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I've got few of these LEDs which are in a 3S3P configuration and rated 9-12V 900-1050mA. I tried to power one of these with this module which I earlier set at 12V and 1A with a multimeter in short circuit.

The problem is that when I connect the LED in series with the multimeter, the latter tells me that is drawing less than 400mA.

So, out of curiosity and willing to fry one LED, I connected it directly to a power supply (a laptop brick, 12V 3A) and the LED doesn't draw more than 600mA. More LEDs in parallel gives me these values:

  • 2 LEDs: 1200mA
  • 3 LEDs: 1600mA

I'd like to push the rated current through the LEDs (so to take them to full brightness), also is weird that seems like they're current-limiting themselves without the need of a constant-current circuit.

Does anyone have any explanation/experience with this behaviour?

Thanks

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  • \$\begingroup\$ sounds like banggood's led's.. no specs. no guarantees and not best quality or price \$\endgroup\$ – Tony Stewart EE75 Nov 3 '16 at 3:27
  • \$\begingroup\$ Have you measured voltage on your power supply under load conditions and are you certain you're in constant current mode? \$\endgroup\$ – K H Sep 15 '18 at 18:28
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You have not mentioned the color of the LEDs. The typical white LEDs that make up modules like yours would take more than 3V to reach maximum rated power. I would say around 3.5V could be a common number.

For a 3.5V white LED, probably around 20% of the voltage drop is due to ESR. A LED (of the same base color) requiring lower voltage and therefore has a corresponding lower ESR would be more efficient.

So there is an unavoidable resistor in series with every LEDs already. If you use the number 12V, 0.6A, 20% ESR then the ESR = 12 * 20% / 0.6A = 4 ohms. These are guesses and may not reflect what you have. But say if these guesses are exactly right, then you would get 1A at 13.6V.

(The simple ESR model breaks down badly at low current).

Rereading what I wrote, I was assuming 4 x 3V. But there are only 3 LEDs in series, so the numbers do not add up.

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  • \$\begingroup\$ You're right I forgot to mention that I'm talking about cool white LEDs. \$\endgroup\$ – Ema Nov 3 '16 at 4:57
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rated 9-12V 900-1050mA

That's a clue that your LED arrays have built in current limiting and are designed to be powered by a voltage, not current.

But if they are rated for 1050mA, why do they only draw 600mA or less? Ask your vendor. Perhaps the 1050mA is for a higher voltage (eg. 14V), or perhaps you were sold sub-standard parts, or the specs are just plain wrong. If you bought them from a website such as eBay or Amazon then just be happy that they work at all!

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  • \$\begingroup\$ It's interesting that they could have built in current limiting, it would actually avoid the hassle of a dedicated circuit; is there a way to confirm that? And yes, I indeed bought them from a chinese vendor on eBay for about $1 each. I guess I should be happy they all work and are pretty bright. \$\endgroup\$ – Ema Nov 3 '16 at 3:15
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    \$\begingroup\$ @ema these were probably lot rejects with excessive ESR or Vf.. best way to use them is a CPU heatsink and thermistor to regulate heatsink temperature to 80'C or pref 60'C using a DC-DC converter with feedback or "wing it" with a desktop ATX 12V supply and surplus CPU heatsink and test temperature with your finger ;) \$\endgroup\$ – Tony Stewart EE75 Nov 3 '16 at 4:42
  • \$\begingroup\$ I will do some testing just out of curiosity, but then I'll end up with your second option, 12V ATX and big heatsink. \$\endgroup\$ – Ema Nov 3 '16 at 5:06
  • \$\begingroup\$ If they have current limiting there should be smd resistors on the board. If not then they are probably bare LEDs which are out of spec, and you just aren't feeding them enough voltage to get the rated power. LED voltage drop decreases as temperature increases, so if you run them on a fixed voltage the current will increase as they warm up. You could add a 1 Ohm 5W resistor in series to provide some current regulation and allow you to check the current draw by measuring the voltage across it (1V = 1A). \$\endgroup\$ – Bruce Abbott Nov 3 '16 at 21:38
  • \$\begingroup\$ Multi-LED assemblies, especially cheap ones that claim high power ratings, tend to have high series resistance. That and the series arrangement of LEDs leads to a wide Vf range over its rated current. There are almost no LEDs out there with built-in current limiting. However, some large array devices do have sufficient resistance that you can effectively run them in constant voltage mode without much concern of them blowing up. \$\endgroup\$ – ajb Sep 4 '18 at 1:40
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What you may be missing is that you are not really connecting the LEDs directly to the power supply- the ammeter acts as a resistor - the more current the more voltage drop.

A constant current circuit- or a substantial series resistance and the associated wasted power is highly recommended if you want to run the LEDs at near maximum current.

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  • \$\begingroup\$ I thought the resistance in an ammeter is so minimal that does not affect the reading as such. What I'm gonna try though, is the same configuration with the current limiting circuit, with and without the ammeter. Then I can use the light meter in my DSLR to measure the amount of light. I won't be able to know the current but at least I can figure out if is increased. \$\endgroup\$ – Ema Nov 3 '16 at 3:25
  • \$\begingroup\$ Current and light output are fairly linear. The resistance of the Ammeter can be fairly high. If you do use one, you should always measure the voltage on the other side of it (the load side). \$\endgroup\$ – mkeith Nov 3 '16 at 3:55
  • \$\begingroup\$ the proper way is to make a 50mV current shunt on ground side to easily measure current and voltage with a common ground for fullscale current, here 1A would be 50mOhm which you can make with a 6ft pair of AWG 16 shorted at one end. then subtract the current shunt from your voltage reading. An ammeter is only as accurate as the user's awareness. It should ! be only 50mV full scale for that meter range , but might not be \$\endgroup\$ – Tony Stewart EE75 Nov 3 '16 at 3:55
  • \$\begingroup\$ @ema if you do not measure V and A properly at the same time, the results are meaningless, the supply could also be oscillating and the meter recording inaccurately \$\endgroup\$ – Tony Stewart EE75 Nov 3 '16 at 4:08

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