Will these LEDs work for throwies?

Question

I'm going to try to build some throwies soon and am ordering some LEDs and other needed things for the task.

It is my understanding that a CR2032 battery is 2.6-3.1 volts.

Well, I have these LEDs picked out.

660 nm wavelength

1.85-2.5V Forward Voltage, at 20mA current

Would this be ok to use without any resistors? I'm ok if LED life is shortened, but I definitely don't want for the LED to just light up and then be burnt out a second later.

@Leon: "LED Throwies are an inexpensive way to add color to any ferromagnetic surface in your neighborhood. A Throwie consists of a lithium battery, a 10mm diffused LED and a rare-earth magnet taped together. Throw it up high and in quantity to impress your friends and city officials." source — stevenvh, Dec 22 '10 at 12:56
Did you consider making improved throwies with photoresistors? This way you'll save some power during daylight. — AndrejaKo, Dec 22 '10 at 13:38
You may be interested in reading up on the spectral response of the human eye or luminosity function. — tyblu, Dec 22 '10 at 15:26

score 8 · Accepted Answer · edited Jun 11 '20 at 15:10

Here is some interesting info on throwies from evilmadscientist.com

[Connecting a 1.7 V LED directly to a CR2032]

Wait-- 107 mA?!--‽

Yes, this is reproducible. (That is to say, we ~~wasted~~ used up another battery just because we didn't believe it either.) But holy cow anyway.

"And they said this was safe?" There are a couple of legitimate concerns here. Lithium coin cells aren't designed to source nearly that much power-- and aren't lithium batteries a fire hazard? And why does my LED-- rated for 25 mA continuous current survive this? I've certainly seen enough LEDs destroyed by overcurrent, and this one was over 25 mA for ten minutes solid. But, and perhaps against my better judgement, I do believe that this actually is safe in practice. With all of the throwies and similar things out there -- don't forget the keychain flashlights -- they just don't seem to be exploding or catching on fire. (Breaking, falling apart, running out of photons, yes-- but those modes of failure are usually not as dangerous.)

[...]

So.... Do you need a resistor? No, not really. As we said, it seems to be reasonably safe without one. Should you use a resistor? Yeah you should, if you want a red, yellow, or orange LED to last more than a day or so. So you can solder a resistor in place, but that somewhat defeats the purpose of easy to assemble throwies.

So to make it not a total pain-in-the rear to add resistors, here's a way to do it without soldering: just twist it.

Source: http://www.evilmadscientist.com/article.php/throw

Please don't just provide a link, it provides little to no information on its own, and worse, if the link dies (fairly common for blogs), your post becomes worthless, and this question is significantly harmed as a result. Copy and quote (and provide attribution) to key parts from the page you link. I copied my post over yours, which I initially overlooked due to it's overwhelming blandness. — Nick T, Dec 22 '10 at 20:58
Enlightening stuff.+1 .This led did not blow and there are measurements to back this up.What about temp extremes? What about unit to unit spreads?What was the resistance of your Amp meter when you measured your LED currents?How far does the LED draw down the battery before it goes out? — Autistic, Apr 14 '16 at 20:49

score 7 · Answer 2 · edited Apr 13 '17 at 12:32

You need a series resistor to limit the LED current. You can find out how to calculate its value here. In this case your problem is that you have little overhead combined with large tolerances in voltages. The voltage difference between Vbat and Vled can vary between 0.1V and 1.25V. A new battery should give you over 3V; I've measured voltages of 3.4V on CR2032s. OTOH, 2.5V for a red LED is rather high, so I would use the 1.85V for my calculation (even that is a bit high!). So, taking 3.1V for the battery and 1.85V for the LED you need a series resistor of (3.1 - 1.85)V / 20 mA = 62 ohm. A E12 value of 56 ohm will do fine.

edit
You probably have seen the battery directly placed between the legs of the LED, no resistor, but this will drain the battery fast.

tyblu · Answer 3 · 2010-12-22T15:18:45.127

Take a peek at the pulsed characteristics for the CR2032. This load may pull down the voltage to 2.3V-2.9V and IR is ~30Ω. Using the "Pulse" data in the 4th graph (bottom right): (2.4V-1.85V)/(20mA)=27.5Ω. Your milage may vary.

Also, looking at the "Spacial Distribution" graph in the LED datasheet, you can see that its viewing angle is only about 20°. If you want it to look brighter despite poor viewing angle (I bet it is often non-ideal when all you're doing is "throwing" them on out-of-reach surfaces), look for an LED with a better viewing angle. While you're at it, you can get high-efficiency ones that only need a few mA for the same lumens. These are often in surface mount packages, so constructing a throwie would be quite different. Examples: SSL-LX203CSRT, LR T68F-U1AA-1-1-Z, LY T68F-T2V2-35-1-Z.

score 1 · Answer 4 · answered Dec 22 '10 at 14:15

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According to the Energizer CR2032 datasheet I'd say the internal resistance of the battery (10 to 40 Ohm) is too low, so you need a series resistor.

answered Dec 22 '10 at 14:15

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score 0 · Answer 5 · answered Dec 22 '10 at 14:10

This is the question I was puzzled myself. In theory, even without resistor it might work as battery and LED have internal resistance, but for lithium batteries it's rather low so most likely LED will operate out of specs (but will probably work if you take a powerful one).

You'd better use series resistor, you may solder SMD one right on LED legs, so it will not increase dimensions. Values to try 10-100 Ohms.

Will these LEDs work for throwies?

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