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I know that you're supposed to wire a resistor with an LED to stop it from being damaged, but I don't know how resistant a resistor to put in. My input voltage will be from a 3.7V Li-Ion battery. The LED says it's 12w, and the LED is red (no info on the led besides that). It has a heatsink. What resistor should I wire up to prevent the LED from being damaged while maximizing it's brightness? If this isn't enough info, let me know what else I need to answer this question. The LED was running fine connected to a 6v (1.5V x 4 aaa batteries) with a resistor in series that was heatshrinked and couldn't be read once I took it off.

This is the LED

Jasmine
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  • Please edit your question to include the make an model number of your LED, or a link to where you bought it or whatever data you can provide about the LED. If nothing else, then you can at least provide a good clear picture of the LED from top and bottom. Somehow, it would be good to determine the forward voltage and max current of the LED before answering your question. – user57037 May 25 '21 at 19:49
  • Is it necessary to know the voltage rating of the LED? I see that on the question you posted, but I don't have that info (I'm reusing an LED) – Jasmine May 25 '21 at 19:50
  • @mkeith I added the photo – Jasmine May 25 '21 at 19:56
  • The board says 94V. That may mean it ran at 94V or its a red herring. Where did it come from? – Passerby May 25 '21 at 20:11
  • Consider that a 1 sq.in metal substrate can only dissipate 1W. Where’s your heatsink? Also consider Ohm’s Law from Pd and V*I drop resistor and derate size 50% – Tony Stewart EE75 May 25 '21 at 20:18
  • @Passerby It came from a lightsaber from ultrasabers.com. I'm reusing it for a project. – Jasmine May 25 '21 at 20:20
  • @TonyStewartEE75 The heatsink threads on – Jasmine May 25 '21 at 20:22
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    @Passerby 94V is something to do with the other compliance marks on that line, A007 appears to be the model number. – Jasen Слава Україні May 25 '21 at 20:34
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    @Passerby - Hi, Regarding: "*The board says 94V. That may mean it ran at 94V or its a red herring.*" That text is the claimed [UL 94](https://en.wikipedia.org/wiki/UL_94) flammability rating for the FR4 material. In this case "94 V-0". It doesn't mean 94 volts :-) Hope that helps. – SamGibson May 25 '21 at 20:37
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    doesn't look like a 12W LED to me: https://leddynamics.com/indus-star-a007-a008 – Jasen Слава Україні May 25 '21 at 20:37
  • @Jasen maybe I'm wrong then. I believe thats what it's marketed as on the Ultrasabers website – Jasmine May 25 '21 at 20:40
  • @Jasen What does the number E320003 on the top refer to then? – Jasmine May 25 '21 at 20:42
  • when I searched it I just found A007 again – Jasen Слава Україні May 25 '21 at 20:51
  • it turns out that it is a 12W LED, it runs at 12V. 1000mA – Jasen Слава Україні May 25 '21 at 20:52
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    @jake the 12w claims seems to be for the multi chip led boards or the rgba light saber with multiple color leds. They have alot of customizations so your saber may have been a simple 2W led. – Passerby May 25 '21 at 20:58
  • @Passerby Gotcha. Is there a "safe" resistance I can add where you'd be confident I wouldn't overheat the LED? I just want to make sure I don't break things – Jasmine May 25 '21 at 20:59
  • The best course would be to set up 2 multimeters and a constant current driver with adjustment. Measure the current and voltage and slowly creep up the current until you see about 2V (since it's red) and the current involved. That will give you what should be the nominal max voltage/current for full brightness. It's going to need a constant current regulator circuit able to support up to 1A. – Passerby May 25 '21 at 22:21
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    Have you considered using a switch-mode supply instead? What about using the driver-board from an LED torch - either from salvaged parts or FleaBay? Here's a 20mm diameter, 10 watt driver for $2.64 australian pesos https://www.ebay.com.au/itm/282637782081 - I'd stat with a 1w or 3w driver. A local store sells torches with a 5 mode, 3W driver for about AUD $6 – enhzflep May 25 '21 at 22:23

3 Answers3

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According to the Industar Luxdrive datasheet the LED is a Cree XP-E. In that datasheet the red and orange-red versions have this current vs voltage curve:-

enter image description here

So you can expect the LED to drop between ~1.9 V and 2.25 V depending on current draw.

