 2006 [1 favorite]

Sorry, I simply don't believe that such films can carry multiple amps of current. Each LED draws a few tens of milliwatts, and they're all in parallel. You're talking 5-10 amps at least. I think that you've invented the world's most expensive fuse here.
posted by Steven C. Den Beste at 6:30 PM on September 2, 2006

scdb, the operating voltage for an LED is a volt or two. 5-10 amperes at that voltage is something on the order of 10 watts.

ITO films are also used as the top transparent electrode for solar cells, which are basically like an LED operating in reverse. depending on the bandgap of the material the voltage is something also on the order of a few volts, and ITO films carry quite a bit more than 10 W in a solar cell application.

finally, if still "just don't believe" it's possible, how do you propose this table was made otherwise?
posted by sergeant sandwich at 7:02 PM on September 2, 2006

if you still
posted by sergeant sandwich at 7:03 PM on September 2, 2006

Sgt. sandwich - That's pretty clever (And it sounds like you have a lot of experience with this stuff.) - But is there any reason somebody couldn't just buy two plates of coated glass, and then charge one positive, and one negative?

Of course, the real fun would be combining two plates and masking in order to make each LED individually addressable, and hooking it up to some sort of display driver...

Mr. Den Beste - I had based my recommendation on this page stating that 'Circuit Substrates' were one of the applications of ITO glass. I haven't found a ref as to it's load-bearing quality, but I'd think that would be a function of layer thickness.
posted by Orb2069 at 7:50 PM on September 2, 2006

How do I think that table was made? I think they're using wires.

Resistance in a series system is additive. You have to deliver two volts or so to the LEDs, but that doesn't mean the whole system voltage drop is two volts. Whatever resistance the thin film has must be used in the calculation

According to this the thickest film quoted has an resistance of 8-12 ohms on a microscope slide. So just for a thumbnail calculation we'll call it 12 ohms for 10 centimeters.

All the LEDs are wired in parallel, and so from your geometry and a wild ass guess as to the size of the table we figure an average run length of half a meter. That's 600 ohms (plus change for the LEDs). Figure 5 amps needed to feed all the LEDs (another wild ass guess, but we're just trying to get an idea of the magnitude here) then the total voltage over the film is 3000 volts and the power dissipated in the film is 15 kilowatts.

Which is obviously not going to happen. What this really means is that using a film like this you'll never be able to deliver enough current to the LEDs to make them glow. The series resistance is about 4 orders of magnitude too high.
posted by Steven C. Den Beste at 7:54 PM on September 2, 2006

Evidently, it's not the case here, but would it be possible to use some tiny neon lamps and an electric field Tesla coil style?
posted by euphorb at 8:16 PM on September 2, 2006

Good Lord. Tesla coils are dangerous! If you can induce enough current with one to light a bulb, what's it going to do to the first kid with braces who sits down at the table?
posted by Steven C. Den Beste at 8:34 PM on September 2, 2006

...or anyone with a piercing! Yeowch!
posted by Steven C. Den Beste at 8:35 PM on September 2, 2006

Response by poster: Thanks for the informative posts! It is very clearly not wires, I probably should have phrased the question better. At the exhibit we were allowed to touch and sit on these two pieces, so I could get as close as I wanted and there were -no- wires connecting the LED's. So obviously there is some kind of clear coating on the glass. This gives me enough information to start trying things at least.

And a few grand for a table like that is perfectly reasonable. I mean, have you seen the prices at Pottery Barn lately?
posted by Ookseer at 8:36 PM on September 2, 2006


i don't understand your math at all. 3kV? 15 kilowatts?

here's my thinking: model each parallel branch of the circuit as a resistor in series with an LED. rearrange the equation here to get Vneeded = RI + VLED.

say the LED voltage you want is 2V, the current is 20 mA, and the resistance is your 600 ohms. i calculate 14 volts for that circuit, and the amount of power dissipated in the resistor is I2R = about a quarter of a watt.

if i have 200 of them in parallel i still only need 14 volts, but the powers add up and now i'm dissipating 50 watts, distributed across the whole surface of the table, which doesn't strike me as a huge deal. and that's only if every single LED were on its own leg of the circuit.

orb - that was actually my first thought, but an axial LED (or a surface mount component stood up on its end) will emit sideways. it still might work but i think it'd also be a lot harder to put together because you'd have to somehow ensure that the top contacts make an electrical connection to the upper glass plate, which sounds tricky.
posted by sergeant sandwich at 9:24 PM on September 2, 2006


you might also look into conducting polymers like polypyrrole, polyanilines and the like. probably cheaper than ITO.
posted by sergeant sandwich at 9:32 PM on September 2, 2006

If you have 200 of them in parallel, at 20 mA each, you're running 4 amps. 4 amps through 600 ohms is 2400 volts and 9.6 kilowatts.

9.6 kilowatts == "explosion and fire"
posted by Steven C. Den Beste at 9:36 PM on September 2, 2006

The problem with the sergeant's drawing is that he's trying to run all the current through only one face of the table.

The calculations I was doing were based on using relatively narrow paths of the film to carry the current.

If you use both faces, more or less making one the source and the other the sink, with the LEDs in between and connecting to both top and bottom, then you've drastically increased the width of the film and the resistance will plummet.

