Does this LED need current limiting? - Page 1
Forum section: electronics / projects

author: skyler
author_rank: Regular Contributor
timestamp: January 31, 2014, 03:57:04 pm
content: I recently ordered this 10 watt LED on electronic goldminehttp://www.goldmine-elec-products.com/prodinfo.asp?number=G19653I want to drive it with a 12 volt battery. I know I need a good heat sink for it, but I don't know how I am going to limit the current to 1 amp.The page states LED operates on 9 to 12VDC. Does this mean with my 12 volt battery, I don't need any current limiter, or do I still need it? If so, what should I use, because I know a resistor won't be practical because of how much power will be wasted.

author: IanB
author_rank: Super Contributor
timestamp: January 31, 2014, 04:04:30 pm
content: It's a raw LED, so it needs current limiting. The reason it is said to need 9-12 V DC is most likely because it has four individual LED elements in series on the die.With a very powerful LED like that you may be able to get away with a small series resistor and/or the internal resistance of an alkaline battery supply to limit the current. But if you want to power it with something like a 12 V lead acid battery you will definitely need some current regulation.

author: DmitryL
author_rank: Regular Contributor
timestamp: January 31, 2014, 04:15:20 pm
content: Quote from: IanB on January 31, 2014, 04:04:30 pmIf the picture is more or less correct, it looks more like 3x3 cluster. 3 LEDs in series and 3 such lines in parallel. Thus, the voltage drop might be 3*3.6VQuoteWhat a bad advice....

author: dannyf
author_rank: Super Contributor
timestamp: January 31, 2014, 04:48:25 pm
content: QuoteThe general rule is that no if you know what you are doing. Otherwise, use a resistor.Having said that, you will find using a switching mode regulator to drive it (at fixed current) is much better.

author: mariush
author_rank: Super Contributor
timestamp: January 31, 2014, 07:44:35 pm
content: If you want to add some protection super cheap, just get a LD1117V (or whatever you find with 1117 in name and adjustable) and a couple of resistors and capacitors :http://uk.farnell.com/stmicroelectronics/ld1117v/v-reg-adj-1-25-15v-1117-to-220/dp/9755829http://uk.farnell.com/stmicroelectronics/ld1117dt/v-reg-ldo-adj-smd-1117-dpak-3/dp/1095395etc just enter 1117 in the search box.It's a simple linear voltage regulator with around 1v drop and 0.8-1A current max and a current limit set at around 0.95-1A (some go up to 1.2A depending on who makes them), so you can set it to output around 9-10v and just rely on the internal current limiting (10v x 1A current = 10w).

author: Dave
author_rank: Super Contributor
timestamp: January 31, 2014, 08:03:45 pm
content: Quote from: dannyf on January 31, 2014, 04:48:25 pmThis. For high power LEDs like this one using resistors is a pure waste of power. A fixed current SMPS will be much more efficient and reliable.

author: jman
author_rank: Contributor
timestamp: February 01, 2014, 05:44:15 am
content: Quote from: skyler on January 31, 2014, 03:57:04 pmYou will need a LED Driver with constant Current, Power, Check datasheet what is the max current IFmax, and also what would be your Supply Voltage? AC Mains?

author: jwm
author_rank: Regular Contributor
timestamp: February 04, 2014, 03:36:44 am
content: A LM317 in current limiting mode is my go to solution for this and easily junk-boxable. I would never use a resistor for a high power LED unless I heavily derate it. Assuming you don't mind the inefficiency. Just a lm317 and a 1.2 ohm resistor will do for a one amp current limited supply.Use your lab power supply in constant current mode set to one amp to test what voltage it is roughly going to use for that too see what size heat sink you will need for the LM317. (Vin - Voltage shown on power supply)*1Amp gives the watts that the LM317 will dissipate, which won't be that much heat sinking needed relative to what your LED will need anyway.

author: skyler
author_rank: Regular Contributor
timestamp: February 09, 2014, 02:22:16 am
content: Quote from: jwm on February 04, 2014, 03:36:44 amThanks, so hooking up my 12v to LM317/1.2ohm and straight to my LED will be all?And I kind-of blew my only bench power supply 4 months ago, it was a nice astron 15 amp 34 volt analog one, but I blew it trying to charge a 150F supercap and scrapped it for parts.

author: sacherjj
author_rank: Frequent Contributor
timestamp: February 09, 2014, 02:37:32 am
content: Quote from: skyler on February 09, 2014, 02:22:16 amIn this situation, you are using the 1.2 ohm resistor to cause the 1.2V drop that the LM317 is looking for and I believe that the full minimum drop is 1.6V. So if your LEDs need more than 10.4V to reach 1A, you might need to go higher than 12V in front of the LM317.

