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I salvaged the LED disk from a defunct LED light bulb. (The problem was an arced transformer, the LEDs are intact.) It has several unmarked white SMD LEDs soldered on it: 5 lines parallel, with 3 LEDs in series on each line.

I want to use this disk in a strobe. My aim is to get the maximum brightness out of it without damaging the LEDs. Is there any way I can determine (vaguely) its nominal current rating, rather than guessing it?

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  • \$\begingroup\$ You look at the datasheet. Otherwise, 20mA is vaguely the right answer. \$\endgroup\$ – PlasmaHH Jul 5 '17 at 14:21
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    \$\begingroup\$ 250 mA is another vague answer. 1 amp might be another. \$\endgroup\$ – Andy aka Jul 5 '17 at 14:23
  • \$\begingroup\$ @PlasmaHH as I said, I have zero information about the LEDs, how could I get a datasheet?... and that's it, I have some white LEDs with 100mA nominal. \$\endgroup\$ – Neinstein Jul 5 '17 at 14:25
  • \$\begingroup\$ You might be able to infer it from the voltage (after transformer) and nominal power rating of the original light bulb. \$\endgroup\$ – pjc50 Jul 5 '17 at 14:27
  • \$\begingroup\$ @pjc50 great idea! Unfortunately the circuit is totally defunct. (blows the fuse...) \$\endgroup\$ – Neinstein Jul 5 '17 at 14:29
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It has several unmarked white SMD LEDs ...

... without marking you cannot get a datasheet and without datasheet there you don't have a chance of finding out the real current.

However there is a possibility of getting a rough impression of the current needed by the LEDs:

Attach a current source and a voltage meter to the LEDs (the current source may be a voltage source with a resistor in series).

Slowly increase the current from 0A until the product of current and voltage is the power you estimate (the nominal power of the bulb multiplied by the efficiency of the transformer; maybe 5W for a 7W bulb).

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VERY approximately:

LED power rating = bulb_ratinf/number_of_LEDs. So if eg 5W bub with 15 LEDs.
LED power ~= 5W/15 = 1/3 Watt per LED.
LED forward voltage is typically ~= 3V.
So LED current ~= LED_Power/3V = 0.333/3 = 111 mA.

This should not be too far from "about right".

Bulb Wattage ratings are expected to be ~= LEDs DC Watts in.

Modern white phosphor LEds usually have quite a low ratio between
i_operating_max_typical and I abs_max - often in the 10% - 20% range.
LED lifetimes increase for I_LED < Imax and usually are acceptably affected for I_LED > Imax by a moderate amount. Operating on the conservative side seems wise.

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    \$\begingroup\$ Nice heuristic. It sounds like rounding down to the nearest "round number" of 100ma would be about right. \$\endgroup\$ – pjc50 Jul 5 '17 at 14:32
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    \$\begingroup\$ @pjc50 At the assumed wattage, yes. It could be anything in say the 2w to 10w range - but I'd guess it was liable to be at the low end. LEDs decrease in efficiency slightly as I rises but can be assumed to have output proportional to current. Increasing current by say 50% produces an increase in brightness almost unfrtrctablr by eye. So running them somewhat below max rating is often a win-win situation. \$\endgroup\$ – Russell McMahon Jul 5 '17 at 15:09
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    \$\begingroup\$ Yes, derating the LEDs somewhat is a good plan. Honestly given the cornucopia of new LED components on the market that let you choose the perfect one for your application, I barely see a point in salvaging them. \$\endgroup\$ – Harper - Reinstate Monica Jul 5 '17 at 15:34
  • \$\begingroup\$ I'm in trouble with which answer should I accept. Though this is a faster way that doesn't need equipment, and I like it, I choose Martin Roseau's because it uses less assumption. Thanks for your answer! \$\endgroup\$ – Neinstein Jul 5 '17 at 16:40
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I got two great answers from @MartinRosenau and @RusselMcMahon. Both answer relies on the wattage of the bulb. They both missed the fact, however, that this is the consumed power by the LEDs AND the series resistor. One could find and measure the resistor, but I used another method.

What I did was using another working bulb of the same type. I eliminated all other light scources, then measured it's brightness from a given distance using a smartphone and this app. Then I used an adjustable supply to find the voltage/current where the brightness of the led disk matches the bulb's from the same distance.

The result is rather accurately 150mA, or 30mA per LED. Pretty beliveable, so the method seems to work.

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    \$\begingroup\$ FWIW, no :-). I noted (for better or worse) "Bulb Wattage ratings are expected to be ~= LEDs DC Watts in." ie if you buy a 5W LED bulb and find that only 3W is supplied to the LED "there will be problem" with consumer satisfaction. || In real world products there IS NO series resistor per se. Rather, they use constant current circuitry and adjust the voltages to suit. There may be very small sense resistors that drop minimal voltage. \$\endgroup\$ – Russell McMahon Jul 7 '17 at 2:34
  • \$\begingroup\$ @RusselMcMahon Hmm, makes sense. And probably explains why I failed to find the series resistor... \$\endgroup\$ – Neinstein Jul 7 '17 at 8:00
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Is there any way I can determine (vaguely) its nominal current rating, rather than guessing it?

the best is to find the model number and get hold of its datasheet;

alternatively, experiment and guestimate.

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    \$\begingroup\$ -1. This is not helpful at all and shows zero effort. As I said in the short description, I don't have any information on the LEDs. Including serial numbers. And "experiment and guestimate", apart from being obivious, is what others already recommended, only they also described their ideas. \$\endgroup\$ – Neinstein Jul 8 '17 at 14:57

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