Figuring out LED specs

I have a bunch of loose LEDs. Some of them are completely clear. Is there anyway that I can figure out the specs for the LEDs? I no longer have the packaging?

i.e. how many volts they use, and their current?

I am afraid if I hook them up to check them out, I will burn them out.

mike c

Reply to
Mike Chambers
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First you have to sort by color. Put the leads directly across a 3V lithium watch battery (no resistor) which will safely light red LEDs, and won't kill them if you reverse-connect them. If it stays dark, stack two lithium cells for 6V; same lack of reverse-voltage concern. Been sorting LEDs this way for some time. As for ultimate current limits per device, well, suck 'em and see.

IRLEDs will be a tad more difficult...

Mark L. Fergerson

Reply to
Mark Fergerson

actually if you put 3V across a red LED, you'll probably fry it. most reds are 1.7V, high brightness are 1.9V. amber, and green are around 2.0-2.1, whites and blues are 3.3-3.6 usually.

as long as you get some current through the LED, u'l be able to tell what color it is. 3V will destroy or damage anything other than blue or white. to be safe, test everything with a 1.5V battery or a well drained 3V. most LEDs wont be full brightness at 1.5V, but 85%-90% will at least glow.

regards, Leon

Reply to
Leon Sorokin

Mike,

You've gotten some conflicting advice here. Just wanted to add another vote to the ones who are saying "put a current-limiting resistor in series with the LED". Do not hook it up directly to a battery or other voltage source without a resistor.

Mark

Reply to
redbelly98

Dumb Luck has always played a large part in my life; I always use batteries from "dead" or otherwise defective watches (I suppose I might have mentioned that), so maybe you're right.

Haven't had any die from testing so far.

Again, Dumb Luck on my part. I'll go with the other suggestion about a current-limited supply and voltmeter if the OP can manage it, but I felt comfortable recommending watch batteries for quick 'n' dirty sorting since I inferred the OP was more interested in not throwing the unknowns away than assembling "proper" test gear. Might be wrong again...

Mark L. Fergerson

Reply to
Mark Fergerson

longer

out.

You've been lucky. Without current limitation (perhaps your battery was weak), pretty much any LED will burn out soon after you exceed the Vf and get it turned on. Granted that a quick hit probably won't do it (I've done dangerous things too ;-) I do find that many LED's are quite durable to this kind of thing, but there are plenty that will just go poof in the blink of an eye.

stack

per

AFAIK, 6V is likely to destroy just about any LED without some kind of current limiting. Most LED's will light dimly if you supply as little as 1 or 2mA.

Reply to
Anthony Fremont

The right way to do this is to send a specific current through the led, and then measure the voltage over the led.

5mA is a safe current.

Use a voltage source which gives significantly more volts than the led.

So a 6 - 12 volt DC source is suitable.

Calculate the resistor you need to put in series with the led to set the current to 5mA.

Assuming a 12 V supply, 12 minus the led voltage which is max 3 or so,

12-3=9Volt.

9V/0.005=1800 = 1k8 resistor.

Ok, so you should connect the power supply 12 V in series with a 1k8 resistor and put the led you want to test in series with the 1k8 resistor.

Then measure the voltage over the led. Check visually if it lights up and in what color.

You have built an "unknown led" tester, from only a battery and a resistor, and a voltmeter.

What happens if you happen to connect the led backwards?

The led has a back voltage and you will find out in a safe way how much it is. The 1k8 resistor limits the current to 5mA in both directions. The led does not work as a lamp when turned the other way around.

So you can try each led twice, once in each direction. That way you find out the polarity of the led too.

If you want to use a higher current, like 10mA, that is okay. Most leds work well at currents up to 20mA.

To raise the current to 10mA you lower the resistor to 1kOhm

--
Roger J.
Reply to
Roger Johansson

That is a 6V lithium battery. It can deliver at least an amp for short periods of time, according to the datasheet:

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Watch batteries are a different thing altogether.

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They are rated to deliver at most 1mA.

--
Regards,
   Robert Monsen
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Reply to
Robert Monsen

Another simple scheme, if you can get a jfet, is to use it as a current limiter:

.--------------. | | | |-+d | | N-Channel JFET | .-g->|-+s | | | | 9V | .-. --- | | | R - | | | | | '-' | | | | '------o | | | .---------. | | | | | Your | | | LEDS | | | | | | | | | | | '---------' | | '--------------' (created by AACircuit v1.28 beta 10/06/04

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The JFET automagically limits the current. Start with a 10k resistor, work your way down to something that makes a normal red LED glow appropriately.

N-JFET with a Vgs(off) of between -1 and -8, like a BF245A would work well. Heck, with that JFET, connecting the gate to the source would limit current to max 6.5mA, since that's the Idss. Using a resistor is a good idea, however, with different JFETs.

--
Regards,
   Robert Monsen
 Click to see the full signature
Reply to
Robert Monsen

For decades, I've had a 9volt battery with a 1K resistor soldered to one terminal. The resistor limits the current, and the lead of the resistor makes it easy to hook up the LED.

It's not only handy to make sure the LED is working (or at least a visible light LED), but handy to get the polarity right.

Michael

Reply to
Michael Black

might

Me too.

Like I said, you've been lucky. I've killed a few small ones by just juicing them for a split second with 5V.

about a

felt

unknowns

The only problem is that there is risk involved when not using a resistor.

have fun

Reply to
Anthony Fremont

Mark mentioned using a lithium. They aren't going to deliver enough current to fry much of anything.

