Illuminating yourself with 10 kilolux of LEDs to combat seasonal affective disorder

Kragen Javier Sitaker, 2013-05-17 (5 minutes)

Lots of people suffer from "seasonal affective disorder", or just think they'd function better, because they spend a lot of their awake time in environments illuminated at 50-500 lux, which is hundreds of times dimmer than full daylight. 10klux is the low end of full daylight; 32klux is the low end of direct sunlight; 130klux is the high end.

What's the minimal amount of equipment I'd need to illuminate my body (including eyes, but other things just in case) at 10klux?

A lux is a lumen per square meter. At 555nm, the most efficient wavelength, a lux is 1.5 mW/m², so 10klux would be 15W/m². I might have a surface area of about 2m², so I'd need about 30W of light, or 20klm. The most efficient light source is probably an LED; red-orange and green LEDs are around 95 lumens per watt, which means I'd need (20klm/(95 lm/W)) = 210 watts of LED power to reach that. Common LEDs are closer to a third of that efficiency, so I'd need about 630 watts, and about US$600 worth of LEDs. Cree Inc. supposedly has 200lm/W white-light room-temperature LEDs available, but Digi-Key only has up to 153lm/W.

Cree's CXA2530-0000-000N00S40E7 ("Cree XLamp CXA2530 19mm white") LED array, in stock at Digi-Key, costs US$25.50 in quantity 1 and delivers 4643 lm at 1.5 amps at 37 volts, for 84 lm/W; it gets above 100 lm/W at half the current. (That's about 15% efficiency: 150mW/W.) Getting 20klm out of these guys would require five of them, costing US$127.50; they could then deliver 23klm at max current (278W), or 15.5klm at optimal efficiency (148W). Linearly interpolating, that gives you 20klm at 226 watts (+ 148 (/ (- 20000 15500) (/ 1 (/ (- 278 148) (- 23000 15500.0))))).

So you could probably cure seasonal affective disorder for the cost of US$130 plus a three-hundred-watt 37-volt power supply (say, US$50), plus 226 watts. You make a chamber of mirrors with all-transparent furniture, stick the brilliant lightbulbs into it (maybe with some frosted glass to make them less harsh), and go sit in it, naked, for some 16 hours a day. (You could imagine that this might cause some other kinds of disorders related to social interaction, though.) 226 watts for 16 hours a day is 151 watts or 1320 kWh/year, or US$132 per year, assuming a 100% efficent power supply.

Given the high power cost, so it probably makes sense to use six of the Cree 19mm LED arrays instead of five to improve efficiency: the cost for the LEDs jumps from US$127.50 to US$153, but the necessary power drops: if the output light from a single array in the 0.8 to 1.5 amp range is 3095lm plus (/ (- 4643 3095) (- 1.5 0.8)) = 2211 lumens per amp, and 20klm is 2860lm per array, we can interpolate to (+ 0.8 (/ (- (/ 20000 6.0) 3095) 2211)) = 0.91 A per array, or (* 37 6 0.91) = 202 W. That means we can get by with a two-hundred-watt power supply and also reduces our power usage by more than 10%, to 1180 kWh/year, US$14 less. The extra array pays for itself in energy savings in less than two years, not even taking into account the likely parts savings from the smaller power supply, the increased resilience to part failure from having more redundant LEDs, and the lower junction temperature.

From the datasheet, it doesn't look like you can significantly improve efficiency by lowering the current further.

20%-efficient photovoltaic panels covering the roof of a 10-meter by 10-meter house in a zone with 300W/m² average day/night insolation would give you 6kW average. If you were powering your LEDs with photovoltaic panels, you could provide light for about 40 people. Of course, during summer daylight hours, it would be more efficient to just put the daylight inside the house directly; then, instead of 20% efficiency at the panels times 15% efficiency at the LEDs, for a total of 3% efficiency, you'd have more like 80% efficiency. Too bad that only works a small amount of the time.

If you only want to illuminate your eye pupils at 10klux, well, your pupils are about 4mm in diameter, so they have a total area of about 25 millionths of a square meter; a quarter of a lumen would suffice to illuminate them to 10klux. You could get that out of 2.5 milliwatts, a hundred thousand times less than what I'm proposing above.

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