author: skyler
author_rank: Regular Contributor
timestamp: February 09, 2014, 02:45:17 am
content: Quote from: sacherjj on February 09, 2014, 02:37:32 amOk i guess I can find a bit larger power supply. I was wanting it to work with my 12v battery on my bicycle

author: sacherjj
author_rank: Frequent Contributor
timestamp: February 09, 2014, 03:37:02 am
content: Hook it up with a 12V battery. Measure the voltage across the 1.2 ohm resistor. If you are getting 1.2V, you are running at 1A.

author: jwm
author_rank: Regular Contributor
timestamp: February 10, 2014, 01:11:41 am
content: Quote from: skyler on February 09, 2014, 02:22:16 amYes, hook it up in the 'current limiting' configuration as shown in the data sheet. You can use a multimeter to measure how it is working. just measure the voltage across the led while it is lit to see what voltage it is getting or the voltage accross the resistor as in, hook your voltmeter leads to each side of the resistor withe the led on and current = voltage shown/reistor value. I recommend using the voltage method rather than directly using the amps feature of your multimeter because it won't blow your meter (well, just its fuse if it's a well designed meter) if you accidentally wire it up incorrectly.If you happen to be reading laser/led forums and see people complain that the lm317 configuration "wastes power" because it has a resistor in it, ignore that. it's complete rubbish and somehow gets repeated a lot and people end up with convoluted and worse supplies that waste the same power and wonder why their expensive laser LEDs burn out. A current limiting linear supply will pass as much power as is needed to get the desired current, if some is dissipated in the resistor it just means the regulator will dissipate exactly that much less to make up the difference. If anything you improve things by split the heating between a resistor and a lm317. for you, 1 amps through 1.2 ohms = 1.2 volts, power = volts * amps so your resistor will dissipate 1.2 watts. If you don't have a power resistor, you can just use multiple resistors in parallel, as in, 3 3.6 ohm 1/2 watt resistors in parallel or 5 1/4 watt 6 ohm ones. That is very conservative, when splitting resistors you can actually dissipate a lot more power than simply summing the wattages because the resistors are separate from each other and have more surface area for air to flow between them. If they are not getting too hot to touch, you are okay.I am surprised you were able to blow your bench power supply, good ones are designed to be bulletproof. Actually it looks like the astron are not lab grade power supplies, but meant to power RF and audio equipment so a different beast that won't have the protections of a lab supply and shouldn't be used for experimenting.You can probably use its case/meters/transformer to make a traditional kit lab supply with your extra LM317s.

author: dannyf
author_rank: Super Contributor
timestamp: February 10, 2014, 02:10:40 am
content: QuoteThat's true, if the LM317 CCS can operate in a different universe,.The LM317 (or similar linear solutions) will dissipate power, (Vin - Vfwd) * Ifwd, where Vin is the input voltage, Vfwd is the diodes' forward voltage and Ifwd is the current going through the LM317 / diodes.Some of that, 1.25v * Ifwd to be exact, is dissipated on the resistor, wth the rest on the LM317. So if we are talking about 12v input, 10v Vfwd on the diodes, and 1amp, we are talking about 2w being dissipated on the LM317 + the resistor.Should the input voltage go to 14v, you are talking about 4w being dissipated. and so on and so forth.That's a lot of power wasted and needed to be dissipated away.

author: jwm
author_rank: Regular Contributor
timestamp: February 10, 2014, 03:30:11 am
content: Quote from: dannyf on February 10, 2014, 02:10:40 amYeah, I was refering to the fact that they seem to think that simply because there is a resistor in the load path with the lm317 current limiter it will inherently waste more power as a whole than some other linear current supply circuit that happens to not have a resistor in the path. I can sort of see how they come to that conclusion without knowing much about regulators. they read 'resistors resist the flow of electricity by dissipating some as heat' then look at the lm317 docs and their eyes glaze over and are just left with the impression that resistors are bad not realizing that the linear regulator effectively is doing the same thing and its feedback makes things work out the same.Like you say, the power wasted by the whole circuit depends on the forward voltage and current and input voltage only. No matter how you slice your supply (assuming it is based on a linear regulator) you still have the same loss, just split up differently. If 9V is what is needed to get an amp through the LED, and the resistor drops 1.2V then the LM317 will stabilize at 10.2V, so a wacky circuit that gets your regulator to stabilize at 9V directly by not using a resistor doesn't help. as it just moves the wastage to the regulator from the resistor.