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If I'm playing with these little toys:

I don't know if I need to put a resistor in series to protect the LED, as it is needed with 5 mm standard LEDs.

Doing some maths, the needed resistor would be calculated using R = (Vs-Vi)/I, being I the desired Intensity current.

As the LED is rated at 10 W, its max intensity would be P = V x I; 10 W = 12 V x I; I = 10 W / 12 V = 0.83 A.

If I want to, let's say, protect the LED, I should apply a lower intensity, e.g. 0.75 A.

Then, going back to the formula to calculate the resistor, R = (12.4 V - 12 V) / 0.83 A = 0.48 ohms ?!

I'm pretty sure my calculations must be bloody and utterly incorrect, because such a ridiculous value for a resistor could not be right... or indeed it is telling me that I don't need that resistor...

  1. Do I actually need a resistor with this kind of LED?
  2. If so, how can I calculate its correct value?
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    \$\begingroup\$ They do make 0.47 ohm resistors, believe it or not. But consider what happens if the LED takes 0.83A at only 9V (or 10V. Picture and text don't agree, which must tell you something about the product). What should the resistor be then? At which point you start to realise why constant current sources for LEDs exist. \$\endgroup\$ – Brian Drummond Oct 26 '20 at 17:18
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    \$\begingroup\$ Since there is no detailed datasheet, you do not know if your LED needs 10V or 12V for it to be at its maximum allowed current. Guess that it is 10V then calculate a current-limiting resistor from your 12.4V supply. How will you cool the LED? \$\endgroup\$ – Audioguru Oct 26 '20 at 17:34
  • \$\begingroup\$ I woudl guess, since the specification (such as it is...) shows an operating voltage range, that the device includes some form of current control to permit it to be operated from the specified voltages without additional series resistors. \$\endgroup\$ – Peter Bennett Oct 26 '20 at 22:20
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As you have guessed, such a low resistor value means that any variation in LED Vf, which will happen as it heats up, will translate into a substantial change in current. Since LED forward voltage decreases as they get hotter, more current leads to more current. It might smoke, or it might not, but you'll have to fiddle with it, measure current when hot, try various resistors, etc...

These LEDs are called COB (chip on board) and are driven with AC to DC constant current drivers that set the current and adjust the output voltage accordingly. The power supply you linked is constant voltage, not constant current. The usual voltage for COBs is 36V, yours is 9-12V which is a bit unusual.

So you need a driver like this one, which can output the proper voltage range while regulating current. Although this one is not exactly the one you need since it outputs a bit too much current. Or you could buy a 36V COB instead, there are a zillion drivers for those. Or you could just get a COB downlight which will save you the trouble of having to bolt it on a heat sink and spend 3 hours looking for stuff that costs $2.

Note these "bargain" COBs usually have garbage-tier Color Rendering Index (CRI) suitable for car parks, plus suspicious tint, usually greenish but expect anything except white. If you want advice on good quality LEDs that look good, ask in the comments.

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  • \$\begingroup\$ Since the web site says "Integrated scheme: 3 in series, 3 in parallel" I suspect that there are three distinct strings. These can be likely composed into a full series string that does work with 36 V. If, instead, I'm wrong about that and the 3 series strings are directly strapped together on the device, then it doesn't make any sense unless there is active circuitry on-board. The reason is because the series strings will NOT share current well, unless there is an active circuit there. \$\endgroup\$ – jonk Oct 26 '20 at 17:40
  • \$\begingroup\$ @jonk At 39c per COB with free shipping, I'm not sure they bother. Here's the same COB for 26 cents with neat photo of bondwires where you can see the parallel wiring. Looks like this LED is all over aliexpress. Another selles specs CRI as "60-85", not suspicious at all... \$\endgroup\$ – bobflux Oct 26 '20 at 17:52
  • \$\begingroup\$ The close-up picture (and some of the pictures showing a close-up when lit) exhibit more than 9 total LEDs. I don't know why they show pictures that cannot conform to a 3x3 arrangement in your link. But they appear to do so. Do you see what I see there? I don't believe I'm seeing an actual 3x3. That said, they do show only two wires used to light one up -- so that says they are series-parallel devices -- and I agree that the pictures suggest that there is no active circuitry on board. \$\endgroup\$ – jonk Oct 26 '20 at 18:58
  • \$\begingroup\$ If no active circuitry and if we assume these things work reasonably well, then the only other possible explanation is that they "bin" the LED strings, jogging things around so that they do a credible job of current sharing. It's possible, I suppose. But unless they have active circuitry or else bin the LED strings for adequate current sharing -- neither thing "easy" -- then I'm kind of flummoxed. It just won't work well. LEDs don't magically share current well at all (terribly, actually) and humans can easily see variations. Something is going on if these work at all. \$\endgroup\$ – jonk Oct 26 '20 at 18:59
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    \$\begingroup\$ Very much appreciate the discussion. I worked for OSRAM doing LED characterization and binning and color balancing (for modules used to make those large outdoor displays.) Two decades back, though. \$\endgroup\$ – jonk Oct 26 '20 at 20:04

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