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I received from a friend a bunch of 1W/3W (??) LED. The thing is, I can't find any specs for them. For sure I have a method to test them and find Vf/If, but is that method any good? I need EE's advice.

Here's how I proceed:

  • first, I use my multimeter in diode mode to get Vf. Here I got 2.4V.
  • I hook up the LED to my CC/CV, set the voltage to a bit more that Vf, here let's say 3.2V.
  • I set the current limit to ~400mA
  • turn it on
  • immediately check the current draw (here 270mA) and set the current limit to that value.

This method is empirical, there must be a better way out there.

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  • \$\begingroup\$ I'd set the power supply to maximum voltage and set the constant current limit to 20 mA to start. If the power supply doesn't have a simultaneous voltage readout, then add a voltmeter across the LED, as well. Then gradually adjust the constant current limit upward until you get the brightness you feel comfortable with, but consistent with, say, the 1 W limit (just multiply your current setting with the measured voltage and keep it under "1".) With that CC locked in, repeat for other LEDs in the bunch to get minimum voltage and maximum voltage observed for the current you settle on, from above. \$\endgroup\$ – jonk Jan 31 at 17:56
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    \$\begingroup\$ If you have a picture of the LED, someone might recognize it. I bought some cheap high power LEDs from China for a project a while back, and it was a nightmare looking for specs. I found a datasheet eventually, but it took a lot of tedious googling. \$\endgroup\$ – Chris Fernandez Jan 31 at 18:28
  • \$\begingroup\$ I can estimate current, voltage and power from a single measurement but you must not apply more power than you can dissipate into a heatsink rated for a 50’C rise of total ‘C/W * W . Which theoretical method would you prefer? Measure chip area? W/sqmm? With wide Rs tolerances or a single deltaV/deltaI=Rs calc and use my 0.5/Rs=Pd max depending on your heatsink? \$\endgroup\$ – Sunnyskyguy EE75 Feb 5 at 14:18
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Faced with the problem I would proceed in a similar manner but set the voltage much higher than the expected Vf and rely on the current limit setting to keep things safe. At 100 mA (for a high-powered LED) you should get a reasonable reading for the Vf.

Next I would monitor the chip temperature as the current is slowly increased letting it soak at some point where I felt things are starting to get hot.

Ultimately, without the specifications you are going to have to determine some safe operating temperature and set the current to the value that maintains that temperature.

Watch out for shifts in colour. That may also be a clue but it may be impossible with white LEDs which use phosphors to generate the longer wavelengths.

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  • \$\begingroup\$ Thanks a lot for your answer. I was not so far from the truth! \$\endgroup\$ – X99 Jan 31 at 18:22
  • \$\begingroup\$ I've always found that if it's tremendously important that you know some spec and you can't find it, especially for a commodity item like an LED, the most reliable way to proceed is to buy one where you know exactly what you're buying. \$\endgroup\$ – Scott Seidman Jan 31 at 19:27
  • \$\begingroup\$ Checking the temperature is a good approach, even if you know the spec. Because temperature is the most important parameter to define the safe current range. \$\endgroup\$ – Fredled Jan 31 at 21:13
  • \$\begingroup\$ longer wavelengths??? I don't think so. "Watch out for shifts in colour"??? How would you suggest that be done? Otherwise a good answer. I would add that a safe temperate is one that does not burn you when you touch it. \$\endgroup\$ – Misunderstood Feb 5 at 2:08
  • \$\begingroup\$ Actually higher temps cause Blue shift or more blue from lower phosphor efficacy at high temp. \$\endgroup\$ – Sunnyskyguy EE75 Feb 5 at 2:53
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If it's to find out whether it's 1W or 3W, it's very easy: 1W leds are usualy 300mA, 3W leds about 1A. They both needs 3V. (if they are white. This changes with colors too) If you have an adjustable current regulator, set it to 300mA and see if the LED shines normally. If it's a 3W led it will be anormally dim. Make sure to protect your eyes by covering the LED with translucid material or wearing sunglass.