A typical 3.7 V Li-ion battery has an operating voltage range of ~4.1 V to 3.0 V under load. If you choose a maximum current of 500 mA with a fully charged battery the voltage difference will be ~ (4.1 V - 2.2 V) = 1.9 V, and the required resistance is ~ 1.9 V / 0.5 A = 3.8 Ω (the nearest 5% series value of 3.9 Ω will be close enough). The resistor will dissipate ~ 1.9 V * 0.5 A = 0.95 W, so it should be rated for at least two watts.

The LED itself will consume ~2.2 V * 0.5 A = 1.1 W, which doesn't sound like much. But the light output will probably be roughly equivalent to a 12 W incandescent bulb with red filter. As the battery runs down the current will drop to ~ (3.0 V - 2.0 V) / 3.8 Ω = ~ 0.26 A, producing about half the light output. Again this sounds worse than it will look, as the human eye is not very sensitive to absolute brightness.

Bruce Abbott
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  • This is the best answer, short of a constant current regulator setup with a pair of multimeters to find the actual usage. – Passerby May 25 '21 at 22:22
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This appears to be a classic 500mA that marketing guy is boasting 1A without and Vf specs (lame) Vf for this 1.2W LED will rise with an ESR of about 0.5 ohm from similar specs of 2.05V @ 500mA

yet I know they handle can 1.5A pulses with an infinite heatsink.

There are better ways to design a CC with a precision cutoff, but this may be adequate for rookies. But a fixed resistor will dissipate more heat than the LED.

This should only be driven with a resistor controlled current limiter such as an LM317 with both LED star and LM317 mounted on a 10W heatsink for 3.7W total heat dissipation to prevent burning your finger.

Design R for 500mA and get a few power resistors for 1.25V/0.5A=R so you can use R//2R for 750 mA then R//R for 1000 mA or a voltage controlled FET to regulate the LM317 . This will start around Vf=2.05 to 2.1V at 25’C and quickly reduce about 4mV/‘C as it heats up so when Vbat =3.1 with a 1.1V drop to 2V it will quickly reduce the current demand as the LED turns off around 1.8V (dim). So get a bunch of 1.25 ohm 1W resistors for the LM317 TO-220 and measure constant current from the voltage drop from Vout to Vadj=1.25V

Tony Stewart EE75
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This web page claims that the LED module wants 12V, Oddly none of the data-sheets mention this.

https://www.ledsupply.com/leds/cree-xlamp-xhp35-high-density

You can't get 12V that from a lithium cell using resistors. You'll need a boost converter.

that page and this datasheet page say they want 11.3V or more just to run at 1/3 power.

Cree datasheet: https://www.ledsupply.com/content/pdf/cree-xlamp-xhp35.pdf

The markings on yours seem slightly different, it might not be a genuine part.

Jasen Слава Україні
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  • It was previously running connected directly to a 6v (4x1.5v aaa) battery pack, with a resistor in series that was heatshrinked and couldn't be read once I took it off :( – Jasmine May 25 '21 at 20:54
  • the data sheet says the LED can take up-to 1A you're probably going to have to measure what voltage is needed for that. you could try a 2.7 ohm resistor with your battery and see what happends – Jasen Слава Україні May 25 '21 at 20:59
  • @jake use a multimeter and measure the resistor if you still have it. – Passerby May 25 '21 at 20:59
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    @Passerby XD duh, I should have thought of that. I'll see if I can find it – Jasmine May 25 '21 at 21:02