Circuit/Battery combos for driving a single 5mm LED
January 18, 2011 9:10 AM   Subscribe

I'd like to use a single 5mm UV LED (3.4Vf, 20-25mA) to charge a daub of glow-in-the-dark paint. It only needs to be lit around 1 minute of every 10 (or 6 seconds of every 60), and only when the (room) lights are out. What battery/circuit would be most economical to operate, and require the least-frequent battery changes?

I'm not a circuit-design wiz, but I can RTFM and I can follow directions. I breadboarded someone else's design using an LM555, but it sucked down batteries (and was probably imperfectly implemented on my part).

What battery would be best for long-life (3 large 1.5V cells? a single specialty 4.5V cell?), and what circuit would maximize this? Realistically, what's an order-of-magnatude estimate of the type of battery life I could expect? 1 month? 6 months? More? Less?

The project allows a lot of wiggle room, and if there is a "sweet spot" to be found by taking it into a direction I've not anticipated, I'm all ears. I can answer any questions about anything I've left unclear. Many thanks.

Also: any suggestions of other online communities that frequently discuss things like this?
posted by jjjjjjjijjjjjjj to Technology (5 answers total)
From an energy efficiency point of view, it will be far more efficient to keep a visible LED lit very dimly, perhaps at a moderately high frequency >100hz but extremely low duty cycle. See but you'd need to change some of th e timing values. (Because the light return of GITD paint is pretty darn low, you'll lose a lot of energy charging it.) You could also get a tritium marker which will glow for a decade or so.
posted by seanmpuckett at 9:23 AM on January 18, 2011

Out of order, but:

Any suggestions of other online communities that frequently discuss things like this?

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What battery would be best for long-life (3 large 1.5V cells? a single specialty 4.5V cell?)

To a first-order approximation, battery life is proportional to size & weight. And often cost.

What circuit would maximize this?

There's a bunch of ways of doing it. Efficiency is often proportional to cost.

You could, relatively cheaply and easily, put together a low power microcontroller that reads a phototransistor every once in a while and "integrates" light state, and turns on your UV LED for a while after an adequate period of dark.

I'd pick a 4.5V battery voltage so you don't have to boost or buck, and use FETs to turn on the light sensor circuit and the LED (when you want it on). The Atmel ATtiny13A could do all that, and it's got pretty good idle/sleep power consumption. I doubt you could do a non-computerized version with as sophisticated (and configurable) behavior, that is equally efficient.
posted by spacewrench at 10:12 AM on January 18, 2011

A 555 is the first thing to come to mind, but it's not the only timing game in town. Obviously, you need a CMOS 555... LMC555 .

Also, in timer design, it's not just the T that counts, but the RC. You want the largest resistors possible for this.

At a duty cycle of 10%, and a current drain for the LED of 20 mA, you have basically a 2 mA net current. Add a milliAmp for the timer, and you have a 3 mA load.

An alkaline D cell has about 12,000 mA/hr capacity. Three of them in series have the same capacity, since you are stacking them, but if you REALLY wanted to change batteries infrequently, use 4 and put them in two parallel strings of 2, for a voltage of 3 Volts and a capacity of 24,000 mA/Hrs.

Ballpark life before battery change... about 15,000/3 = 5000 hours. (Not all the energy in the batteries is recoverable.)

I was going to suggest using a constant current driver for your LED, but decided you can get by with the output from the 555. Use a sink drive (i.e., anode of the LED to V+ and a current limiting resistor... ballpark size is ((3V-2V)-.2V)/.020 = 40 ohms. Use a visible light at first to verify timing and brightness, then replace it with the UV part.
posted by FauxScot at 10:22 AM on January 18, 2011

BTW.... the 3V is the supply voltage (nominal), the 2V is the LED forward voltage (ballpark) and the .2V is the Vce sat of the 555 output transistor. The .020 is the 20 mA drive current.

This doesn't give you a lot of headroom on the decay of the battery voltage as they weaken, but the LMC555 is good to 1.5V and the discharge curve of the alkalines is pretty flat.

A smaller battery with higher terminal voltage (3.6V) is the lithium thional chloride battery from Tadiran. It's a little pricier, but available in an AA size and / or D size! You'd have similar operation from one battery. The Tadiran D-cell has 17,000 mA/hr capacity at 3.6V.
posted by FauxScot at 10:29 AM on January 18, 2011

I'm going to take this from a completely different angle.

First, you should take a good look at your paint. Rather than the underperforming zinc sulfides in the days of yore, much better (brighter, longer-lasting) paints exist today. Some of them, if you are not particular about the color, last for twelve hours on a "charge." This is all a function of color, how bright it needs to be, and so forth.

Second, as a safety measure, exactly how ultra is your violet? By which I mean, would this particular LED display in the visible spectrum at all? If not, it is rather easy to burn your retinas. We evolved for blackbody curves wherein our pupils snap shut to protect us and expect that any significant amounts of UV light come with lots of regularly perceivable light as a signal to tighten up. If the UV light is truly beyond visible range to a great degree, you can injure your eyes rather easily.

This is not to say you should not do it, only that you might want to consider putting a cone around your UV LED such that all of the light heads towards the paint and none towards a bystander's eyes. Alternatively, you could "frame" your UV LED with a UV-proof glass or glass-with-coating.
posted by adipocere at 1:01 PM on January 18, 2011

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