Why is my LED's Vf higher than the specs?

Cemoi

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France
Hello,
I've just bought my first power LED: a red-orange Luxeon I.
I'd like to power it from two AA NiMH in series (so Vbat ranges from 2.7 V to 2.4 V for most of the runtime).
According to the datasheet for this LED Vf should be around 2.3V @ 70mA, and 2.6V @ 200mA.
So I tried connecting my LED directly to the batteries (Vbat=2.7V), expecting the current to be around 200mA or slightly more.
But I measured only 70mA, and the actual Vf was 2.6V.

So I tried with three NiMH in series, i.e. Vbat= 4V. I put a variable resistor just in case, to avoid any damage to the LED.
From the datasheet Vf=2.5V @ 150mA.
So doing the easy maths, Vbat-Vf=1.3V, so to get 150mA I need a 9 ohm resistor. But with this setup I measured only slightly more than 100mA.
Then I tried to adjust the resistor value until I got the expected 150mA: it turned out to be 3.2 ohms, and Vf=2.8V instead of 2.5V from the specs.
Now completely removing the resistor, the current only reaches 194mA, and Vf=3.3V whereas the specs say it should be 2.6V @ 200mA.
Next I put back the variable resistor (still with 3 NiMH) and adjusted it until the current was 70mA, like in my first attempt with 2 NiMH. I then measured Vf and found 2.33V, i.e. a much lower value than when I connected the LED to 2 NiMH and got 70mA. Strange.

How come I find such a difference with the Vf/I chart in the specs?
How come I can't get more than 194mA when driving it direct from a 4V battery set?
Why do I find different values for Vf at the same current, depending on Vbat?

At this stage I should mention that I had trouble soldering the leads on the LED star pads (it was my first time:() so I may have overheated the LED. Could this be the reason for these off-specs values?
 

baterija

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How come I find such a difference with the Vf/I chart in the specs?
I'm assuming you are looking at the graphs and estimating values (either that or I missed where it had numbers for 70 and 200ma.) Note that it says "average forward current" in the graphs. So for a large sample of emitters that's average. It shows you the general shape of the curve relating Vf and I but that graph doesn't speak to how big the deviation in Vf is.

If you look at the numbers for 350ma in the chart earlier it has Min Vf 2.31 Typical Vf 2.95 and Max Vf 3.51. Doing the math the bottom is 21% lower than the typical and the max is 19% higher than the typical. Call it roughly +-20% deviation from the "typical" value.

Looking back at the chart for 200ma it's right around the 2.6 you quoted, maybe 2.7. Assuming the percentage of deviation is about the same at lower current, that give you 2.9-3.0 max Vf at that current. So yours appears to fall within spec, just towards the higher end of the distribution.

Luxeon lottery isn't just for tints. :(
 

jeffosborne

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Hi Cemoi - soldering on those stars is always a challenge because the contacts are on top of aluminum - taking heat away whilst you are trying to apply heat! I use a 45-watt iron for the stars. And I am sure to pre-tin the star contact and the wire end before soldering them together. I doubt that you damaged your emitter while soldering. Also, for your voltage readings, put your meter probes right on the star contacts for most accurate readings. Hey, what are you going to do with your new LED? Cheers, Jeff O.
 

ellerbro

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Jul 10, 2007
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How come I can't get more than 194mA when driving it direct from a 4V battery set?

I'm guessing you had a poor solder connection or the batteries are worn out (high internal resistance). Vbat should not have dropped from 4 to 3.3V when putting out only 194mA. That means there were about 3.5 ohms in the current path, either in the battery or the wiring/connections. I recommend against doing that again. A 4V source will most likely fry that LED. Look at the I-V graph in the datasheet.

Why do I find different values for Vf at the same current, depending on Vbat?

Vbat has nothing to do with it. Vf depends primarily on current and temperature. Maybe you didn't let the LED's temperature stabilize? I can't think of why Vf would be different if the temperature and current were the same.
 

Cemoi

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soldering on those stars is always a challenge because the contacts are on top of aluminum - taking heat away
I might get a bare emitter next time (I thought stars would be easier to handle for a newbie), especially as i noticed that many optics are designed to be placed on bare emitters rather than stars.
What are the easiest to solder emitters(i.e. not tiny terminals)?