But I don't know how far. But it would have to drop a hell of a lot, because voltage and power are both linearly proportional to the resistance.
posted by Steven C. Den Beste at 9:43 PM on September 2, 2006

steven please decide whether you are talking about series circuits or parallel circuits. you're right that it'd take 2.4kV to push 4A through a single 600-ohm resistor, but we're talking about two hundred 600-ohm resistors in parallel. that has an equivalent resistance of 3 ohms. 4A2 * 3 ohms = 48 watts.
posted by sergeant sandwich at 9:48 PM on September 2, 2006

If you use both surfaces of the table, then you do have parallel conductors. Especially if you line the edge of both plates with foil so that you have a good solid conductor all around the edge for the interface.
posted by Steven C. Den Beste at 10:04 PM on September 2, 2006

One of the comments on one of the pages the original question links to leads to Sun-Tec, a Swiss company which sells (among other things) plastic sheets containing embedded LEDs (bare dice, it sounds like) and coated with "transparent conductive oxide". It sounds like they're putting one transparent electrode on each side of the sheet, with the LEDs in the middle. They're intended to be laminated between glass exactly like the table in the link.

The Sun-Tec data sheet doesn't mention exactly what oxide they're using, nor do they mention its resistance-per-square.
posted by hattifattener at 1:16 AM on September 3, 2006

Sheet resistivity is measured in ohms per square (ohms/sq). What this means is that a sheet 1 cm x 1 cm will have the same resistance from edge to edge as a sheet 100 cm x 100 cm.

A sheet 1 cm x 100 cm will have 100 times the resistance between the narrow edges because the length is 100 times the width. Conversely, across the wide edges, the resistance will be one-tenth, because the length is one tenth the width.

This is the same principle used in semiconductor design. As you decrease feature size, as long as the length of wires and width of wires decrease by the same ratio, resistance is unchanged. (This principle breaks down below 100 nm due to second order effects like mean free path of electrons, not an issue here.)

The thicker the ITO film, the lower the resistance, but also the lower the transparency. An ITO film with 80% light transmission will be about 10 ohms/sq. An ITO film with 90% light transmission will be about 100 ohms/sq.

If the power source is 48V and all LEDs are wired in parallel, then the resistance for the current path would be in the neighborhood of 4600 ohms to limit current to 10 milliamps in each LED. But this would mean that total power for 278 LEDs would be about 130 watts for each panel. It think the glass panel would get pretty toasty.

A more efficient method would be to combine series and parallel so that series strings of 10 or so LEDs were connected in parallel. This would cut total power down to about 13 watts.

You would have to be very clever in your routing of the conductors so that all paths were exactly the same length. Otherwise some LEDs would be brighter than others. An interleaved anode and cathode scheme such as the sergeant depicted would achieve this. You could also fine tune things by varying the width of the conductors in accord with the ohms/sq principle. This is exactly how the defogger conductors on a car's rear window are designed.
posted by JackFlash at 1:25 AM on September 3, 2006 [1 favorite]

I'll come in as also questioning wtf Steven C. Den Beste is going on about.

Wire the 600-ohm traces and LEDs in parallel, and assuming that these traces can handle the 20 mA, which seems likely (but I haven't checked), the only place you could run into a problem is the power/ground rails. (In sandwich's picture, the verticals.) A power rail the width of the rails connecting the LEDs and connected to the power supply at one end would burn up at that end as 4 amperes flow through it. So you need to make the power/ground traces very big (perhaps impractically so depending on the material), with a nice big copper connection to the power supply.

Perhaps more practically, the table has to be supported, so if a via from the invisible conductor to the bottom of the glass to connect to the table support isn't too hard, just have a lot of them and use metal in the table supports as power/ground rails.

Did you see where the power was connected to the table?

If you wanted to get fancy you could do RGB LEDs for most any color you care. You'd need some sort of controller, rather than just a power supply, and more complicated traces for four-terminal devices.

Consider not buying from Digikey unless you can't find LEDs you like elsewhere - in general they have very wide selection, good service, etc., but higher prices. Since you'd just be buying this one thing in bulk you'd probably save a bit of money shopping around.
posted by TheOnlyCoolTim at 1:48 AM on September 3, 2006

Resistance of a trace depends on the volume resistivity of the material and the length, width, and thickness of the trace. Fix the thickness and material, and you can begin speaking of surface resistivity, which is defined as the ratio of current density across the width of a square, to the voltage applied across the length of a square. In this way, the size of the square is not important, and you can quote resistivity per dimensionless square. To find the resistance of a trace, first find how long the trace is relative to its width - which is the length in units of squares, sort of - and multiply by the resistivity in ohms/square. ITO coatings are available in 8 to 1500 ohms/sq.

The LED pattern on the table is fairly sparse, and each table is about 2' x 6' with 278 LEDs. The electrical contact is probably where the legs meet the glass. Assuming a single layer of conductor, It is very hard to say what trace pattern they might follow to get power to each position on the table. However, given the 48V supply, you might guess that they are arranged in 28 strings of 10 LEDs. If each string of LEDs was connected with it's own unique trace, the maximum length possible would be about 8' (from one leg, out 2' to the edge, and back 6' to the other leg - 96"). The trace width could be pretty wide.. At 2" the maximum trace length would be 48 squares, for 400ohms per trace (at 50mA per LED you need 20V+10xLED voltage) and you dissipate 2.4W per string, or 67W for the table.

Well, 2" wide traces might be impractical (it could get narrower for short lengths without any substantial effect, but still..), and there will be an additional resistance at the interface from LED to trace (because LEDs are very small compared to 2"), but overall there really isn't any problem building the circuit on one layer. You just need a sensible layout.
posted by Chuckles at 2:25 AM on September 3, 2006

Well, that watt estimate is a bit off, I don't actually know what 20V+10xLED voltage is. White LEDs seem to be about 3.6V, which means you would probably have to go to strings of 8 LEDs..

Whatever.. The system seems completely practical.
posted by Chuckles at 2:33 AM on September 3, 2006

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