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  • \$\begingroup\$ Not true: If it's a 3W led it will be abnormally dim. A 3W LED @ 300 mA will look very much the same as a 1 W LED at 300 mA. This is true only sometimes: They both needs 3V. Most current white (and blue) LEDs Vf run about 2.7-2.9V. \$\endgroup\$ – Misunderstood Feb 5 at 4:13
  • \$\begingroup\$ @Misunderstood Sorry, but I have to confirm and reiterate what I wrote: If a 3W led requires 900mA in otpitmal operations, with 300mA it will be much less bright than at 900mA. However you are right that a 3W led with 300mA will emits as much or almost as a 1W LED with the same amount of amperes. But much less than a 3W led with 900mA. That's what I mean. \$\endgroup\$ – Fredled Feb 6 at 23:05
  • \$\begingroup\$ About voltage: Mentioned voltages are "typical". You can apply slightly less or slightly more (but not too much) than indicated. It's of course better not to exceed indicated voltage in absence of current regulation. \$\endgroup\$ – Fredled Feb 6 at 23:11
  • \$\begingroup\$ My point is using brightness is not a very effective method. Temperate is the key. No matter the characteristics of the LED and the PCB it is mounted on, temperature is the single most important factor. You really do not need the datasheet in most cases after the LED is selected. You design your PCB for the most effective thermal management you know how, and adjust the current by temperature. Whether the LED is 1 W or 3 W does not matter either. They will both have similar luminous intensity and temperature. The variable is thermal management design of the PCB. \$\endgroup\$ – Misunderstood Feb 8 at 19:15
  • \$\begingroup\$ @Misunderstood I agree with you but the question here was how to guess the specification of a LED. Not how to power it in the best way. I agree that temperature is the key to define voltage and/or current provided to the LED. But to find out whether it's 1W or 3W, observing the light intensity is the easiest way IMO. \$\endgroup\$ – Fredled Feb 10 at 0:30
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Eventually you will need a constant current driver to keep them from over heating, drawing even more current, and burning up. A simple one is this, and there are plenty more.

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  • \$\begingroup\$ I like the current regulator you linked in. With a potentiometer it would be a good setup to test Vf. For high power LEDs, not so useful. If using a CC driver, how would it draw even more current? Would not the current be constant? Sometimes there are better solutions than a CC (e.g. resistor). A CC will not always keep them from over heating either. \$\endgroup\$ – Misunderstood Feb 5 at 2:42
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I use my formula Vf=Vt +If*Rs

I use some low current for Vt like 10% of rated DC current at some known temp. for silicon, ~Vt=0.55 , for Red/Yellow UB , Vt= 1.7 to 1.8 , for BLue/White Vt =2.7 to 2.8V (low for high power diodes)

  • 2.8 +If*Rs (+/-50%)=Vf for low power devices

  • or Vf=2.7 +If * Rs =Vf +/-10% for high power

  • where Rs internal anode and bulk resistance Rs=0.5/Pmax +/-50% depending quality of LED and heatsink @ Tj= 85’C for If>10%

So for your results 3.2=2.8 +0.27A*Rs gives Rs= 1.5 Ohm or 2/3W. +/- 50%

  • or Vf=2.7+If * Rs for If >=1A white LED Thus at 1A this LED is 2.7+1A * 0.5 Ohms = 3.2 V +/- 10% which agrees with the datasheet 2.9 to 3.5V at some temp.

  • The VI typ curve shows Vf drops 0.1V from 1000mA to 800mA thus 0.1/0.2=0.5ohm so Pd=1/Rs= 2W +/- 50% or more depends greatly on your max Tj you allow above your heatsink

    • where 3.5W worst case @1000mA in the datasheet with heatsink say of +50{‘C/W} of PCB copper area of 3 sq.in.
    • Results in a Tjc+Tca rise of (50+6.5 ’C/W) * 3.5W = 200’C rise !! This tells you your junction cooling design is no good and you need Alum clad board or you cannot use 3.5W thus my formula of 2W+/- 50% is accurate for a typical design.
    • Yet a good thermal cooling designer can limit the junction rise to 50’C and the others fry their good LEDs ignoring Ohm’s Law for my Rs vs Pd formula and thermal resistance.