Also, for your voltage readings, put your meter probes right on the star contacts for most accurate readings.
I'm afraid when doing so it is hard not to look at the LED, which I prefer to avoid after this clear warning

what are you going to do with your new LED?
DIY bike lights.
 

Cemoi

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I'm guessing you had a poor solder connection or the batteries are worn out (high internal resistance)
I checked all my batteries internal resistance, and found that some were higher than others (up to 0.7 ohms for the worst ones, vs. 0.3 ohms for the best ones). Then I tried again with the best ones, but it made very little difference.
Vbat should not have dropped from 4 to 3.3V when putting out only 194mA. That means there were about 3.5 ohms in the current path, either in the battery or the wiring/connections
I tried again with two NiMH only, and eventually found the reason for this very high Vf. I noticed that my LED was much brighter when connecting it directly to the batteries, WRT when I measured the current using my DMM. To check this I made a small circuit using a switch allowing me to include or not my DMM in the battery+LED circuit. And I could notice a huge difference in brightness depending on the presence of the DMM in the circuit.

Next step was to use two multimeters: this one to measure the current, and a second one (an old but good quality analog one) to measure Vf at the same time. This gave a Vf value which matches perfectly the datasheet: 2.3V @ 70mA.
Swapping the multimeters gave almost identical values: 2.23V @ 60mA.

In fact in my first measurements I was measuring Vf values which did not correspond to the actual current values found when connecting my DMM in the circuit.

So my LED is OK :) and its Vf seems on par with its specs.

But my question is now: how come my DMM affects the current in the circuit so much? AFAIK multimeters, when used to measure currents, have a very low resistance which modifies only marginally the overall circuit current. FYI my DMM is a good one (a Fluke 75).
What is the good method to properly measure LED currents?
 

HKJ

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What is the good method to properly measure LED currents?

Keep the meter in the 10A range, to reduce the voltage drop.
Or get a DC current clamp.

Your can also measure the voltage after the current meter, a voltmeter will not draw any significant current (In this case).
 

faklya

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In fact in my first measurements I was measuring Vf values which did not correspond to the actual current values found when connecting my DMM in the circuit.


I was trying to point out the same problem. See this thread. https://www.candlepowerforums.com/threads/201190&page=2

Do you have any idea, how could I measure current and the appropriate Vf with only 1 DMM and battery(or perhaps using resistors)
 

Cemoi

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Keep the meter in the 10A range, to reduce the voltage drop.
Thanks HKJ, unfortunately the fuse of my 10A range is blown out.

But in the meantime, I had checked my DMM specs and saw that its "typical full scale burden voltage" is 6mV/mA, so when inserted into a circuit it is equivalent to a 6Ω resistor*, which makes a big difference and fully explains
the biased values I got for my current, and apparently shifted values for Vf.

*I could confirm this value by measuring the voltage drop across my DMM, using my analog multimeter.

So I started all my measurements again, by measuring the voltage drop across a 1Ω resistor in series with the LED, which gives me a direct reading of the current since in this case 1mV corresponds to 1mA. The current I measure this way matches pretty well what I expect from the I/Vf curve in the datasheet.
 

Illum

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S26H 0014679
3207LXHLMH1D (the last digit might also be a zero, not sure).

I had asked in this thread how to read such a code, but no reply for the Luxeon.

Thanks for the link to the Bining and Labeling document, I'll read it.

the post important part of the bin numbers are the first four digits representing the flux rating [1 letter], tint[2 letters, 1 letter with one number, or possibly 2 numbers], voltage [1 letter], and color [1 letter]. on white LEDs, the color code is usually left off, so it looks something like UX0M instead of UX0MW

just for reference, the last part: LXHL-MH1D indicates its a luxeon star, lambertian [or high dome] and its red-orange, which the color designation in the first four digits should indicate an H

according to http://home.comcast.net/~theledguy/bin_codes/index.htm
Bin codes explained said:
Flux.......Min Lumens............Max Lumens
M................13.9...........................18.1
N................18.1........................... 23.5
P.................23.5.......................... 30.6
Q................30.6........................... 39.8
R.................39.8.......................... 51.7
S.................51.7...........................67.2
T.................67.2...........................87.4
U.................87.4..........................113.6
V................113.6.........................147.7
W...............147.7.........................192.0
X................192.0.........................249.6

So S means your luxeon when driven to spec gives between 51.7-67.2 lumens

I'm not sure how to go about designating the "26" but the other parts should be adequately explained on that link:grin2:
 
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