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I went to my local electronics store and asked for a powerful LED because I wanna make a small flashlight. They gave me this thing, but the problem is that the stores in my area don't have any datasheets or anything of the like. When I asked about the current or forward voltage they just told me they don't know and that I should try several resistors and stick to the one that doesn't burn the LED... (basically trial and error because they don't bother with that datasheet stuff...).

So... perhaps could somebody here give me an idea as to what kind of resistor I should use with this thing?

I am using 2 3-volt lithium batteries (CR2016) to provide 6 volts. The guys at the store know that the maximum voltage that the LED can withstand is 3V because they burned 2 LEDs in front of me when I asked about the voltage...

Here are the front and back views of this thing. The metal thingy is supposed to be a dissipator:

Front view Back view

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    \$\begingroup\$ Your local electronics store sounds kinda crappy \$\endgroup\$ – Funkyguy Jul 19 '14 at 20:35
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    \$\begingroup\$ Looks like one of these 1W LEDs. I'd get a 1KΩ pot, attach it in series as a rheostat and start reducing it's resistance while monitoring the temperature. \$\endgroup\$ – Evan Jul 19 '14 at 20:57
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    \$\begingroup\$ Next time, buy a component only if it has a datasheet. No datasheet => No sale. Might also relay this to your store. \$\endgroup\$ – Nick Alexeev Jul 19 '14 at 21:06
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    \$\begingroup\$ Open box, undocumented surplus. Any electronics store that doesn't have some is boring. \$\endgroup\$ – Passerby Jul 19 '14 at 21:44
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    \$\begingroup\$ While it is fair to recommend that electronic parts be purchased only if supported by datasheets, ground reality, especially for hobbyists in geographies like India, is that the choice is between buying whatever is available at electronic part markets, or not attempting the hobby at all. The vast majority of shops in the Lamington Road electronics market (Mumbai, India) for instance, are run by traders who don't have the slightest interest in the specifics of what they sell, let alone maintain part numbers and datasheets. Yet they sell thousands of parts a day each. All countries aren't equal. \$\endgroup\$ – Anindo Ghosh Jul 20 '14 at 9:52
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I would say you want to get a handful of the LEDs and test them to destruction.

You will need to get or build a variable constant current source. This can be made with an LM317 or similar adjustable regulator - google for LM317 constant current. The current resistor should be a wire wound low ohmage rheostat ideally to take the current, plus a tiny (though high wattage) zero-stop resistor.

Power the LED with the constant current source set to its minimum and monitor the current value with a DMM.

Slowly increase the current until the LED starts to illuminate. Measure the voltage across the LED for your \$V_F\$. Slowly increase the current more and more, noting down the current value with each adjustment.

When the LED burns out the last current reading you took will be the absolute maximum current for the LED. Do not, I repeat do not run the LED at that current, as you are right on the cusp of burning it out.

Get a second LED and replace the blown one. Now adjust the current to just below where it burned out before, and this time slowly reduce the current down until you see the LED visibly reduce in brightness. This is the current that you don't really want to drop below as the LED won't be properly bright.

Now pick a value roughly half way between the two and calculate a resistor (or better yet a constant current source or sink) accordingly.

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Without a datasheet answering your question is not difficult, but impossible.

Forward voltage may well be in the 3V range, forward current may be about anything between 50mA and 1A (not much more than 1A I guess).

To avoid burning your LED you should use a bench power supply, regulate it on about 5V and limit the current to 10mA or so, connect the LED and, measuring the voltage, slowly tune up the current. When voltage starts to rise slowly, well that's where your led should work (a little less maybe).

Assuming you don't have a bench supply you should assume 3V of forward voltage drop, then use the formula: $$R_{LED}=\frac{V_{CC}-V_{LED}}{I_{LED}}$$ where:

  • \$V_{CC}\$ is your source voltage, i.e. 6V
  • \$V_{LED}=3\$V
  • \$I_{LED}\$ should start from some 50mA and increase to a maximum that you must discover

For example, for \$I_{LED}=50\$mA you should grab a \$60\Omega\$ resistor (it does not exist actually), make your circuit and monitor the LED. If it does not get much hot and you want to increase luminositi, go for a higher current changing the resistor.

Please note that the resistor might dissipate quite a lot of power, the formula is the old \$P=RI^2\$ so in our case \$P=150mW\$, that's not much. If you go for a current of \$200\$mA (that might be too low) you get R\$=15\Omega\$ and \$P=600\$mW, that's too much for an ordinary \$\frac{1}{4}\$W resistor.

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Your LED looks like it is just a simple 1W LED These are fairly common and require about 3.3V and can sink, at most, 350mA

For any resistor, you'll need a pretty hefty one since it'll have to be able to handle the amount of power you're shoving through it. Each answer is good for the specified current that the LED will consume.

The size resistor you need depends on your voltage source but since you say you are using a 6V source, you should use around a 10 ohm reistor, but it needs to have a power rating of at least 2W to be able to dissipate the power, see calculation:

These LEDs can be bought cheaply and they are already mounted on its proper circuit board.

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  • \$\begingroup\$ No problem. I use it all the time b/c I don't feel like doing the math, although it is important to know. \$\endgroup\$ – Funkyguy Jul 19 '14 at 23:20
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From the data sheet it appears that CR2016's are rated for 90 mAh with a 30k ohm load, and their internal resistance is high enough that you won't have to worry about burning out the LED, no matter what.

On the down side, the battery's life will be very short.

Assuming a drop of about 3.5 volts for the LED with the cells sourcing 3 volts each and an internal resistance of about 40 ohms for both of them means that the current into the LED will be:

$$I_{LED}=\frac{V_{BAT}-V_{LED}}{R_{BAT}} = \frac{6\ V-3.5V}{40\Omega\ } = 62.5\ mA $$

For a capacity of 90 mAh, that means a life of about an hour and a half, best case. It'll be worse than that, though, because the cells are rated at 90 mAH with a 100 microampere - not a 63 milliampere - load.

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The maximum current will depend on how well you can get heat out of the package. (Heat dissipated = diode voltage * current) You should check if that back pad is connected to either of the diode leads. If it is you'll need to be aware of the heat sink voltage. (if not things are easier. I'm assuming you'll use a heat sink.) Other's have given you lots of information on how the calculate the current. If you're not sure, start with 1k ohm and work down from there. I'm don't know what's a reasonable max. temperature, but ~60C should be OK. (Warm to the touch, but it doesn't burn you.)

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  • \$\begingroup\$ Thank you guys, these are really useful ideas. I'm using a 10 Ohm 1/2 watt resistor and he LED doesn't blow up but it does get a little warm. I'll see if using a 2 watt resistor (if any in this place...) helps. \$\endgroup\$ – A R Jul 19 '14 at 23:09

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