This require more detail to prove, so check your design tolerances and thermal resistance.

For S x P arrays it would be, \$V_f=2.8*S+S/P*If*Rs\$ Then heatsink should be Rca= Watts/deg C to to be less than Pmax/40 deg C for a good design., better if poss.

Ref for more thermal details.

If you don’t know the SxP array , choose S from Vf/3 and then estimate P and so Rs and Pd will be for the array vs individual.

If you want to use a Vcc then add to Rs for desired If and chose Vcc less than 1 LED Vf drop above string for choosing Rs for efficiency and Pd.

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  • \$\begingroup\$ That's interesting. Could you add a worked example for a known LED, showing how your formula gets an answer close to the one given on that LED's datasheet? Thanks. -- I also wasn't sure what you meant by "If>10%" 10% of what? Again, if you can include this in the worked example, that would help with the explanation. Thanks again. \$\endgroup\$ – SamGibson Jan 31 at 18:39
  • \$\begingroup\$ 10% rated If. The diode is saturated and bulk Rs is > quadratic reducing junction R \$\endgroup\$ – Sunnyskyguy EE75 Jan 31 at 18:47
  • \$\begingroup\$ Sam search my other LED ESR answers or try and LED spec yourself !f you like. I’ve done this hundreds of times \$\endgroup\$ – Sunnyskyguy EE75 Jan 31 at 18:52
  • \$\begingroup\$ Thanks for adding the calculations for the OP. Unfortunately the formula didn't seem to work very well on the first (Cree) LED I found on the Farnell website. (This is why I asked for a worked example against a datasheet - I don't know if the problem is with the formula, or the datasheet, or my interpretation etc.) Anyway, as you suggested, I'll keep trying when I have more time. \$\endgroup\$ – SamGibson Jan 31 at 19:30
  • \$\begingroup\$ what tolerance error? And p/n? If it is an array you need to factor per LED \$\endgroup\$ – Sunnyskyguy EE75 Jan 31 at 19:44
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This method is empirical, there must be a better way out there.

Without a datasheet where the manufacturer did the empirical experiments for you, there is no other way.

A better way is to use an LED IV analyzer.

I use my multimeter in diode mode to get Vf. Here I got 2.4V.

You need two multimeters to do this. One to provide the current (in ohms mode) the other to measure the voltage.

If the Vf is indeed 2.4V (not likely) it must be a red, orange, or yellow LED. More likely your meter uses 2.4V to test. LEDs have a higher Vf than other types of diodes.

set the current limit to ~400mA
immediately check the current draw (here 270mA) and set the current limit to that value.

This is puzzling. With this procedure the current should have gone to 400 mA and stayed there. Why did it only go to 270 mA? What limited it to 270 mA? Possibly the Vf is more than the voltage your were using.

If you are indeed using a CC/CV driver then the voltage should not matter, it will adjust itself. However you do want to start out with a lower current (e.g. 300 mA or less).

Bottom Line

This is a waste of time.
1. You should not buy LEDs that do not have a datasheet.
2. How much will knowing the forward voltage help? Hint: Not much.

Even if you have a datasheet there is not parameter in the datasheet for your PCB and its thermal management. The datasheet will not say what the temperature is for any given current.

Either way, datasheet or not, you must not allow the LED to get too hot. How hot is hot? Check the datasheet. Get the max Tjunction and the junction to case thermal resistance calculate the thermal resistance for your PCB and heatsink to estimate the temperature. This is the simplest app note I know of to help you do that: Thermal Design By Insight, Not Hindsight

OR

Use the low tech empirical method I use. No Vf required.

Touch the PCB near the LED.
If it burns, the temperature is too high.
I prefer a temperature so that I can put my finger on the PCB for an extended amount of time without much, if any, discomfort.

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