Jon

Reply to
Jonathan Kirwan

Have you looked at the battery specs for a CR2016 or CR2025 or CR2032?

I don't think it would be possible to burn out a red LED with one.

Alkaline? Yup -- that would be bad. But these lithium button batteries used in watches just aren't designed to deliver appreciable currents.

Jon

Reply to
Jonathan Kirwan

If you are willing to use 6 AA alkalines, you can arrange things so that you use a current limited driver:

,-------+---------- VP2 (+) | | | --- ~ LED --- \ / ~ under B1 - --- test

6V --- 4 | - AA +---------- VP2 (-) | | | |/c Q1 +-----| 2N2222 | |\e (NPN) | | --- +---------- VP1 (+) B2 - | 3V --- 2 \ R1 - AA / 120 Ohms | \ | | '-------+---------- VP1 (-)

If you now place the LED into the circuit, it will limit the current to it. Typically, below 19mA. If you now measure the voltage for VP1 (setting your test leads as indicated) you can compute the current through the LED as:

(VP1/120)

If you measure the voltage for VP2 (again, as indicated), you will have already measured the operating voltage of your LED. It's just VP2, at the above computed current.

The 6V of battery B2 should allow this to work for a wide variety of LED colors.

You can adjust the current to something else by changing R1. For example, by using a 100 Ohm for more current or a 180 OHM or 270 Ohm for less current.

You can also achieve something very similar with a single 1.5 AA alkaline battery, using a hand-wired transformer (easily done with a couple of feet of magnet wire on a tiny toroid) and two resistors and a capacitor and a simple

1N4148 diode. If you are interested, I can post that, too.

Jon

Reply to
Jonathan Kirwan

Forgot to mention that it also needs the NPN transistor. Oh, well.

Jon

Reply to
Jonathan Kirwan

3V

LEDs,

was

Honestly I haven't. It makes sense that being able to deliver high current supply wasn't a primary design consideration, long shelf life was.

That may be, they're pretty tough (mostly).

batteries used in

I'm actually kinda surprised that they won't supply much current because I have an SLR 35mm camera that uses a lithium battery (2cr5). Besides physical size, does anyone know what the big difference is that allows it to output, what must amount to, several amps of current when it recharges the flash?

Reply to
Anthony Fremont

The CR2016, according to Energizer's data sheet, is about 38 Ohms rising offscale past 120 Ohms shortly after 70mAh of use. They point out that it can be used in pulse applications and supply about 6.5mA.

Assume that 38 Ohms is fixed, independent of current being supplied (it's not, it gets worse when you pull more), then if your LED needs 2V and the battery is supplying a peak of about 2.9V (about right for most of its life) you will get about 0.9V/38ohms or slightly less than 24mA. Which is about right for 2V on a red LED (the old style red LED is about 1.55V+21*I, which in this case is about

2.05V at 24mA -- about 2V.)

I doubt the CR2016 would supply even that 24mA, though.

Now, if you look at the fatter CR2032, it's all the same. The spec shows about

38 ohms, again. They suggest a slightly higher pulse current as an example, 7.3mA, but the device is otherwise very similar except for its greater energy storage.

None of these things deliver the energy very quickly.

The 2CR5 you mention is NOT a button battery. And it is designed to deliver a

20mA continuous current, an ability to actually provide 1.5A continuous, and with peaks of up to 3.5A!! -- as compared to a button battery's design for about 0.1mA and ... maybe up to 0.5mA. That's quite a healthy difference. Notice that the 2CR5 weighs some 38 grams as compared to about 2 grams for the button batteries.

Comparing those two families is TOTALLY crazy-minded!!

Jon

Reply to
Jonathan Kirwan

Now, do all that with the JFET using only a 1.5V AA battery, two resistors, a capacitor, an 1N4148 diode and a simple, hand-wound toroidal. ;)

V+ V+ V+ | | | | | | | )| .|( (about 50" of | )| T1 |( magnet wire | )| |( for both | + )|. |( windings) --- ,' | - B1 | | D2 R2 --- \ +---|>|---+---/\/\---, - / R1 | 1N4148 | 220 | | \ 2200 | | | | / | | | | | |/c Q1 | --- ~ | '--------| 2N3904 --- C1 \ / ~ | |>e --- 100uF --- LED | (B1 > 1V) | | | | | | | gnd gnd gnd gnd

Jon

Reply to
Jonathan Kirwan

I agree with Jon, some time ago I worked with some Red, white and blue LEDs using CR2016 and CR2032 (very common batteries in LED torches) The batteries won't fry an LED as the batteries can't supply enough current.

It's common to use the CR2032 for red ones, and two CR2016 for white and blue ones, I found though that the last option will end with the LED's life faster.

It's important to remember the LED's are current divices, so I think that the simpliest thing is to use a resistor to limit the current below 20mA.

Reply to
memo_electronico

Here's a circuit for testing LEDs that works very well.

Use a +5 volt source. Any power supply will do the trick.

Connect a 150-ohm resistor in series with the LED.

If you look at an LED from above, you'll note the circular shape; the flat spot indicates the cathode.

So....

Cathode of LED to - terminal of power supply. Anode of LED to 150-ohm resistor. Other end of resistor to + terminal of power supply.

Turn on power supply, and you're in business!

If you get an LED that won't light up, there's a possibility that it's an infra-red LED and its light won't be visible. Some video cameras can see it, though, so try looking at it through one.

Reply to
Matt J. McCullar

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